CSAT Solved Papers/ 2023/Q24

2023 CSAT — Q24

Quant Logical & quantitative reasoning 2.5 marks Medium

If 7⊕9⊕10=87 \oplus 9 \oplus 10 = 8, 9⊕11⊕30=59 \oplus 11 \oplus 30 = 5, 11⊕17⊕21=1311 \oplus 17 \oplus 21 = 13, then what is the value of 23⊕4⊕1523 \oplus 4 \oplus 15?

  1. A 6 Answer
  2. B 8
  3. C 13
  4. D 15

Worked rationale

Fit a rule to all three anchors before computing. Try the sum of the three operands, then its digit sum:

  • 7+9+10=26⇒2+6=87 + 9 + 10 = 26 \Rightarrow 2 + 6 = 8 ✓
  • 9+11+30=50⇒5+0=59 + 11 + 30 = 50 \Rightarrow 5 + 0 = 5 ✓
  • 11+17+21=49⇒4+9=1311 + 17 + 21 = 49 \Rightarrow 4 + 9 = 13 ✓

All three match: a⊕b⊕c=a \oplus b \oplus c = (sum of the digits of a+b+ca + b + c).

Apply to 23⊕4⊕1523 \oplus 4 \oplus 15:

23+4+15=42⇒4+2=6.23 + 4 + 15 = 42 \Rightarrow 4 + 2 = 6.

Answer: (a) 6.

Why the other options miss

  • B
    solved the wrong question: reuses the first anchor’s output, or applies a "a−b+ca - b + c"-type rule that happened to give 88 once but fails the other anchors.
  • C
    the half-finished answer: stops at the raw sum’s structure (echoing the third anchor’s 1313) or forgets to take the digit sum of 4242.
  • D
    an arithmetic slip: returns an operand (1515) or mis-adds 23+4+1523+4+15, never reducing to the digit sum.

Specialist insight

The discipline on “define-the-operator” items is to test one candidate rule against every anchor — a rule that fits two but not three is wrong. “Sum, then digit-sum” is the workhorse pattern (CSAT loves digit sums); note the third anchor’s 49→1349 \to 13 proves the output need not be a single digit — it is the digit sum, full stop, not “repeated until one digit.” Mistaking it for the digital root would give 49→449 \to 4 and break the anchor.

The trap, in one line

Rule == digit-sum of (a+b+c)(a+b+c): 49→1349\to 13 (not 44) confirms it; 23+4+15=42→6⇒23+4+15=42\to 6 \Rightarrow (a).

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