Caselets

Caselets In caselets data are given in the form of paragraph. No charts/graphs are provided with the data. Caselets vary considerably in length, in the amount of information contained, in different sentences and paragraph. While reading a caselets it s always advisable to underline the important fact and figures and if necessary make your own table/chart/graphs for solving the questions. In CAT Caselets can be asked either/both in Quants and Data Interpretation section and it s similar to the Reading Comprehension part in English Usage section. The best way to mastery caslets is to practice

Example 1: Directions for Q. 1 to 5: Refer to the following information and the answer the following questions. People Power Corporation presently employs three Managers (A, B and C) and five recruitment agents (D, E, F, G and H). The company is planning to open a new office in San Jose to manage placement of software professionals in the US. It is planning to relocate two of the three managers and three of the five recruitment agents to the office at San Jose. As it is an organization which is highly people oriented the management wants to ensure that the individuals who do not function well together should not be made as a part of the team going to the US. The following information was available to the HR department of People Power Corporation. • • •

•

Managers A and C are at each others throat and therefore cannot be sent as a team to the new office. C and E are excellent performers in their own right. However, they do not function together as a team. They should be separated. D and G have had a major misunderstanding during the last office picnic. After the picnic these two have not been in speaking terms and should therefore not be sent as a team. D and F are competing for a promotion that is due in another 3 months. They should not be a team.

Q1. If D goes to the new office which of the following is (are) true? I. C cannot go II. A cannot go III. H must also go (a) I only (b) II and III only (c) I and III only (d) I, II and III

2. If A is to be moved as one of the Managers, which of the following cannot be a possible working unit? (a) ABDEH (b) ABFGH (c) ABEGH (d) ABDGH 3. If C and F are moved to the new office, how many combinations are possible? (a) 4 (b) 1 (c) 3 (d) 5 4. Given the group dynamics of the Managers and the recruitment agents, which of the following is sure to find a berth in the San Jose office? (a) B (b) H (c) G (d) E 5. If C is sent to the San Jose office which member of the staff cannot go with C? (a) B (b) D (c) G (d) F ANSWERS: 1. (c) 2. (d) 3. (b) 4. (a) 5. (b)

Example 2 Ghosh Babu took voluntary retirement in Dec. 1991 and received a certain amount of money as retirement benefits. On Jan 1, 1992, he invested the entire amount in shares. At the end of the month, he sold all his shares and realised 25% profit. On Feb 1, he reinvested the entire amount in shares which he sold at the end of the month at a loss of 20%. Again, he invested the entire amount on Mar 1 in a new company. At the end of the month, he sold the new company to a friend and realised a profit of 20% in the process. He invested the entire amount in shares on Apr 1, which he sold at the end of the month for Rs. 1,08,000 incurring a loss of 10%. 1. What is the amount of retirement benefits received by Ghosh Babu? a) Rs. 1,08,000 b) Rs. 1,25,000 c) Rs. 1,20,000 d) Rs. 1,00,000

2. The percentage profit received by Ghosh Babu between Jan 1 and Apr 30 is: a) 8.00% b) 15.00% c) - 10.00% d) None of these 3. The amount of loss incurred by Ghosh Babu based on his operation in Apr 1992 is: a) Rs. 25,000 b) Rs. 12,000 c) Rs. 20,000 d) Rs. 8,000 4. The maximum amount invested by Ghosh Babu in any one month was in: a) January b) February c) March d) April Answers: 1. d Let the amount received by Ghosh Babu in Dec. 1991 be Rs. x, as retirement benefits: Therefore, investment in the month of Jan 1992 = 100 Profit of 25% at the end of Jan 1992. Hence, investment in the month of Feb 1992 = 125 Loss of 20% at the end of Feb 1992 Hence, investment in the month of March 1992 = 100 Profit of 20% at the end of March 1992 Hence, investment in the month of April 1992 = 120 Loss of 10% at the end of April 1992 Therefore the amount left at the end of April 1992 = 108 Amount at the end of April 1002 = Rs. 1,08,000 Therefore, simply equating figures, he would have started with Rs 1,00,000 2. a % Profit between Jan 1 and Apr 30 = (1.08x - x/x) X 100 3. b Investment in the month of April = Rs. 1,20,000 Amount received at end of April = Rs. 1,08,000 Therefore, Loss = Rs. 12,000 4. b Maximum amount invested by Ghosh Babu is in the month of February = Rs. 1,25,000

Reminder Theory This chapter is incomplete

1. 2. 3. 4. 5.

