CSAT Solved Papers/ 2025/Q76

2025 CSAT — Q76

Quant Arithmetic & numeracy 2.5 marks Medium

Let p+q=10p + q = 10, where p,qp, q are integers.

Value-I == Maximum value of p×qp \times q when p,qp, q are positive integers.

Value-II == Maximum value of p×qp \times q when p≥−6, q≥−4p \ge -6,\ q \ge -4.

Which one of the following is correct?

  1. A Value-I < Value-II
  2. B Value-II < Value-I
  3. C Value-I = Value-II Answer
  4. D Cannot be determined due to insufficient data

Worked rationale

With a fixed sum p+q=10p + q = 10, the product pq=p(10−p)pq = p(10 - p) is a downward parabola, maximised when the two parts are as equal as possible.

Value-I (p,qp, q positive integers): the closest equal split is p=q=5p = q = 5, giving

pq=5×5=25.pq = 5 \times 5 = 25.

Value-II (p≥−6, q≥−4p \ge -6,\ q \ge -4, integers, p+q=10p+q=10): the same vertex p=q=5p = q = 5 is inside this feasible region (5≥−65 \ge -6 and 5≥−45 \ge -4), and the parabola’s maximum is at the vertex. So the constraint does not push the optimum away:

max⁡pq=5×5=25.\max pq = 5 \times 5 = 25.

(Moving toward the boundaries only decreases the product, e.g. p=−6,q=16⇒−96p=-6, q=16 \Rightarrow -96.)

Both maxima equal 2525.

Answer: (c) Value-I = Value-II.

Why the other options miss

  • A
    wider range ≠ bigger max: imagines the wider range p≥−6p\ge-6 allows a larger product, not seeing the parabola peaks at the interior point 5,55,5 common to both.
  • B
    the constraint doesn’t bind: thinks the extra constraint restricts the max below 2525, but (5,5)(5,5) is feasible under both.
  • D
    gave up instead of computing: treats the two conditions as incomparable rather than evaluating each maximum explicitly.

Specialist insight

For a fixed sum, the product is maximal at equal parts — the vertex p=q=5p=q=5 gives 2525. The whole item tests whether you notice that the second condition’s wider bounds (p≥−6,q≥−4p\ge-6, q\ge-4) are a red herring: the maximum sits at the interior point (5,5)(5,5), which both feasible sets contain, so the bounds never bind and both maxima coincide. Don’t assume a looser constraint yields a bigger maximum — locate the vertex, check it lies in the region, and compare. Here both land on 2525.

The trap, in one line

For fixed sum 1010, pqpq peaks at the interior p=q=5p=q=5 (=25=25), feasible under both conditions — so the wider bounds don't change it: Value-I == Value-II.

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