CAT 2004 QA Question 25 · Arithmetic - Percentage

c = 8
∴ a + b = 92   ...(1)

Also, $\frac{2b}{a+2b+3c}\ge 0.2$

⇒ $\frac{2b}{92-b+2b+24}\ge 0.2$

⇒ 2b ≥ 23.2 + 0.2b

⇒ $b\ge \frac{232}{18}$

⇒ b ≥ 13   ...(2)

Also, as group A carries at least 60% of the total marks,

$\frac{a}{a+2b+3c}\ge 0.6$

⇒ $\frac{a}{a+2(92 - a)+24}\ge 0.6$

⇒ aa ≥ 0.6a + 110.4 - 1.2a + 14.4

⇒ 1.6a ≥ 124.8

⇒ a ≥ 78  ...(3)

From (1), (2) and (3)
⇒ b can be 13 or 14.

Hence, option (c).

Alternately,
Consider options.

For c = 8 and b = 11, a = 81. The total marks = 127.
Group B has 17.32% marks, which is not possible.

For c = 8 and b = 12, a = 80. The total marks = 128.
Group B has 18.75% marks, which is not possible.

For c = 8 and b = 13, a = 79. The total marks = 129.
Group B has 20.15% marks and group A has 61.24% marks, which is possible.

For c = 8 and b = 14, a = 78. The total marks = 130.
Group B has 21.53% marks and group A has 60% marks, which is possible.

For c = 8 and b = 15, a = 77. The total marks = 131.
Group B has 22.9% marks and group A has 58.77 marks, which is not possible.

Hence, option (c).