Basic Concepts Fermat’s Theorems Euler’s Theorems Chinese Remainder Theorem Discrete Logarithms

Examples : 1. What is the remainder of (7777...upto56 digits)/19 ?

Froum post : http://www.cat4mba.com/node/2971#comment-1378 The basic funda is Any number a repeated N times is divisible by P where P is a prime number and N is the recurring decimal of P For example: Recurring decimal of 7 is 6 since 1/7 = 0.142857 ( Repeats after six digits) So 111111 (6 times) is divisible by 7 222222 (6 times) is divisible by 7 555555(6 times) is divisible by 7 Similarly for 19 the recurring decimal is 18 So 111111, 111111, 111111 (18 times) is divisible by 19 And 777777,777777,777777 (18 times) is divisible by 19 => 7777(54 times) is divisible by 19 Now 7777( 56 times) = 7777( 54 times) 0 0 + 77 =>required reminder = reminder of 77 divided by 19 = 1 Now Big Question How to find the recurring decimal for a prime number? From my experience I found the number P-1 is always a multiple of recurring decimal and this information is enough to solve most of the questions. For example 11 , 1/11 = .0909 ; recurring decimal =2 and p-1=10 is a multiple of 2 13, 1/13 = 0.0769230; recurring decimal = 6 and p-1=12 is a multiple of 6. Notes: 1. I have never tried to search/prove the above.

2. There are few exception like number 3 and 5 and I would love to know that if there z any more

================================== Chinese Remainder Theorem is used to find integers x that simultaneously solve the following equation x = a1 (mod m1) x = a2 (mod m2) ..... x = an (mod mn) For example : A number when divided by 4 gives a reminder 3, when divided by 3 gives a reminder 2 and when divided by 5 gives a reminder 4. Find the number METHOD1 (Simple Logic) x =3 (mod 4 ) ------- (eq1) x =2 (mod 3) -------- (eq2) x = 4 (mod 5) ------- (eq3) The values of x which satisfy equation 1 are 7, 11, 15 and etc. Our first target is to find the value of x from this series which is divisible by both 3 and 5 (i.e by 15) The first number in the series that is divisible by 15 is 15. So lets take n1 = 15. Now we‘ll follow the same process for eq (2) The series is 5, 8, 11, 14 . . . . . The first number which is divisible by both 4 and 5 (4. 5 = 20) and gives a reminder 2 when divided by 3 is 20 (n2). Now we‘ll follow the same process for eq (3) The series is 9, 14, 19. . . . . The first number which is divisible by both 4 and 3 (4. 3 = 12) and gives a reminder 4 when divided by 5 is 24 (n3). Now the number which satisfy all the above three equation is N = n1 + n1 + n3 = 15 + 20 + 24 = 59. N = 59 + (4 * 3 * 5)p where p is any number

METHOD2 (Stepwise Calculation) Let x = a1 (mod m1) x = a2 (mod m2) ..... x = an (mod mn) Step 1: Calculate M = m1 x m2 x m3. . . . . x mn Step 2:

Calculate Z1 = M/m1 Z2 = M/m2 Z3 = M/m3 . . . . . . . . . .. . . .. Zn = M/mn Step 3: Calculate Y1 where Z1Y1 = 1 (mod m1) Y2 where Z2Y2 = 1 (mod m2) . . . . . . . . . .. . . .. Yn where ZnYn = 1 (mod mn)

