Qqad-_2007

Question: 1 The equation |x-1| - |x-2| + |x-4| = m has exactly n real solutions for some real m. Then which among the following relations between m and n can not be true? (a) m/n = 3/5

(b) m = n

(c) m/n = 3/2

(d) m/n = 5/3

(e) m = n-1

Solution: Since 1, 2, 4 are the critical point, we divide the domain into 4 regions; 1) when x > 4 |x-1| - |x-2| + |x-4| = (x-1)-(x-2)+(x-4) = m => x-3 = m => m>1 2) when 2 < x <=4 |x-1|-|x-2|+|x-4| = (x-1)-(x-2)-(x-4) = m => 5-x = m => 1 <= m < 3 3) when 1 < x <=2 |x-1|-|x-2|+|x-4| = (x-1)+(x-2)-(x-4) = m => x+1 = m => 1 < m <= 3 4) when -Infinity < x <=1 |x-1|-|x-2|+|x-4| = -(x-1)+(x-2)-(x-4) = m => 3-x = m => 2 <= m <= Infinity Clearly for n=4 we have 2 < m < 3; for n=3 we have m = 2,3; for n=2 we have 1 < m < 2; for n=1 we have m=1; for n=0 we have 1 > m. Looking at the choices (a) m = 2.4 and n = 4 satisfy (b) m = n = 3 satify (d) m = 10/3 and n = 2 satisy (e) m = -1 and n = 0 satisy Hence, choice (c) is the right answer. Question # 2 When Katrina get's Swiss chocolcates, she swings in delight, equal to her total chocolates at that time. For instance, if Katrina gets 3 chocolates, then 7 chocolates and then 3 chocolates again, she at first makes 3 swings, then she makes 3+7 = 10 swings and then she makes 3+7+3 = 13 swings, making a total of 3+10+13 = 26 swings. If all of Katrina's Swiss chocolates are in a group of either 3 or 7 and Katrina makes 99 swings during the process, in how many different ways can she get the chocolates in that process ? (a) 3

(b) 6

(c) 5

(d) 2

(e) none of the foregoing

Solution: If a1, a2, a3, ... are the numbers in which Katrina gets chocolates in order then Katrina's total swings will be n*a1 + (n-1)*a2 + .... + an where a1, a2 etc. take the values 3 or 7. Given the fact n*a1 + (n-1)*a2 + .... + an = 99; if all a1, a2 etc are 3 then 3n(n+1)/2 = 99 => n(n+1) = 66, so we know n can atmost be 7. Taking a1, a2 as 7 we can see that n is greater than 4. OK, if it's 7a1 + 6a2 + ... + a7 = 99 then because 3 and 7 are each of the form 4k-1 we see 7a1 + 6a2 + ... + a7 = 99 reduces to 4*(7k1 + 6k2 + ...+ k7) = 99 + (1 + 2 +... +7) = 99 + 28, which is not possible. Similarly it's not for n=5. For 6 it's possible as 99 + (1 + 2 +...+ 6) = 120 is div by 4. Now you are left with 6k1 + 5k2 + ...+ k6 = 30 where k1,k2 ... takes either 1 or 2

(k1, k2, k3, k4, k5, k6) = (1, 2, 1, 2, 1, 2), (2, 1, 1, 1, 2, 2), (1, 1, 2, 2, 2, 1), (1, 2, 2, 1, 1, 1), (2, 1, 1, 2, 1, 1) Hence, choice (c) is the right answer.

Question # 3 Let f(x ,y) be a function satisfying f(x ,y)=f(2x + 2y, 2y -2x) for all x , y. Define g(x)=f(2^x , 0). What is the minimum positive integer p if g(x+p)=g(x) for all x? (a) 4

(b) 6

(c) 8

(d) 12

(e) none of the foregoing

Solution: f(x ,y)=f(2x+2y,2y-2) = f(8y ,-8x ) f(8y,-8x) = f (8(-8x), -8(8y)) = f(-64x , -64y) = f(-64(-64x),-64(-64y)) = f(2^12 x ,2^12 y) = f(x,y) => f(x,0)=f(2^12 x,0) so , f(2^x,0) = f(2^(x+12) ,0) = > g(x) = g(x+12) Hence, choice (d) is the right answer.

Question # 4 Consider a triangle PQR so that PQ=4, PR=5 and QR=6. Similarly, we have a point S so that QS=5, RS=4, and PQSR is a paralleogram (not an isosceles trapezium). We draw the angle bisector of < P, which hits QS at T, and the angle bisector of
(b) 4v6/ 3 (c) 2v3

(d) 9/2

(e) none of the foregoing

Solution: Letting PA /PT = PR / PR + QT
Since the angle bisector of the vertex angle of an isosceles triangle bisects its base, TU =1/2 *6 =3 Hence, choice (a) is the right answer.

Question # 5 Three positive integers a, b, and c are consecutive terms in an arithmetic progression. Given that n is also a positive integer, for how many values of n below 1000 does the equation a^2 - b^2 - c^2 = n have no solutions? (a) 458

(b) 493 (c) 524

(d) 559 (e) 596

Solution: Let a = b+d, and c = b-d => n = b(4d - b) put b = 4d - 1, it gives n = (4d - 1)*1 = 4d - 1 = 3,7,11,...,999 for d = 1,2,...,250 put b = 4d - 2 it gives n = 8d - 4 = 4,12,...,996 for d = 1,2,...,125 put b = 4d -3, n = 12d - 9 which is of the form 4k - 1 and already counted put b = 4d - 4 it gives n = 16d - 16 = 16,32,...,992 for d = 2,3,...,63 it can be seen that b = 4d - 3, 4d - 5,... etc. give these already counted values of n. => total n are 999 - 250 - 125 - 62 = 562. The answer is not in the choices! For more rigorous mathematical proof please refer innocent_123's post # 214 in the discussion thread. Question # 6 20 teams of five archers each compete in an archery competetion. An archer finishing in kth placecontributes k points to his team, and there are no ties. The team that wins will be the one that has the least score . Given that, the 1st position team’s score is not the same as any other team,the number of winning scores that are possible is ? (a) 236

(b) 237 (c) 238

(d) 239 (e) none of the foregoing

Solution: The teams’ scores must sum to 1 + 2 + . . . + ..+99+100 =5050. The winning score must be no larger than 1/20 *5050 = 252.5 and is at least 1 + 2 + 3 + 4 + 5 = 15. However, not all scores between 15 and 252 inclusively are possible because all teams must have integer scores and no team can tie the winning team. If the winning score is s, the sum of all teams’ scores is at least s + 19(s + 1) = 20s +19, so solving gives s <= 251. Hence, 251 − 15 + 1 = 237 winning scores are possible. Hence, choice (b) is the right answer

Question # 7 P(x) is a third degree polynomial and the coefficients of P(x) are rational. If the graph of P(x) touches the x-axis, then how many rational roots does P(x) = 0 have? (a) 0

(b) 1

(c) 2

(d) 3

(e) none of the foregoing

Solution: Since P(x) touches x-axis => P(x)=0 has a repeatable root and is real. If the repeatable root is irrational and is a+vb then the non-repeatable root is k-2vb where k is a rational number. WHY? => P(x) = p*(x-a-vb)^2*(x-k+2vb) which means the constant term of P(x) is irrational, a contradiction. Thus, the repeatable root is rational. And so is non-repeatable. Alternate Method: Let P(x) be x^3 − ax^2 + bx − c. Let m be a root of P(x) = 0 and P′(x) = 0 m^3 − am^2 + bm − c = 0 … (1) and 3m^2 − 2am + b = 0 … (2) => am^2 − 2bm + 3c = 0 … (3) => m = (ab – ac)/2(a^2 – 3b) is rational; also 2m + n = a (rational) => n is also rational => all the three roots of P(x) = 0 are rational. Hence, choice (d) is the right answer

Question # 8 QQAD team decides to go on vacation for 8 days trusting their NL software system. The problems are fed into the system and date timer set in advance for each 8 days for the questions to be delivered to the subscribers daily. However, the software follows a weird rule. It doesn't always deliver the NL daily in those 8 days, but never misses 3 consecutive deliveries. How many possible ways are there for the NLs delivery in those 8 days? (a) 162

(b) 138

(c) 117

(d) 176

(e) 149

Solution: 8 NLs can be delivered in 8C8 = 1 way; 7 NLs can be delivered in 8C7 = 8 ways; 6 NLs can be delivered in 8C6 = 28 ways. 5 NLs can be delivered in 8C5 - (8-3+1) = 50 ways (6 is subtracted as

it's no. of ways of 3 consecutive misses). 4 NLs can be delivered in 8C4 - 20 (with 3 consecutive misses) - 5 (with 4 consecutive misses) = 45 ways. 3 NLs can be delivered in 16 ways. WHY? 2 NLs can be delivered in 1 way - on 3rd and 6th day. Thus making a total of 149 ways of delivery in those 8 days. Alternate Method: On each day NL can take 2 states - it either delivers or it doesn't. Let f(n) be the possible ways for n days with given conditions. => f(1) = 2, f(2) = 2^2, f(3) = 2^3-1 (1 subtracted as we can't have 3 misses). Let n > 3. Then, NL either delivers on day 1 or it doesn't. When it does it can have f(n-1) ways from there; on day 2 also it either delivers or it doesn't, when it does it can have f(n-2) ways from there. Now on day 3, it has to deliver as we have missed first 2 days, thus after delivering on day 3 it can have f(n-3) ways => f(n) = f(n-1)+f(n-2)+f(n-3) => f(4) = 13, f(5) = 24, f(6) = 44, f(7) = 81 and f(8 ) = 149 Hence, choice (e) is the right answer

Question # 9 Moiz has a machine into which he can put any number of one rupee coins. If he inserts n rupees, the machine returns 2n rupees. Each time he uses the machine, however, he must insert more money than he did on the previous use. If he starts with exactly Rs 1 and use the machine once, he will have Rs 2. On his next use of the machine, he is forced to insert Rs 2 yielding Rs 4, and on his third use of the machine, he can insert either Rs 3 or Rs 4 yielding a total of Rs 7 or Rs 8. The largest 2 digit integer Z such that it is impossible to obtain exactly Z rupees with the machine, starting with Rs 1 is (a) 10

(b) 65

(c) 96

(d) 97

(e) none of the foregoing

Solution: Note that Rs10 is an impossible amount because after four uses of the machine Moiz must have at least Rs 11. Now it is easy to check that all amounts from 11 to 20 rupees can be obtained. Suppose, by way of contradiction, that some value larger than Rs 10 is impossible and let m be the smallest number exceeding 10 such that m rupees cannot be obtained. Then m > 20. If m is even, write m = 2k and note that k > 10. Thus k rupees are possible and he never inserted as much as k rupees to get it. He can thus insert all k rupees to get 2k = m rupees. If m is odd, write m = 2k + 1 with k ¸ 10. Then k + 1 rupees is possible and he never inserted as much as k rupees to get it. He can thus insert k rupees yielding a total of 2k + 1 = m rupees. This shows that m rupees is possible, contradicting the choice of m. Thus no value larger than Rs 10 is impossible to obtain. Hence, choice (a) is the right answer

Question # 10 Rani draws a square ABCD of side 1 unit. She then draws 10 straight lines connecting A to each of 11 equally spaced points lying internally on CD (including C

and D). What is the total area (in unit square) of all the possible triangles that can be formed? (a) 5

(b) 11/2

(c) 11

(d) 23/2

(e) none of the foregoing

Solution: Clearly area of each of the smallest triangles is 1/20. Let the 9 internal points be M, N, O .... With AD as one side, the area of all triangles = 1/20 [1+2+..10] = 55/20 With AM as one side, the area of all triangles = 1/20[1+2+..9] = 45/20 Similarly for others... SO total area of all such triangles including triangle ABC is 1/40[1*2+2*3+..9*10+10*11] +1/2 =11+1/2 = 23/2 square units. Hence, choice (d) is the right answer

Question # 11 PQR is an acute angled triangle with perimeter 60 cm. S is a point on QR. The circumcircles of triangles PQS and PSR intersect PR and PQ at T and U respectively such that ST = 8 cm and SU = 7 cm. If
(b) 3/4

(c) 14/17

(d) 5/6

(e) none of the foregoing

Solution: Given that QU X 24 = 8 X 15 => QU = 5 So, UP = 19, similarly TR X 21 = 7 X 15 => TR = 5. So, TP = 16 Therefore PT/PU = 16/19 Hence, choice (a) is the right answer

Question # 12 How many two digit or three digit positive integers in base 6 are there such that if 0 in inserted between the units and the tens digit, the multiple of the original number is obtained? Solution: (a) 12

(b) 15

(c) 16

(d) 18

(e) none of the foregoing

Solution: Let AB be the 2 digit number => in base 6 it's 6A+B => 36A+B is divisible by 6A+B; for B=0, A can take 5 values from 1 to 5. 14 and 23 satisfy the given property (check yourself!) => 7 two digit numbers are possible. Let ABC be the three digit number then 216A+36B+C is divisible by 36A+6B+C; for C=0 we have 5*6 = 30 possibilities. Thus, in all 37 such two or three digit numbers are possible. Hence, choice (e) is the right answer Question # 13 Apple had to study Mathematics from his teacher Orange. As usual, Apple was sleeping in the class. Immediately he heard a thundering voice. "All roots of this equation are real as well as positive in nature". Apple woke up from deep slumber. He hurried to copy the 10 degree equation written on the board but could copy only the first two terms written on the blackboard before Orange sir wiped it all. Apple however remembered that the constant term was 2. He noted down the equation as 2x^10 - 20x^9………. + 2 = 0. Apple was very sure that if someone would tell him the sum of all the coefficients of all the powers of x in the equation, he would solve it anyhow. He asked Mango about the same. The answer which Mango correctly gave was (a) -1024 foregoing

(b) 0

(c) 1024

(d) Mango himself was confused

(e) none of the

Solution: Let the roots be a1, a2,….., a10 then a1 + a2 + …. + a10 = 10 , and (a1.a2….a10)=1 Now since all the roots are real and positive in nature We can say that (a1+a2+….+a10)/10>=(a1.a2….a10)^1/10 For real numbers, AM >= GM but here we find that AM = GM hence a1 = a2 = … = a10 = 1 So our equation actually is 2(x-1)^10=0 and then the sum of all the coefficients of all the powers of x in the equation is 0. Hence, choice (b) is the right answer

Question # 14

The sum of all the possible values of integral n such that (n^2-2n) ^ (n^2+47) = (n^2-2n) ^ (16n-16) is (a) 16

(b) 17

(c) 18

(d) 19

(e) none of the foregoing

Solution: Equate the bases to 0 and 1 and -1, we get n = 0, 1, 2 .. out of which 0 is rejected Equate the powers to get n=7,9 Thus, we have 1+2+7+9=19 as the answer Hence, choice (d) is the right answer

Question # 15 The total number of 7 digit positive integers whose digits are in increasing order (not necessarily strictly) is (a) 15C7

(b) 6! + 7.4!

(c) 14C6

(d) 7! - 4.6!

(e) none of the foregoing

Solution: Consider arranging 7 X and 8 Y in a row. We can do this in 15!/(8!*7!) ways. Replace each X with (1+the number of Ys prior to it), and now remove all Ys. We will have the desired numbers! Hence, choice (a) is the right answer

Question # 16 The 150 contestants of Miss India 2010 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ...). This process is carried out repeatedly until there is only the winner left. What is the number of Miss India 2010? (a) 93

(b) 48

(c) 119

(d) 38

(e) 140

Solution: Suppose a number is at nth position. First time n/3, [n/3] + 1, [n/3] + 1 integers less than n are eliminated for n = 3k, 3k+1, 3k+2. So, nth number's new position will be 2n/3, eliminated, [2n/3] for n = 3k, 3k+1, 3k+2. Working this backwards it's like if a number is at nth position now, earlier it must have been at 3n/2 or [3n/2] + 1. So, what is at 1st position now (winner) was earlier at position [3/2] + 1 = 2.

Moving back one more step what was at position 2, even earlier was at position 3*2/2 = 3. Going on like this the only positions that can end up at first position are 1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315...(These are positions, not actual values) And when I have moved back enough steps to get back the original series (1,2,3,4,5,...), positions become equal to actual values cause in the actual series nth number is at nth position. So these are the actual winning contestant numbers 1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315... subject to how many total contestants are there. Hence, choice (e) is the right answer

Question # 17 Let 3 statements be made (P) 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10 (Q) The remainder on dividing 16! + 89 by 323 is q (R) 46C23 leaves remainder r on division by 23 Then p+q+r equals (a) 12

(b) 19

(c) 26

(d) 33

(e) none of the foregoing

Solution: The notation a % b = c means that when a is divided by b, c is the remainder that is obtained. (P) Check for p = 9; 10^12 % 13 = 1 by Fermat's theorem => 10^18 % 13 = 10^6, since 10^3 % 13 = -1 + 13 => 10^18 % 13 = 1 => 10^18 - 10^9 + 1 leaves remainder (1-(-1)+1=3) by p = 9; Check yourself the remainder when p = 8. When p = 7, as 10^12 % 13 = 1 => 10^14 % 13 = 9; also 10^6 % 13 = 1 => 10^7 % 13 = 10. Thus p = 7 satisfies our condition. (Q) 323 = 17*19. (p-1)! + 1 is divisible by p when p is prime => 16! % 17 = -1 + 17. Also 18! % 19 = 1 + 19 If 16! % 19 = x => 18! % 19 = 17*18x => 306x % 19 = 18 => 2x % 19 = 18. Thus, 16! leaves the remainder 16 by 17 and 9 by 19 => It leaves 237 by 323 => q = 3 (R) 2nCn = (nC0)^2 + (nC1)^2 + ... + (nCn)^2; when n is prime, each of the term in RHS except 1st and last is divisible by n => r = 2 Hence, choice (a) is the right answer Question # 18 From a cylinder of height 3 cm and radius 2 cm , two identical cones each of height 2 cm are to be cut such that they have the maximum volume. The volume of a cone will be (in cubic cm) (a) (2/3)*pi (b) (5.12/3)*pi (c) (6.16/3)*pi (d) (7/3)*pi (e) none of the foregoing Solution:

The two cones must be symmetrically selected with their bases on opposite bases of the cylinder respectively and also the two cones must touch each other. Let the radius of base of each cone be x and let y be the perpendicular distance of the point of tangency of the two cones from the altitude of either cones. Then, 2x +2y = 2*2=4 If the height of cylinder is h, then we have (2h/3 -h/2)/y = (2h/3)/x =>y/x =1/4 Hence, x+x/4 =2 => x = 8/5. So volume of the cone = pi/3 (8/5)^2 *(2/3)*h = (5.12/3) *pi cubic cm. Hence, choice (b) is the right answer

The question that was 19th Last Year Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of (a) 18%

(b) 20%

(c) 21 %

(d) 23%

(e) Cannot be determined

Let the CP of two vodkas be Rs 100 and Rs 100x and individual profit in Rs on them being A and B. => (A+2B)/3 = 10/100*(100+200x)/3 and (2A+B)/3 = 20/100*(200 + 100x)/3. solving we get A = (70+20x)/3 and B = (20x-20)/3 => profit percentages on each is (70+20x)/3 and (20x-20)/3x. When they are increased to 4/3 and 5/3 times respectively and mixed in the ratio 1:1 we get total profit % as (4/3*100*(70+20x)/3 + 5/3*100x*(20x-20)/3x)/(100+100x) = 100*(20x+20)/(100+100x) = 20 => choice (b) is the right answer. The question that was 20th Last Year N girls and 2N boys played a chess tournament. Every player played every other player exactly once. The boys won 7/5 times as many matches as the girls (and there were no draws). Then which among the following is definitely false? (Assume 1 point for a win and 0 for a loss) (a) Boys pocketed prime number of points against girls (b) Girls always won twice or more matches than boys won against them (c) The sum of the scores of top 3 individual players was not between 25 and 33 (d) The sum of the scores of top 3 individual players was 69 (e) none of the foregoing Total number of matches among boys were 2nC2, among girls were nC2 and between boys and girls were n*2n. Please note that 2nC2 + nC2 + 2n^2 = 3nC2. Assume 1 point for a win and 0 for a loss => Girls pocketed nC2 points amongst themselves and boys pocketed 2nC2 points among themselves. Let boys take k points from their matches against girls => girls take 2n^2 - k from

their matches gainst boys. => 2nC2 + k = 7/5*(nC2 + 2n^2 - k), solving we get 8k = n(5n+1). for n = 3, k = 6. For n = 8, k = 41, For n = 11, k = 77. (a) can be true as for n = 8, k = 41. (b) can be true as can be seen for for n = 3, 8, 11, ... (c) is true as for n = 3, top 3 can score 8+7+6 = 21 points and for n = 11, when 33 matches are played top 3 will always score more than 16+15+14 = 45. (d) is true, For n = 11, we can have the top 3 score as 23+23+23 = 69 => choice (e) is the right answer.

The question that was 21st Last Year Amrutesh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Amrutesh walks according to the following plan: He moves along the altitude-tothe-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Amrutesh walks? (a) 48/25

(b) 12/5

(c) 12

(d) 15

(e) none of the foregoing

Let M be the endpoint of the altitude on the hypotenuse. Since we are dealing with right triangles, triangle MAC ~ triangle ABC, so AM = 12/5. Let N be the endpoint he reaches on side AC. Triangle MAC ~ trangle NAM,So , MN/AM =4/5 . This means that each altitude that he walks gets shorter by a factor of 4/5 . The total distance is thus (12/5) /(1- 4/5) =12 => choice (c) is the right answer. The question that was 22nd Last Year The minimum possible value of the largest of ab, 1-a-b+ab, and a+b-2ab if 0 <= a <= b <=1 is (a) 4/9

(b)1/9

(c)5/9

(d)1/3

(e) none of the foregoing

Let s = a + b, p = ab, so a and b are (s+/- root(s^2-4p))/2 . Since a and b are real , s^2 - 4p>=0 . If one of the three quantities is less than or equal to 1/9, then at least one of the others is at least 4/9 by the pigeonhole principle since they add up to 1. Assume that s-2p < 4=9, then s^2 - 4p < (4/9 + 2p)^2 - 4p , and since the left side is non-negative we get 0<= p^2 -(5/9) p +4/81 =(p1/9)(p-4/9). This implies that either p<=1/9 or p>=4/9 , and either way we're done. This minimum is achieved if a and b are both 1/3, so the answer is 4/9 => choice (b) is the right answer The question that was 23rd Last Year Let f be a factor of 120, then the number of positive integral solutions of xyz = f is

(a) 160

(b) 240

(c) 320

(d) 480

(e) none of the foregoing

Let k be such that k = 120/f. Then, the number of positive integral solutions of xyz = f is same as that of number of positive integral solutions of xyzk=120=2^3.3.5 We can assign 3 and 5 to unknown quantities in 4*4 ways. We can assign all 2 to one unknown in 4C1 ways, to two unknowns in (4C2)(2) and to three unknown in 4C3 ways. Hence, the number of required solutions =4*4*[4C1 + (4C2)(2) +4C3] =4*4*20 =320 => choice (c) is the right answer.

