CSAT Solved Papers/ 2022/Q60

2022 CSAT — Q60

Quant Arithmetic & numeracy 2.5 marks Medium

When $70\%$ of a number $x$ is added to another number $y$ , the sum becomes $165\%$ of the value of $y$ . When $60\%$ of the number $x$ is added to another number $z$ , then the sum becomes $165\%$ of the value of $z$ . Which one of the following is correct?

A z < x < y Answer
B x < y < z
C y < x < z
D z < y < x

Worked rationale

Translate each sentence into an equation.

First: $0.7x + y = 1.65y \Rightarrow 0.7x = 0.65y \Rightarrow x = \dfrac{0.65}{0.70}\,y = \dfrac{13}{14}\,y$ . Since $\tfrac{13}{14} < 1$ , we get $x < y$ .

Second: $0.6x + z = 1.65z \Rightarrow 0.6x = 0.65z \Rightarrow x = \dfrac{0.65}{0.60}\,z = \dfrac{13}{12}\,z$ . Since $\tfrac{13}{12} > 1$ , we get $x > z$ , i.e. $z < x$ .

Combine: $z < x < y$ .

Answer: (a) z < x < y.

Why the other options miss

B

gets a comparison backwards: inverts the second relation, placing $z$ above $x$ .
C

wrong reading of the sentence: reads “sum is $165\%$ ” as ” $x$ is $165\%$ ,” flipping both comparisons.
D

half right, half reversed: gets $z < x$ but flips $x$ vs $y$ .

Specialist insight

Reduce each statement to $x$ in terms of the other variable: $x = \tfrac{13}{14}y$ (so $x<y$ ) and $x = \tfrac{13}{12}z$ (so $x>z$ ). The single number $13$ versus the denominator $14$ or $12$ decides each inequality’s direction. Express everything through $x$ to chain the order cleanly: $z < x < y$ .

The trap, in one line

$x = \tfrac{13}{14}y$ (so $x<y$ ) and $x = \tfrac{13}{12}z$ (so $x>z$ ) $\Rightarrow z < x < y =$ (a).

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