Step 4: Calculate x where x = a1 y1 z1 + a2 y2 z2 + a3 y3 z3+ . . . . . . . Example: x =3 (mod 4 ) ------- (eq1) x =2 (mod 3) -------- (eq2) x = 4 (mod 5) -------- (eq3) Step 1: Calculate M = m1 x m2 x m3. . . . . x mn = 60 Step 2: Calculate Z1 = M/m1 = 15 Z2 = M/m2 = 20 Z3 = M/m3 = 12 . . . . . . . . . .. . . .. Zn = M/mn Step 3: Calculate Y1 where Z1Y1 = 1 (mod m1) 15Y1 = 1 (mod 4) Calculate it by trial and error method For Y1= 1 ; 15 = 3 (mod 4) For Y1= 2 ; 30 = 2 (mod 4) For Y1= 3 ; 35 = 1 (mod 4) So Y1=3 satisfy the condition Similarly Calculate Y2 where Z2Y2 = 1 (mod m2) 20Y2 = 1 (mod 3) => Y2 = 2 ......... Calculate Y3 where Z3Y3 = 1 (mod m3) 12Y3 = 1 (mod 5) => Y3 = 3 .. . . .. Yn where ZnYn = 1 (mod mn)

Step 4: Calculate x where x = a1 y1 z1 + a2 y2 z2 + a3 y3 z3+ . . . . . . . X = 3 * 3 * 15 + 2 * 2 *20 + 4 * 3 *12 = 135 + 80 + 144 = 359 359 = 59 mod (60) And the number in generalized form = 59 + 60P

EXAMPLES 1. find out the last non zero digit of 37! ANSWER: From the post http://www.cat4mba.com/node/5310 I have come across this question many times in many places so thought of writing a detail solution. Please go through it carefully and revert back if you have any problem at any step. Initially it might seem lengthy but once you have the expertise it wont take more than couple of minutes. STEP1 Find out the prime factors in the factorial 37! = 2

34

x5

8

x3

17

x7

5

x 11

3

x 13

2

x 17

2

x 19 x 23 x 29 x 31 x 37

STEP2 37! Contains 8 zeros at the end as we have 8 fives in 37!. So we can get the first non-zero digit by dividing 37! With 109 but that is not an easy. In stead of doing that we ‘ll first divide 37! With 108 and then we ‘ll try to find out the reminder of the new number when divided by 10. (Try to clearly understand what is the difference between above two) STEP3 37!/108 = ( 234 x 58 x 317 x 75 x 113 x 132 x 172 x 19 x 23 x 29 x 31 x 37 )/108 = 226 x 317 x 75 x 113 x 132 x 172 x 19 x 23 x 29 x 31 x 37 = N STEP4 We ‘ll find the reminder of N divided by 10. 226 = 2 x 225 = 2 x 25 = 2 x 2 = 4 317 = 3 x 316 = 3 x 34 = 3 x 1 = 3 75 = 7 x 492 = 7 x (-1)2 = 7 x 1 = 7 113 = 1 132 = 9 172 = 72 = 9 So reminder of N divided by 10 reduces to 4 x 3 x 7 x 1 x 9 x 9 x 19 x 23 x 29 x 31 x 37

Now multiply left to 2 digits get the reminder divided by 10 and then proceed to the right 4 2 4 6 4 6 8 2 2

x x x x x x x x x

3=2 7=4 9=6 9=4 19 (or 9)= 6 3( or 23) = 8 9=2 1=2 7=4

And that’s the answer 4

Euler Function In number theory, the totient ( also called phi) Φ (n) of a positive integer n is defined as the number of positive integers less than or equal to n that are co prime to n. For example, Φ(9) = 6 since the six numbers 1, 2, 4, 5, 7 and 8 are co prime to 9. A few first values: Φ (1)=1, Φ (2)=1, Φ (3)=2, Φ (4)=2, Φ (5)=4, Φ (6)=2, Φ (7)=6, Φ (8)=4, Φ (9)=6, Φ (10)=4, Φ (11)=10, Φ (12)=4, Φ (13)=12, etc., do not appear to follow any law. But there is a formula discovered by Euler to which helps in calculating these numbers. According to Euler

