CSAT Solved Papers/ 2024/Q69

2024 CSAT — Q69

Quant Data sufficiency 2.5 marks Medium

A Question is given followed by two Statements I and II. Consider the Question and the Statements.

A person buys three articles $p$ , $q$ and $r$ for Rs. $50$ . The price of the article $q$ is Rs. $16$ which is the least.

Question: What is the price of the article $p$ ?

Statement-I: The cost of $p$ is not more than that of $r$ .

Statement-II: The cost of $r$ is not more than that of $p$ .

Which one of the following is correct in respect of the above Question and the Statements?

A The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone
B The Question can be answered by using either Statement alone
C The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone Answer
D The Question cannot be answered even by using both the Statements together

Worked rationale

Set up the fixed facts first: $p + q + r = 50$ and $q = 16$ , so $p + r = 34$ . Since $q = 16$ is the least, $p \ge 16$ and $r \ge 16$ .

Statement-I alone ( $p \le r$ ): with $p + r = 34$ and $p \le r$ , we get $p \le 17$ . Combined with $p \ge 16$ , $p$ lies in $[16, 17]$ — e.g. $p = 16$ (then $r = 18$ ) and $p = 17$ (then $r = 17$ ) both fit. Not unique.

Statement-II alone ( $r \le p$ ): symmetrically $r \le 17$ so $p \ge 17$ ; with $p \le 18$ (from $r \ge 16$ ), $p$ lies in $[17, 18]$ — e.g. $p = 17$ or $p = 18$ . Not unique.

Both together ( $p \le r$ and $r \le p$ ): force $p = r$ . With $p + r = 34$ , $p = r = 17$ — a single value. Check: $p = r = 17 \ge 16 = q$ , and $q$ is the least ✓. Unique: $p = 17$ .

Answer: (c) The Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2024 CSAT Q69 — Data sufficiency — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

thought a statement was enough when it wasn’t: assumes " $p \le r$ " already squeezes $p$ to a single value, forgetting it only bounds $p \le 17$ and leaves $16$ and $17$ both open.
B

read an inequality as an equality: reads each inequality as an equality, treating “not more than” as “equal to.”
D

missed that the two bounds combine to an equality: misses that the two “not more than” statements combine into $p = r$ , which the sum $p + r = 34$ then pins exactly.

Specialist insight

The engine here is the classic squeeze: " $p \le r$ " and " $r \le p$ " are weak alone (each only caps one side around the midpoint $17$ ), but together they are an equality $p = r$ , and a linear sum closes the problem. Spot that the midpoint of $p + r = 34$ is $17$ and that each statement merely says ” $p$ is on its side of $17$ ” — only both, asserting $p$ sits exactly at $17$ , decide it. The ” $q = 16$ is the least” clause is the guardrail ensuring $p, r \ge 16$ so no degenerate split sneaks in.

The trap, in one line

Each "not more than" alone only bounds $p$ around the midpoint $17$ of $p+r=34$ ; only both together give $p=r=17$ $=(c)$ — don't read one inequality as an equality.

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