The question that was 24th Last Year Apple, Bombardier, Chat.sun and Doomsayer are to compile this year's Quant Question A Day. They can finish this work together in a certain number of integer days. However, they work two in a day and it is found that the compilation is completed when (Apple, Bombardier), (Bombardier, Chat.sun) and (Chat.sun, Doomsayer) worked for respectively 5, 9 and 4 days or 7, 6 and 5 days. They could not have all together done the work in (a) 8 days

(b) 9 days

(c) 10 days

(d) 11 days

(e) none of these

Let Apple, Bombardier, Chat.sun and Doomsayer do 1/a ,1/b,1/c and 1/d parts of work per day respectively . Then , 5(1/a +1/b) +9(1/c +1/b) +4(1/c +1/d) =1 ; 7(1/a +1/b) +6(1/c +1/b) +5(1/c +1/d) =1 and 1/a +1/b + 1/c +1/d = 1/n (say) =>12( 1/c +1/b) = 1-3/n which implies n>3 and 4( 1/a +1/b) = 1- 7/n which implies n>7 . Consequently, 1/c +1/d = 1/n - ( 1/a +1/b) = 1/n -1/4(1-7/n) =(11-n)/4n which implies n<11; So, from the given options n=11 is not possible => choice (d) is the right answer. The question that was 25th Last Year Through T, the mid-point of the side QR of a triangle PQR, a straight line is to meet PQ produced to S and PR at U, so that PU = PS. If the length of UR = 2 cm, then the length of QS is (a) 3/2 cm

( b) 2 cm

(c) 5/2 cm

(d) 3 cm

(e) none of the foregoing

If QS=UR=2

=> choice (b) is the right answer. The question that was 26th Last Year A new Ducati is designed for the Indian market such that its mileage at a particular speed follows a certain relationship with that speed. Also , the speed decreases linearly with the mass of the rider while the petrol consumption per km increases linearly with the mass of rider .Ideally , when the mass of the rider is negligible, the speed is 100km/hr and mileage is 100km/l .When the speed of the Ducati is 50 km/hr , the mileage is 50 km/l.When the speed of the Ducati is 75 km/hr , the mileage will be (a) 60 km/l determined

(b) 67 km/l

(c) 72 km/l

(d) 75 km/l

(e) Cannot be

Speed = 100-k1*mass Petrol Consumption per KM = c2+k2*mass or Mileage = 100/(1+k2*mass) When speed = 50KMPH, K1*mass = 50 when mileage = 50KMPL, k2*mass = 1 k1 = 50*k2 When speed = 75kmph, k1*mass = 25 => k2*mass = 0.5 mileage = 100/1.5 = 67km/l => choice (b) is the right answer.

The question that was 27th Last Year A function V(a, b) is defined for positive integers a, b and satisfies V(a, a) = a, V(a, b) = V(b, a), V(a, a+b) = (1 + a/b) V(a, b). The value represented by V(66, 14) is ? (a) 364

(b) 231

(c) 455

(d) 472

(e) None of the foregoing

V(66,14 ) = V(14,66 )=(33/26)* V(14,52) =(33/26) * (26/19)* V(14, 38) = (33/19) *(19/12)* V(14, 24) = (33/12) *(12/5)* V(14,10) = (33/5)* (7/2)*V(10, 4) = (231/10)* (5/3) *V(4, 6) = (77/2)* 3 *V(4, 2) = (231/2)*2* V(2, 2) = 231*2 = 462 => choice (e) is the right answer.

The question that was 28th Last Year A cargo ship circles a lighthouse at a distance 20 km with speed 1500 km/h. A torpedo launcher fires a missile towards the ship from the lighthouse at the same speed and which moves so that it is always on the line between the lighthouse and the ship. How long does it take to hit? (a) 37.7 secs

(b) 56.57 secs

(c) 75.43 secs

(d) 94.29 secs

(e) 113.14 secs

Let M be the position of the ship at the moment the missile is fired. Let V be the point a quarter of the way around the circle from M (in the direction the ship is moving).Take the point at which Lighthouse is situated to be U. Then the missile moves along the semi-circle on diameter UV and hits the ship at V. To see this take a point T on the quarter circle and let the line UT meet the semi-circle at R. Let Z be the center of the semicircle. The angle VZR is twice the angle VUR, so the arc VT is the same length as the arc VR. Hence also the arc UR is the same length as the arc MT. so time needed= 10 *pi*3600 /1500 =75.43 => choice (c) is the right answer.

The question that was 29th Last Year --------------------------------------------Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers? (a) 24

(b) 23

(c ) 22

(d) 20

(e)21

Each angle in degrees is 180(p-2)/p. 180-{360}/{p}=k, so 360/p has to be an integer. Factors of 360 = 2^3.3^2.5^1 So there are 4.3.2 solutions, but we exclude 1 and 2, because p>= 3 So, 24 -2 =22 => choice (c) is the right answer.

The question that was 30th Last Year Consider a pair (x,y) of natural numbers satisfying x + y^2 + g^3 = xyg, where g is the greatest common divisor of x and y .Then , how many such pairs are possible? (a) 2

(b) 3

(c) 4

(d) 5

(e) 6

Put x = Mg, y = Ng, so that M and N are coprime. Then: M + N^2g + g^2 = MNg^2. So g must divide M. Put M = M'g, then M' + N^2 + g = M'Ng^2. So M' = (N^2 + g)/(Ng^2 - 1). Hence M'g^2 = N + (N+g^3)/(Ng^2 - 1). So (N + g^3)/(Ng^2 - 1) is an integer. If g = 1, (N+1)/(N-1) can only be an integer for N = 2 or 3. That gives the solutions (x, y) = (5, 2) and (5, 3). So assume g > 1. Since (N + g^3)/(Ng^2 - 1) is an integer and positive, we must have N + g^3 >= Ng^2 - 1, so N <= (g^3 + 1)/(g^2 - 1). If g = 2, then N <= 3. Then N = 1 gives the solution (x, y) = (4, 2), N = 2 gives (N + g^3)/(Ng^2 - 1) non-integral and hence no solution, N = 3 gives the solution (x, y) = (4, 6). So assume g > 2. Then (g^3 + 1)/(g^2 - 1) < g + 1. Hence N ≤ g. Hence M' = (N^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(g^2 - 1), since N ≥ 1 and hence M' < 2 (since g > 2). So M' = 1. So N is a root of the quadratic N^2 - g^2N + g+1 = 0. But g^4 - 4(g+1) > g^4 - 4g^2 + 1 = (g^2 - 1)^2 and < g^4, so g^4 - 4(g+1) cannot be a square and hence N cannot be integral. So there are no solutions with g > 2 => choice (c) is the right answer.

The question that was 30th Last Year --------------------------------------------Consider a pair (x,y) of natural numbers satisfying x + y^2 + g^3 = xyg, where g is the greatest common divisor of x and y .Then , how many such pairs are possible? (a) 2

(b) 3

(c) 4

(d) 5

(e) 6

Put x = Mg, y = Ng, so that M and N are coprime. Then: M + N^2g + g^2 = MNg^2. So g must divide M. Put M = M'g, then M' + N^2 + g = M'Ng^2. So M' = (N^2 + g)/(Ng^2 - 1). Hence M'g^2 = N + (N+g^3)/(Ng^2 - 1). So (N + g^3)/(Ng^2 - 1) is an integer. If g = 1, (N+1)/(N-1) can only be an integer for N = 2 or 3. That gives the solutions (x, y) = (5, 2) and (5, 3). So assume g > 1. Since (N + g^3)/(Ng^2 - 1) is an integer and positive, we must have N + g^3 >= Ng^2 - 1, so N <= (g^3 + 1)/(g^2 - 1). If g = 2, then N <= 3. Then N = 1 gives the solution (x, y) = (4, 2), N = 2 gives (N + g^3)/(Ng^2 - 1) non-integral and hence no solution, N = 3 gives the solution (x, y) = (4, 6). So assume g > 2. Then (g^3 + 1)/(g^2 - 1) < g + 1. Hence N ≤ g. Hence M' = (N^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(Ng^2 - 1) <= (g^2 + g)/(g^2 - 1), since N ≥ 1 and hence M' < 2 (since g > 2). So M' = 1. So N is a root of the quadratic N^2 - g^2N + g+1 = 0. But g^4 - 4(g+1) > g^4 - 4g^2 + 1 = (g^2 - 1)^2 and < g^4, so g^4 - 4(g+1) cannot be a square and hence N cannot be integral. So there are no solutions with g > 2 => choice (c) is the right answer. The question that was 31st Last Year

LMN is a triangle. MO is the angle bisector. The point P on LM is such that >
Now take triangles LMN and LPO (LM / MN ) = ( LO / ON ) ..... angle bisector theorem But ON=OP ... Given ==>> ( LM / MN ) = ( LO / OP ) And , > Both Triangles are similar.

==>> > x=180 - (7/5)x ==>> x=
While required angle = > >
=> choice (e) is the right answer.

The question that was 32nd Last Year --------------------------------------------N! is defined for non negative integers as N!=N*(N-1)*(N-2)* ...3*2*1 . The number of positive integers which divide (2^5 )! are (a) 2^13.3^3.5^2 of the foregoing

(b) 2^8.3^2.5^2

(c) 2^11.3^2.5

(d) 2^8.3^3.5^3 (e) None

32! is divisible by [32/2] + [32/4] + [32/8] + [32/16] +[32/32] = 31 powers of 2, and [32/3] + [32/9] + [32/27] = 14 powers of 3. It is divisible by 5^7, 7^4, 11^2, 13^2, 17 , 19, 23, 29, 31. In other words, 32! = 2^31 .3^14 . 5^7 .7^4 .11^2 •13 ^2 •17•19.23 .29.31 So it has (31+1) (14+1)(7+1)(4+1)(2+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1) = 2^13 . 3^3 . 5^2 => choice (a) is the right answer.

The question that was 33rd Last Year --------------------------------------------The sum of the infinite series 1/3 + 2/21 + 3/91 + 4/273 + ... is given by which of the following?

(a) 1/4

(b) 1/2

(c) 3/4

(d) 1

(e) 3/2

General term T(n) = n/(n^2-n+1)(n^2+n+1) = 1/2 * (1/(n^2 - n + 1) - 1/(n^2 + n + 1)) Hence sum = 1/2 *(1 - 1/3 + 1/3 - 1/7 + 1/7 - ........) = 1/2 *(1) = 1/2 => choice (b) is the right answer. The question that was 34th Last Year --------------------------------------------Consider a positive integer Z which when represented in decimal base,does not end in zero. Z and the number obtained by reversing the digits of Z are both multiples of seven. Let the number of such Zs in the set {10, 11, 12... 998, 999, and 1000} be K. Then k is given by which of the following? (a) 13

(b) 14

(c) 15

(d) 16

(e) None of the foregoing

There is only one 2-digit number, namely 77. Indeed, if 10a + b is (where a, b are integers 1–9), then 7 divides 10a + b and 7 divides 10b + a. Thus, 7 divides{ 3(10a + b) − (10b + a)} = 29a − 7b = a + 7(4a − b), so that 7 divides a, and similarly 7 divides b, so we’d better have a = b = 7. There are 16 3-digit numbers. First consider the 12 palindromic ones (ones where the hundreds and units digits are the same): 161, 252, 343, 434, 525, 595, 616, 686, 707, 777, 868, and 959. Now consider the general case: suppose 100a + 10b + c is, where a, b, c are integers 1–9. Then 7 divides 100a + 10b + c and 7 divides 100c + 10b + a, so 7 divides (100a+10b+c)−(100c+10b+a) = 99(a−c), and so 7 divides a−c. In order for this not to result in a palindromic integer, we must have a − c = +/- 7 and, moreover, both 100a + 10b + a and 100c + 10b + c must be palindromic integers. Consulting our list above, we find 4 more integers: 168, 259, 861, and 952. So K= 1+16 = 17 => choice (e) is the right answer.

The question that was 35th Last Year ---------------------------------------------

Shravya invests some amount of money in a firm M .This amount grows upto 5000 in 2 years and upto 5500 in 3 years on R% compound inetrest . Then she goes to another firm N and borrows Rs 7000 at a compound interest of R%. At the end of each year she pays back Rs 3000 to firm N. Then, the amount she should pay to firm N at the end of 3 years to clear all the dues is? (a) 2387

(b) 2550

(c) 2667

(d) 2477

(e) None of the foregoing

CI of 3rd year =5500-5000 =500 Rate, R = 500*100/5000 =10% At the end of 3 years she pays = 7000[1+11/100]^3 - 3000[(1+10/100)^2 + (1+10/100)] = 2387 => choice (a) is the right answer.

The question that was 36th Last Year --------------------------------------------If |(a − b + c)(b − c + a)(c − a + b)| = 15 ,where |x| has its usual meaning , then the possible number of ordered integral triplets (a,b,c) are (a)36

( b)33

(c)66

(d)60

(e) None of the foregoing

Take b=a+k (where k is any integer) We boil down to |(c^2-k^2)*(2a+k-c)|=15 Cases can only be out of the following : (+/-1,+/-15) ,(+/-15,+/-1) , (+/-3,+/-5) , (+/-5,+/-3) (c,k) can be - > (+/-1,0) ,(0,+/-1),(+/-4,+/-1),(+/-1,+/-4),(+/-8+/-7),(+/-7,+/-8 ),(+/-2,+/-1),(+/1,+/-2) ,(+/-3,+/-2),(+/-2,+/-3)

And each case for (c,k) will yield 2 values of a and for a given a and c , b is uniquely determined by k Total c,k from above is (2+2) +(4+4)+(4+4)+(4+4) +(4+4) =36 Since for each (c,k) we get two values of a . Hence total (a,b,c) =2*36 =72 => choice (e) is the right answer.

The question that was 37th Last Year --------------------------------------------Consider a scalene triangle PQR.Points S, T and U are selected on sides QR, PR, and PQ respectively. The lines PS, QT, and RU meet at point Z. If area(PUZ) = 126, area(UQZ) = 63, and area(RTZ) = 24, the area of triangle PQR is ? (a) 324

(b) 351

(c) 360

(d) 364

(e) 378

Consider triangle PQT and PQZ and PZT The area of PQZ = 1/2 QZ *h = 189 the area of PZT= 1/2 ZT *h =x so area of PZQ /area of PZT = 189/x = QZ/ZT ----------(1) Then take triangle QTR, similarly area of QZR / area of RTZ = y /24 = QZ/QT ------(2) from 1 and 2---- 189/x = QZ/ZT = y /24 Then take triangle PUR and URQ and apply the same procedure so we get the rel 24 +x /126 = y /63 Solve and get the value of x and y.. So get total as 351 => choice (b) is the right answer.

The question that was 38th Last Year --------------------------------------------There are three runners viz, Nishant , Deepak and Mohit who jog on the same path. Nishant goes jogging every two days. Deepak goes jogging every four days. Mohit goes jogging every seven days. If it’s the first day that they started this routine, what is the total number of days that each person will jog by himself in the next seven weeks? (a) 12

(b) 13

(c) 14

(d) 15

(e)16

Nishant jobs a total of 25 days, Deepak a total of 13 days, and Mohit a total of 7 days. We know that Nishant jogs on days 1,3,5,7,.....49, and Deepak jobs on days 1,5,9,.....49, and Mohit jogs on days 1, 8, 15, 22,...43. Obviously when Deepak jogs, Nishant will always be jogging also, so Deepak jogs 0 days alone. Nishant jogs 25-13=12 days without Deepak also jogging, but Mohit jogs on 4

odd numbered days, 2 of which Deepak also jogs, giving us 12-4+2=10 days that Nishant jogs alone. Nishant jogs on every odd numbered day, so Mohit jogs alone only on even numbered days. Because there are 4 odd numbered days, as stated above, there are 3 even numbered days, so Mohit jogs alone on 3 days. Answer: Nishant: 10 Deepak: 0 Mohit: 3 => choice (b) is the right answer.

The question that was 39th Last Year --------------------------------------------Consider a polynomial function P(y) =y^3+2y^2+5 and one another polynomial function, Q(y)=y^4 -3y^2+2y+1. Let there be two more functions S(y) and T(y), that satisfy gcd(P(y), Q(y) ) =S(y)*P(y) + T(y)*Q(y). Now consider a function M(y) = T(y) - S(y). Then, the product of all the roots of M(y) is given by (a) 42

(b) 48

(c) 54

(d) 56

(e) None of the foregoing

First stage division gives ... Q(y) = P(y)(y-2) + y^2 -3y +11 y^2 -3y +11 = Q(y) - (y-2)P(y) And then from the second stage, P(y) = (y^2-3y+11)(y+5) + 4y-50 = {Q(y) - (y-2)P(y)}(y+5) + 4y-50 From the third stage of division, 4(y^2 -3y +11)= (4y-50)(y+19/2) +519 =>519 = 4[Q(y) - (y-2)P(y) ] - [(y^2+3y-9)]P(y) - (y+5) Q(y)] [y+19/2] =>519 = 4 Q(y) - (4y-8 )P(y) - (y^2+3y-9)[y+19/2] P(y) + (y+5) Q(y) [y+19/2] =-[4y-8+y^3+19y^2/2 +3y^2 +57/2y -9y-171/2]P(y) + [y^2+19/2y +5y+95/2 +4]Q(y) =-P(y)/2 [2y^3 + 25y^2 +47y -187] + Q(y)/2[ 2y^2+29y+103]

=>519 = [-(2y^3 + 25y^2 +47y -187)/2]P(y) + [(2y^2+29y+103)/2]Q(y) So, T(y) - S(y)= (2y^3 +27y^2 + 76y -84)/2 = y^3 +27/2y^2 + 38y -42 So, product of roots =42 => choice (a) is the right answer. The question that was 40th Last Year --------------------------------------------A circle passes through the vertex C of rectangle ABCD and touches its sides AB and AD at P and Q respectively. If the distance from C to the line segment PQ is equal to 4 units, then the area of the rectangle ABCD in sq. units (is) (a) 20 (b) can not be determined foregoing

(c) 16

(d) greater than 20 (e) none of the

Let M be the feet of perpendicular from C to PQ. Now, by alternate segment theorem we have => < CQM = < CPB and < CPM = < CQD. Thus, right triangles CQM and CPB are similar and also CPM and CQD. Thus, CQ/CP = CM/CB and CP/CQ = CM/CD. Thus, CB.CD = CM^2 = 16. => choice (c) is the right answer.

The question that was 41st Last Year Maximums needed to calculate the volume of a rectangular room. He multiplied the length and the breadth correctly but the breadth had been incorrectly jotted down, it was one-third larger than what it should have been. To compensate for this, he reduced the height by one-third, then multiplied it on. He figured this was okay since the breadth was equal to the height. He then found his volume was off by 20 m^3. What was the actual volume? (a) 160

(b) 120

(c) 210

(d) 159

(e) None of the forgoing

Actual Volume = L*B*H. Given, L*B*H - L*4/3B*2/3H = 20 => 1/9*LBH = 20. => choice (e) is the right answer.

The question that was 42nd Last Year --------------------------------------------Which among the following have the same graph? I) y = x-2

II) y = (x^2-4)/(x+2)

III) y(x+2) = x^2-4

IV) y = (√x-√2)*(√x+√2)

(a) I and III only the forgoing

(b) II and IV only

(c) I, III, IV

(d) II and III only

(e) none of

y = x-2, will be a straight-line with slope 45 deg and making intercept of 2 and -2 on X and Y axis respectively. II) Same as I except for a breakpoint at x = -2. III) at x = -2, we will also have a line parallel to the Y axis. IV) Graph is restricted to first and fourth quadrants only. => choice (e) is the right answer.

The question that was 43rd Last Year --------------------------------------------All the digits of a 50 digit positive number are 4 except for the nth digit. If the number is divisible by 13 for some choice of that nth digit, then how many possible values can n have? (a) 17

(b) 21

(c) 25

(d) 33

(e) none of them

The seed of 13 is 4 which means to check the divisibility by 13 of a n digit number, multiply the last digit by 4 and add the result to the initial (n-1) digits and see if it is div by 13 e.g. 182 -> 18+2*4 = 26. Thus, 182 is div by 13. 1001 -> 100 +1*4 = 104 -> 10 + 4*4 = 26, hence 1001 is div vy 13. It's also the same thing for 182 as saying that 2*(4^2) + 8*4 + 1 is div by 13 or for 1001 1*(4^3) + 1 is div by 13 - this comes from recurrence. Note that we can find the seed of every prime number. The seed of 7 is 5, the seed of 17 is 12, the seed of 19 is 2. If 444...x...444 is div by 13 => 4*(4^49 + 4^48 + ... + 1) + (x-4)*4^(n-1) is div by 13, where nth digit is x and not 4. 4*(4^49 + 4^48 + ... + 1) = 4/3*(4^50-1). By Fermat's theorem 4^12 = 1 mod (13) => 4^50 = 3 mod (13). Thus, 4*(4^50 - 1) leaves remainder of 8 when div by 13. If 4/3*(4^50-1) leaves x by 13 => 3x%13 = 8 => x = 7. Thus 4/3*(4^50-1) leaves remainder 7 when div by 13. Now, 7 + (x-4)*4^(n-1) is div by 13. when n = 1, 7+(x-4) has to be div by 13, we have no such x when n = 2, 7+4*(x-4) has to be div by 13, we have no such x when n = 3, 7+3*(x-4) has to be div by 13, we have x = 6 when n = 4, 7+12*(x-4) has to be div by 13, we have no such x when n = 5, 7+9*(x-4) has to be div by 13, x = 9 when n = 6, 7+10*(x-4) has to be div by 13, x = 2

when n = 7, 7+(x-4) has to be div by 13, we have no such x and the whole cycle of 6 repeats from here. In each cycle of 6 we have 3 desired n. Till n = 50 we have 48*3/6 + 0 = 24 such numbers. => (e) is the right answer.

