CSAT Solved Papers/ 2025/Q77

2025 CSAT — Q77

Quant Arithmetic & numeracy 2.5 marks Hard

Consider a set of 1111 numbers:

Value-I == Minimum value of the average of the numbers of the set when they are consecutive integers ≥−5\ge -5.

Value-II == Minimum value of the product of the numbers of the set when they are consecutive non-negative integers.

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

Value-I — minimum average of 1111 consecutive integers ≥−5\ge -5. The average of consecutive integers is their middle term. To minimise it, start as low as allowed: −5,−4,…,5-5, -4, \dots, 5. Their sum is 00 (symmetric about 00), so the average is

011=0.\frac{0}{11} = 0.

Any set starting above −5-5 has a larger middle term, so the minimum average is 00.

Value-II — minimum product of 1111 consecutive non-negative integers. Non-negative means ≥0\ge 0. The smallest such set is 0,1,2,…,100, 1, 2, \dots, 10, whose product is

0×1×⋯×10=0.0 \times 1 \times \cdots \times 10 = 0.

Since all terms are ≥0\ge 0, no product can be negative, and 00 is achievable (any set including 00), so the minimum product is 00.

Both values equal 00.

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

Why the other options miss

  • A
    ignored the floor: thinks the average can go negative (e.g. forgets the ≥−5\ge -5 floor or mis-centres the set), making Value-I <0< 0.
  • B
    missed the zero: overlooks that the consecutive non-negative set includes 00, so the product is 00, not some positive minimum.
  • D
    gave up instead of computing: treats “minimum” as ill-defined rather than computing each extremum directly.

Specialist insight

Two short extremum reads that both land on 00. Value-I uses the fact that the average of consecutive integers is the middle term, minimised by starting at the floor −5-5, where the set −5..5-5..5 is symmetric and sums to 00. Value-II hinges on the zero-product trap: “consecutive non-negative integers” must include 00 (the smallest non-negative), so the product collapses to 00 — the minimum possible for a non-negative product. The reward is recognising both floors give 00; the decoys come from forgetting the ≥−5\ge -5 bound or the presence of 00 in the product.

The trap, in one line

Min average is the middle of −5..5=0-5..5 = 0; min product of consecutive non-negatives includes 00, giving 00 — both are 00, so equal.

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