Q257 / 270 · NDA 2026 (Phase 1) · Mathematics

Passage: Let $A$, $B$, $C$ and $D$ be mutually exclusive and exhaustive events and $\dfrac{P(A)}{2}=\dfrac{P(B)}{3}=\dfrac{P(C)}{5}=\dfrac{P(D)}{8}$.

Question: What is $\dfrac{[2P(A)+3P(B)]}{[4P(C)+5P(D)]}$ equal to ?

  • A
    $\dfrac{13}{18}$
  • Correct answer
    B
    $\dfrac{13}{60}$
  • C
    $\dfrac{4}{21}$
  • D
    $\dfrac{5}{28}$

Solution

Let $\dfrac{P(A)}{2} = \dfrac{P(B)}{3} = \dfrac{P(C)}{5} = \dfrac{P(D)}{8} = k$

Then $P(A) = 2k$, $P(B) = 3k$, $P(C) = 5k$, and $P(D) = 8k$

Since $A$, $B$, $C$, and $D$ are mutually exclusive and exhaustive events, $P(A) + P(B) + P(C) + P(D) = 1$

$2k + 3k + 5k + 8k = 1$

$18k = 1 \Rightarrow k = \dfrac{1}{18}$

We need to evaluate $\dfrac{2P(A) + 3P(B)}{4P(C) + 5P(D)}$

Substituting the values in terms of $k$:

$\dfrac{2(2k) + 3(3k)}{4(5k) + 5(8k)} = \dfrac{4k + 9k}{20k + 40k} = \dfrac{13k}{60k} = \dfrac{13}{60}$

Answer: $\dfrac{13}{60}$

Hence, the correct answer is option: 'B'.

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