Quantitative Question # 043 -----------------------------------------------------Consider two cones of heights 1 and 8 units having the same base radii. It is found that their height is increased by x keeping their vertex angle unchanged, their volume becomes equal. Then x equals (1) 2/3

(2) 4/3

(3) 8/3

(4) 16/3

(5) none of these

Suppose the radii of two cones to be r initially and r1 and r2 after increasing height by x since the vertex angle are unchanged r/1 = r1/(1+x) => r1 = r(1+x) r/8 = r2/(8+x) => r2 = r(8+x)/8 now the volumes are equal => 1/3*pi*r1*r1*h1 = 1/3*pi*r2*r2*h2 => [r(1+x)]^2 * (1+x) = [r(8+x)/8]^2*(8+x) => x = 4/3 => Choice (2) is the right answer 44. Let a, b, c be positive reals. (I) and (II) are independent statements. (I) Minimum value of a^3/4b + b/8c^2 + (1+c)/2a is p (II) a + b + 2c = 8 and a^2 + b^2 + 2c^2 = 25. Maximum possible value of c is q. Then which among the following is |p-q|? (a) 5/2

(b) 9/4

(c) 1/2

(d) 3/8

(e) none of the foregoing

(I) a^3/4b + b/8c^2 + (1+c)/2a = a^3/4b + b/8c^2 + 1/4a + 1/4a + c/2a. For positive reals AM >= GM and the equality occurs when all the numbers are equal. Thus, a^3/4b + b/8c^2 + 1/4a + 1/4a + c/2a >= 5(a^3/4b*b/8c^2*1/4a*1/4a*c/2a)^1/5 = 5/4. Thus, p = 5/4 when a^3/4b = b/8c^2 = 1/4a = c/2a. (II) a+b=8-2c; a^2+b^2 >= 1/2*(a+b)^2 = 1/2(8-2c)^2 => 25-2c^2 = 1/2(8-2c)^2 => 1/2 <= c <= 7/2. Thus, q is 7/2 when a=b. => (b) is the right answer. 45. Let F be be mini 4X4 chessboard => it has 16 fields in all. In how many ways is it possible to select two fields of F such that the midpoint of the segment joining the centres of the two fields should also be the centre of a field? (1) 15

(2) 18

(3) 24

(4) 32

(5) none of these

Solution: The fields of the centre of the fields of the chessboard can be assigned coordinates. Let the centre of the 1st field in the left most bottom corner be (0, 0) => the centres of the fields are of the form (x, y) where 0 <= x, y <= 3. The coordinate of the midpoint of the segment connecting the points (a,b) and (c,d) are (a+c)/2 and (b+d)/2, and these coincide with the coordinates of the centre of a field if and only if a+c and b+d are both even. The point (a,b) may be the centre of any field out of the 16 fields of the mini chessboard. (a,b) having been selected, the number c can be any of the 2 numbers that have the same parity as a, and independently of that, d can also have 2 different values. In order to make the chosen points different, the number of the possible pairs (c,d) is 1 less than 2.2=4. Buthe order of the two selected points does not matter => the number of the appropriate number pairs is 1/2*(4*4)*3 = 24. If our question was on NXN square, then it could have been solved in a jiffy with the above logic. For 6X6 the answer is 144. For 8X8, the answer is 480. => Choice (3) is the right answer 46. All India Pagalguy Meet of year 2008' is in June (which has 30 days), but Allwin forgot which day, so he asked around. Rohit said that the date was an odd number; Apurv claimed it was greater than 13. Divya declared it was not a perfect square, while Sonam swore it was a perfect cube. Finally, Grand-ma tells Allwin the date was less than one-fourth her (Grand-ma's)age, which Allwin knew to be 68. Yesterday Allwin learned that only one of them had told the truth! If the date of the All India PG Meet is D (numerical value), then (a) D is uniquely determinable (b) D can have exactly 2 values (c) D doesn't exist (d) D has atleast 4 values (e) none of the foregoing

The question says that only one of them spoke the truth! Let's see if Rohit spoke the truth or not. If he has then the date D is an odd number <= 13 (since Apurv must have lied), and D > 17 (Grandma must have lied too in this case). Contradiction, and hence we have no such value of D. If Apurv spoke the truth, then D > 13 and D >= 17 (Grandma must have lied). Since, Rohit and Divya also lie => D is a even perfect square. Not possible as D <= 30. Please check yourself that if Divya or Sonam speak truth then no value of D exists. Let's check if Grandma has spoken the truth or not. If she has then D < 17. Rohit lies => D is even, Divya lies => D is a perfect square, Sonam lies => D is not a perfect cube => D can be 4 or 16 but as Apurv also lies => D = 4 is the only possibility => choice (a) is the right answer

47. Let m be the largest positive term of an harmonic progression whose first two terms are 2/5 and 4/9. A real number r satisfying m/2-1/n < r <= m+1/n, for every positive integer n, is best described by:

(a) 1 < r < 5 foregoing

(b) 2 < r <= 4

(c) 1 < r <= 5

(d) 2 <= r <= 4

(e) none of the

If 2/5 and 4/9 are the first two terms of the harmonic progression then 5/2 and 9/4 are the terms of the corresponding AP => common difference of the AP is -1/4 => we want 5/2 + (n-1)*(-1/4) as positive as well as minimum. Thus, n = 10 and the AP term is 1/4. m = 4. Please note that our HP consists of 10 terms only. Now, 2-1/n < r <= 4+1/n for all positive intger n. Putting n = 1 we get 1 < r <= 5. Putting n = 2 we get 3/2 < r < 9/2 etc. If r is more than 4 then r <= 4 + 1/n fails for some n (e.g. if r = 4.01 then n = 101). If r is less than 2 then 2-1/n < r fails for some n (e.g. if r = 1.99 then n = 100) Note that 1/n > 0 always for any choice of positive integer n. Thus, 2-1/n < 2, and 4 <= 4+1/n is true for all postive integers n. => the best description of r contains 2 and 4 as well besides the range (2, 4). => choice (d) is the right answer.

The number of real roots of the equation |1 - |x|| - (1.01)^(1.01x) = 0 is/are (a) 1

(b) 2

(c) 3

(d) 0

(e) none of these

Solution: At x = 0 we have a solution. Let f(x) = |1 - |x|| - (1.01)^(1.01x) f(1) < 0, f(3) > 0 => we have odd number of solutions between (1, 3) but f(x) is increasing in (1, 3) => we have just 1 solution in (1, 3) f(-1) < 0 and f(-2) > 0 and f(x) is decreasing in (-2, -1) => 1 solution in this interval also. Also, draw the graphs of y = |1 - |x|| and y = (1.01)^(1.01x) and see they intersects at 3 points. => Choice (3) is the right answer 48. The set S has 5 elements. In how many ways can one select two (possibly identical) subsets of S whose union is S? (a) 32

(b) 63

(c) 64

(d) 93

(e) 122

Let the subsets of S be A and B. For each element in S we have three choices (it can belong to either of A, B or both). That gives each pair of subsets twice except for the case A = B = S. Hence, we can select 2 subsets in (3^5 + 1)/2 ways. => choice (e) is the right answer. 48. The number of real solutions to |1 - |x|| - (1.01)^(1.01x) = 0 is (a) 1

(b) 2

(c) 3

(d) 0

(e) none of these

At x = 0 we have a solution. Let f(x) = |1 - |x|| - (1.01)^(1.01x) f(1) < 0, f(3) > 0 => we have odd number of solutions between (1, 3) but f(x) is increasing in (1, 3) => we have just 1 solution in (1, 3) f(-1) < 0 and f(-2) > 0 and f(x) is decreasing in (-2, -1) => 1 solution in this interval also.

Also, f(1000) < 0 and f(3) > 0 and f(x) in (3, 1000) is x - 1 - (1.01)^(1.01x) which is a decreasing function in this interval => we have one more root in (3, 1000) => 4 roots in all Ö choice (e) is the right answer.

48. Twinkle tells Raveena that she has got 3 kids and 2 of these kids are twins, and also that their ages are all integers. She tells Raveena the sum of the ages of her kids and also the product of their ages. Raveena says that she has insuficient information to determine the ages, but one possibility is that the twins are a prime number of years old. If Twinkle’s twins are teenagers and their age is not prime, then the sum of the ages of her kids is (a) a prime number (b) is greater than 43 (c) a composite number (d) exactly 2 of the foregoing (e) still undeterminable Let x,and y be the ages of Twinkle’s children. Since the twins are teenagers, x is an integer with 13 <=x <=19. Furthermore, Let p, p and q be the solution that Raveena discovered with the twins having prime age p. Since Raveena knows the sum of the ages, we must have 2x + y = 2p + q. Similarly, she is given the product of the ages, so x^2y = p^2q. Multiply the first equation by p^2 and replace p^2q by x^2y to obtain p^2(2x + y) = p^2(2p + q) = 2p^3 + p^2q = 2p^3 + x^2y, and thus y(x−p)(x+p) = y(x^2−p^2) = 2p^2(x−p). Since x is different from p, we can divide by x−p and get y(x + p) = 2p^2. In particular, the integer x + p divides 2p^2. Since p is a prime, the positive divisors of 2p^2 are 1, p, p^2, 2, 2p and 2p^2. Furthermore, x + p > 1, 2, p and x + p not equal to 2p since x not equal to p. This leaves x + p = p^2 or x + p = 2p^2 and hence x = p^2 − p or x = 2p^2 − p. Since p is a prime and 13 <=x <= 19, the only possibility here is p = 3 and x = 2p^2 − p = 15. Using (x + p)y = 2p^2, it then follows that y = 1. Thus Twinkle’s children have ages 15, 15 and 1. Note that 2x + y = 31, so 2p + q = 31 and q = 25. As a check, we observe that (15)^2•1 = 3^2 * 25. Ö choice (a) is the right answer.

49. The equation |x-1| - |x-2| + |x-4| = m has exactly n real solutions for some real m. Then which among the following relations between m and n can not be true? (a) m/n = 3/5

(b) m = n

(c) m/n = 3/2

(d) m/n = 5/3

(e) m = n-1

Since 1, 2, 4 are the critical point, we divide the domain into 4 regions; 1) when x > 4 |x-1| - |x-2| + |x-4| = (x-1)-(x-2)+(x-4) = m => x-3 = m => m>1 2) when 2 < x <=4 |x-1|-|x-2|+|x-4| = (x-1)-(x-2)-(x-4) = m => 5-x = m => 1 <= m < 3 3) when 1 < x <=2 |x-1|-|x-2|+|x-4| = (x-1)+(x-2)-(x-4) = m => x+1 = m => 1 < m <= 3 4) when -Infinity < x <=1 |x-1|-|x-2|+|x-4| = -(x-1)+(x-2)-(x-4) = m => 3-x = m => 2 <= m <= Infinity Clearly for n=4 we have 2 < m < 3; for n=3 we have m = 2,3; for n=2 we have 1 < m < 2; for n=1 we have m=1; for n=0 we have 1 > m. Looking at the choices (a) m = 2.4 and n = 4 satisfy (b) m = n = 3 satify (d) m = 10/3 and n = 2 satisfy (e) m = -1 and n = 0 satisfy Ö choice (c) is the right answer. 50. Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 Choose 2 Choose 3 Choose 4 Choose 5

if the question can be answered by X only if the question can be answered by Y only if the question can be answered by either X or Y if the question can be answered by both X and Y if the question can not be answered by combining X and Y also

If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x. (X) [a[k/(2 - a)] + a/2] = p (Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even Solution: Put 2-a = m => 0 < m < 1, we will show that [(2-m)k/m] = [(2-m)[k/m] + (2-m)/2] [(2-m)k/m] = [2k/m] - k, since k is an integer. Let us take this case by case. Case 1: when k/m is an integer = i [(2-m)k/m] = 2i - k, and [(2-m)[k/m] + (2-m)/2] = [2i -mi + 1-m/2] = [2i -k + 1-m/2] = 2i - k. Case 2: when i-1/2 < k/m < i, where i is an integer [(2-m)k/m] = 2i - 1 - k, [(2-m)[k/m] + (2-m)/2] = [2i - m/2 - k] = 2i - 1 - k Case 3: when i < k/m < i + 1/2, where i is an integer Prove yourself here that [(2-m)k/m] = [(2-m)[k/m] + (2-m)/2] => X is sufficient for all k Now Y adds 1/2 in X and depending on the value of a[k/(2-a)], we can have multiple (two) possible value of [ak/(2 - a)] for q. => Choice (1) is the right answer 51. When Katrina get's Swiss chocolcates, she swings in delight, equal to her total chocolates at that time. For instance, if Katrina gets 3 chocolates, then 7 chocolates and then 3 chocolates again, she at first makes 3 swings, then she makes 3+7 = 10 swings and then she makes 3+7+3 = 13 swings, making a total of 3+10+13 = 26 swings. If all of Katrina's Swiss chocolates are in a group of either 3 or 7 and Katrina makes 99 swings during the process, in how many different ways can she get the chocolates in that process ? (a) 3

(b) 6

(c) 5

(d) 2

(e) none of the foregoing

If a1, a2, a3, ... are the numbers in which Katrina gets chocolates in order then Katrina's total swings will be n*a1 + (n-1)*a2 + .... + an where a1, a2 etc. take the values 3 or 7. Given the fact n*a1 + (n-1)*a2 + .... + an = 99; if all a1, a2 etc are 3 then 3n(n+1)/2 = 99 => n(n+1) = 66, so we know n can atmost be 7. Taking a1, a2 as 7 we can see that n is greater than 4. OK, if it's 7a1 + 6a2 + ... + a7 = 99 then because 3 and 7 are each of the form 4k-1 we see 7a1 + 6a2 + ... + a7 = 99 reduces to 4*(7k1 + 6k2 + ...+ k7) = 99 + (1 + 2 +... +7) = 99 + 28, which is not possible. Similarly it's not for n=5. For 6 it's possible as 99 + (1 + 2 +...+ 6) = 120 is div by 4.

Now you are left with 6k1 + 5k2 + ...+ k6 = 30 where k1,k2 ... takes either 1 or 2 (k1, k2, k3, k4, k5, k6) = (1, 2, 1, 2, 1, 2), (2, 1, 1, 1, 2, 2), (1, 1, 2, 2, 2, 1), (1, 2, 2, 1, 1, 1), (2, 1, 1, 2, 1, 1 Ö choice (c) is the right answer.

52. Let the equation 8^(x/(x+2)) = 6/3^x has n roots m(1), m(2), ..., m(n) such that m(1) >= m(2) >= ... >= m(n). Then the numerical value of m(1) - m(2) - ... - m(n) is (a) 2+3log2/log3 (b) 2-3log2/log3 (c) 3+2log2/log3 (d) 3-2log2/log3 (e) none of the foregoing The eqn becomes 2^(3x/(x+2)) . 3^x = 2*3 =6, or, 2^(3x/(x+2) - 1) . 3^(x-1) = 1. So, (2x - 2)/(x+2). log2 + (x-1).log3 = 0 => either x-1 = o => x = 1 or,x+2 = -2log2/log3, so, x = -2 -2log2/log3 So, m1 - m2 = 1-(-2 - 2log2/log3) = 3+2log2/log3 Ö choice (c) is the right answer. 53. Concentric circles radii 1, 2, 3, ... , 100 are drawn. The interior of the smallest circle is colored red and the annular regions are colored alternately green and red, so that no two adjacent regions are the same color. The total area of the green regions divided by the area of the largest circle is (a) 1/2 (b) 51/100 (c) 101/200 (d) 50/101 (e) none of the foregoing Total Area of Green Regions = pi*(100^2-99^2+98^2-97^2+....+2^2-1^2) Area of the largest circle = pi*100^2 =>Required ratio = (100^2-99^2+98^2-97^2+....+2^2-1^2)/100^2 = {(100+99)(100-99)+(98+97)(98-97)+.....+(2+1)(2-1)}/100^2 = (199+195+191+...+7+3)/100^2 = 5050/100^2 = 101/20 Ö choice (c) is the right answer 54. A biologist catches a random sample of 60 fish from a lake, tags them and releases them. Six months later she catches a random sample of 70 fish and finds 3 are tagged. She assumes 25% of the fish in the lake on the earlier date have died or moved away and that 40% of the fish on the later date have arrived (or been born) since. What does she estimate as the number of fish in the lake on the earlier date? (a) 420 (b) 560 (c) 630 (d) 720 (e) 840 Let no. of Fishes on earlier day = x 60 fishes are tagged, out of which 25% have died/moved remaining fishes that are tagged = 45

No. of fishes on day of second observation = 5/4x No. of fishes caught = 70 No. of Fishes tagged = 3 = 4.2857% (3/70) Total No. fishes on Second day of Observation = 45/.042857 (45*70/3) = 1050 No of Fishes on earlier day = x = 4/5*1050 = 840 Ö choice (e) is the right answer. 55. The cost of 20 oranges and 1 kg of apple is Rs 60 while their selling price is Rs 72. The cost of 10 apples and 1 kg of orange is Rs 50 while their selling price is Rs 60. Given that the profit percent on the sale of two fruits are different, then the sum of the selling price of 5 oranges and 3 apples and the cost price of 6 kg oranges and 5 kg apples (is) (a) can not be determined (b) Rs 318 (c) Rs 375 (d) Rs 384 (e) none of the foregoing 20 Oranges and 1KG of apple = 12 Apples and 1.2 KG of Oranges (50*1.2) If the profit %ages are different, then it must be clear that the ratio of apples : Oranges in the 60Rs. sale and the 72Rs. sale must be the same. Hence, we can safely say that 1.2 KG of oranges = 20 Oranges and 1KG of apples = 12 apples. SP of 20 Oranges and 12 Apples = 72 => SP of 5 Oranges and 3 apples = 18 CP of 1.2KG of Oranges and 1KG of Apples = 60 => CP of 6KG Oranges and 6KG apples = 300

=> choice (b) is the right answer. 56. A word is a combination of 8 letters, each either A or B. Let x and y be 2 words differing in exactly 3 places. How many words differ from each of x and y in at least 5 places? (a) 38

(b) 48

(c) 58

(d) 68

(e) 78

Exp-X--_ _ _ * * * * * Y--_ _ _ * * * * * Case 1. Lets assume that X and Y are same in the * positions. So our number now can have anything in _ positions if we assign a different alphabets in *. S0, 2*2*2 =8 Case 2: Let’s say only 4 positions are satisfied. In that case 1 will have to come from the first 3 positions.

There are 6 ways to do that and since there are 5 positions where we can change the alphabet we get 5*6=30 So total sum=38 Ö choice (a) is the right answer. 57. Squares WXYZ, XPQR, PWMN are drawn externally on the sides of a triangle WXP. The line segments YZ, QR, MN, when extended, form a triangle W'X'P'. Find the area of W'X'P' if WXP is an equilateral triangle of side length 2. (a) 6 +9(3)^1/2 (b)12 + 13(3)^1/2 (c) 9 + 13(3)^1/2 (d) > 12 + 9(3)^1/2 (e)None of the foregoing Area of triangle W'X'P' is area of three squares + area of equi triangle +area of 3 Quads..calc area of squares and triangle WXP is no prob....area of quad is area of the two triangles formed in that quad...we get area of one such triangle as 2(3)^1/2..(using trigonometry...)there are six such triangles (contained in the three quads) and hence the total area is Area of three squares = 12 (4*3) + area of equilateral triangle WXP = 3^1/2 + Area of all six triangles is 12(3)^1/2 => total area is 12+13(3)^1/2 Ö choice (b) is the right answer. 58. The following sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17 ... has one odd number followed by two evens, then three odds, four evens, and so on. What number is the 2003rd term? (a) 3942 (b) 3943 (c) 3944 (d)3945 (e) None of the foregoing 1,--2,4,--5, 7, 9,-- 10, 12, 14, 16,-- 17,19,21,23,25 ... 1. n(n+1)/2 63(64)/2 = 2016. i.e from 1st to 63rd set there will be in total 2016 terms. 2. Each set will cover (2n-1) consecutive integers. Therefore 63 set will cover 125 consecutive numbers, 62nd set will cover 123 numbers.. so on. therefore 1+3+5+....+125 = 63*63 = 3969 consecutive integers are covered by the end of 63rd set. 3. 63rd set will contain 63 odd numbers, 3969 as the 2016th term. 3969, 3967, 3965, 3963, 3961, 3959, 3957, 3955, 3953, 3951, 3949, 3947, 3945, => choice (b) is the right answer.

59. Each day, Chetna is either happy or sad. If she is happy one day, then four times out of five she is happy the next day. If she is sad one day, then she is sad the next day one time out of three. In the long run, what are the chances that Chetna is happy on any given day? (a) 8/15 (b) 15/13 (c) 10/13 (d) 4/15 (e) None of the foregoing Use Bayes Theorem. It states that P(A/B) = [P(B/A) * P(A)] / P(B) where : P (B/A) : Probability of B ( sad ) given A(happy) = 2/3 ( Since she is sad 1 times out of 3 if she is sad one day - implies she is happy 2 times out of 3) P (A) = Probability of being happy P(B) = Probability of being sad P(A/B) : Probability of A (happy) given B = 1/5 ( similar logic as above ) this gives 2/3 = (1/5*P(A)) / P(B) or, P(A) / P(B) = 10/3 Now, We have the ratio of P(A) and P(B) as 10:3 So, P(A) = 10x and P(B) = 3x Here, P(A) + P(B) = 1 (Since Chetna is either happy or sad every day) So,10x+3x =1 or 13x = 1 Implies that, P(A) = 10x/13x = 10/13 and P(B) = 3x/13x = 3/13 or,P(Happy) = P (A) = 10/13 Ö choice (c) is the right answer.

60. Let the net profit when pencil, pen and eraser are sold respectively at 20%,30% and 40% profits be 10% more than that when they are sold respectively at 40%,20% and 30% profits. Let K be the net profit when they are sold respectively at 30%, 40% and 20% profits. Then the least value of K will be? (a) 10% (b) 15% (c) 20% (d) 25% (e) None of the foregoing Let x, y, z be the prices of pencil, pen and eraser .2x+.3y+.4z = 1.1(.4x+.2y+.3z) 7z +8y = 24x => z+y = 3x + z/8 ...(1) K = (0.3x+0.4y+0.2z)/(x+y+z) = 0.2+ (0.1x+0.2y)/(x+y+z) = 0.2+ (0.7x - 1.4z/8 )/(4x+z/8 ) equating 0.7x to 1.4z/8, we get z = 4x but from equation (1), 24x > 7z and also, after simplifying the above fraction, since the Nr. of the fraction is less than the Dr., the value of the fraction increases with x. Least K = 0.2 + (4.9Z/24 - 1.4z/8 )/(7z/6+z/8 ) = 0.2 + 0.7/31 ~ 22.26% Ö choice (e) is the right answer.

61. If in the expansion of (1+x)^a (1–x)^b, the coefficient of x and x^2 are 3 and –6 respectively, then 'a' is (a) 6

(b) 9

(c) 12

(d) 24

(e) None of the foregoing

(1 + x)^a(1 - x)^b = (1 + aC1.x + aC2.x^2 + ..)(1 - bC1.x + bC2.x^2 - ..) Co-effn of x: (a-b) = 3...(1), Co-effn of x^2: (b(b-1)/2 + a(a-1)/2 -ab) = -6...(2) =>(a-b)^2 - (a+b) = -12 (a+b) = 21, Hence a = 12 => choice (d) is the right answer. 62. On a regular Chess Board, the numbers of rectangles such that in each one of them the number of Black Houses is equal to the number of White Houses are (a) 896

(b) 1036

(c) 1156

(d) 796

(e) 856

Only those rectangles with even number of houses satisfy this condition. Total ways in which length = 1 can be selected ->8 Total ways in which length = 2 can be selected ->7 ... ... Total ways in which length = 8 can be selected ->1 So, Total ways in which all different length can be selected = 8+7+6+5+4+3+2+1 same holds for Breadth. So, Total no. of rectangles (by Multiplication Theorem) = (8+7+6+5+4+3+2+1)*(8+7+6+5+4+3+2+1) = 36^2=1296 Total no. of rectangles with odd number of houses = (8+6+4+2)*(8+6+4+2) =400 => the no. of rectangles satisfying the above condition is 1296-400 =896 Ö choice (a) is the right answer. 63. Let V(t) =V(1-t)+V((1-t^2)^0.5), for all 0
Ö choice (d) is the right answer. 64. Let PQRS be a convex quadrilateral with PQ=QR=RS with PR not equal to QS and T be the intersection point of its diagonals. If PT=ST then sum of the angles QPR and QSR in degrees? (a) 30

(b) 60

(c) 90

(d) 120 (e) can not be determined

Let U be the reflection of R through PS. Thus, <(UPS) = <(TPS) = <(TSP) => PU || SQ. => PQSU is an isosceles trapezium. So < PQS+< PRS=< PQS+< PUS=180 < TPS+< TSP=< QTP=< SQR+< PRQ so < QPR+ < RSP=2*(< SQR+< PRQ) ...... (1) < QPS+< RSP=360-< PQR-< QRS=180-< SQR-< PRQ ....... (2) from (1) and (2) < SQR+ < PRQ=60 Ö choice (b) is the right answer. 65. In a class there are 100 students. A division of the students in n sections is good if: 1) the sections have different numbers of students 2) for any partition of one of the sections in 2 smaller sections, among the (n+1) sections you get 2 with the same number of students (any section has at least 1 student). The positive difference between the maximal and minimal possible value of n such that the division is good is (a) 2

(b) 3 (c) 7

(d) 8 (e) none of the foregoing

Well, the language of the question wasn't perfect and people interpreted this question in different manners. Some assumed "one of the sections" as - some one, while others assumed it as each one. Then, "you get 2 with the same number", some assumed at least 2, while some assumed exactly 2. We give the answer of all possible 4 cases. Case 1: when the question meant each one + at least 2 Max = [1, 2, 3, ..., 12, 22]. Min = [1, 3, 5, ..., 19]. Thus, answer is 13-10 = option (a). Case 2: when the question meant each one + exactly 2 Max = [1, 3, 5, ..., 19]. Min = [1, 3, 5, ..., 19]. Thus, answer is 10-10 = option (e). Case 3: when the question meant some one + exactly 2 Max = [1, 3, 4, ..., 12, 24]. Min = [1, 3, 96]. Thus, answer is 12-3 = option (e). Case 4: when the question meant some one + at least 2 Max = [1, 2, 3, 4, ..., 12, 22]. Min = [1, 3, 96]. Thus, answer is 13-3 = option (e).

