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CAT 2022 Slot 2 — QA Question 16

PercentageEasy

In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was:

Answer & solution

  • A

    60288

  • B

    50240

  • C

    40192

  • 62800

Solution

Easy

Let registered voters be 100x100x, so 80x80x votes are cast. One candidate gets 30%30\%; the rest is split 1:2:31:2:3. Identify the winner and runner-up, set their gap to 25122512, solve for xx.

1

Set up vote counts:

Cast=80x,C1=30%×80x=24xRemaining=80x−24x=56x split 1:2:3\begin{aligned} &\text{Cast}=80x,\quad C_1=30\%\times 80x=24x\\ &\text{Remaining}=80x-24x=56x \ \text{split }1:2:3 \end{aligned}
2

Split the remaining 56x56x:

C2=16(56x)=28x3,C3=26(56x)=56x3,C4=36(56x)=28xC_2=\tfrac{1}{6}(56x)=\tfrac{28x}{3},\quad C_3=\tfrac{2}{6}(56x)=\tfrac{56x}{3},\quad C_4=\tfrac{3}{6}(56x)=28x
3

Winner vs runner-up. Largest is C4=28xC_4=28x; next is C1=24xC_1=24x (since 56x3≈18.7x<24x\tfrac{56x}{3}\approx 18.7x<24x). Set the gap:

28x−24x=2512 ⇒ 4x=2512 ⇒ x=62828x-24x=2512\ \Rightarrow\ 4x=2512\ \Rightarrow\ x=628
4

Registered voters:

100x=100×628=62,800100x=100\times 628=62{,}800
Registered voters=62800\text{Registered voters}=\mathbf{62800}
CAT 2022 Slot 2 QA Q16: In an election, there were four candidates and 80% of the registered voters casted their votes. One of the can — Solution | TheCATExam