CSAT Solved Papers/ 2022/Q45

2022 CSAT — Q45

Quant Arithmetic & numeracy 2.5 marks Medium

Five friends $P, Q, X, Y$ and $Z$ purchased some notebooks. The relevant information is given below:

$Z$ purchased $8$ notebooks more than $X$ did.
$P$ and $Q$ together purchased $21$ notebooks.
$Q$ purchased $5$ notebooks less than $P$ did.
$X$ and $Y$ together purchased $28$ notebooks.
$P$ purchased $5$ notebooks more than $X$ did.

If each notebook is priced Rs. $40$ , then what is the total cost of all the notebooks?

A Rs. 2,600 Answer
B Rs. 2,400
C Rs. 2,360
D Rs. 2,320

Worked rationale

Solve the chain of relations:

From (2) and (3): $P + Q = 21$ and $Q = P - 5 \Rightarrow 2P - 5 = 21 \Rightarrow P = 13$ , $Q = 8$ .
From (5): $P = X + 5 \Rightarrow X = 13 - 5 = 8$ .
From (1): $Z = X + 8 = 16$ .
From (4): $X + Y = 28 \Rightarrow Y = 28 - 8 = 20$ .

Total notebooks:

P + Q + X + Y + Z = 13 + 8 + 8 + 20 + 16 = 65.

Total cost $= 65 \times 40 = 2600$ rupees.

Answer: (a) Rs. 2,600.

Why the other options miss

B

an arithmetic slip: gets $60$ notebooks (e.g. drops the $Z = 16$ to $11$ ), $60 \times 40$ .
C

an arithmetic slip: a one-notebook slip ( $59$ ) from mis-solving $X$ or $Y$ .
D

a relation read backwards: mis-reads relation (1) or (5) direction, undercounting by two.

Specialist insight

Anchor on the only fully self-contained pair — equations (2) and (3) pin $P$ and $Q$ outright — then cascade: $P\to X \to Z$ and $X\to Y$ . Every other quantity is one substitution away. Convert to cost only at the very end ( $65 \times 40$ ); converting early invites a units slip.

The trap, in one line

$(P,Q,X,Y,Z) = (13,8,8,20,16)$ , total $65$ books $\times 40 =$ Rs. $2600 \Rightarrow$ (a).

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