66. Let a, b, c be distinct non-zero integers such that -5 <= a, b, c <= 5. How many solutions (a, b, c) does the equation 1/a + 1/b + 1/c = 1/(a+b+c) have? (a) 72

(b) 120

(c) 192

(d) 240

(e) none of the foregoing

1/a + 1/b + 1/c = 1/(a+b+c) reduces to (a+b)(b+c)(c+a) = 0; if a+b = 0 then we can chose b in 10 ways (integers from -5 to 5 excluding 0) and c in 10-2 = 8 ways (since a, b and c are different). Also, b+c = 0, and c+a = 0 gives 2 more 80 each such solutions => option (e).

67. One day Vikram was out bicycling. After entering a one-way tunnel and after having ridden one-fourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr? (a) 32

(b) 36

(c) 40

(d) 48

(e) none of the foregoing

Let d=distance truck is in front of tunnel entrance, L=length of tunnel, x=Vikram's speed. Case 1: Vikram turns around and heads for entrance, a distance of L/4. Vikram and truck get to entrance at same time T1=d/80=(L/4)/x. Case 2: Vikram streaks for exit, a distance of 3L/4. Vikram and truck get to exit at same time T2=(L+d)/80=(3L/4)/x. Solving both equations for x and setting them equal, x=(L/4)*80/d=(3L/4)*80/(L+d) After simplifying, d=L/2, hence x=40. => option (c).

68. A square sheet of paper ABCD is so folded that B falls on the mid point of M of CD. The crease will divide BC in the ratio (a) 3 : 2

(b) 5 : 3

(c) 2 : 1

(d) 9 : 4

(e) none of the foregoing

Point B is coincided with the midpoint of CD(i.e., M). Let the crease cut the side BC at N. Assuming the length of the sides as 2a, CM=MD=a. Let NC=x, making BN=2a-x. Now when B is coincident with M, a rt triangle is formed: triangle BNC(or MNC, as B now lies on M) Applying Pythagoras' theorem, (2a-x)^2= x^2+a^2 solving for x we get: x=3a/4 therefore 2a-x=5a/4 BN:NC=5a/4:3a/4 or 5:3. => option (b).

69. Five containers each have one litre of 10%, 20%, 30%, 40%, 50% milk in them. By mixing the contents of just two of the containers, one litres each of w%, x%, y%, z% and 42% (22 < w <= x <= y <= z < 42) milk is prepared, contents of no two containers are mixed twice in different proportions to get different concentrations. Also each original container is used exactly twice in making new mixtures. If the containers are mixed only in the multiples of 100 ml then z+y-x-w (is) (a) can not be determined

(b) 4

(c) 16

(d) 20

(e) none of the foregoing

Let’s start with 42% milk preparation. 10% milk can be mixed with 50% milk to produce 42% milk but on doing so we are left with 800 ml of 10% milk which can’t give any possible combination with others to produce milk more than 22% milk. Only 30% milk in quantity 400 ml with 50% milk in 600 ml gives us a 42%

combination. Now we are left with 600 ml of 30% and 400ml of 50%.Now since we are combining 600 and 400 ml of 2 hence all the solution have to be broken in two parts each of 400 ml and 600 ml. Proceeding further from here, we find only 1 valid combination so that all the values are satisfied in the given range. 400ml of C + 600 ml of E ---> [120 + 300] ---> 420 out of 1000 ---> 42% 400ml of E + 600 ml of A ---> [200 + 60 ] ---> 260 out of 1000 ---> 26% 400ml of A + 600 ml of D ---> [40 + 240] ---> 280 out of 1000 ---> 28% 600ml of B + 400 ml of D ---> [120+ 160] ---> 280 out of 1000 ---> 28% 400ml of B + 600 ml of C ---> [80 + 180] ---> 260 out of 1000 ---> 26% Hence, z+y-x-w = 28+28-26-26 = 4 => option (b).

70. Akshay and John started to dig a canal 10 km long. It is calculated that if Akshay takes off for 3 days, then John has to dig 1 km more and if John takes off for 4 days, Akshay has to dig n km more. Which among the following is true? (a) If Akshay and John do the work without absence and Akshay digs 6 km, then the work gets completed in a little over 7 days (b) n = 4/3 (c) If the work gets completed in 7.5 days, when Akshay and John dug without any absence, then Akshay dug 5 km (d) all of the foregoing (e) none of the foregoing Let W be the amount of work done in digging. Let Akshay, John do W/a and W/b parts of work per day respectively. Let N, p, q be the number of days in which the work can be done if Akshay, John do the work together without absence, with Akshay absent for 3 days, with John absent for 4 days respectively. => NW/a + NW/b = W, (p-3)W/a + pW/b = W and qW/a + (q-4)W/b = W. Also, pW/b - NW/b = W/10 and qW/a - NW/a = nW/10. => p = b/10 + N, q = an/10 + N => (b/10 + N-3)/a + (b/10 + N)/b = 1; (na/10 + N)/a + (na/10 + N-4)/b = 1 => (b/10 - 3)/a + 1/10 + N(1/a + 1/b) = 1 and n/10 + (na/10 - 4)/b + N(1/a+1/b) = 1, but N(1/a + 1/b) = 1 => b/10 - 3 + a/10 = 0; and n/10(1/a + 1/b) = 4 => n = 4/3 and a+b = 30 This answer is part (b) Part (a) If Akshay digs 6 km then John digs 4 km => Amount of work done by them is in the ratio 3:2 => b:a = 2:3 but a+b = 30 => a = 12, b= 18 => work is completed in 1/(1/12 + 1/8 ) = 7.2 days Part (c) 1/a + 1/b = 2/15 and a+b = 30 => a = b = 15 => Akshay dug 5 km. Hence, choice (d) is the right answer

71. Consider a set P= {1,2,3...,11,12} of natural numbers. We define another set Q such that it contains no more than one out of any three consecutive natural numbers. How many subsets Q of P including the empty set are possible? (a) 114

(b) 117

(c) 129

(d) 136

(e) 130

Let f(n) be number of subsets of {1,2, .....n} which contain no more than one out of any three consecutive natural numbers. then f(n) = f(n-1) + f(n-3) Logic is to include f(n-1) subsets and add n to all the subsets for f(n-3) to get all possible subsets for n with f(1) =2, f(2) = 3 , f(3) =4. Solving we get f(12)=129. => option (c). 72. PQRST is a cyclic pentagon with QR = RS = ST.The diagonals PR and QT intersect at X.Y is the foot of the altitude from X to PQ. We have XP = 25,XS = 113, and XY = 15. The area of triangle PQT is approximately (a)1544

(b)1600

(c) 1648

(d)1560

(e) None of the foregoing

Pythagoras gives PY = 20. We draw QS and PS, and construct the altitude XZ to PS, with Z on PS, and altitude XX' to PT, with X' on PT. Because QR = RS = ST, angles QPR, RPS, and SPT are congruent. Because Z is on PS, triangles XYP and XZP are congruent by AAS, so XZ = 15 and ZP = 20, from which Pythagoras gives ZS = 112, implying PS = 132. Let k = option (a). 73. Maalchand had 100 pieces each of articles A and B. A was cheaper in price than B. Fearing police raid, Maalchand decides to sell all the pieces in two days. On the first day, he sold A and B at 10% and 30% profit respectively. Although, he made a net profit of 25%, the articles did not sell much and was not more than 25 pieces of each type. On second day, A and B were sold at 30% and 10% profit respectively. In the process, all the articles of Maalchand were sold, but his net profit for the second day was reduced to 20%. Which among the following is not true? (a) the cost price of each piece of article A was less than 18% cheaper than each piece of article B

(b) the total sale on the first day was 19.5% of the total sale (c) the net profit on the total sale was 20.9% (d) all of the foregoing (e) none of the foregoing Let 'a' be the price per item and 'x' be the no: of article sold of article A on the first day. Similarly let 'b' be the price per item and 'y' be the no: of article sold of article B on the first day. so, 1.1ax+1.3by=1.25(ax+by) => 3ax = by------------------(1) also for the second day, (100-x)*1.3a+(100-y)*1.1b=1.2[(100-x)a+(100-y)b] =>[(100-x)a = (100-y)b--------------------(2) from (1) and (2) y = 300x/(2x+100) since x,y<=25; only one integer value satisfies this i.e. x=10 and y=25, from (1) we get, 30a = 25b a = .833b--------------------------(3) so, ((b-a)/b)*100 = 16.7% so option (a) is true. Total sale on the first day = 11a+32.5b = 50a Total sale on the second day = 117a+82.5b = 117a+99a = 216a ratio of sale on first day to second day = 50/256 = 19.5% Hence (b) is also true. Also total sale = (216a+50a) = 266a, total cost = 100a+120a = 220a, hence net profit percent = (46/220)*100 = 20.9% hence (c) is also true. => option (e) is the correct answer. 74. What is the smallest positive integer k for which there are at least 11 even and 11 odd positive integers m so that (m^3 +k)/(m+2) is an integer? (a) 268

(b) 448

(c) 638

(d) 858

(e) none of the foregoing

Notice that m^3 + 8 is divisible by m+2. Therefore, k−8 must be divisible by m+2 for the expression to be an integer. If f is a factor of k − 8, m = f − 2 is a corresponding suitable m; we then need f >=3 to make m > 0. Thus k − 8 must have twelve each odd and even factors including 1 and 2. To make the number of odd and even factors equal in order to minimize k, the power of 2 in the prime factorization of k− 8 must be 1. Suppose the prime factorization of k− 8 is then 2^1•3^a •5^b •7^c •11^d (larger prime factors will clearly not minimize k. Then (p+1)(q+1)(r+1)(s+1)>= 12. To minimize k, p >=q>= r>= s. We then examine values of (k-8)/2 to determine the best (p, q, r, s). 3 • 5 • 7 • 11 = 1155, 3^2 • 5 • 7 = 315. Moving any more factors into smaller primes involves multiplying by 3^2/7 or 3^2/5 (or subsequent larger powers of 3), which increases the value. Therefore k− 8 = 2 •3^2 • 5 • 7, so k= 638. => option (c) is the correct answer.

75. Every week the Pagalguy.com magazine publishes a list of the Top 20 contributors on Pagalguy.com. If the order is never the same in any two consecutive weeks, and no contributing member ever regains any lost popularity (i.e. no contributor rises in ranking once he/she starts to drop in ranking) how many consecutive weeks could the same twenty contributors remain on the Top 20 list? (a) 199

(b) 191

(c) 190

(d) 171

(e) none of the foregoing

If you look at the initial and the final weeks of the rankings, during this time, the contributor initially in the top position can drop at most 19 spots to position 20; the second ranked contributor can drop at most 18 spots to position 19, and so on. This can take at most 19 + 18 + ... + 2 + 1 = 190 weeks. Therefore there are at most 191 weeks in which the same contributors can appear on the charts. You can achieve this maximum, by switching only two contributors each week. => option (b) is the correct answer. 76. Four persons go to a birthday party. They leave their top-coats and hats in the lounge and pick them while returning back. The number of ways in which none of them picks up his own top-coat as well as his own hat is p. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q. The number of ways in which a person picks up someone else’s top-coat and yet someone else’s hat is r. Then p+q+r is (a) 117

(b) 104

(c) 113

(d) 108

(e) none of the foregoing

The no. of ways in which the 1st person doesn't pick up his own top-coat is 3. The 2nd person (whose top-coat the 1st one picked) can pick someone else's top-coat in 3 more ways. For the 3rd and 4rth person we have just 1 choice. Thus, in all when none among the 4 pick his/her own topcoat is 9. Thus p = 9*9 = 81. A person among the 4 who picks his own top-coat as well as hat can be selected in 4 ways. The other 3 falls in the category [not own top-coat or not own hat] = not own top-coat + not-own hat (not won top-coat and not own hat) = 2*6 + 2*6 - 2*2 = 20. Thus, q = 80. As explained in the 1st case, each person can pick someone else's top-coat in 9 ways. To each of these 9 ways we have 2 ways where he picks yet someone else's hat. Thus, r = 9*2 = 18 => option (e) is the correct answer. 77. (i) The sum of real numbers x and y is 1. The maximum value of xy(x^3 + y^3) is p. (ii) Let the equation [n/2] + [n/4] = n has q possible solutions, where [n] denotes the greatest integer less than or equal to n e.g. [3.21] = 3. What is the value of p*q? (a) 1/4

(b) 3/16

(c) 1/3

(d) 1/2

(e) none of the foregoing

(i) If n is positive then [n/2] + [n/4] can at the max be 3n/4. Thus, no solutions for positive n. For non-negative n, n = 0 is a trivial solution. For n < -5 LHS > RHS. For -5 <= n < 0, we have 3 solutions, n = -2, -3, -5. Thus p = 4. (ii) xy(x^3 + y^3) = xy(1-3xy) = 1/3*(3xy*(1-3xy)). Thus, 3xy + (1-3xy) is constant => max of 3xy*(1-3xy) occurs when 3xy = 1-3xy => xy = 1/6 => q = 1/12

=> option (c) is the correct answer. 78. A car gives the mileage of 60km/L, 50km/L and 40km/L when driven at the speeds of 40km/hr, 50km/hr and 60km/hr respectively. Assume that each car is driven only at the three speeds mentioned above. The car is driven for 3 hours using 2 Litres of petrol. The distance covered by the car (a) > 120 km (b) < 120 km of the foregoing

(c) = 120 km

(d) can not be determined

(e) none

Let the distance traveled with 40km/hr, 50km/hr, 60km/hr is A, B and C respectively. => A/40 + B/50+ C/60 = 3 and A/60 + B/50 + C/40 = 2. Eliminating A, we get 2B/5 + 5C/6 = 0 => B=C=0 => A = 120 km => option (c) is the correct answer. 79. Let f(x) be an algebric expression of odd degree (> 1). If the degree of f(x) + f(1-x) is at least two less than the degree of f(x) and the coefficients of x and x^2 are equal in magnitude but opposite in sign in the given expression, then the highest possible degree of f(x) is (a) 9

(b) 13

(c) 5

(d) 7

(e) none of the foregoing

Taking f(x) = ax^n + bx^n-1 + cx^n-2 + .... , where the third last coefficient and the second last coefficient are equal in magnitude but opposite in sign and n is odd. f(x) + f(1-x) has at least first 2 terms as 0 => 2b + na = 0. Thus, we can have such f(x) for any n, thus n-> Infinity. => option (e) is the correct answer. 80. 100 ex-students of the same batch meet for alumni meet in the campus of a Bschool. The benches of Fr Prabhu Hall in the B-school are arranged in a rectangle of 10 rows of 10 seats each. All the 100 have different salaries. Each of them asks all his neighbours (sitting next to, in front of, or behind him, i.e. 4 members at most) how much they earn. They feel a lot of envy towards each other: a person is content with his salary only if he has at most one neighbour who earns more than himself. What is the maximum possible number of ex-students who are satisfied with their salaries? (a) 72

(b) 32

(c) 48

(d) 64

(e) none of the foregoing

Let us represent the ex-students with a square grid of 10x10 points, and label each point with the salary of the ex-student. Have a look at the figure in the document by following the undermentioned link. Let us draw arrows between neighbouring points such that the arrow is directed from the smaller to the larger number: www.pagalguy.com/forum/814259-post1030.html (Satisfied alumni are marked in green, and dissatisfied ones in red.) Let a be the number of satisfied EX-STUDENT's sitting in the corners, b the number of those sitting at the sides of the square, and c the number of those sitting inside. The number of arrows is 180. There is at most one arrow originating at any satisfied EXSTUDENT, and there will be at least one point where no arrow originates, the EX-STUDENT with the largest salary (obviously satisfied). Hence the number of arrows originating at satisfied EXSTUDENT's is at most a+b+c-1. There are at most (4-a).2 arrows from the 4-a dissatisfied EX-STUDENT's in the corners, at most

(32-b).3 from the 32-b dissatisfied EX-STUDENT's along the sides, and at most (32-b).3 from those (64-c).4 sitting inside. The total number of arrows is thus 180 <=(a+b+c-1)+(4-a).2+(32-b).3+(64-c).4, that is, a+2b+3c <=179. The one with the lowest salary out of the 36 EX-STUDENT's around the circumference is necessarily dissatisfied, thus a+b<=35. It is also obvious that a <=4. By adding the inequalities, we have 3(a+b+c)=(a+2b+3c)+(a+b)+a <=179+35+4=218, that is, a+b+c <=72. Hence, the number of satisfied EX-STUDENT's cannot be greater than 72. The diagram shows the case when there are exactly 72 EX-STUDENT's who are content with their salaries. => option (a) is the correct answer.

81. Consider two triangles PRS and PQS.Let two circles circumscribe these two triangles and their radii be 25 and 25/2.Also,PQRS is a rhombus whose area is equal to A. Then the value of A is (a) 200

(b) 125

(c) 375

(d) 400

(e) 325

Let O be the point of intersection of diagonals PR and QS,and T be the point of intersection of PR and the circumcircle of triangle PQS.Extend SQ to meet the circumcircle of triangle PRS at U. PO.OT=QO.OS and SO.OU=PO.OR Let PR=2a and QS=2b.Because PT is a diameter of the circumcircle of triangle PQS, and SU is a diameter of the circumcircle of triangle PRS, the above equalities can be rewritten as a(25-a)=b^2 and b(50-b)=a^2 Hence, a=2b.It follows that 50b=5b^2=>b=10,a=20.Thus area of PQRS=1/2 .PR.QS=2ab=400 => option (d) is the correct answer.

82. The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...up to infinity is (a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these Each term in the summation is n/(n^4 + n^2 + n). Had each term be n/(n^4 + n^2 + 1), then the summation would become 1/2(1 - 1/3 + 1/3 - 1/7 + .... up to infinity) = 1/2. Note that n^4 + n^2 + 1 can be factored as (n^2+1)^2 - n^2 = (n^2+n+1)*(n^2-n+1). n/(n^4 + n^2 + n) < n/(n^4 + n^2 + 1) for all n > 1. Thus, our answer is < 1/2. All options among the first 4 are >=1/2. => option (e) is the correct answer. 83. The value of 1/(1+1^2+1^4)+ 2/(2+2^2+2^4) + 3/(3+3^2+3^4) + ...up to infinity is (a) 1/2 (b) 2/3 (c) 1 (d) 3/2 (e) None of these

Each term in the summation is n/(n^4 + n^2 + n). Had each term be n/(n^4 + n^2 + 1), then the summation would become 1/2(1 - 1/3 + 1/3 - 1/7 + .... up to infinity) = 1/2. Note that n^4 + n^2 + 1 can be factored as (n^2+1)^2 - n^2 = (n^2+n+1)*(n^2-n+1). n/(n^4 + n^2 + n) < n/(n^4 + n^2 + 1) for all n > 1. Thus, our answer is < 1/2. All options among the first 4 are >=1/2. => option (e) is the correct answer. 84. The number of permutations of the set {1, 2, 3, 4} in which no two adjacent positions are filled by consecutive integers (increasing order) (a) is a prime (b) is a composite number divisible by 3 (c) does not exceed 17 (d) is at most 19 (e) None of the foregoing The permutations where there are consecutive numbers are 1234, 1243, 1342, 1423, 2134, 2341, 2314, 3124, 3412, 3421, 4123, 4312, 4231 => option (a) is the correct answer.

85. When Divya is one year younger than Twinkle will be when Divya is half as old as Twinkle will be when Divya is twice as old as Twinkle is now, Twinkle will be three times as old as Divya was when Twinkle was as old as Divya is now. One of them is in her teens and the ages are natural numbers. What is the sum of the ages? (a) 42

(b) 44

(c) 48

(d) 52

(e) 54

DIVYA TWINKLE w-1 3z v w 2y 2v X(present ages) Y (present ages) Z (some time in past) x (some time in past) Here 2y-x = 2v-y, v = (3y-x)/2 Similarly, w = (5y-3x)/2 ; z = 2x-y or 3z = 6x-3y and w-1 = (5y-3x-2)/2 Now,{(5y-3x-2)/2} – x = (6x-3y) – y =>13y = 17x+2 Here only at x = 19 and y =25 satisfies for one is a teenager. => option (b) is the correct answer.

86. Let there be Z number of rational triplets (g, h, i) for which g, h, i are the roots of x^3 + gx^2 + hx + i = 0. Then which of the following is the value of Z ? (a)0 (b)1 (c)2 (d)3 (e) Cannot be determined We require (1) g + h + i = -g, (2) gh + hi + ig = h, and (3) ghi = -i. From (3), either I = 0, or gh = -1. If i = 0, then (1) becomes h = -2g, and (2) becomes h(g - 1) = 0. Hence either g = h = 0, or g = 1, h = -2. So assume i ≠ 0, and gh = -1. (1) becomes i = - h – 2g. Substituting in (2), we get: -1 - (2g+ h)(g + h) = h, so -g^2 –2g^4 + 3g^2 - 1 = -g, or 2g^4 – 2g^2 - g + 1 = 0. So g= 1, or 2g^3 + 2g^2 - 1 = 0 …. The first possibility gives g = 1, h = -1, i = -1. Suppose g = m/n is a root of 2g^3 + 2g^2 - 1 = 0 with m, n relatively prime integers. Then 2m^3 + 2m^2n - n^3 = 0. So any prime factor of n must divide 2 and any prime factor of m must divide 1. Hence the only possibilities are g = 1, -1, 1/2, 1/2, and we easily check that these are not solutions. So 2g^3 + 2g^2 - 1 = 0 has no rational roots. => option (c) is the correct answer. 87. Let x, y, z be prime numbers in the arithmetic progression such that x > y > z. Which among the following is always true? (a) (b) (c) (d) (e)

x – z is divisible by 12 x + y is divisible by 8 xz - 1 is divisible by 7 atleast two of the foregoing none of the foregoing

Take x = 7, y = 5 and z = 3 rules out first 4 options. However, if z were > 3 then x - z is always divisible by 12. Check yourself for that. => option (e) is the correct answer.

88. What is the minimum value of the expression 6x^2+3y^2-4xy-8x+6y+2? (a) -13/7

(b) -2

(c) -8/5

(d) -7/3

(e) none of the foregoing

The expression 6x^2+3y^2-4xy-8x+6y+2 can be written as (14x^2 -12x - 3)/3 + 3(y - (4x-6)/6)^2 To minimize (y - (4x-6)/6) = 0 and 14x^2 -12x - 3 should be minimum. 14x^2 -12x - 3 is minimum for x = 3/7 => y = -5/7. Alternatively: Differentiate F(x, y) = 6x^2+3y^2-4xy-8x+6y+2 = 0 w.r.t to x 1st and then y. Solve the 2 equations you get and the values of x and y give you the answer. => option (a) is the correct answer.

