CSAT Solved Papers/ 2024/Q77

2024 CSAT — Q77

Quant Logical & quantitative reasoning 2.5 marks Medium

If in a certain code, ‘ABCD’ is written as $24$ and ‘EFGH’ is written as $1680$ , then how is ‘IJKL’ written in that code?

A $11880$ Answer
B $11240$
C $7920$
D $5940$

Worked rationale

Infer the rule from the two anchors, confirm it on both, then apply it.

Use the alphabet positions $A=1, B=2, \dots, Z=26$ .

‘ABCD’ $\to 1 \times 2 \times 3 \times 4 = 24$ ✓
‘EFGH’ $\to 5 \times 6 \times 7 \times 8 = 1680$ ✓

The code is the product of the alphabet positions of the four letters. For ‘IJKL’, the positions are $I=9, J=10, K=11, L=12$ :

9 \times 10 \times 11 \times 12 = 11880.

(Compute in steps: $9 \times 10 = 90$ , $90 \times 11 = 990$ , $990 \times 12 = 11880$ .)

Answer: (a) $11880$ .

Why the other options miss

B

an arithmetic slip: a multiplication error in $990 \times 12$ (e.g. $990 \times 12$ mis-carried), producing a near-miss product.
C

counted one letter off: shifts the block by one letter (uses $H,I,J,K = 8 \times 9 \times 10 \times 11$ ) or otherwise mis-indexes the starting position $I = 9$ .
D

an arithmetic slip: drops a factor — e.g. computes $9 \times 10 \times 11 \times 6$ or halves the true product through an arithmetic slip.

Specialist insight

The two anchors are deliberately consecutive-letter blocks, which lets you test the simplest rule first — product of positions — and confirm it on both before trusting it. Resist the urge to look for additions or place-value tricks: $24$ and $1680$ are exactly $4!$ and $5\cdot6\cdot7\cdot8$ , an unmistakable product signature. Once verified twice, the answer is pure multiplication; the only risk left is an arithmetic slip, so multiply in clean steps ( $90 \to 990 \to 11880$ ) and keep the running product visible.

The trap, in one line

The code is the product of alphabet positions ( $I,J,K,L = 9\cdot10\cdot11\cdot12 = 11880$ ) $=(a)$ — verify the rule on both anchors and multiply in steps to avoid a slip.

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