CSAT Solved Papers/ 2025/Q36

2025 CSAT — Q36

Quant Logical & quantitative reasoning 2.5 marks Medium

What is XX in the sequence 24, X, 12, 18, 36, 9024,\ X,\ 12,\ 18,\ 36,\ 90?

  1. A 18
  2. B 12 Answer
  3. C 9
  4. D 6

Worked rationale

Read the multipliers between known consecutive terms from the right:

1218 (×1.5),1836 (×2),3690 (×2.5).12 \to 18 \ (\times 1.5), \quad 18 \to 36 \ (\times 2), \quad 36 \to 90 \ (\times 2.5).

The ratio rises by 0.50.5 each step. Extending the same rule backward to the left:

24X (×0.5),X12 (×1).24 \to X \ (\times 0.5), \quad X \to 12 \ (\times 1).

So X=24×0.5=12X = 24 \times 0.5 = 12, and the check X×1=12×1=12X \times 1 = 12 \times 1 = 12 ✓ matches the third term. The full multiplier chain is ×0.5, ×1, ×1.5, ×2, ×2.5\times 0.5,\ \times 1,\ \times 1.5,\ \times 2,\ \times 2.5.

Answer: (b) 12.

Why the other options miss

  • A
    reached for the wrong rule: assumes an additive pattern (differences) rather than a multiplicative one, interpolating XX between 2424 and 1212 as a midpoint-ish 1818.
  • C
    an arithmetic slip: applies the wrong multiplier (24×0.37524 \times 0.375 or a mis-stepped ratio), landing on 99.
  • D
    reached for the wrong rule: reads the multiplier as ×0.25\times 0.25 (24624 \to 6), over-shrinking the first step.

Specialist insight

The fastest series-decoding move is to read the operation between adjacent terms rather than the terms themselves: here the clean arithmetic progression of multipliers (1.5,2,2.51.5, 2, 2.5, step 0.50.5) is far more visible than any pattern in 12,18,36,9012, 18, 36, 90. Extend that progression leftward (1.0,0.51.0, 0.5) to recover the hidden term. A CSAT series rewards the student who classifies the rule (additive? multiplicative? second-difference?) in one pass and then runs it both directions — guessing a “nice” interpolated value (a) is the planted trap.

The trap, in one line

The pattern is multiplicative with ratios stepping by 0.50.5 (×0.5,×1,×1.5,\times0.5, \times1, \times1.5, \dots), so X=24×0.5=12X = 24 \times 0.5 = 12 — not an additive midpoint.

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