89. Allwin likes to talk only in integer numbers. So much that he rounds off everything including his course average points to the nearest integer. For example,

89.34 is 89 and 99.54 is 100, and 115.5 is 116. Allwin always calculates the average (real) on the cumulative points so far. After his 75 points in Finance, his rounded average drops by 1 point. Next, after 83 points in strategy paper, his rounded average further plummets down by 2 points. Which among the following is not true? (a) The minimum possible number of courses is less than 15 (b) The maximum possible number of courses is not more than 50 (c) The minimum possible current rounded average is 95 (d) Either of 126 or 127 can be the current rounded average (e) none of the foregoing An + 75 = (A - d1)*(n+1); An + 75 + 83 = (A - d1 - d2)*(n+2) -> A is the current average. Let's simplify the expression above: 75 + d1 = A - nd1 and 79 + (d1 + d2) = A - n/2(d1 + d2) => n*(d1 - d2) = 8 + 2d2 => (n*d1 – 8)/*(n+2) = d2 The fact to be noted here is d1 > d2 as adding 75 will bring down the average more than adding 83. Let's look at the range of both d1 and d2 now. d2 will lie in [1+x, 3-y] where both x and y -> 0 => x + y > 2 => d1 will lie in [1+a, 2-b] where both a and b -> 0 => a < x, 1 > y - b, a+b > For n = 10, we have (5d1 - 4)/6 = d2, for extreme values of d1 < 2 we have d2 < 1 => n > 10. For n = 11, we have (11d1 - 8)/13 = d2; but d2 > 1 => d1 > 21/11. After this we require to do some quick calculations. See yourself that An = 1083 satisfies our conditions where d1 is approx. 1.95 Hence, choice (a) as well as choice (c) is correct. The expression in pink tells us we can have n > 50 and d1 > d2. Hence, choice (b) is false. No need to check further for choice (d) as we have our answer already. For completion sake we can verify that current round average 127 is obtained when to start with there are 3653 points from 28 course current round average 126 is obtained when to start with there are 4521 points from 35 courses Hence, option (b) is the right answer L 90. Let P(x) = nx + a where n and a are integers with n > 0. If the solution to 2^ P(x) = 5 is x = log10/log8, then 4n-3a is (a) 11 (b) 12

(c) 13

(d) 14

(e) none of the foregoing

Taking log on both the sides of 2^P(x) = 5 we get P(x) = log5/log2. x = log10/log8 = 1/3(1 + log5/log2) => n = -3 and a = 1. Hence, option (e) is the right answer

L91. For each positive integer p, let S(p) denotes the sum of the digits of p. For how many values of p is p+ S(p) + S(S(p)) = 2007? (a) 1

(b) 2

(c) 3

(d) 4

(e) none of the foregoing

Four numbers satisfy the given condition: 1977, 1980, 1983, 2001. numbers must be of the form of 19ab or 2001 (obvious) case 1: a+b<10

above equation is 1900+10a+b+(10+a+b)+(a+b+1)=2007 12a+3b=96 solution is a=8;b=0 1980 case 2: a+b>10 equation is 1900+10a+b+(10+a+b)+(a+b+2-10)=2007 that is 12a+3b=105 solutions are a=8;b=3 1983 a=7;b=7. 1977 Hence, option (d) is the right answer --------------------------------------------------------------------------------------------------------------------------------------L92. For each positive integer p, let S(p) denotes the sum of the digits of p. For how many values of p is p+ S(p) + S(S(p)) = 2007? (a) 1

(b) 2

(c) 3

(d) 4

(e) none of the foregoing

Four numbers satisfy the given condition: 1977, 1980, 1983, 2001. numbers must be of the form of 19ab or 2001 (obvious) case 1: a+b<10 above equation is 1900+10a+b+(10+a+b)+(a+b+1)=2007 12a+3b=96 solution is a=8;b=0 1980 case 2: a+b>10 equation is 1900+10a+b+(10+a+b)+(a+b+2-10)=2007 that is 12a+3b=105 solutions are a=8;b=3 1983 a=7;b=7. 1977 Hence, option (d) is the right answer

L93. Let 3 statements be made (P) 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10 (Q) The remainder on dividing 16! + 89 by 323 is q (R) 46C23 leaves remainder r on division by 23 Then p+q+r equals (a) 12

(b) 19

(c) 26

(d) 33

(e) none of the foregoing

The notation a % b = c means that when a is divided by b, c is the remainder that is obtained. (P) Check for p = 9; 10^12 % 13 = 1 by Fermat's theorem => 10^18 % 13 = 10^6, since 10^3 % 13 = -1 + 13 => 10^18 % 13 = 1 => 10^18 - 10^9 + 1 leaves remainder (1-(-1)+1=3) by p = 9; Check yourself the remainder when p = 8. When p = 7, as 10^12 % 13 = 1 => 10^14 % 13 = 9; also 10^6 % 13 = 1 => 10^7 % 13 = 10. Thus p = 7 satisfies our condition.

(Q) 323 = 17*19. (p-1)! + 1 is divisible by p when p is prime => 16! % 17 = -1 + 17. Also 18! % 19 = 1 + 19 If 16! % 19 = x => 18! % 19 = 17*18x => 306x % 19 = 18 => 2x % 19 = 18. Thus, 16! leaves the remainder 16 by 17 and 9 by 19 => It leaves 237 by 323 => q = 3 (R) 2nCn = (nC0)^2 + (nC1)^2 + ... + (nCn)^2; when n is prime, each of the term in RHS except 1st and last is divisible by n => r = 2 Hence, option (a) is the right answer

L94. In a sports tournament, the number of European teams is 9 more than the number of Asian teams. Each team plays the other exactly once, the winner gets 1 point and the loser gets nothing. There were no draws during the tournament. If the total points earned by European teams in 9 times the total points earned by the Asian teams. What is the maximum possible score of the best Asian team? (a) 9

(b) 11

(c) 21

(d) not be uniquely determined

(e) none of the foregoing

Let the no. of Asian teams be x. Therefore no. of European teams will be x+9. Matches can be divided into three groups: 1) Those played b/w two Asians 2) Those played b/w two Europeans 3) Those played b/w an Asian and a European In the 1st group no. of points won (by Asians) = x(x-1)/2 In the 2nd group no. of points won (by Europeans) = (x+9)( x+8 )/2 Let k be the points Asians gain in the 3rd group. Therefore those won by Europeans = x(x+9) - k Using the given conditions we have 9*[x(x-1)/2 + k] = (x+9)( x+8 )/2 + x(x+9) - k This gives 3x^2 - 22x + 10k - 36 = 0 => Factorize it to get the following: (3x+5)(x-9) = -10k - 9 Since RHS is negative, LHS too should be negative. Hence, x lies between 0 and 9. On checking, we find that the only integer solutions of the above equation occur when either x=8 and k=2 or when x=6 and k=6. The maximum no. of matches an Asian team can win if x = 6 and k = 6 is 5 (won from other 5 Asian teams in 1st Group) + 6 (If the same Asian team is the one which wins all the Asian-European matches in 3rd Group) = 11 and if x = 8 and k = 2 it is 7 + 2 = 9 So the maximum no. of points an Asian team can have is 11. Hence, option (b) is the right answer L95. Given p <= 4 is a positive real . Let A be the area of the bounded region enclosed by the curves y = 1 - |1-x| and y = |2x-p|. Then which among the following best describes A? (a) 0 < A <= 1/2 <= 1/3

(b) 1/6 <= A <= 1/4

(c) 0 < A <= 1/3

(d) 0 < A < 2/3

(e) 1/6 <= A

Case 1: 0 < p <= 1 The area formed is of a triangle with vertices (p/3, p/3), (p/2, 0) and (p, p). Thus, the area is p^2/6 sq. units. Case 2: 1 <= p <= 3 The figures formed is a quadrilateral (p/3 ,p/3),(p/2,0), ((p+2)/3 , (4-p)/3) and (1,1). Thus, the area formed is 1/3 - 1/6*(p-2)^2 Case 3: 3 <= p <= 4

The area formed is the image of case 1. Area = (4-p)^2/6 Thus, min(A) = 0 at p = 4 and max(A) = 1/3 at p = 2 Hence, option (c) is the right answer L96.If x, y, z are natural numbers such that y ≠ 1 and (x-y√3)/(y-z√3) is a rational number. Then which among the following is always true? (a) x^2 + 4z^2 is composite prime (d) both (a) and (b)

(b) x^2 + y^2 + z^2 is non-prime (e) none of the foregoing

(c) x^2 + 4z^2 is

(x-y√3)/(y-z√3) = (x-y√3)*(y+z√3)/(y^2 - 3z^2) = (xy +√3(xz - y^2) -3yz)/(y^2 - 3z^2) is rational => y^2 = xz. Let y = xr, z = xr^2 => x^2 + 4z^2 = x^2*(1+4r^4) = x^2*(2r^2 + 2r+1)(2r^2-2r+1). If x = 1 then r ≠ 1; Thus, x^2 + 4z^2 can be written as product of 2 natural numbers > 1 which are x(2r^2 + 2r+1) and x(2r^2 - 2r + 1). Similarly, x^2 + y^2 + z^2 = x(r^2 + r + 1)*x(r^2 - r + 1) and can be written as product of 2 natural numbers > 1. Hence, option (d) is the right answer L97. Let three 3-digit integers abc, bca, cab when divided respectively by 10, 8 and 3 leave the same remainder 1. If 2 < b < 8, then a = (Part B) A four digit number is formed by using 5, 6, 8 and 9 (each number being used once). How many of them will be divisible by 7? (a) 3

(b) 4

(c) 5

(d) can not be determined

(e) none of the foregoing

(Part A) abc leaves remainder of 1 when divided by 10 => c = 1. bca leaves remainder of 1 on dividing by 8 => 4b+2c+a = 8m+1 and similarly a+b+c = 3n+1. On solving we get 3b = 8m -1 - 3n => m = 2, 5. Also 4b+a = 39 means b > 7 => m = 5 rejected => b = 5-n => a+b = 3n => a = 4n-3 but b > 2 => n < 3. But 0 < a < 10 => n = 2 and hence a = 3. Hence, choice (a) is the right option (Part B) The concept of seeds can again be applied to this problem. The seed of 7 is 5 => if abcd is div by 7 then a+5b+25c+125d is div by 7 => (a-d) + 2(2c-b) is div by 7. (a-d) + 2(2c-b) can take 0, 7, 14, 21. If (a-d) + 2(2c-b) were 21, we have just 1 solution i.e. c = 9, b = 8, a = 6, d = 5. If (a-d) + 2(2c-b) were 0, we have just 1 solution i.e. c = 5, b = 9, a = 6, d = 8. if (a-d) + 2(2c-b) were 7 => 2c-b=3 && a-d = 1 OR 2c-b=2 && a-d=3. It is easy to see one solution in each case. Similarly, we will have 2 solutions when (a-d) + 2(2c-b) = 14. Check yourself! Thus, in all 6 possible such numbers. Hence, choice (e) is the right option

L98. (Part A) Three articles are sold at the profit of 20%, 25% and 40% and the net profit comes out to be 30%. Had the article sold at 25% been sold at 15% profit and the article sold at 40% been sold at its cost price then the profit on the sale of three articles would have been (Part B) Let the cost price of an article A is as much more than another article B as much that of B is more than that of another article C whose cost price is more than that of fourth article D by the same amount. If A, B, C, D are sold at 10%, 15%, 30% and 5% profit respectively, then the net profit percent on the sale of four articles together is (a) 5% (b) 10% foregoing

(c) 15%

(d) can not be determined

(e) none of the

(Part A) From 1st condition, C3 = C1 + 0.5C2 => new profit = (0.2C1+0.15C2)*100/(2C1+1.5C2) = 10% Hence, choice (b) is the right option (Part B) If Cd is the Cp of D and x be the common difference in CPs, then net profit = (60Cd+90x)/(4Cd+6x) = 15% Hence, choice (c) is the right option L99. (Part A) Two triangles of sides (2,2,a) and (3,3,b) are inscribed in a circle. If a^4+36b^2 = b^4 +16a^2, then what is the ratio of a/b? (Part B) RST is a triangle and U, V, W are the feet of perpendiculars dropped from R, S, T on sides ST, TR and RS respectively. If SU=2, SW=3 and TV=4, then UV is (a) 9/4

(b) 7/3

(c) 7/2

(d) 3/2

(e) None of the foregoing

(Part A solution ) a^4+36b^2 = b^4 +16a^2 => a^2*(a^2-16) = b^2*(b^2-36) => a^2/b^2 = (36-b^2)/(16a^2)=>a/b = root((36-b^2)/(16-a^2))-------------------(A) Also , area of the triangle with side (2,2,a) = (a*2*2)/(4*R) = (a/4) * root(4*2^2 - a^2) {Where R is the circumradius} => R = 4/root(16 - a^2)------------(1) Also from the second triangle with sides(3,3,b) we get R = 9/root(36-b^2)------------------(2) Equating (1) and (2) we get,(36-b^2)/(16-a^2)= 81/16 Using (A) we have a/b =9/4 Hence, choice (a) is the right option (Part B) Triangles TVU & TSR are similar, 4/(TU+2)=UV/(3+RW) Triangles SUW & SRT are similar, 2/(3+RW) = 3/(2+TU)

Hence UV=8/3 Hence, choice (e) is the right option L100. (Part A) Consider the series S=1+ 1/2^2 + 1/3^2 + 1/4^2 +... +1/n^2.The value of S for any n is less than which of the following? (Part B) If x,y,z are positive real numbers representing the sides of a triangle,then (x^2+y^2+z^2)/(xy+yz+xz) is necessarily less than which of the following? (a) 1.80

(b) 2.0

(c) 1.85

(d) 1.95

(e) None of the foregoing

(Part A) The roots of sinx/x are n*pi,viz +/- pi.+/-2pi and so on. Hence, sinx/x = (1-x/pi) (1+x/pi) (1-x/2pi) (1+x/2pi) (1-x/3pi) (1+x/3pi)... Coeff of x^2 here= -(1/pi^2 + 1/4pi^2 + 1/9pi^2 + ...) But we also know that, sinx = x-x^3/3! +... Here coeff of x^3 is -1/6 Hence, -(1/pi^2 + 1/4pi^2 + 1/9pi^2 + ...) =-1/6 =>1+1/4+1/9+... = (pi^2)/6 Hence, all the choices from choice (a) to choice (d) are the right option (Part B) As x,y and z are sides of the triangle hence, x-y < z ----------------(1) z-x < y ----------------(2) y-z < x ----------------(3) squaring and adding all this three equation we get: (x-y)^2 + (z-x)^2 + (y-z)^2 < x^2 + y^2 + z^2 => 2*(x^2 + y^2 + z^2) - 2*(xy + yz + zx) < x^2 + y^2 + z^2 => x^2 + y^2 + z^2 < 2*(xy + yz + zx) =>(x^2+y^2+z^2)/(xy+yz+xz) < 2 Hence, choice (b) is the right option L101. (Part A) An insect starts at vertex A of a certain cube and is trying to reach at vertex B, which is opposite A, in 5 or fewer steps where a step consists travelling along an edge from one vertex to another. The insect will stop as soon as it reaches B. The number of ways in which the insect can achieve its objective is (Part B) A committee of 5 is to be chosen from a group of 9 people. How many ways it can be done if Vikram and kaizen must serve together or none at all, and Sunit and Pratyush refuse to serve with each other? (a) 40

(b) 36

(c) 41

(d) 30

(e) 48

L102. (Part A) Let f(x+1, y+1) = f(x, y) for all integers x and y. If f(x, 0) = x then f(0, 4) is (Part B) Let f(x+1, y) = 1 + f(y, x) for all real x and y. Then, f(x+2, y+2) - f(x, y) = (a) -4

(b) 0

(c) 4

(d) 2

(e) none of the foregoing

(Part A) f(0, 4) = f(-1, 3) = f(-2, 2) = f(-3, 1) = f(-4, 0) = -4 Hence, choice (a) is the right option (Part B) f(x+2,y+2) - f(x,y) = 1+f(y+2,x+1) - f(x,y) = 2+f(x+1,y+1) - f(x,y) = 3+f(y+1,x) - f(x,y) = 4+f(x,y) - f(x,y) = 4 Hence, choice (c) is the right option

L103. (Part A) The painting of a hotel can be done by 10 painters in 5 days. But, in reality there were few dabblers in the group of 10 painters and therefore the work took a little over 6 days for completion. A painter worked twice as fast as the dabbler and used 25% less paint than a dabbler used. If 5 litres extra paint was used than expected, then the total paint used in painting the hotel was (in litres) (Part B) A, B and C can do a work together in a certain number of days. If A leaves after half the work is done, then the work takes 4 more days for completion, but if B leaves after half the work is done, the work takes 5 more days for completion. If A takes 10 more days than B to do the work alone, then in how many days can C alone do the work? (a) 45 (b) 52.5 (c) 60 (d) 65 (e) none of the foregoing (Part A) Let the number of painters be x, then dabblers is 10-x Now x/50 + (10-x)/ 100 = slightly less than 1/6 . (2x+10-x)/100 = slightly less than 1/6 x+10 = slightly less than 100/6 x= slightly less than 6.67 Since x has to be integer nearest integer is 6. That means the number of dabblers = 4. Let p be the amount of paint used by each dabbler, so by each painter = .75p Extra paint used = .25p * 4 = 5 litres So p = 5 litres.

Dabblers will be using 5*4 = 20 litres of paint. Now as the painters are working at twice the efficiency so they will use twice the amount of 75% of paint used by dabbler in the same time. i.e. paint used by each painter = 2*(.75)*5 = 7.5 litres. Hence paint used by 6 painters = 7.5*6 = 45 litres So total paints used = 45 + 20 = 65litres Hence, choice (d) is the right option (Part B) Let A alone can do the work in a days, B alone in b days and C alone in c days. Let A, B and C together complete the work in x days. So, 1/a+1/b+1/c = 1/x -----------(1) Then half of the work they will complete together in x/2 days. When A leaves, Rest half of the work is done by B and C together, (2/b+2/c) = 1/{(x/2)+4} Using (1) , 2*(1/x-1/a) = 1/{(x/2)+4} Solving we get, a = [x*{(x/2)+4}]/4----------------(2) On similar lines b = [x*{(x/2)+5}]/5 ------------------(3) Given that a = b+10. [x*{(x/2)+4}]/4 = [x*{(x/2)+5}]/5 + 10 Solving we get x = 20. From (2) and (3) we get a = 70 and b = 60 Put these values in (1) we get c = 52.5 days. Hence, choice (b) is the right option L104. (Part A) A regular hexagon ABCDEF is inscribed in a circle with radius r. Another regular hexagon is formed (lying fully inside the circle with radius r) with 2 of its vertices internally on AB and another 2 on the circle. What is the ratio of the sides of the smaller hexagon to the larger hexagon? (Part B) In circle O, chords AB and AC are drawn so that AB = AC. Chord PQ is drawn intersecting AB and AC at points D and E respectively. D lies internally between E and P. Given AD = EC = 1/3 and DB = AE = 1/4, what is the maximum value of |PD QE|? (a) 1/18

(b) 1/12

(c) 1/13

(d) 1/12

(e) none of the foregoing

(Part A) Please refer sunit's post #122 here http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007quantitative-questions-day.html Hence, choice (c) is the right option (Part B) By chord theorem; PD*QD = AD * BD = (1/3)*(1/4) = 1/12 ----------------(1) QE*PE = AE * EC = (1/4)*(1/3) = 1/12 ----------------(2) from (1) and (2) we get, PD*QD = QE*PE => PD*(QE+ED) = QE*(PD+ED) => PD*QE + PD*ED = QE*PD + QE*ED

=> (PD - QE) * ED = 0 when ED not equal to zero |PD - QE| = 0 when ED = 0 ; |PD - QE| will take its maximum value. If ED = 0 then D and E coincide which is not possible as it is given that AD = 1/3 and AE = 1/4 so D and E cannot coincide. so ED cannot be equal to zero hence only possible value of |PD-QE| is 0 . hence maxm. value of |PD-QE| = 0. Hence, choice (d) is the right option L105. (Part A) How many integral values of p are there for which the inequality 3 - |x-p| > x^2 is satisfied by at least one negative x? (Part B) How many integral values of p are there for which the inequality x^2 + px + p^2 + 6p < 0 is satisfied for all x in (1, 2)? (a) 3

(b) 4

(c) 5

(d) 6

(e) none of the foregoing

(Part A) If x > p then 3-x+p > x^2 => p > -13/4. If x < p then 3-p+x > x^2 => p < 3. Thus, p lies in (-13/4, 3) Hence, choice (d) is the right option (Part B) If f(x) = x^2 + px + p^2 + 6p then f(1) < 0 && f(2) < 0 => p lies in ((-7-√45)/2, -4+2√3)) Hence, choice (d) is the right option

L106. (Part A) A tower stands on a horizontal plane. From a mound 14 m above the plane and at a horizontal distance of 48 m from the tower, an observer notices a loophole and finds that the portions of the tower above and below the loophole subtend equal angles. If the height of the loophole is 30 m, then the height of the tower is (Part B) Two straight roads OA and OB intersect at O. A tower is situated within the angle formed by them and subtends angles of 45˚ and 30˚ at the points A and B where the roads are nearest to it. If OA = 81 m and OB = 17 m, then what is the height of the tower? (a) 56

(b) 64

(c) 78

(d) 92

(e) data insufficient

(Part A) Please refer post #148 here http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007quantitative-questions-day.html Hence, choice (c) is the right option

(Part B) Please refer post #145 here http://www.pagalguy.com/forum/quantitative-questions-and-answers/23791-cat-2007quantitative-questions-day.html Hence, choice (a) is the right option

L107. (Part A) Divya works for Google that claims only 4 working days. In the 1st week of joining, she went for movies after 4 working days. On Friday, she noticed that the fraction p of the line is in front of her, while 1/q of the line was behind her. On Saturday, the same fraction p of the line was in front of her, while 1/(q+1) of the line was behind her. On Sunday, the same fraction p of the line was in front of her, while 1/(q+2) of the line was behind her. For how many values of p is this possible? (a) 3

(b) 1

(c) 2

(d) no such value

(e) infinitely many

(Part B) What is the sum of the reciprocals of all the numbers in the form (2^a)*(3^b), where a and b are non-negative integers? (a) 3

(b) 3.5

(c) 2√3

(d) √3 + 1

(e) none of the foregoing

(Part A) Let p = (x-1)/x for some natural number x. Let q = 1/(x+y), Divya herself represents the line => 1 ((x-1)/x + 1/(x+y)) = y/x(x+p) => x(x+p)/y is the length of the line. To make it an integer we can make x to be a multiple of 6. For q = x+1, we want to get an integer for y=1, 2, 3. More generally, for x(x+1), x(x+2)/2, and x(x+3)/3 to be integers we require x to be a multiple of 6. On Friday, there are 36k^2+6k people in the queue, on saturday 18k^2+6k and on Sunday 12k^2+6k. Hence, choice (e) is the right option (Part B) Let the asked sum be x => 1+(1/2+1/4+1/8+...)+(1/3+1/9+1/27+...)+1/6(x) = x => 1+1+1/2+1/6(x) = x => x = 3 Hence, choice (a) is the right option The question was that was 109TH Last Year --------------------------------------------What is the greatest integer n for which there exists a simultaneous solution x to the inequalities k < x^k < k+1, k = 1, 2, 3, ..., n? (a) 2 (b) 3 (c) 4 (d) 6 (e) none of the foregoing 1 < x < 2, 2 < x^2 < 3 is satisfied by all x in (√2, √3), if we have one more inequality with the former two i.e. 3 < x^3 < 4 is satisfied by all x in (3^1/3, 4^1/3). If we also have 4 < x^4 < 5; then 4 simultaneous are satisfied for x in (3^1/3, √√5). If we also have 5 < x^5 < 6, there will be no x satisfying 5 equations as 3^1/3 > 6^1/5 (243 > 216). Hence, choice (c) is the right option

The question was that was 110TH Last Year --------------------------------------------If three unequal positive real numbers p, q, r are in GP and q-r, r-p, p-q are in HP, then the value of p+q+r is independent of (a) p (b) q (c) r (d) all of the foregoing (e) none of the foregoing since p,q,r are in GP => q^2 = p*r also since q-r,r-p and p-q are in HP So 1/q-r,1/r-p and 1/p-q are in AP => 2(q-r)(p-q) = - (r-p)^2 => p^2 + r^2 - 2q^2 +2 pq +2qr-4pr=0 => p+q+r = -3q => data is inconsistent with the given information Hence, choice (e) is the right option

The question was that was 111TH Last Year --------------------------------------------(Part A) A candy company makes 5 colours of jelly-beans which comes in equal proportions. If Divya grabs a random sample of 5 jellybeans, what is the probability that she gets exactly 2 distinct colours?

