A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly?

Question

A student got twice as many sums wrong as he got right. If he attempted 48 sums in all, how many did he solve correctly?

1. A. 32
2. B. 16 (Correct answer)
3. C. 18
4. D. 24

Correct Answer

Option B — 16

Detailed Solution & Explanation

The correct answer is 16.

Key Points

• Let the number correct be r. Wrong answers are 2r, so r + 2r = 48.
• Therefore 3r = 48, giving r = 16 correct.
• Check: 16 correct and 32 wrong sum to 48, and 32 is exactly twice 16.

Additional Information

• Whenever quantities are in a fixed ratio, express them as multiples of one variable — here 1 : 2, so parts of 3.
• The ratio shortcut: correct : wrong : total = 1 : 2 : 3, so correct = (1/3) × 48 = 16.
• Reading "twice as many wrong as right" backwards is the usual error and yields 32.