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CAT 2019 Slot 2 — QA Question 16

RatioEasy

In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11 : 10 : 3. Then Anjali's score exceeded Rama's score by

Answer & solution

Correct answer: 32

  • A

    26

  • 32

  • C

    24

  • D

    35

Solution

Easy

Let the revised scores carry the ratio 11:10:311:10:3 (Anjali : Mohan : Rama). Subtract 6 from each to recover the original scores, then apply Rama's original condition (Rama=112(Mohan+Anjali)\text{Rama}=\tfrac{1}{12}(\text{Mohan}+\text{Anjali})) to solve for the scale factor.

1

Set up revised and original scores. Revised: Anjali =11x=11x, Mohan =10x=10x, Rama =3x=3x. Each rose by 6, so subtract 6 for the originals.

Anjali0=11x−6,Mohan0=10x−6,Rama0=3x−6\begin{aligned} &\text{Anjali}_0=11x-6,\quad \text{Mohan}_0=10x-6,\quad \text{Rama}_0=3x-6 \end{aligned}
2

Apply Rama's original condition. Originally Rama scored one-twelfth of (Mohan + Anjali).

3x−6=112[(11x−6)+(10x−6)]⇒ 12(3x−6)=21x−12(multiply by 12, combine)⇒ 36x−72=21x−12⇒ 15x=60 ⇒ x=4\begin{aligned} &3x-6=\frac{1}{12}\big[(11x-6)+(10x-6)\big]\\ &\Rightarrow\ 12(3x-6)=21x-12 \quad\text{(multiply by 12, combine)}\\ &\Rightarrow\ 36x-72=21x-12\\ &\Rightarrow\ 15x=60\ \Rightarrow\ x=4 \end{aligned}
3

Compute the excess. Revised Anjali =11x=44=11x=44, revised Rama =3x=12=3x=12.

Anjali−Rama=44−12=32\begin{aligned} &\text{Anjali}-\text{Rama}=44-12=32 \end{aligned}
Anjali exceeds Rama by 32\text{Anjali exceeds Rama by }32

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CAT 2019 Slot 2 QA Q16: In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, — Solution | TheCATExam