(Part B) A bag contains 5 jelly-beans and it is not known how many of them are blue in colour. If 2 jelly-beans are drawn and found to be blue, then the probability that all the jelly-beans in the bag are blue is (a) 2/5 (b) 20/63 (c) 8/625 (d) 1/2 (e) 2/21 (Part A) Let there be A,B,C,D,E number of candies with five colours So total ways is number of solutions of the equation A+B+C+D+E = 5 which is 9C4 = 126 when the candies are of exactly two different colors than Total number of ways = Number of ways of selecting two colors out of five x ( nunmber of solutions of equation X+Y = 5 , where X, Y are natural numbers ) = 5C2 x 4C1 = 40 Hence Probability = 40/126= 20/63 Hence, choice (b) is the right option (Part B) This is a case of conditional probability:

Required probability = (probability that 2 jelly beans drawn being blue when all are blue)/((probability that 2 jelly beans drawn being blue when all are blue) + (probability that 2 jelly beans drawn being blue when 4 are blue) + (probability that 2 jelly beans drawn being blue when 3 are blue) + (probability that 2 jelly beans drawn being blue when 2 are blue)) = 1/(1+ 4/5.3/4 + 3/5.2/4 + 2/5.1/4) = 1/2 Hence, choice (d) is the right option The question was that was 112TH Last Year --------------------------------------------(Part A) XYZ is a triangle with <X = 60 and incenter K. E lies on YZ with 3 YE = YZ. The point J lies on XY and KJ is parallel to XZ. Then the ratio of
(Part B) T and B are points on the sides LM and GM respectively of triangle LMG such that LT = k, TM = 6, MB = 20, BG = k, GL = 25. If triangle TMB and quadrilateral LTBG have equal areas, what is k? (a) 1

(b) 2

(c) 3

(d) 4

(e) None of the foregoing

(Part A) Consider XYZ to be an equilateral triangle. produce JK to meet YZ at L(say). then YE=EL=LZ now ang.YJL=ang.YXZ=60 deg. so triangle JYL is equilateral and JE is median. so ang.YJE=1/2ang.YJL=30 deg. also TK being angle bisector of ang.XYZ,ang.JYK=30 deg. so reqd ratio is 1. Hence, choice (a) is the right option (Part B) Let 120/((20+k)(6+k)) = 1/2 Ö (k+30)(k-4) = 0 => k = 4 Hence, choice (d) is the right option The question was that was 113TH Last Year

--------------------------------------------Let's consider all irreducible fractions below 1, of type p/q where q =3007 and p/q is positive .If S denotes the sum of all such fractions denoted by p/q ,the the value of S is best represented by (a) 1001 (b) 2002 (c) 1234 (d) 1405 (e) None of the foregoing A fraction a/b is called irreducible iff a and b are coprime--GCD(a,b)=1 3007 has prime factors 31 and 97 3007=31x97 so in series sum where we obtains multiple of 31 and 97--that nos we should not take sum of all nos (1+2+3+...+3006)/3007=1503 sum of multiples of 31--->(31+62+...+2976)/3007=48 sum of multiples of 97---->(97+194+...+2910)/3007=15 ans. is==1503-(48+15)=1503-63=1440 Hence, choice (e) is the right option The question was that was 114TH Last Year --------------------------------------------If we choose 4 people randomly,find the probability (in %) that none of their birthdays fall on the same day of the week (a) 23

(b) 35

(c) 38

(d) 27

(e) None of the foregoing

We can choose 4 different days out of 7 days in 7C4 ways. Now these 4 days can be given to 4 people in 4! ways. So possible ways are = 4! * 7C4 Total no of ways is 7*7*7*7(all 4 can have any day as B'day) = 7^4 Probability Percentage is [(4! * 7C4)/7^4]*100 = 35 Hence, choice (b) is the right option The question was that was 115TH Last Year --------------------------------------------Pratyush sold few articles each for the same amount either making 10% profit or 10% loss on them. (PART A) If he breaks even on the total sale,then the least number of articles sold by Pratyush is (a) 18 (b) 19 (c)20 (d) 21 (e) None of the foregoing (Part B) If he makes 5% profit on the total sale,then the least number of articles sold by Pratyush is (a) 14 (b) 15 (c) 16 (d) 17 (e) None of the foregoing (Part A)

Let CP of the first type of article be C1 and that of the other type be C2. SP is same for both that is S. let total number of articles be n of which x is of type C1 and (n-x) of type C2. C1 = (10/9)*S and C2 = (10/11)*S Then according to question: x*C1 + (n-x)*C2 = n*S x*(10/9)*S + (n-x)*(10/11)*S = n*S=> (n/x) = (20/9) So,the minimum possible integer value of n is 20. Hence, choice (c) is the right option (Part B) (21/20)*[x*C1 + (n-x)*C2 ] = n*S (n/x) = 14/3 So, the minimum possible integer value of n is 14. Hence, choice (a) is the right option The question was that was 116TH Last Year --------------------------------------------Arvind and Swarnali are team mates and they can do a project in certain number of days.If Arvind is on holiday for x days,then they take y more days to complete the work while if Swarnali is on holiday for x days,then they take z more days to complete the same project.Then

(a)y, x/2,z are in AP (b) y, x/2,z are in GP determined (e) None of the foregoing

(c)1/x=1/y+1/z

(d) Cannot be

Let Aravind complete 1/a portion of the work in a day and Swarnali 1/b portion of the work in a day Aravind on holiday for x days implies loss of (1/a)*x fraction of the work This loss is compensated by arvind and swarnali in the coming extra y days so ((1/a)+(1/b))*y =(1/a)*x Similarly, for Swarnali's holiday of x days ((1/a)+(1/b))*z= (1/b)*x Adding the two equations ((1/a)+(1/b))*(y+z)=x*((1/a)+(1/b)) implying y+z=x x/2=(y+z)/2 Hence y,x/2,z are in AP Hence, choice (a) is the right option The question was that was 117TH Last Year --------------------------------------------GOI directive proposed formation of more new IIMs. Fearing lack of good faculty and geographical distances as disadvantageous, IIMs decide to have one for all campus. The School of Planning and Architecture proposed 2 new projects for the new Housing campus. In each project, the campus is designed to have several identical dorimiatory buildings, with the same number of 1 bedroom apartment in each building. In the first project there are 12096 apartments in total. There are 8 more buildings in the second project than the the first and each building has more

apartments which raises the total of apartments in the project to 23625. Let N be the number of buildings the second project require. Then N is divisible by (a) 5 (b) 7 (c)9 (d) at least two of the foregoing (e) none of the foregoing 12096 = (2^6)*(3^3)*7 and 23625 = (5^3)*(3^3)*7; well you don't have to be a genius to observe this; how numbers move in tandem is best demonstrated by this example. For 12096, the last 3 digits tell us it's div by 8, thus 12096/8 = 1512, again 512 is div by 8; for 23625 the last 3 digits tell it's div by 125. So divide! 12096/n < 23625/(n+8). Now, comes a bit of data manipulation. n must divide 12096 and n+8 should divide 23625. And n has to be odd => n = 3^3 and # of buildings 2nd project requires is 5*7 =3^3 + 8 = 35 Hence, choice (d) is the right option The question was that was 118TH Last Year --------------------------------------------(Part A) A rectangle of the greatest area is inscribed in a trapezium ABCD, one of whose nonparallel sides AB (8 cm long) is perpendicular to the base, so that one of the rectangle's sides lies on the larger base of the trapezium. The base of the trapezium are 6 and 10 cm long, respectively. What is the area of the rectangle in sq. cm? (Part B) A rectangle ABCD is inscribed in a circle with a diameter lying along the line 3y = x + 10. If A and B are the points (-6, 7) and (4, 7) respectively, then the area of the rectangle is (a) 40

(b) 48

(c)60

(d) 80

(e) none of the foregoing

(Part A) Let one of the sides of the rectangle be on AB and be of length x then other side which is on BC will be (10 - x/2). Area = x*(10 - x/2) This is maximum when x = 10 But x cannot be 10 as the side AB is of length 8 cm so maximum area is 8*6 = 48 cm^2. Hence, choice (b) is the right option (Part B) The centre O is equidistant from A and B. Thus if O = (a, (a+10)/3) then (a+6)^2 = (a-4)^2 => a = -1 and O = (-1, 3). From AB i.e y = 7 its distance is 4 units => BC = 8 units and area = 80 sq. units. Hence, choice (d) is the right option

The question was that was 119TH Last Year --------------------------------------------(PART A) The number of values of z for which the equation zx^2 + (z-7/2) x + 1/(16z)=0 has a unique solution is

(PART B) How many distinct positive integral solutions of the equation a+b+c+ab+bc+ac =89 are possible? (a) 2 (b)3 (c)5 (d) 6 (e) none of the foregoing (Part A) For the solution to be unique D = 0 => (z-7/2)^2 = 1/4 => z = 3, 4. Hence, choice (a) is the right option Part B) a+b+c+ab+bc+ca = 89 => (90-ab)/(a+b+1) = c+1 => 90-ab must divide a+b+1 => 90+a+a^2 must divide (a+b+1). Let's enumerate a bit now. For a = 1, we must have 92 div by b+2 => b = 2. For a = 2, we must have 96 div by b+3, but 96 = (2^5)*3 => b = 3, 5, 9, 13, 21 For a = 3 we must have 102 = 2.3.17 div by (4+b), thus no new solution For a = 4 we must ahve 110 = 2.5.11 div by (5+b) => b = 5, 6 For a = 5 we must have 120 div by (b+6) => b = 6, 9 For a = 6 we must have 132 = 2^2.3.11 div by (7+b), thus no new solution And we stop here. WHY? Thus, in all 10 unordered solutions. Hence, choice (e) is the right option The question was that was 120TH Last Year --------------------------------------------Each question is followed by two statements A and B. Answer each question using the following instructions Choose 1 if the question can be answered by A alone Choose 2 if the question can be answered by B alone Choose 3 if the question can be answered by A and B combined Choose 4 if the question can be answered by either A or B Choose 5 if the question can be answered by neither A and B

(PART I) A N-sided regular polygon was inscribed in a circle. The consecutive vertices of the polygon were numbered from 1 to N. If vertex number n lies on one end of the circle’s diameter, then the other end of the diameter lies on vertex number 3n+1. What is the value of N? A) 2N -4 = n2 B) 14 < N < 22

(PART II) A set S = {a, b, c, d} consists of prime numbers between 20 and 100, such that no two elements of S have any digit in common. What is the value of a+b+c+d? A) The difference of some pairs of elements in S is prime

B) If a > b > c > d, then the number of distinct ways in which the elements of S can be chosen is 14 (Part I) According to question As vertex number n and 3n +1 are opposite ends of the diameter hence, number of sides of the polygon is 2n+2n+2 = 4n + 2 = N or, N-2 = 4n =>2N-4 = 8n---------------(1) A) says that 2N-4 = n^2---------------(2) from (1) and (2) we have N=34 So question can be answered by A) B) 14 < N < 22 We have N = 4n+2, => N = 18 is the only possibility Hence, choice (4) is the right option (Part II) A) gives 2 set (23, 47, 59, 61) and (47, 59, 61, 83) B) is redundant Hence, choice (5) is the right option The question was that was 121st Last Year --------------------------------------------Part (A) Let X be the set of non-negative integers. Let functions f: X -> X such that x f(y) + y f(x) = (x + y) f(x^2 + y^2) for all x, y. Then which among the following is always true? (1) f(x) >= 0 none of these

(2) f(x) = f(1)

(3) f(0) = 0

(4) Atleast 2 of the foregoing

(5)

PART (B) For all positive integers n, let nf(n+1) = n^2 + f(n). If f(1) is a positive integer and f(n) is an integer only for 1 <= n <= 6, then the minimum value of f(1) is (1) 25 (2) 49 (3) 81 (4) 121 (5) none of these (Part A) Putting x = 0 we get y f(0) = y f(y^2), so f(y^2) = f(0) for all y. That strongly suggests f is constant Obviously any constant function satisfies the condition. Suppose f(x) < f(y) and neither x nor y is zero, then (x + y) f(x) < x f(y) + y f(x) < (x + y) f(y). Hence, f(x) < f(x^2 + y^2) < f(y). But that is impossible, because we could repeat the argument to get an infinite number of distinct values between f(x) and f(y). But we know that f(1) = f(0). Hence f(x) = f(1) for all x > 1. Thus, f is constant. Also since, X is non-negative and f: X->X, f(x) >= 0 Hence, option (1) and option (2) are always true. Hence, choice (4) is the right option (Part B) nf(n+1) = n^2 + f(n). put n = 1 we get, f(2) = 1 + f(1). put n =2 we get, f(3) = 2 + f(2)/2 = {5 + f(1)}/2

put n = 3 we get, f(4) = 3 + f(3) /3 = { 23 + f(1) } / 6 put n = 4 we get, f(5) = 4 + f(4) /4 = { 119 + f(1) } /24 put n = 5 we get , f(6) = 5 + f(5) / 5 = { 719 + f(1)}/ 120 put n = 6 we get , f(7) = 6 +f(6)/6 = { 5039 + f(1)}/720 f(n) is an integer only for 1 <= n <= 6 so f(7) need not be an integer so minimum such value can be 121. because for 1 every number will be an integer. Hence, choice (4) is the right option The question was that was 122nd Last Year --------------------------------------------Part (A) Salman does pushups in sets each consisting either of 6, 9, or 20 pushups. By doing 2 sets of 6 pushups, he can do 12 pushups. But he cannot do 13 pushups, as one cannot get 13 by using 20,6 or 9. What is the maximum number of pushups that Salman cannot do?

PART (B) A gardener was given a Rs. 100 note by his landlord and asked him to buy exactly 100 flowers from the nursery by spending all the money. A piece of Daffodil costs Rs15. Canterbury Bells cost Re 1, and Marigold are 25 paise each. He was asked to buy at least one flower of each of the three varieties. How many Canterbury Bells did the gardener buy from the nursery? (1) 39 (2) 43 (3) 45 (4) 57 (5) None of the foregoing (Part A) 9mod6=3 . Hence all multiples of 3 greater than 6 can be arrived at 20mod6=2. Hence all numbers >25 which leave remainder of 2 when divided by 3 can be arrived at 40mod6=1. Hence all numbers >43 which leave remainder of 1 when divided by 3 can be arrived at Hence, choice (3) is the right option (Part B) Let the number of Daffodils be 'd' Canterbury be 'c' and marigold be 'm' Then according to question 15d + c + (.25)m = 100 or, 60d + 4c + m = 400 --------------- (1) also, d+c+m = 100 -------------------(2) subtracting (2) from (1) we get, 59d + 3c = 300 or, c = 100 - (59d / 3) As c can only be an integer and d has to be atleast 1 only d =3 satisfies. so c = 41 and m = 56. so answer is 41 Hence, choice (5) is the right option The question was that was 123TH Last Year ---------------------------------------------

ABC is a triangle with AB = 14, BC = 10 and CA = 6. D and E are points on BC and CA respectively such that CD = 3 and CE = 2.5. A line passing through C and the point of intersection of AD and BE cut the side AB at F. Then AF = (a) 5 (b) 5.25 (c) 6 (d) 6.25 (e) 7.5 CD = 3, DB = 7, AC = 6 and AB = 14. => CD/DB = AC/AB => AD is the angle bisector of angle A. Similarly: CE/EA = BC/AB => BE is the angle bisector of angle B. So CF is the angle bisector of angle C AF/FB = 6/10 anf AB = 14. => AF = 5.25. Hence, choice (b) is the right option The question was that was 126TH Last Year --------------------------------------------Part (A) A sphere is wrapped completely by the minimum amount of single piece of paper. What percent of the paper will go wasted?

PART (B) A hemisphere is wrapped completely by the minimum amount of single piece of paper. What percent of the paper will go wasted? (1) 40

(2) 47

(3) 55

(4) 60

(5) 67

(Part A) For wrapping the sphere, the paper should be a circle of diameter 2r*pi => area of the paper will be pi*(pi*r)^2 = 10r^2*pi as pi^2 =~ 10. The surface area of the sphere is 4r^2*pi => wastage is 60% => Choice (4) is the right answer (Part B) The paper will be a circle of diameter pi*r + r. The surface area of hemisphere is 3r^2*pi => wastage is 55% Ö choice (3) is the right answer. The question was that was 127TH Last Year --------------------------------------------The number of integral x < 4 such that 2^|x+2| - |2^(x+1) - 1| = 2^(x+1) + 1 are (1) 2 (2) 3 (3) 4 (4) 6 (5) none of these The equation is not satisfied for x = -2 but satisfied for x = -1. For integral x > -1 i.e. x >= 0 . 2^(x+2) - 2^(x+1) + 1 = 2^(x+1) + 1..

=> 2^(x+2) - 2*2^(x+1) = 0. => 2^(x+2) - 2^(x+2) = 0. which is always true. For integral x < -2. i.e x <= -3. 2^(-(x+2)) - 1 + 2^(x+1) = 2^(x+1) + 1. => 2^(-(x+2)) = 2. This is true only for x = -3. Therefore all values x<4 = -3,-1,0,1,2,3. =>Choice (4) is the right answer The question was that was 128TH Last Year --------------------------------------------Let a(n) = (n+9)!/(n-1)! for each positive integer n. If k is the smallest integer with the rightmost nonzero integer a(k) is odd. The right most nonzero digit of a(k) is (1) 1

(2) 3

(3) 5

(4) 7

(5) 9

a(k) = k*(k+1)*(k+2)*....*(k+9) The given condition would occur when the powers of 2 and 5 are the same in a(k) or the power of 5 is greater than the power of 2. Take any 10 consecutive positive integers. 5 of them will be even. And exactly 2 of them will be div by 5. The smallest highest power of 2 that will be div by the product of these 10 consecutive numbers is 1+2+1+3+1 (three of them will be div by 2, one by 4 and one by 8) => we can have k near 5^7 as another factor of 5 will occur with a number ending with zero. But 5^7 = 78125. Checking that no consecutive number div by 16 comes anywhere and just 2 are div by 4, we get k = 78117 and hence the last non-zero gigit of a(78117) is (1*3*7*9) => 9 Hence, choice (5) is the right option The question was that was 129TH Last Year --------------------------------------------Divya and Raveena can do a work alone exactly in 20 and 25 days respectively. However, when they work together, they do 25% more work than is expected. If they work for a few days alone and for few days together (both being integers only), then the work could not have been completed in exactly (1) 10 days (2) 14 days (3) 16 days (4) 17 days (5) either none or atleast 2 of these Let divya work alone for a days Raveena work alone for b days and they work together for c days then a/20 + b/25 + 9c/80 = 1 => 20 a + 16b + 45 c = 400 => here when c=4 ; a= 7 and b = 5 it satisfies the above equation . 5+4+7 = 16 is possible . when c=4 ; a= 3 and b = 10 it satisfies the above equation . 3+4+10 = 17 is possible a+b+c = 10 or 14 is not possible. Hence, choice (5) is the right option

The question was that was 130TH Last Year --------------------------------------------A graph has p points. The degree of a point in the graph is the number of other points it is connected to by edges. Each point has degree atmost 3. If there is no edge between two points then there is a third point joined to them both. What is the maximum possible value of p? (1) 12

(2) 7

(3) 10

(4) 9

(5) 8

Take any point A. A is joined to at most three points, call them B, C, D. Any other point P must be joined to one of B, C, D. Each of B, C, D is joined to at most two other points. So there are at most 1 + 3 + 3•2 = 10 points in all. But is 10 possible? Take points A, B, C, D, E, F, G, H, I, J and edges AB, AC, AD, BE, BF, CG, CH, DI, DJ, EH, EJ, FG, FI, GJ, HI. It is easy to check that every point has degree 3. Now take any two points not joined by an edge. For A,E or A,F both are joined to B. Similarly, both A,G and A,H are joined to C, and both A,I and A,J are joined to D. Both B,C and B,D are joined to A. B,G and B,I are joined to F. B,H and B,J are joined to E. Similarly, C,D are joined to A. C,E and C,I are joined to H. C,F and C,J are joined to G. Similarly, D,F and D,H are joined to I. D,E and D,G are joined to J. E,F are joined to B. E,G are joined to J. E,I are joined to H. F,H are joined to I. F,J are joined to G. G, I are joined to F. H, J are joined to E. Hence, choice (4) is the right option The question was that was 131st Last Year --------------------------------------------Which one of the following is the remainder when x + x^7 + x^16 + x^37 is divided by x^4 - x? (1) 4x (2) 2x(x-1) (3) x^2 + 2x (4) x(x-1) (5) none of these x + x^7 + x^16 + x^37 is divided by x^4 - x => x*{1 + x^6+ x^15 + x^36 is divided by x^3 - 1}=4x (using factor theorem) If the question had mentioned that x is an integer then remainder is 4x for x > 1. For x = -1 remainder is 0 and for x < -1 remainder is x^4 + 3x. In polynomial division where nothing about x is mentioned, don't put integer values to derive the answer. Hence, choice (1) is the right option The question was that was 132nd Last Year --------------------------------------------Three squares each of area 16 sq.units have their one diagonal along a common line. Let the area common to first and second square be 9 sq. units while that common to second and third be 4 sq. units.The area (in sq. units) common to first and third squares will be

(1) 0

(2) 1

(3) 2

(4) 3

(5) none of these

As the area common between Ist and IInd square is 9 and the diagonal lie on the same line so the the length of line common to the diagonal of both the square will be (3*V2) as the common area is also a square only with side 3.Also as the area common between IInd and IIIrd square is 4 and the diagonal lies on the same line so the the length of line common to the diagonal of both the square will be (2*V2) . Now if we take square I , II and III in the order from left to right, then length of line common to the diagonal of Ist and IIIrd square will be (V2) and so the common area is 1. =>Choice (2) is the right answer The question was that was 133rd Last Year --------------------------------------------Three vessels A, B and C have different concentrations of alcohol. If the contents of pairs of vessels (A, B), (B, C) and (C, A) are mixed, then the concentrations of alcohol become 30%, 40% and 50% respectively. If the concentration of alcohol in vessel A in % is 30 less than twice the concentration in % in vessel B, then the concentration of vessel C is (1) 42% (2) 45% (3) 48% (4) can not be determined (5) none of these Let L1, L2, L3 be the amount of alcohol in vessels A, B and C while n1%, n2%, n3%be their respective concentration. Given that n1 = 2n2 - 30. Also, (n1L2+n2L1)/(L1+L2) = 30 => L1/L2 = (30-n2)/(n1-30). Similarly, L2/L3 = (40-n3)/(n2-40) and L3/L1 = (50-n1)/(n3-50). solving we get n3 = 45% but that gives n1 > 50 and n2 > 40 which is inconsistent with the given data. Hence, choice (4) is the right option The question was that was 134th Last Year --------------------------------------------Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is (1) 2 (2) 4 (3) 3 (4) 1 (5) none of these A fractiox = 10(11-n)/(n-1); y = 10(12-2n)/(n-1) ; z = 10(13-3n)/(n-1) Hence n = 2 and n = 3 satisfy. Hence 2 distinct solutions. Hence, choice (1) is the right option The question was that was 135th Last Year ---------------------------------------------

