Qqad Test 1 Answers

QQAD, Practice Test 1 Answers prepared by the students of QQAD

Instructions: 1) The duration of this test is 50 minutes and the test is meant to be taken in one-go without any break(s). 2) This test has 25 questions. Each question carries +4 marks on answering correctly. 3) Each wrong answer attracts penalty of negative 1 mark. 4) Use of slide rule, log tables and calculators is not permitted. 5) Use the blank space in the question paper for the rough work.

Keys to 1st Practice Test 1 -> B 2 -> B 3 -> D 4 -> A 5 -> A 6 -> E 7 -> E 8 -> D 9 -> C 10 -> C 11 -> B 12 -> D 13 -> B 14 -> E 15 -> D 16 -> A 17 -> C 18 -> E 19 -> C 20 -> A 21 -> E 22 -> A 23 -> B 24 -> A 25 -> D

(1) Sarah and Neha start running simultaneously 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 reduces to half of their current speeds. If the length of the circular track is 1500 m, how many times will Sarah and Neha meet on the track? (A) 6

(B) 9

(C) 11

(D) 7

(E) 8

Solution: (by Implex) 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

=> choice (B) is the right answer (2) A cone of radius 1 unit and height 2 units just fits inside a cylinder with their axis perpendicular. The radius of the cylinder in unit is (A) 1

(B) 1.25

(C) 1.5

(D) 2

(E) more than 2

Solution: (by Implex) Triangle ABC be the cross section of cone and circle K be the cross section Of Cylinder Let AM be perpendicular to BC clearly BC=2( diameter of cone) AM=2 (Height of cone) Let O be the center of K , so O lies on AM In right traingle OBM , OB=r, BM=1 and OM=AM-AO=2-r r^2=1+(2-r)^2 r=1.25

=> choice (B) is the right answer (3) There are a certain facts about the software engineers working with an IT firm. Total number of software engineers is 70 out of which 30 are females. 30 people are married. 24 software engineers are above 25 years of age. Out of all married software engineers, 19 are above 25 years, of which 7 are males. 12 males are above 25 years and overall 15 males are married. How many unmarried females are there which are above 25? (A) 12

(B) 8

Solution: (by Implex)

(C) 7

(D) 0

(E) 11

15 males are married, so 15 females are married now 7 males above 25 are married. so (19-7)=12 females above 25 are married given that 12 males are above 25 years so 24-12 =12 females are above 25 years so all females above 25 years are married this means 0 unmarried female above 25

=> choice (D) is the right answer

DIRECTIONS for Questions 4 and 5: Each question is followed by two statements X and Y. Answer each question using the following instructions: Choose A Choose B Choose C Choose D Choose E

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 be answered by neither X and Y

(4) Let p(x) = x^2 + 40. Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number? (X) p(i) – p(j) is not a composite number (Y) p(2i) + p(2j) is a composite number Solution: (by Implex) p(i) – p(j) is not a composite number =>i^2-j^2 is a prime as i ,j are positive integers and i >j, ( i^2-j^2) can't be 1 =>(i+j)(i-j)= prime so i-j=1 let p be the prime so i=(p+1)/2 j=(p-1)/2 clearly p is not 2 hence all p is odd p(i) + p(j)=80 +(p^2+1)/2 now p^2=6k+1 ( can be easily proved) [ p(p-1)(p+1) is divisible by 6 now p is a prime so p^2-1=6k p^2=6k+1, for any prime p greater than 3] therefore p(i) + p(j)=80 +(p^2+1)/2 becomes 80+(6k+2)/2=81+3k=3(27+k) so not a prime => can be answered by using X now, p(2i) + p(2j) is a composite number 4(i^2+j^2+20) is composite now i and j can be anything can't make any conclusions

=> choice (A) is the right answer

(5) What is the length of the side AB of triangle ABC? (X) AB <= AC = 2, and area of triangle ABC is 2 (Y) Exactly two sides have integer length Solution: Area = ½*(AB).(AC).sin A => 2 = ½*AB.2.sinA => ABsinA= 2.

But AB <=2 and sin A <= 1 => AB = 2. => choice (A) is the right answer (6) The numerical value of f(1/10) + f(2/10) + …. + f(9/10), where f(x) = 9^x/(3+9^x) is (A) 10/3

(B) 4

(C) √10

(D) 5

(E) 9/2

Solution: (by Implex, thebornattitude) f(x) = 9^x/(3+9^x)=1-3/(3+9^x) now look at this 3/(3+9^x) +3/(3+9^(1-x))= [9+9+3(9^x++9^(1-x))]/[9+9+3(9^x++9^(1-x))]=1 f(1/10) + f(2/10) + …. + f(9/10)=9-4 -3/(3+sqrt(9))=9-4-1/2=9/2

=> choice (E) is the right answer

(7) A blackboard bears a half-erased mathematical calculation exercise: 2 3 ? 5 ? + 1 ? 6 4 2 ------------4 2 4 2 3 In which number system was this calculation performed? (A) 4

