CSAT Solved Papers/ 2021/Q55

2021 CSAT — Q55

Quant Arithmetic & numeracy 2.5 marks Hard

A person $P$ asks one of his three friends $X$ as to how much money he had. $X$ replied, “If $Y$ gives me Rs. $40$ , then $Y$ will have half of as much as $Z$ , but if $Z$ gives me Rs. $40$ , then three of us will have equal amount.” What is the total amount of money that $X$ , $Y$ and $Z$ have?

A Rs. 420
B Rs. 360 Answer
C Rs. 300
D Rs. 270

Worked rationale

Let $X, Y, Z$ be the three amounts.

Condition 1 (” $Y$ gives $X$ Rs. $40$ , then $Y$ has half of $Z$ ”):

Y - 40 = \tfrac12 Z.

Condition 2 (” $Z$ gives $X$ Rs. $40$ , then all three are equal”): after the transfer $X$ has $X+40$ , $Y$ unchanged, $Z$ has $Z-40$ , and these are equal:

X + 40 = Y = Z - 40.

From Condition 2: $Y = X + 40$ and $Z = Y + 40 = X + 80$ .

Substitute into Condition 1:

(X + 40) - 40 = \tfrac12 (X + 80) \;\Rightarrow\; X = \tfrac{X+80}{2} \;\Rightarrow\; 2X = X + 80 \;\Rightarrow\; X = 80.

So $X = 80,\ Y = 120,\ Z = 160$ , total $= 80 + 120 + 160 = 360$ .

Check: Cond 1: $Y-40 = 80 = \tfrac12(160)$ ✓. Cond 2: $X+40 = 120 = Y = Z-40$ ✓.

Answer: (b) Rs. 360.

Why the other options miss

A

an arithmetic slip: a sign or transfer-direction error in one condition shifts the total upward.
C

misread the setup: applies the ” $+40$ / $-40$ ” transfers to the wrong people, breaking the equal-amount condition.
D

the relationship flipped: reads ” $Y$ will have half of $Z$ ” as ” $Z$ has half of $Y$ ,” inverting Condition 1.

Specialist insight

Money-transfer puzzles are cleanest when you express everything through one variable. Condition 2 chains $Y$ and $Z$ off $X$ ( $Y = X+40$ , $Z = X+80$ ) in one stroke; Condition 1 then becomes a single equation in $X$ . The two traps are (i) the transfer accounting — only the giver and receiver change — and (ii) the direction of “half of” ( $Y-40 = \tfrac12 Z$ , not the reverse). Write the post-transfer amounts explicitly before equating.

The trap, in one line

$Y=X+40$ , $Z=X+80$ , and $Y-40=\tfrac12 Z \Rightarrow X=80$ , total $80+120+160 = 360 \Rightarrow$ (b).

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