Let X = {1, 2, 3, 4}. How many functions f: X->X are there satisfying (f.f)(x) = x for all x in X? (1) 13

(2) 15

(3) 14

(4) 16

(5) none of these

If f:X -> X where X = {1,2,3,4} such that f.f (x) = x, then f is a bijection. There are 24 bijections from X to X. Of these only 10 are such that f.f (x) = x for all x in X. In giving these maps explicitly we will use a cyclic notation for convenience, e.g. (1 2) stands for the map 1 -> 2, 2 -> 1, 3 -> 3, 4 -> 4 Note that in this cyclic notation 3 and 4 are not mentioned which means that they are mapped to themselves i.e. they are fixed. The maps are : (1 2), (1 3), (1 4), (2 3), (2 4), (3 4) and also (1 2) (3 4) (1 3) (2 4) (1 4) (2 3) Note that (1 2) (3 4) means 1 -> 2, 2 -> 1, 3 -> 4, 4 -> 3 And lastly the identity map which gives us a total of 10 maps satisfying the given condition. Hence, choice (5) is the right option The question was that was 8th Last Year --------------------------------------------QQAD team decides to go on vacation for 8 days trusting their NL software system. The problems are fed into the system and date timer set in advance for each 8 days for the questions to be delivered to the subscribers daily. However, the software follows a weird rule. It doesn't always deliver the NL daily in those 8 days, but never misses 3 consecutive deliveries. How many possible ways are there for the NLs delivery in those 8 days? (a) 162

(b) 138

(c) 117

(d) 176

(e) 149

On each day NL can take 2 states - it either delivers or it doesn't. Let f(n) be the possible ways for n days with given conditions. => f(1) = 2, f(2) = 2^2, f(3) = 2^3-1 (1 subtracted as we can't have 3 misses). Let n > 3. Then, NL either delivers on day 1 or it doesn't. When it does it can have f(n-1) ways from there; on day 2 also it either delivers or it doesn't, when it does it can have f(n-2) ways from there. Now on day 3, it has to deliver as we have missed first 2 days, thus after delivering on day 3 it can have f(n-3) ways => f(n) = f(n-1)+f(n2)+f(n-3) => f(4) = 13, f(5) = 24, f(6) = 44, f(7) = 81 and f(8 ) = 149 Hence, choice (e) is the right option

The question was that was 137th Last Year --------------------------------------------3 Honda Civics, 4 Pajeros, and 5 BMWs are to be parked in front of a dude's house for whom girl's family is coming to see him. In how many ways the cars be parked such that no 2 Honda Civics are next to each other, no Pajero is adjacent to a BMW, and all the cars are being parked? (1) 129

(2) 135

(3) 152

(4) 153

(5) 165

1H2H3H4 Consider that the Hondas occupy the positions marked as H. Now the rest can be filled in certain ways First consider only slots 1,2,3 to be filled We get (3+4)*3c2 possible ways = 21 ways For 2,3,4 it’s the same Hence here also, no. of ways = 21 If only 2 and 3 are occupied by the cars We have 2 ways Now consider all 4 slots to be filled No. of ways = no. of ways in which p occupies (1 slot + 2s lots + 3 slots) = (3+3)*4c1 + (3*4*4c2) + 3* 4c3 = 24+ 72+12 = 108 ways total = 108+21+21+2 = 152 ways Ö Choice (3) is the right answer The question was that was 138th Last Year --------------------------------------------Within a 5X5 table, a box is marked at the intersection of second row and third column. How many rectangles formed by the boxes of the table do not contain the marked box? (1) 129 (2) 135 (3) 152 (4) 153 (5) 165 The rectangle can be defined without ambiguity by its upper left and lower right vertices. To contain the marked box, the upper left vertex must be in a row with a number less than or equal to 2 and in a column with number less than or equal to 3. The lower right vertex must be in a row with a number greater than or equal to 2 and in a column with a number greater than or equal to 3. Thus, there are 2.3 = 6 different positions for the upper left vertex and there are 4.3 = 12 different positions for the lower right vertex. Therefore, there are 2.3.4.3 = 72 rectangles containing the marked box. Ö choice (4) is the right answer

The question was that was 139th Last Year --------------------------------------------The angular deviation of a tower CD at a place A due south of it is 60˚, and at a place due west of A, the elevation is 30˚. If AB = 3 km, then the height of the tower is (1) 3√3/2√2 (2) 3√6/4 (3) 2√3 (4) 2√6 (5) none of these If the height of the tower is h, then tan 60 = CD/AC= h/AC tan 30 = CD/BC =h/BC Also triangle ABC is a right angled one , hence , 3^2+AC^2 = BC^2 or, 9^2 +h^2/3 = 3h^2.. or, h=3v3/2v2 Ö Choice (1) and (2) are the right answer

The question was that was 140th Last Year --------------------------------------------S is a region bounded by 5|y|=2(x-2) and x=7.There is an infinite plane mirror of negligible width at x=-3 perpendicular to the x-axis.Consider a region T bounded by x(x+6)=16-y^2 for x<=-3 and the mirror.Find the area of the region contained within y=+/-2 and x=+/-14 which does not include S,T or their images as formed by the mirror. (a)13.46

(b)15.75

(c)52.73

(d)53.875

(e)None of the foregoing

A fractiox = 10(11-n)/(n-1); y = 10(12-2n)/(n-1) ; z = 10(13-3n)/(n-1) Hence n = 2 and n = 3 satisfy. Hence 2 distinct solutions. Hence, choice (1) is the right option The question was that was 141st Last Year --------------------------------------------If the remainder when 281^508 is divided by 153 is P, and p^(46P) when divided by 83 gives remainder Q.Then sum of the digits of Q is (a)1 (b)3

(c)7

(d)11

(e)None of the foregoing

281^508 by 9 gives remainder 16 and by 17 16 => 281^508 by 153 gives remainder 16. 16^82 when divided by 83 gives remainder as 1 => 16^(46*16) mod 83 == 16^80 mod 83.

If 16^80 = x mod (83) => 256x by 83 gives 1 => (249+7)x by 83 gives 1 => 7x by 83 gives 1 => x = 12 Ö Choice (2) is the right answer The question was that was 142nd Last Year --------------------------------------------To offset the increase in price of sugar and rice, either Sargam has to reduce the consumption of sugar by 20% or rice by 25%. How much percent consumption of sugar Sargam must reduce if she reduced the consumption of rice by 10%? (1) 12% (2) 12.5% (3) 15% (4) 16% (5) can not be determined Let the initial prices of sugar and rice be x and y per kg and their consumption be a and b per kg respectively. Also, let there been an increase of 100p1% and 100p2% in their prices respectively. Then, ax+by = 0.8(1+p1)ax + (1+p2)by and ax+by = (1+p1)ax + 0.75(1+p2)by. Let ax+by = (1-k)(1+p1)ax + 0.9(1+p2)by where 100k% is reduction in consumption of sugar when consumption of rice is reduced by 10%. Solving we get k = 0.12 Ö Choice (1) is the right answer The question was that was 144th Last Year --------------------------------------------Consider the following system of equations: r + x = l ; l + p = n ; n + r = k ; r = 8 ; x + p + k = 30 The value k is (1) 17 (2) 22

(3) 23

(4) 15

(5) 11=>

Add the given 4 equations to get 2(x+p)+2r = 30 r= 8 (given) hence, x+p =7 => k = 23 Ö Choice (3) is the right answer The question was that was 143rd Last Year --------------------------------------------Let the cost of 3 apples and 4 oranges be Rs 21. If Anjli can buy at most 4 apples and 3 oranges in Rs 20, then the maximum amount that could be left with Anjli will be about (1) Rs 1 (2) Rs 1.50 (3) Rs 2 (4) Rs 2.50 (5) none of these 3x+4y = 21 and let r be the residual left after 4 oranges and 3 apples => r = 20 (4x+3y). Also r < x and r < y.

Solving we get 17/11 < x. Take x = 1.55 => y = 4.0875. r comes out to be little over Rs 1.50 but less than Rs 1.55 which is the cost of an apple. Ö Choice (2) is the right answer The question was that was 145th Last Year --------------------------------------------Kaizen and Warrior are 1 km apart when they decide on phone to meet after some time. Warrior starts moving at 60 degrees to the line joining them initially and at the same time Kaizen starts moving at an angle of 45 degrees to the line joining them initially. It is known that Warrior and Kaizen move at a constant velocity which is 10m/s in case of Warrior. Both of them reach the meeting point at the same time.After the meeting they retrace the path that they took to reach the point in order to go back from where they came. Assuming that they retrace as soon as they meet each other, what is the time taken(in mins) by Warrior to cover the whole journey? (1) 2.44 (2) 2.86

(3)9.107

(4) 10.274

(5)None of the foregoing

Option (1) incase both start moving towards each other => angles of the triangle in degrees are 45, 60 and 75. Option (3) incase warrior moves away from kaizen and kaizen moves towards warrior => angles of the triangle in degrees are 45, 120 and 15. Use sine rule of the triangle to find the length travalled by Warrior. In another case when kaizen moves away from warrior and warrior moves towards kaizen, it is not possible for them to meet. => Choice (1) and (3) is the right answer Quantitative Question # 146 (Solved by vineetvijay,kondapalli,cloudsonfire,vfactor) -------------------------------------------------------Let f(a+x) - f(a-x) = pf(2a) + qf(2x) for all real x and non-zero f(2a). Which of the following is always true? (1) p+q = 0 these

(2) q = 1

(3) both (1) and (2)

(4) either (1) or (2)

(5)none of

Solution: Given f(a+x) - f(a-x) = pf(2a) + qf(2x). Putting x = 0 we get, pf(2a) + qf(0) = 0. Putting x=a in the original equation we get f(2a) - f(0) = pf(2a) + qf(2a) => f(0) = (1p-q)f(2a) => (p+q)(1-q) = 0 => either p+q = 0 or q = 1. Ö Choice (4) is the right answer Quantitative Question # 147 (Solved by Superstar, Marshall, warrior) --------------------------------------------------------

Twenty metres of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle in meters be, if we wish to have a flower bed with the greatest possible surface area? (1) 2√2

(2) 2√5

(3) 5

(4) 4√2

(5) none of these

Solution: Let the sector be of angle x Area of Sector, A = x/360*pi*r^2 Circumference of the sector = 20 => x/360*2*pi*r +2r= 20 => 2A/r+2r=20 => A= r(10-r) Ö Choice (3) is the right answer Quantitative Question # 148 (Solved by kekambus,sanyo,andy_84) -------------------------------------------------------A car travels downhill at 72 kmph (kilometers per hour), on the level at 63 kmph, and uphill at only 56 kmph The car takes 4 hours to travel from town A to town B. The return trip takes 40 minutes more. What is the distance between the two towns in kilometers? (1) can not be determined

(2) 191

(3) 255

(4) 273

(5) none of these

Solution: Let the total distance travelled downhill, on the level, and uphill, on the outbound journey, be x, y, and z, respectively. The time taken to travel a distance s at speed v is s/v. Hence, for the outbound journey x/72 + y/63 + z/56 = 4 While for the return journey, which we assume to be along the same roads x/56 + y/63 + z/72 = 14/3 Multiplying both equations by the least common multiple of denominators 56, 63, and 72, we obtain 7x + 8y + 9z = 4 · 7 · 8 · 9 9x + 8y + 7z = (14/3) · 7 · 8 · 9 Now it is clear that we should add the equations, yielding 16(x + y + z) = (26/3) · 7 · 8 · 9 Therefore x + y + z = 273; the distance between the two towns is 273 kms. Ö Choice (4) is the right answer Quantitative Question # 149 (Solved by Rohit.sinha, pallavgarg2003) --------------------------------------------------------

Triangle ABC is right-angled at A. D is a point on AB such that CD = 1. AE is the altitude from A to BC. If BD = BE = 1, what is the length of AD? (1) 2^1/3 - 1 these

(2) (5^1/2 - 1)/2

(3) (5^1/2 + 1)/4

(4) 2^1/2 - 1

(5) none of

Solution: Let AD = x, CE = y, and < ABC = t. Let AE and CD meet at F. Since BCD is isosceles, < BCD = t. Hence < CFE = 90° − t, and so < DFA= 90° − t. Since also < FAD = < EAB = 90° − t, DFA is isosceles, and so DF = AD = x. Hence CF = 1 − x. Triangles ABE and CFE are similar, as each contains a right angle, and < ABC = < ECF. Hence y/(1 − x) = 1/(1 + x), and so y = (1 − x)/(1 + x)(1) Triangles ABC and ABE are similar, as each contains a right angle, and < ABC = < ABE. Hence (1 + x)/(1 + y) = 1/(1 + x), and so (1 + x)^2 = 1 + y. Substituting for y from (1), we obtain (1 + x)^2 = 1 + (1 − x)/(1 + x) = 2/(1 + x). Hence (1 + x)^3 = 2. Ö Choice (1) is the right answer Quantitative Question # 150 (Solved by Koush) -------------------------------------------------------Let three positive integers p, p^2 + 2, p^3 + 2 be given. Which among the following is always true? (1) All the three numbers are prime for atleast 2 values of p (2) Exactly 2 of these numbers are perfect squares for some p (3) The product of 2 of these numbers can be expressed as a 6 digit number in the base of the third number (4) atleast 2 of the foregoing (5) none of these Solution: If p is prime > 3 then it is always of the form 6k+/-1 => p^2 + 2 is always div by 3 => for p = 3 only all the 3 numbers are prime => (1) is false p^2 < p^2 + 2 < (p+1)^2 => only possibility is that p and p^3 + 2 can be both perfect squares. If p is a perfect square then it is of the form 4k or 4k+1 => p^3 + 2 is of the form 4k+2 or 4k+3 => (2) is false

The product (p^2 + 2).(p^3 + 2) has highest power as 5 and hence in base p can be written as a 6 digit number => (3) is true Ö Choice (3) is the right answer Quantitative Question # 151 (Solved by Kekambus, Marshall) -------------------------------------------------------Each of the 10 digits from 0 to 9 are used exactly once altogether to form 3 positive integers. One of these numbers is the sum of the other two. What is the difference between the largest possible and the smallest possible of the largest of these 3 numbers? (1) 4995 (2) 5775 (3) 6858 (4) 7632 (5) none of these Solution: Since the sum has at least as many digits as either of the summands, the sum must have at exactly 4 digits. A 4-digit sum can arise either as the sum of two three-digit numbers or as the sum of a 4-digit and a 2-digit number. In the former case, the sum must exceed 1000 and be less than 2000 and, in the latter case, it must be at least 2000. Thus, the smallest possible sum must be obtained by adding two three-digit numbers to get a 4-digit sum. Since, the digits of the sum are all distinct, the smallest sum is atleast 1023. Since, 589+437 = 1026, the smallest sum is atmost 1026. We may assume that each digit in the first summand exceeds the corresponding digit in the second summand. The only possibilities for a lower sum are 5pq + 4rs = 1023, 6pq + 3rs = 1024, 6pq + 3rs = 1025. None of these work. For the largest sum, let the first summand have 4-digits and the second two. The hundereds digit of the first summand is 9 and the thousands digit of the sum exceeds the thousand digit of the first summand by 1. Since 5987 + 34 = 6021, the largest sum is atleast 6021. The only possibility to consider for a larger sum are 79ab + cd = 80ef, 69ab + cd = 70ef, 59ab + cd = 60ef. None of these work. Thus, max is 6021 and the min is 1026. Ö Choice (1) is the right answer Quantitative Question # 152 (solved by amit_adobe, andy_84) -------------------------------------------------------In the quadrilateral ABCD, AD = DC = CB, and < ADC = 100˚, < ABC = 130˚. Then the measure of < ACB (is) (1) 20˚ (2) 30˚ (3) 50˚ (4) can not be determined (5) none of these Solution: A, B, C will lie on a circle with centre at D (as the angle subtended by the arc at the centre i.e. 260˚ is twice subtended at the circle i.e. 130˚ ) => in triangle DAC, < DAC = < DCA = 40˚. Let < ADB = 2x => < ACB = x, and let < BDC = < CDB = y => 2x+y = 100˚, and 2y+x = 140˚. => Choice (1) is the right answer

Quantitative Question # 154 (solved by ... ) -------------------------------------------------------The remainder when 100*(99^10) is divided by 100*99 + 1 is (1) 0 (2) 1 Solution:

(3) 100

(4) 9900

(5) none of these

Let n = 99 => 100*99 + 1 = n^2 + n+1. Also the dividend is n^11 + n^10. Note that n^2 + n + 1 = 0 gives roots (w, w^2) of unity. And w^2 + w + 1 = 0, w^3 = 1. Also w^11 + w^10 = -1 and w^22 + w^20 = -1 => n^11 + n^10 + 1 is divisible by n^2 + n + 1. => Choice (4) is the right answer Quantitative Question # 154 (solved by krsh.vik) -------------------------------------------------------Two kinds of rice are mixed in the ratio 1:2 and 2:1 and then they are sold fetching the profit of 10% and 20% respectively. If they are mixed in equal ratio and the individual profit percents on them are increased 4/3 and 5/3 time respectively, then the profit % would be (1) 18 Solution:

(2) 20

(3) 21

(4) 25

(5) none of these

Let the profit be 100p1% and 100p2%. Given (p1c1 + 2p2c2)/(c1 + 2c2) = 0.1 and (2p1c1 + p2c2)/(2c1 + c2) = 0.2 => c1(p1-0.3) = p2c2 && 3(p1c1+p2c2) = 0.5c1 + 0.4c2. => (4/3p1c1 + 5/3p2c2)/(c1+c2) = 1/5 => Choice (2) is the right answer Quantitative Question # 155 -------------------------------------------------------For how many intergral k does the inequality log2 + log (2x^2 + 2x + 7/2) >= log(kx^2 + k) possesses at least one solution? (1) 6

(2) 8

(3) 11

(4) 5

(5) 10

Solution: log (4x^2 + 4x + 7) >= log(kx^2 + k) => for 4x^2 + 4x + 7 >= kx^2 + k, we must have atleast one real solution for k > 0. => (k-4)x^2 -4x + (k-7) <= 0 Multiply both the sides by (k-4). => LHS of the inequality becomes ((k-4)x - 2)^2 + (k-8 )(k-3). For k >=4 we have ((k-4)x - 2)^2 + (k-8 )(k-3) <= 0 ... we have k = 4, 5, 6, 7, 8 For k < 4 we have ((k-4)x - 2)^2 + (k-8 )(k-3) >= 0 ... we have k = 1, 2, 3 => Choice (2) is the right answer

Quantitative Question # 156 (solved by sanyo, v-factor, krish_mec, de_profundis) -------------------------------------------------------Five former beauty contest winners Sushmita, Aishwarya, Diana, Lara, Priyanka were placed 1 to 5 in a contest with no ties. One prediction was that the result would be the order Sushmita, Aishwarya, Diana, Lara, Priyanka. But no contestant finished in the position predicted and no two contestants predicted to finish consecutively did so. For example, the outcome for Diana and Lara was not 1, 2 (respectively), or 2, 3, or 3, 4 or 4, 5. Another prediction was the order Lara, Sushmita, Priyanka, Diana, Aishwarya. Exactly two contestants finished in the places predicted and two disjoint pairs predicted to finish consecutively did so. Who finished immediately next after Priyanka? (1) Sushmita to determine

(2) Aishwarya

(3) Diana

(4) Lara

(5) Impossible

Solution: Lets ignore the first set of conditions and examine the second set. Since two disjoint sets are valid - that means that the following are valid consecutive sets. LS (12) and PD (34) LS (12) and DA (45) SP (23) and DA (45) However, since exactly two of them are in the right position, it logically means that exactly one pair of one of the three pairs above is in the correct order and positon while the other is in the correct order but wrong position. This means that the correct order and position can only be in position 12 or 45 (23 and 34 is not valid because then the other valid pair cannot be shifted to another valid position). I know this reasoning can be confusing but read it a couple of times and it will make sense. That means - either: 1) LS (12) is correct and PD (34) is repositioned to PD (45). This way we get LSAPD - SA is consecutive which invalidates condition (1) hence false. 2) LS (12) is correct and DA (45) is repositioned to DA (34). This way we get LSDAP - D has the same position as condition (1) hence false. 3) DA (45) is correct and SP(23) is repositioned to SP (12). This way we get SPLDA - S has the same position as condition (1) hence false. 4) DA (45) is correct and LS(12) is repositioned to LS (23). This way we get PLSDA - This is valid. => Choice (4) is the right answer Quantitative Question # 157 (solved by sharang, varun_nakra, kekambus) -------------------------------------------------------Let ABC be a triangle and D be the midpoint of AC. Point E lies internally on BD such that BE = 2. Also, AB = 3, BC = 4 and < AEC = 90˚. Then AC =

(1) √21 - 1 these

(2) √20 - 1

(3) √18 - 1

(4) √24 - 1

(5) none of

Solution: Let AD = x = CD => ED = x => BD = x+2 using Appollonius theorem 3^2 + 4^2 = 2{(2+x)^2 + x^2} => x = √21 - 2 => Choice (5) is the right answer Quantitative Question # 158 (solved by frigid sapphire, kaizen_2007, sandeep jha, sanyo, jbadri, mejogi, Me@Biz, Mits@Bits) -------------------------------------------------------Let the product of four consecutive integers be a five-digit integer pq0pq, where p and q are single digit positive integers. Then p+q is (1) 4

(2) 6

(3) 8

(4) 9

(5) none of these

Solution: The five digit number is pq*1001 = pq*7*11*13 => pq = 24 as the product is 11*12*13*14 => p+q = 6 => Choice (2) is the right answer Quantitative Question # 159 (solved by varun nakra) -------------------------------------------------------A person is said to be n years old, where n is a non-negative integer, if the person has lived at least n years and has not lived n+1 years. At some point in time, Anupam is 4 years old and Nilesh is three times as old as Shrikant. At some other time, Shrikant is twice as old Anupam, and Nilesh is 5 times as old as Anupam. At yet another time, Nilesh is twice as old as Shrikant and Anupam is Y years old. There are different possibilities of what Y can be. The largest possible Y is in the range (1) [15, 18] of these

(2) [21, 25]

(3) [27, 32]

(4) [35, 39]

(5) none

Solution: Let each f1 lies in [0, 1) and each y1 lies in (0, 1) At some point, Anupam has lived 4 + f1 years, Nilesh has lived n + f2 years and Shrikant has lived s + f3 years where n and s are integers satisying n = 3s. At another time, say a + f4 years later where a is an integer, Shrikant is twice as old as Anuapm .... => Anupam would have lived 4 + a + f1 + f4 years, he will be 4 + a + y1 years old. Similarly, Nilesh would be n + a + y2 years old and Shrikant would be s + a + y3 years old. => s + a + y3 = 2(4 + a + y1) and n + a + y2 = 5(4 + a + y1)

