CSAT Solved Papers/ 2021/Q76

2021 CSAT — Q76

Quant Statement validity 2.5 marks Hard

The difference between a 22-digit number and the number obtained by interchanging the positions of the digits is 5454.

Consider the following statements:

  1. The sum of the two digits of the number can be determined only if the product of the two digits is known.

  2. The difference between the two digits of the number can be determined.

Which of the above statements is/are correct?

  1. A 1 only
  2. B 2 only
  3. C Both 1 and 2 Answer
  4. D Neither 1 nor 2

Worked rationale

Let the digits be aa (tens) and bb (units). The number is 10a+b10a + b, its reverse 10b+a10b + a, and the difference is

∣(10a+b)−(10b+a)∣=9 ∣a−b∣=54  ⇒  ∣a−b∣=6.|(10a + b) - (10b + a)| = 9\,|a - b| = 54 \;\Rightarrow\; |a - b| = 6.

Statement 2: the difference of the digits is ∣a−b∣=6|a - b| = 6, fully determined. Correct.

Statement 1: the sum a+ba + b is not fixed by ∣a−b∣=6|a-b| = 6 alone — e.g. (7,1),(8,2),(9,3)(7,1), (8,2), (9,3) give sums 8,10,128, 10, 12. But if the product abab is known, then

(a+b)2=(a−b)2+4ab=36+4ab,(a + b)^2 = (a - b)^2 + 4ab = 36 + 4ab,

which pins a+ba + b. So the sum is determinable only once the product is supplied — exactly Statement 1’s claim. Correct.

Answer: (c) Both 1 and 2.

Why the other options miss

  • A
    missed a case: accepts the product-condition for the sum but somehow doubts ∣a−b∣|a-b|, even though 9∣a−b∣=549|a-b| = 54 fixes it directly.
  • B
    missed the algebraic link: gets the digit difference but rejects Statement 1, not seeing that (a+b)2=(a−b)2+4ab(a+b)^2 = (a-b)^2 + 4ab needs the product.
  • D
    wrong formula: mishandles 9∣a−b∣=549|a-b| = 54, breaking both deductions.

Specialist insight

The reverse-difference identity 9∣a−b∣9|a-b| instantly gives ∣a−b∣=6|a-b| = 6 (Statement 2). Statement 1 is a subtler “what’s determinable” claim: with the difference known, the sum still floats across (7,1),(8,2),(9,3)(7,1),(8,2),(9,3), and the missing link is the product — via (a+b)2=(a−b)2+4ab(a+b)^2 = (a-b)^2 + 4ab. The discipline is to separate “always determined” (the difference) from “determined only with extra data” (the sum). Both statements are precisely calibrated and both hold.

The trap, in one line

9∣a−b∣=54⇒∣a−b∣=69|a-b| = 54 \Rightarrow |a-b| = 6 (S2 ✓); sum needs the product via (a+b)2=36+4ab(a+b)^2 = 36 + 4ab (S1 ✓) ⇒\Rightarrow (c).

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