If P denotes '+, Q means '÷', R means ‘−’ and S means ‘×’, then which of the following equations is correct?
- (a)16 S 5 R 10 P 4 Q 3 = 9
- (b)16 P 5 R 10 S 4 Q 3 = 19
- (c)16 Q 5 P 10 S 4 R 3 = 12
- (d)16 S 5 Q 10 P 4 R 3 = 9
Answer
Why
Correct — D.
Rule: turn each letter into its sign, then do × and ÷ before + and − (BODMAS).
16 S 5 Q 10 P 4 R 3
= 16 × 5 ÷ 10 + 4 − 3
= 80 ÷ 10 + 4 − 3
= 8 + 4 − 3 = 9 → option (d).
Why the others are wrong
- (a)16 S 5 R 10 P 4 Q 3 = 9 — 16 × 5 − 10 + 4 ÷ 3 = 80 − 10 + 1⅓ = 71⅓, not 9. It is the keyed equation with Q and R exchanged, so it reaches 9 only if R means ÷ and Q means −.
- (b)16 P 5 R 10 S 4 Q 3 = 19 — 16 + 5 − 10 × 4 ÷ 3: the product and quotient go first, 40 ÷ 3 = 13⅓, so the value is 21 − 13⅓ = 7⅔, not 19.
- (c)16 Q 5 P 10 S 4 R 3 = 12 — 16 ÷ 5 + 10 × 4 − 3 = 3.2 + 40 − 3 = 40.2, nowhere near 12.
Concept
A symbol-substitution item takes two steps. First rewrite the equation with the real signs, then evaluate it by BODMAS: × and ÷ first, working left to right, then + and −.
Rewrite all four options before judging any. Three of them here do not even come out whole: 71⅓, 7⅔ and 40.2.
Option (a) is option (d) with Q and R exchanged. Mix up those two letters and (a) comes to 9 while (d) does not, which is why the code has to be applied letter by letter.
Key facts
- P = +, Q = ÷, R = − and S = × in this item.
- BODMAS: division and multiplication before addition and subtraction, each worked left to right.
- 16 × 5 ÷ 10 = 8, and 8 + 4 − 3 = 9.
Study next
Common traps
- Swapping the meanings of Q and R, which makes option (a) come to 9
- Reading a letter as the sign it begins, such as S for subtraction, when here S means ×
19 Jan 2026, 11:00 AM, Reasoning Q.17 recodes the signs themselves ('+' means division) and asks for a value: 36 − 2 + 4 × 10 ÷ 5 becomes 36 × 2 ÷ 4 + 10 − 5 = 23.
At 17 Sep 2025, 16:00, Reasoning Q.6 letters stand for the signs as they do here, with P for addition, and a bracketed expression works out to 7.
Related PYQs
No directly related past PYQ was found.