At yet another time, say b + f5 years later after Anupam lived 4 + f1 years, Anupam is 4 + b + y4 years old and n + b + y5 = 2(s + b + y6). This equation and n = 3s imply b = n - 2s + y5 - 2y6 = s + y5 - 2y6. => s = a + 8 + 2y1 - y3 and 3s = 4a + 20 + 5y1 - y2. => m = 4m - 3m = 12 + 3y1 - 4y3 + y2 Combining the above we get Y = 16 + 3y1 - 4y3 + y2 + y5 - 2y6 + y4, since each y(i) lies between 0 to 1 => T <= 16 + 3 + 1 + 1 + 1 = 22. Thus, Anupam's age is 4.6, Shrikanth's is 16 and Nilesh's is 48.6 Next: Anupam's age is 12, Shrikant's is 22.4 and Nilesh's is 55 Next: Anupams age is 22.1, Shrikant's is 33.5 and Nilesh's is 66.1 Alternate solution from Varun Let the ages be of N ,S and A in this order 3[x] + f2 ; [x] + f4 ; 4 + f6 5[y] + f1 ; 2[y] + f3 ; [y] + f5 2[z] + f7 ; [z] + f8 ; [Y] + f9 Now we can equate the differences We have [y] = (3f3 + f2 + 2f6) - (f1 + 3f4 +2f5) + 8 [x] = ( 4f3 + f2 + 3f6) - (3f5 + 4f4 + f1) + 12 [Y] = 2[y] + (2f8 + f1 + f5) - (2f3 + f7 + f9) [z] = 3[y] + (f8 + f1) - (f3 + f7) Now [x] - [y] = f3 + f6 - f5 - f4 + 4 which shud be an integer.So in order to maximise [Y] nd hence maximise [y]. one can see that f4 nd f5 are to be subtracted(in the value of [y] nd [x] both) so we shall make it minimum which is by putting each of them to be zero. So f4 = f5 = 0 nd f3 + f6 shud be an integer..Again for max[Y] we shall make it 1 nd not 0 hence f3 + f6 = 1 hence [y] = (3f3 + f2 + 2f6) - (f1) + 8 [x] = ( 4f3 + f2 + 3f6) – (f1) + 12 again we can Put f1=0 [y] = (3f3 + f2 + 2f6) + 8 [x] = ( 4f3 + f2 + 3f6) + 12 now Put f6 = 1-f3 [y] = (f3 + f2 ) + 10 [x] = ( f3 + f2 ) + 15 Now f3+f2 has to be an integer which shud be 1 only for the max value of [y]..so [y] max becomes 11 Nd [Y] = 2[y] + (2f8 ) - (2f3 + f7 + f9) [z] = 3[y] + (f - (f3 + f7) Now put f7=f9=0

[Y] = 2[y] + (2f8 ) - (2f3) [z] = 3[y] + (f - (f3) Thus [Y] = 22 + (2f8 ) - (2f3) Now f8-f3 shud be an integer which could only be 0 nd nothing else coz if it is 1.in that case f8=1+f3 which is Not possible,so f8=f3 hence forth we have the constants as f1=f4=f5=f7=f9=0 ; f3+f2=1 ;f6+f3=1 and f8=f3 Hence [Y]max becomes 22. => Choice (2) is the right answer

Quantitative Question # 160 (solved by varun nakra) -------------------------------------------------------There are 7 lakes in Lakeland. They are connected by 10 canals so that one can swim through the canals from any lake to any other. How many islands are there in Lakeland? (1) 8

(2) 6

(3) 5

(4) 7

(5) 4

Solution: Consider the planar graph whose vertices are lakes, whose edges are the canals, and whose faces are the islands. Since V-E+F = 2, V=7 and E = 10 => F = 5. But one of the faces is an outer face and not an island. Ö Choice (5) is the right answer Quantitative Question # 161 (solved by kekambus, varun nakra, mejogi, sanyo, krish_mec ) -------------------------------------------------------The four numbers a < b < c < d can be paired in exactly 6 different ways. If each pair has a different sum and if the four smallest sums are 1, 2, 3, 4 then what is the sum of all the possible values of d? (1) 4

(2) 35/6

(3) 15/2

(4) 7

(5) 11

Solution: Case1: a+b < a+c < b+c < a+d = 1 < 2 < 3 < 4 => d = 4 Case2: a+b < a+c < a+d < b+c = 1 < 2 < 3 < 4 => d = 3+1/2 => Choice (3) is the right answer Quantitative Question # 162 (solved by samsing, varun nakra) --------------------------------------------------------

A test has exactly 10 questions and is either answered in True or False. If Deepika answers 5 questions "true" and five "false", her score is guaranteed to be at least 4. How many answer keys are there for which this is true? (1) 12

(2) 13

(3) 16

(4) 19

(5) 22

Solution: Suppose that either 9 or 10 of the questions have the same answer. Then no matter which 5 questions Deepika picks to have this answer, she will be right at least 4 times. Conversely, suppose that there are at least two questions with each answer; we will show that she can get a score less than 4. By symmetry, assume there are at least five questions who answer is true. Then if we label five of these false, not only will Deepika get these 5 wrong, but she will also have answered all the false questions with true, for a total of at least 7 incorrect. There are 2 ways for all the questions to have the same answer and 20 ways for one question to have a different answer from the others, for a total of 22 ways. => Choice (5) is the right answer Quantitative Question # 163 (solved by kekambus, sanyo, swarnali, vivekr) -------------------------------------------------------If p, q, n are positive reals such that 3 = n^2(x^2/y^2 + y^2/x^2) + n(x/y + y/x). What is the maximum value of n? (1) 2

(2) (1 + √5)/2

(3) (-1 + √7)/2

(4) 5/2

(5) (-1 + √5)/2

Solution: To maximize n we must minimize (x/y + y/x) => x/y + y/x = 2 when x = y => 3 = 2n^2 + 2n. => Choice (3) is the right answer Quantitative Question # 164 (solved by andy_84, anupsu) -------------------------------------------------------The numerical value of infinite series 1/(3^2 + 1) + 1/(4^2 + 2) + 1/(5^2 + 3) + ... is (1) 13/36

(2) 5/16

(3) 7/24

(4) 1/3

(5) none of these

Solution: The series reduces to 1/3((1/2-1/5) + (1/3-1/6) + (1/4-1/7) + (1/5-1/8) + ...) = 1/3(1/2 + 1/3 + 1/4) => Choice (3) is the right answer Quantitative Question # 165 (solved by Marshall, doomsayer, v-factor, kekambus) -------------------------------------------------------There are six sticks with lengths 2, 4, 4, 10, 22, 37. How many isosceles trapeziums can be constructed, each time using all six sticks?

(1) 2

(2) 3

(3) 4

(4) 5

(5) none of these

Solution: Combinations 1. (4,4,(22+10+2),37) => (4,4,34,37) 2. (4,4,(22+10),(37+2)) => (4,4,32,39) 3. (10,(2+4+4),22,37) => (10,10,22,37) => Choice (3) is the right answer Quantitative Question # 165 (solved by Me@Biz, junoonmba) -------------------------------------------------------By giving one rubber free with 4 pencils, it means that a discount of 10% is given on the sale of pencils. Then by giving 1 pencil free with 6 rubbers, it means that a discount of x% is given on the sale of rubbers. Then x (approximately) equals (1) 57%

(2) 61%

(3) 64%

(4) 73%

(5) none of these

Solution: Let P and R be the selling price of a pencil and a rubber => .9(4P+R) = 4P => 9R = 4P Similarly, x/100(6R+P) = 6R => x = 8/11*100 =~ 73% => discount given is 3/11*100 =~ 27% => Choice (5) is the right answer Quantitative Question # 167 (solved by getneonow, krsh.vik) -------------------------------------------------------The base 6 representation of 0.33333..... is (1) 3/5

(2) 2/3

(3) 1/3

(4) 3/4

(5) none of these

Solution: Please refer the discussions thread for the answer. Ö Choice (3) is the right answer Quantitative Question # 168 (solved by nebuchadnezzar, srivatsa, arvindva) -------------------------------------------------------Let in a triangle ABC, AD, BE and CF be the altitudes intersecting at H. If AH = 3, AD = 4, BH = 2 then BE = (1) 3

(2) 3.5

(3) 4

(4) 4.5

(5) none of these

Solution: Please refer the discussions thread for the answer. => Choice (2) is the right answer

Quantitative Question # 169 (solved by kekambus, its_vaibhav, doomsayer, targetABC) -------------------------------------------------------Divya and Raveena decide to play the following game. They begin with S = 0. They take turns picking up the number from the set {1, 2, 3, 4, 5, 6}. On each turn any of the 6 numbers can be chosen. The number is added to S and S is then replaced by the sum. The 1st person who choses a number which when added to S gives 40 is the winner. Divya begins by picking 6. Let p be the number which is now the best choice for Raveena Let a and b be integer solution to 17a + 6b = 13. Let q be the smallest possible postive value of a-b The numerical value of pq is (1) 13

(2) 36

(3) 39

(4) 78

(5) 102

Solution: Let 6 be the choice for Raveena. Now 28 is left out of 40 , for any no. from 1 to 6 that Divya picks Raveena can pick a number to make it a 7 in two turns. And since 28=7*4 she would ensure that she wins no matter what Divya picks.Hence p = 6 Also q= minimum positive value of a-b = 17 for a = 5 and b = -12 So, p*q=102 Ö Choice (5) is the right answer Quantitative Question # 170 (solved by vivekr, krish_mec, Rohit.Sinha) -------------------------------------------------------There are z digits in the decimal expression of the natural number N,while there are y digits in the decimal expression of N^3. Then which of the following cannot be equal to y+z? (a) 20

(b) 26

(c) 35

(d) 45

(e) none of these

Solution: As per the question, 10^(z−1) <= N < 10^z =>10^(3z−3)<= N^3 < 10^3z. So, y can be {3z − 2, 3z − 1, 3z} and y+z can be {4z-2 ,4z-1,4z) Clearly all the natural numbers except those of type 4k+1 satisfy the value for y+z. So, 35 cannot be the value. Ö Choice (d) is the right answer

Quantitative Question # 171 (solved by nipun_jain) -------------------------------------------------------I was at a restaurant for lunch the other day with some QQAdites. The bill came and I wanted to give the beautiful waiteress a multiple of 100 number of rupees, with the difference between what I give her and the bill being the tip. I always like to tip between 10 and 15 percent of the bill (ofcourse to a good looking damsel only). But if I give her a certain number of rupees (mutiple of 100), the tip would have been less than 10% of the bill, and if instead I give her 100 rupee more, the tip would have been more than 15% of the bill. The exact bill was a whole number. Let Rs PQRS be the largest possible amount of the bill where P, Q, R and S are each single digit numbers. Then P+Q+R+S is (1) 14

(2) 15

(3) 17

(4) 22

(5) none of these

Solution: Please refer the discussions thread. Ö Choice (1) is the right answer Quantitative Question # 172 (solved by arvindva, sanyo, doomsayer, kekambus) -------------------------------------------------------Two cars A and B started from P and Q respectively towards each other at the same time. Car A was travelling at a speed of 54km/h but due to some problem reduced its speed by 1/3rd after travelling for 60 minutes. Car B was travelling at a speed of 36km/h. Had the technical problem in car A had arisen 30 minutes later, they would have met at a distance which is (1/30*PQ) more than towards Q than where they met earlier(PQ > 120km). Anothet car C starts from P, 90 minutes after car B started at Q, and car C travels towards Q with a speed of 36km/h, at what distance from P will cars B and C meet? (1) 63 km

(2) 54 km

(3) 40.5 km

(4) 36 km

(5) none of these

Solution: Two cars meet at mid-point of PQ if they were at same speed. If the starting point of A shifts by l1 towards Q and that of B by l2 towards Q, where l2 < l1), the meeting point shifts by (l1 - l2)/2 towards Q. => meeting point shifts by (27-18)/2 km towards Q => PQ = 30*(4.5) km = 135 km B covers 3/2*36 = 54 km from Q when C starts. Cars B and C are (135 - 54)km apart when C starts => PR = 81km. If C and B meet => they meet in the mid-way as they have same speed. Ö Choice (3) is the right answer Quantitative Question # 173 (solved by vikas) --------------------------------------------------------

ABCD is a trapezium with AB parallel to CD. M is the midpoint of AD, <MCB = 150˚, BC = 4 and MC = 12. The area of ABCD (is) (1) 16

(2) 24

(3) 16√3

(4) can not be determined

(5) none of these

Solution: Extend CM to meet BA at R. Triangles MRA and MCD are congruent => RC = 24. Area of triangle CRB = area of trapezium ABCD = 1/2*24*4*sin150 = 24 Ö Choice (2) is the right answer Quantitative Question # 173 (solved by warrior, kekambus) -------------------------------------------------------14 students are seated on chairs in a 3X5 array such that the middle seat is unoccupied. A teacher wishes to reassign the seats so that each student moves exactly one place (either to his/her left, right, straight back, or straight forward) . In how many ways is the reassignment possible? (1) 13

(2) 29

(3) 0

(4) 1

(5) 27

Solution: Colour each seat alternately with green and black starting with green => 8 green seats are occupied and 6 blacks are occupied. We wish to make 8 blacks occupied in the reassignment, but only 7 are available. Ö Choice (3) is the right answer Quantitative Question # 175 (solved by kaizen_2007, kekambus, sandeep jha) -------------------------------------------------------A point on a circle inscribed in a square is 1 and 2 units from the two closest side of the square. What is the area of the square? (1) 36

(2) 80

(3) 100

(4) can not be determined

Solution: Please refer the discussions thread. Ö Choice (3) is the right answer Quantitative Question # 176 (solved by doomsayer, Arlen) --------------------------------------------------------

(5) none of these

Amirchand is selling some articles. Amirchand is offering a discount of 33.33% if one pays by credit card. Amirchand has marked on the article in such a way that after giving a discount, he still manages to get a profit of 25%. Garibchand uses false weighing balance and decieves Amirchand by 20% and he also pays the amount by credit card. If Garibchand gives the same article to another customer at 40% discount on marked price of Amirchand and Garibchand has a profit of Rs 20, then what is the cost price of the article in rupees? (1) 160

(2) 200

(3) 240

(4) 300

(5) none of these

Solution: CPa = x => MPa = 15x/8. Using false weighings the MPg = 15x/8 but actual MPa = 15x/8*6/5 = 9x/4. Credit card payment means Garibchand shells out 5x/4 from his pocket for what he actual should have 3x/2. Here, he is already in profit by Rs x/4. 40% discount on 15x/8 = 9x/8 is SPg. => profit = 9x/8 - 5x/4 = -x/8 Garibchand made a profit of Rs 20 = -x/8 + 3x/2 = 11x/8 => x = Rs 160/11. Alternate interpretation The values of MP: SP: CP: Actual MP: MP after deceiving: SP paid by Garibchand = 15: 10: 8: 15: 12: 8 If Garibchand sells the article at 40% discount on MP, he is selling at 9 parts on the ratio scale. On the same ratio scale the difference between SP and CP is Rs 1 and actual is Rs 20. Ö Choice (1) is the right answer Quantitative Question # 178 (solved by kekemabus, warrior, Rohit.sinha, mehtamanas, getneonow, Arun Paliwal, pallavgarg2003) -------------------------------------------------------The area bounded by the region |x| + |y| + |x+y| <= 2 is (1) 2

(2) 3

(3) 2√3

(4) 4

(5) none of these

Solution: The figure formed will be a hexagon. Please refer the discussions thread for the solution. The approach is to form cases and then plot the graph quadrant-wise. Ö Choice (2) is the right answer

Quantitative Question # 178 (solved by vivekr, sandeep jha, Arun Paliwal) -------------------------------------------------------A set of 3 distinct elements which are in arithmetic progression is called a fundootrio. What is the largest number of fundoo-trios that can be subsets of a set of 15 distinct real numbers? (1) 42

(2) 45

(3) 49

(4) 75

(5) none of these

Solution: Let X be one of the elements. What is the largest number of trios that can have X as middle element? Obviously, at most max(b,a), where b is the number of elements smaller than X and a is the number larger. If n = 15, then the no. is at most 0 + 1 + 2 + ... + 6 + 7 + 6 + ... + 1 + 0 = 49. Ö Choice (3) is the right answer Quantitative Question # 179 (solved by deoarshi) -------------------------------------------------------How many real numbers r exist such that the roots of the equation x^2 + rx + 6r = 0 are both integers? (1) 6

(2) 10

(3) 4

(4) 8

(5) none of these

Solution: r(r-24) is a perfect square => (r-12)^2 = p^2 + 144, a perfect square. If r > 0, put r-12 as k => (k-p)*(k+p) = 144 = 72*2 = 36*4 = 24*6 = 18*8 = 12*12 => k = 37, 20, 15, 13, 12 => r=49, 32, 27, 25, 24. When r is non-positive, Let -r = t => t(t+24) is a perfect square. We repeat the same process (t+12)^2 is a perfect square, put t+12 as k => t = 25, 8, 3, 1, 0 => r = -25, -8, 3, -1, 0. Hence, 10 solutions in all. Ö Choice (2) is the right answer Quantitative Question # 180 -------------------------------------------------------Each of the following 3 statements (S) is followed by an inference (I). Answer which pair of statement and inference is not true. (A) S: kekambus found 3 different solutions to the equation f(x - 2/x) = 0 => I: He could have definitely found 1 more different solution.

(B) S: Doomsayer decided to solve a problem where a, b, c are non-zero real numbers such that a^2 + b^2 + c^2 = 1 and a(1/b + 1/c) + b(1/c + 1/a) + c(1/a + 1/b) = -3 => I: Exactly 3 distinct values of a+b+c exist.

(C) S: Arvind and Vivek start running simultaneosly from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduce to half of their current speeds => I: If the length of the circular track is 1500m, Arvind and Vivek meet on the track more than 9 times. (1) only A

(2) only C

(3) B and A

(4) C and B

(5) none of these

Solution: (A) If x - 2/x = a => x^2 - ax - 2 = 0 => D = a^2 + 8 => the equation has unequal roots for the same value of a => the number of roots of f(x - 2/x) if finite are always even. (B) If a+b+c is x and ab+ac+bc = y then x^2 = 1 + 2y and xy = 0 => x = 0, 1, -1 (C) Total distance Shrikant can cover = 10/60*(15 + 15/2 + 15/4 + ...) = 5 km Total distance Sachin can cover = 10/60*(25 + 25/2 + 25/4 + ...) = 25/3 km First time they cover 750 m and subsequently they cover a distance of 1500 m to meet. The total distance they cover together is 40/3 km. Number of meetings possible is 1 + (40000/3 - 750)/1500 = 9.4 => They can meet only 9 times and not more. Ö Choice (2) is the right answer Quantitative Question # 180 -------------------------------------------------------Each of the following 3 statements (S) is followed by an inference (I). Answer which pair of statement and inference is not true. (A) S: kekambus found 3 different solutions to the equation f(x - 2/x) = 0 => I: He could have definitely found 1 more different solution. (B) S: Doomsayer decided to solve a problem where a, b, c are non-zero real numbers such that a^2 + b^2 + c^2 = 1 and a(1/b + 1/c) + b(1/c + 1/a) + c(1/a + 1/b) = -3 => I: Exactly 3 distinct values of a+b+c exist.

(C) S: Arvind and Vivek start running simultaneosly from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduce to half of their current speeds => I: If the

length of the circular track is 1500m, Arvind and Vivek meet on the track more than 9 times. (1) only A

(2) only C

(3) B and A

(4) C and B

(5) none of these

Solution: (A) If x - 2/x = a => x^2 - ax - 2 = 0 => D = a^2 + 8 => the equation has unequal roots for the same value of a => the number of roots of f(x - 2/x) if finite are always even. (B) If a+b+c is x and ab+ac+bc = y then x^2 = 1 + 2y and xy = 0 => x = 0, 1, -1 (C) Total distance Shrikant can cover = 10/60*(15 + 15/2 + 15/4 + ...) = 5 km Total distance Sachin can cover = 10/60*(25 + 25/2 + 25/4 + ...) = 25/3 km First time they cover 750 m and subsequently they cover a distance of 1500 m to meet. The total distance they cover together is 40/3 km. Number of meetings possible is 1 + (40000/3 - 750)/1500 = 9.4 => They can meet only 9 times and not more. Ö Choice (2) is the right answer Quantitative Question # 182 (vineetvijay, doomsayer) -------------------------------------------------------At Pizza-Hut pizzas are made only on an automatic pizza-making machine. The machine continually makes different sorts of pizzas by adding different sorts of toppings on a common base. The machine makes the pizzas at the rate of 1 pizza per minute. The various toppings are added to the pizza in the following manner. Starting from every pizza, every fifth pizza is topped with pepperoni, every seventh with olive and baby corn, every eigth with mushroom, and the rest with chesse and tomatoes. The machine works for 13 hours per day without any breaks in between. How many pizzas per day are made with cheese and tomatoes as topping? (1) 418

(2) 438

(3) 458

(4) 478

(5) 498

Solution: In 13 hours exactly 780 pizzas are made out of which we need all those numbers which are neither multiples of 5 nor 7 nor 8. A U B U C = A + B + C - A/B - A/C - B/C + A/B/C Multiples of 5 are 780/5 = 156 Multiples of 7 are [780/7] = 111 Multiples of 8 are [780/5] = 97 Multiples of 5 and 7 are [780/35] = 22 Multiples of 5 and 8 are [780/40] = 19 Multiples of 7 and 8 are [780/56] = 13

Multiples of 5, 7 and 8 are [780/270] = 2 Since, numbers are counted from 1st pizza => A U B U C = 156 + 112 + 98 - 23 - 20 - 14 + 3 = 312 => answer is 468 and not among the choices. Quantitative Question # 183 (krishp) -------------------------------------------------------Let ABCDE be a regular pentagon and M be a point inside the pentagon such that <(MBA) = <(MEA) = 42˚. Then <(CMD) equals

(1) 48˚

(2) 60˚

(3) 69˚

(4) 78˚

(5) 111˚

Solution: We will prove that the triangle MCD is equilateral. Let M' be a point inside ABCDE such that M'CD is equilateral. Then triangles CM'B and DM'E are both isoscles having 2 sides equal. <(M'CB) = <(DCB) - <(DCM') = 48 degrees, and by symmetry <M'DE = 48 degrees => <(M'BC) = <(M'EA) = 66 degrees. It follows <(M'BA) = <(M'EA) = 42 degrees => M and M' coincide. Hence option (2) is the right choice. Quantitative Question # 184 (answered by aravindva,sandeep jha,kaizen_2007) -------------------------------------------------------Let a^2 + 7b^2 + ab = 106 and a^2 - 2b^2 + ab = 30. Find the value of |2a+b|.

(1) 13

(2) 14

(3) 15

(4) 16

(5) None of the foregoing

Solution: a^2 + 7b^2 + ab + 3*( a^2 - 2b^2 + ab )= 106+3*30=196 =>4a^2+b^2+4ab=196 =>(2a+b)^2=14^2 =>|2a+b|=14 Hence option (2) is the right choice. Quantitative Question # 185 (answered by none) -------------------------------------------------------A right circular cone has base radius 1. The vertex is V. C is a point on the circumference of the base. The distance VC is 3. A particle travels from C around the cone and back by the shortest route. Its minimum distance from V is

(1) 5/4

(2) 2

(3) 5/3

(4) 3/2

(5) None of the foregoing

Solution: The key is to cut the cone along VC and unroll it to give the sector of a circle. If the sector is bounded by VC and VC' and the arc CC', then evidently the particle travels along the straight line CC'. It is closest to V when it is at the midpoint M.. The arc CC' has length 2π, so the
(2) 4

(3) 7

(4) 9

(5) 1

Solution: Please refer the discussions thread for the solution Hence, option (4) is the right choice. Quantitative Question # 187 -----------------------------------------------------How many ordered triplets (x, y, z) of positive integers less than 10 is the product xyz divisible by 20? (1) 72

(2) 120

(3) 90

(4) 102

(5) none of these