(B) 9

(C) 8

Solution: (by Implex) 23A5B 1c642 ------42423 clearly number system is greater than 6 so we get 7, 8, 9, now clearly B=1 now the moment B is 1, there is no carry gives 5+4=9, 9=2(mod7) hence number system is 7

(D) 5

(E) 7

=> choice (E) is the right answer (8) Given points P1, P2, P3, …, P7 on a straight line, in the order stated (not necessarily evenly spaced). Let P be an arbitrary point selected on the line and let s be the sum of undirected lengths PP1, PP2, PP3, …, PP7. Then s is smallest if and only if the point P is (A) midway between P1 and P7 (B) midway between P2 and P6 (C) midway between P3 and P5 (D) at P4 (E) at P1 Solution: Let P1, P2, P3, …, P7 be points in a co-ordinate plane such that P1 = (0, 0), P2 = (a, 0), P3 = (b, 0), …., P7 = (f, o) such that a < b < … < f. Let X be arbit point on P1P7 such that X = (x, o) where a <= x <= f. We want to minimize s = x + |x-a| + |x-b| + |x-c| + |x-d| + |x-e| + |x-f| If X = P4 then s = c + (c-a) + (c-b) + 0 + (d-c) + (e-c) + (f-c) = d+e+f-a-b If X = P1 then s = a + 0 + (b-a) + (c-a) + (d-a) + (e-a) + (f-a) = d=e+f-2a > d+e+f-a-b For choice (A) -> X = (a+f)/2, (B) -> X = (b+e)/2, (C) -> X = (c+d)/2 we see that none is < d+e+f-a-b => choice (D) is the right answer (9) A 33 rpm record which normally plays for 30 minutes was inadvertently started at 45 rpm, then switched to 33 rpm when mistake was realized. Altogether the record played for 26 minutes. For how many minutes was it playing at 45 rpm?

(A) 9

(B) 10

(C) 11

(D) 8

(E) 12

Solution: (by Implex) 33*30=45x+33(26-x) 33.4=12x x=11

=> choice (C) is the right answer (10) Let S be a 6 element set. Then the number of pairings (3 pairs) of S is (A) 9

(B) 12

(C) 15

(D) 20

Solution: {(1, 2), (3, 4), (5, 6)} is one pairing of 3 pairs. Keeping (1, 2) as constant we can have {(3, 5), (4, 6)} or {(3, 6), (4, 5)}  for (1, 2) we have 3 set of pairs to map.

(E) 27

 For (1, x) where x varies from 2 to 6 we will have in all 5*3 = 15 pairing. => choice © is the right answer Directions for questions 11 to 12 : The cost of 1 kg of sugar is Rs 20 while the cost of 1 litre of pure milk is Rs 15. Sweetened milk is prepared by adding a fixed amount of sugar in a litre of milk. (11) If the cost of sweetened milk is Rs 15 per kg, then it can be concluded that the weight of x litre pure milk is y kg more than a litre of sweetened milk where (x, y) is (A) (4, 5)

(B) (4, 3)

(C) (5, 3)

(D) (2, 1)

(E) (3, 2)

Solution: (by Implex) Let the weight of pure milk be a kg/liter assume b kgs of sugar is added to 1 liter of pure milk so now cost is 15+20b=15(a+b), 3+b=3a now y=ax-a-b=a(x-1)-3a+3 if we put x=4 y is 3 (x-1=3)

=> choice (B) is the right answer (12) If the cost of 1 litre of sweetened milk is Rs 16 and its weight is 1.25 kg, then the weight of 1 litre of pure milk is (A) 1.05 kg

(B) 1.04 kg

(C) 1.10 kg

(D) 1 kg

(E) none of these

This is sitter; => choice (D) is the right answer Directions for questions 13 to 14 : A point X is taken on a circle having its centre at O and a chord is drawn at an angle 50 ˚ to OX clockwise to cut the circle at Y. Again a chord is drawn at an angle 50 ˚ to OY clockwise and this process continues until a chord in subsequent process intersects the circle at X. (13) After how many revolution(s) of the circle, the process is completed? (A) 1

(B) 2

(C) 4

(D) 8

(E) 9

Solution: (by Implex) <XOY=80, similarly
=> choice (B) is the right answer (14) At how many points the chords drawn during the process intersect each other inside the circle? (A) 12

(B) 6

(C) 7

(D) 8

(E) 9

Solution:(by Implex) clealry the first 4 chords won't intersect the fifth chord cuts the first teh sixth cuts the first and second seventh cuts second and third eighth cuts third and fourth ninth cuts fourth and fifth so 9 internal intersections

=> choice (E) is the right answer (15) If is a nonzero integer and is a positive number such that the median of the set ?

what is

Solution: Clearly, 0 < b < 1 => a < -1 (always). Thus, a < 0 < b < 1 < 1/b => choice (D) is the right answer

(16) A wooden cube unit on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ? (A) 4

(B) 3

Solution: (by Implex) Total cubes is 6n^3 colored cubes is (1/4)6n^3

(C) 2

(D) 6

(E) none of these

now only the six outer faces are colored = 6n^2 => 6n^2= (1/4)6n^3 => n=4

=> choice (A) is the right answer (17) Katrina, her brother, her son and her daughter are chess players (all relations by birth). The worst player’s twin (who is one of the four players) and the best player are of opposite sex. The worst player and the best player are of the same age. Who is the worst player? (A) Katrina (B) Her brother be determined

