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Hope you liked the first article on Quick Quant. Have a look at these five questions and see if you can crack this set in five minutes.

Q.1 Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose? (CAT 2005)
(1) 5
(2) 10
(3) 9
(4) 15

Q.2 The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n-1)(n-2)…3.2.1 is not divisible by n is (CAT 2003)
(1) 5
(2) 7
(3) 13
(4) 14

Q.3 Let g(x) = max (5 – x, x + 2). The smallest possible value of g(x) is (CAT 2003)
(1) 4.0
(2) 4.5
(3) 1.5
(4) none of these

Q.4 If pqr = 1 then 1/(1+p+q^(-1) ) + 1/(1+q+r^(-1) ) + 1/(1+r+p^(-1) ) is equivalent to (CAT 2002)
(1) p+q+r
(2) 1/(p+q+r)
(3) 1
(4) p^(-1)+q^(-1)+r^(-1)

Q.5 In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. If was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is (CAT 2005)
(1) 200
(2) 216
(3) 235
(4) 256

These questions can be solved using simple logic and conceptual understanding. We’ll go through each of them now.

Q.1 Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose? (CAT 2005)

Solution:

Let’s say, these 6 people are: E1, E2, E3, F1, F2, F3

E1 and F1 can communicate with each other as one of them is bi-lingual.

E2 and E3 individually call E1 and share their secrets with E1. (2 calls)
F2 and F3 individually call F1 and share their secrets with F1 (2 calls)

Once this is done, E1 and F1 call each other and share secrets with each other so that they know all secrets now (1 call)

E1 now calls E2 and E3 respectively and share others’ secrets with E2 and E3 including Frenchmen secrets (2 calls)
F1 now calls F2 and F3 respectively and share others’ secrets with F2 and F3 including Englishmen secrets (2 calls)

In total, 9 calls are required to achieve this.

Q.2 The number of positive integers n in the range 12 ≤ n ≤ 40 such that the product (n-1)(n-2)…3.2.1 is not divisible by n is (CAT 2003)

Solution:

The only way a factorial is not divisible by the next natural number is when the number is prime.
For example: 2! Is not divisible by 3, 4! Is not divisible by 5, 6! Is not divisible by 7 … and so on.

So the question is as simple as finding the number of prime numbers between 12 and 40 = 13, 17, 19, 23, 29, 31, 37. Answer is 7.

Q.3 Let g(x) = max (5 – x, x + 2). The smallest possible value of g(x) is (CAT 2003)

Solution:

We are trying to find the minimum of value of the maximum of two values. By equating the two, we get

5 – x = x+ 2
3 = 2x
x = 1.5

Hence, g(x) = max(5–1.5, 1.5+2) = 3.5

Tip: Don’t mark the answer is 1.5 in a hurry. That’s the value of x and not the final answer.

Q.4 If pqr = 1 then 1/(1+p+q^(-1) ) + 1/(1+q+r^(-1) ) + 1/(1+r+p^(-1) ) is equivalent to (CAT 2002)

Solution:

As nothing has been mentioned about the nature of p, q, r. Let p = q = r = 1
On substituting, we get:

1(1 + 1 + 1^-1) + 1/(1 + 1 + 1^-1) + 1/(1 + 1 + 1^-1) = 1/3 + 1/3 + 1/3 = 3/1 = 1

Substituting the same values of p, q, r in the options, we have our options as: 3, 1/3, 1, 3. Hence, the answer is Option 3.

Q.5 In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. If was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is (CAT 2005)

Solution:

We just need to find the number of boys and girls first.
All girls: 45 games: nC2 = 45 hence, n = 10
All boys: 190 games: nC2 = 190, n = 20
10C1 * 20C1 = 200.

Tip: The first part of the question can be cracked in a matter of seconds like this. When you have a match of two players and every player is playing one match with every other player, and the number of matches is given, multiply the number of matches by 2 and try to find two consecutive numbers who give that as their product. Here, we had 45. 45*2 = 90. 90 is product of two consecutive numbers 9 and 10. So you have 10 players. Similarly, 190*2 = 380. 380 is product of 19 and 20. So 20 players.

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