A student attempted 48 sums. He got twice as many sums wrong as he got right. How many sums did he solve correctly?

Introduction / Context:
This algebra word problem checks ratio translation into equations. The statement “twice as many wrong as right” indicates a 2:1 ratio between wrong and right answers, with a fixed total of 48 attempts.

Given Data / Assumptions:

• Total attempted = 48.
• Wrong = 2 * Right.

Concept / Approach:
Let R be the number correct (right). Then wrong = 2R. Since total = right + wrong, we get a single linear equation in R.

Step-by-Step Solution:
Let R = number right, W = number wrong.Given W = 2R and R + W = 48.Substitute: R + 2R = 48 → 3R = 48 → R = 16.Therefore, he solved 16 sums correctly.

Verification / Alternative check:
Compute wrong: W = 2 * 16 = 32; total = 16 + 32 = 48. Checks out.

Why Other Options Are Wrong:

• 12, 18, 20, 24: Do not satisfy the ratio W = 2R with total 48 when tested.

Common Pitfalls:

• Interpreting “twice as many wrong as right” backwards (setting R = 2W).
• Arithmetic mistakes when dividing by 3.

Final Answer:
16