(C) Her son (D) Her daughter (E) Can not

Solution: (by Implex) If Katrina is worst player , the best player is female , but this defies other condition that the best player and worst player are of same age now if her brother is worst player best player is then her son. this means katrina her son and her brother are of same age. not possible again if her son is worst player, the best player is male so he must be katrina's brother now if we assume that her son and her brother are of same age this can happen. all other cases are impossible

=> choice (C) is the right answer (18) An elastic string laying along the interval [-2, 2] on the x-axis is stretched uniformly and displaced so that it lays along [3, 9]. What is the new location of the point of the string which was formerly at x = 1? (A) 8

(B) 6

(C) 4.5

(D) 4

(E) 7.5

Solution: (by thebornattitude) length of string previously= 4 now = 6, so unit length of string will now become= 6/4= 3/2.....(i) now part of string at x= 1 means 3 units from starting of string.... means now it will be at 3x3/2 units from starting = 4.5 units from starting.. starting is at x=3.. hence now it will shift to x= 3+4.5 = 7.5...answer

(19) A round table has radius . Six rectangular place mats are placed on the table. Each place mat has width and length as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length . Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is ?

Solution: (by thebornattitude) Refer post 88 here http://www.pagalguy.com/forum/quantitative-questions-andanswers/32067-cat-2008-qqad-practice-test-9.html => choice (C) is the right answer

(20) The number of values of k for which the roots of the equation kx^3 + 2x^2 – 3x + 1 = 0 are in harmonic progression is (A) 0

(B) 1

(C) 2

(D) 3

(E) more than 3

Solution: (by Implex) if we put y=1/x we will get an equation which has roots in AP y^3-3y^2+2y+k=0 let roots be a-b ,a, a+b => 3a=3 => a=1 Also, a(a-b)+a(a+b)+(a^2-b^2)=2 3-b^2=2 b^2=1 k=-a(a^2-b^2)=-1(1-1)=0 but for this value of k we get one of the roots as zero ! which makes the root of original equation -> infinity! => No solution Hence, choice (A) is the right answer

(21) Vineet attends an IPL game in Delhi and estimates that there are 50,000 fans in attendance. Rahul attends an IPL game in Bangalore and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:

i. The actual attendance in Delhi is within of Vineet's estimate. ii. Rahul's estimate is within of the actual attendance in Bangalore. To the nearest 1,000, the largest possible difference between the numbers attending the two games is (A) 10,000

(B) 11,000

(C) 20,000 (D) 21,000

(E) 22,000

Solution: Actual attendance in Delhi can at least be 45, 000 Actual attendance in Bangalore can at most be 60, 000/(0.9) = 66, 666 => choice (E) is the right answer Directions for questions 22 to 25 : A Latin square of order n is an arrangement of n symbols in n rows and n columns such that each symbol appears exactly once in each row and each column. For example, two Latin squares of order three are shown below. 0 1 2

1 L 2= 0

2 0 1

1 2 0

0 M1= 2

2 0 1

Two Latin squares of order n are said to be orthogonal if, upon superimposition of one on the other, each of the n² possible ordered pairs of symbols appears in exactly one cell. For example, the Latin squares L and M, as shown above, are orthogonal since superimposing M on L, we get the structure: 01 12 20

10 21 02

22 00 11

Where in each cell the first entry comes from L and the second entry comes from M. Note that each of the nine possible ordered pairs 00, 01, 02, 10, 11, 12, 20, 21, 22, appears exactly once in each cell. This shows that L and M are orthogonal.

(22) The number of distinct Latin squares of order three, with symbols 0, 1, 2 each of which is orthogonal to both L and M shown above is (A) 0

(B) 1

(C)

2

(D)

3

(E) 6

(23) The numbers of distinct Latin squares of order four, with symbols 0, 1, 2, 3 that can be formed by completing the following incomplete structure equals 0

2 1 2 3

(A) 0

(B) 1

(C)

3

(D)

6

(E) 24

(24) Consider the following two Latin squares of order four: 0 1 2 3

1 0 3 2

2 3 0 1

3 2 1 0

x q 2 1

y r 3 0

z s 1 2

w t 0 3

Here each of x, y, z, w, q, r, s, t belongs to the set {0, 1, 2, 3}. If the two Latin squares are orthogonal, then x equals (A) 0

(B) 1

(C)

2

(D)

3

(E) can not be determined

(25) Suppose we have two Latin squares P and Q which are orthogonal. Both P and Q are of order n(>2) and both of them have the symbols 0, 1, . . . , n – 1. Now suppose, a new square R is formed from Q by replacing the symbols 0, 1, 2 in Q by 1, 2, 0 respectively and keeping the other symbols in Q unchanged. Then the P and R are orthogonal Latin squares (A) For no choice of n (> 2) (C) Only for even n (> 2) (E) None of the above

(B) Only for odd n (> 2) (D) For every n (> 2)