CAT 2023 Slot 1 QA Question 10

"In an examination, the average marks of 4 girls and 6 boys is 24. Each of the girls has the same marks while each of the boys has the same marks. If the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy, then the number of possible distinct integer values of the total marks of 2 girls and 6 boys is...?"

The options are A) 19, B) 21, C) 22, D) 20. The correct answer is supposed to be B, but We analyze why.

First, let's break down the problem. We have 4 girls and 6 boys. The average marks of all 10 students is 24. So, the total marks for all of them together is 10 * 24 = 240.

Let the marks of each girl as G and each boy as B. Since all girls have the same marks and all boys have the same marks, the total marks can be written as 4G + 6B = 240.

We also have a condition that the marks of any girl is at most double the marks of any boy, but not less than the marks of any boy. So, that translates to G ≤ 2B and G ≥ B. So, B ≤ G ≤ 2B.

Our goal is to find the number of possible distinct integer values of the total marks of 2 girls and 6 boys. Let's denote this total as T. So, T = 2G + 6B.

We need to find how many different integer values T can take, given the constraints on G and B.

We have the equations:

We can simplify equation 1. Let's divide both sides by 2:

2G + 3B = 120

So, 2G = 120 - 3B

Therefore, G = (120 - 3B)/2

Since G must be an integer (because the problem mentions integer values for total marks, so I assume G and B are integers), (120 - 3B) must be even. So, 3B must be even because 120 is even. 3B is even only if B is even because 3 is odd. So, B must be even.

Let B = 2k, where k is an integer. Then, G = (120 - 3*(2k))/2 = (120 - 6k)/2 = 60 - 3k.

So, G = 60 - 3k and B = 2k.

Now, applying the constraints:

B ≤ G ≤ 2B

Substituting G and B:

2k ≤ 60 - 3k ≤ 4k

Let's split this into two inequalities:

Solving the first inequality:

2k + 3k ≤ 60 5k ≤ 60 k ≤ 12

Solving the second inequality:

60 ≤ 4k + 3k 60 ≤ 7k k ≥ 60/7 ≈ 8.571

Since k must be an integer, k ≥ 9.

So, k can take integer values from 9 to 12, inclusive.

Let me list the possible values of k:

k = 9, 10, 11, 12

Now, Computing G and B for each k:

For k=9: B = 18 G = 60 - 27 = 33

Check constraints: 18 ≤ 33 ≤ 36. Yes, 33 is between 18 and 36.

For k=10: B = 20 G = 60 - 30 = 30

Check: 20 ≤ 30 ≤ 40. Yes.

For k=11: B = 22 G = 60 - 33 = 27

Check: 22 ≤ 27 ≤ 44. Yes.

For k=12: B = 24 G = 60 - 36 = 24

Check: 24 ≤ 24 ≤ 48. Yes, since G can be equal to B.

So, we have four possible pairs of (G, B): (33,18), (30,20), (27,22), (24,24).

Now, we need to find the total marks T = 2G + 6B for each of these pairs.

Computing T for each:

T = 233 + 618 = 66 + 108 = 174

T = 230 + 620 = 60 + 120 = 180

T = 227 + 622 = 54 + 132 = 186

T = 224 + 624 = 48 + 144 = 192

Wait, that's only four values: 174, 180, 186, 192. But the options are 19,21,22,20. Hmm, that can't be right because the number of possible T values is 4, which isn't among the options. So, I must have made a mistake.

Wait, maybe I oversimplified.

I assumed that G and B are integers because the total marks T is an integer, but actually, the problem doesn't explicitly state that G and B are integers. However, since the total marks of 2 girls and 6 boys must be an integer, and since 2G + 6B must be integer, but G and B could be fractions. But the problem says "distinct integer values of the total marks," so T must be integer, but G and B could be non-integers as long as T is integer.

Wait, but in the initial equation, 4G + 6B = 240, which is an integer. If G and B are fractions, their combination must result in an integer. But perhaps it's easier to assume that G and