Example: 12 = 22 • 3, i.e. Φ (12) = 12 • (1 - 1/2) • (1 - 1/3) = 12 • 2 / (2 • 3) = 4. Reminder Theorem

If Φ(n) is the eulers numbr of integer n, the reminder obtained when mΦ(n) is divided by n is 1. Properties 1. The sum over the Euler function values Φ (d) of all divisors d of an integer number n

exactly gives n. E.g. for n=12: Φ (1) + Φ (2) + Φ (3) + Φ (4) + Φ (6) + Φ (12) = 1 + 1 + 2 + 2 + 2 + 4 = 12 2. if GCD(a,n) = 1, then aΦ(n) = 1 (mod n). For example, Φ (12)=4, so if gcd(a,12) = 1, then a4 = 1 (mod 12). 3. Φ (m1m2) = Φ (m1) Φ (m2) for co prime m1 and m2 Reference 1. http://www.cat4mba.com/node/3359#comment-http://cat4mba.com/files/Euler.png1355 4. From Nishit's comment @ http://www.cat4mba.com/node/3359?page=4 Euler's Function for number N e(N) = N(1 - 1/p1)(1 - 1/p2)(1 - 1/p3) . . . . . . Where p1, p2, p3, . . . , pn are prime factors of N. For example if N=60 = 2 x 2 x 3 x 5 then e(N) = 60 (1-1/2) (1-1/3) (1-1/5) = 60 x 1/2 x 2/3 x 4/5 = 16 I think Its wrong to calcualte like

60(1-1/2)(1-1/2)(1-1/3) (1-1/5)

Another way of finding it 60 = 22 x 31 x 51 So e(60) = 21 x(2-1) x 30 x (3-1) x 50 x (5-1) = 16

i.e if N = Xa Yb Zc then

e(N) = Xa-1(X-1) Yb-1(Y-1) Zc-1(Z-1)

Fermat's little Theorem

If p is a prime then for any integer a we have ap = a modulo p.

i.e. If p is a prime and n is an integer then np–n is divisible by p. Example : 7 is a prime so n 7 – n is divisible by 7 . For n = 2 : 2 7 – 2 = 128 – 2 = 126 is divisible by 7 Questions Q1 : what is the reminder when 1139 is divided by 19 Q2. Find the reminder when 591 is divided by 91 Q3. Find the following reminders a. 757575 is divided by 37 b. 2100 is divided by 101 c. 20 51 97 is divided by 17 Corollary :

nq – n is divisible by q where q is a prime number or product of two prime numbers. Important Points: 1. If P be a prime number such that ap – bp is divisible by p, then it is also divisible by p2 2. If an integer n is greater than 2, then the equation an + bn = cn has no solutions in non-zero integers a, b, and c. (Fermat’s Last Theorem) 3. If p be prime and a is prime to p, then a(p-1) – 1 is multiple of p. Submitted by praveen_84 on Fri, 2008-02-01 01:02. •

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commented on: Fri, 2008-02-01 14:31

OSO points: 76 Location: Comments : 14

Q1 answer As per fermat's little theorem (1119 - 11)/19 = k

=> 1119 - 11 = 19k => 1119 = 19k + 11 (squaring it )

=> 1138 = some multiple of 19 + 121 So reminder of 1138 divided by 19 is 7 => reminder of 1139 divided by 19 is 77 = reminder of 1139 divided by 19 is 1

Others a. (any number x + 1) N when divided by x gives reminder 1. b. (any number x ) N when divided by (x + 1) gives reminder 1 when N is even and reminder x when N is odd. c. xn + anis exactly divisible by (x + a) if n is odd, but not if n is even d. xn - an is divisible by (x + a) if n is even but not if n is odd e xn - an is always divisible by (x - a)