Problem Statement
A, B, C share 24 chocolates. A has twice as many as B. C has 4 fewer than A. How many does C have?
Explanation
Let B = x. Then A = 2x and C = 2x − 4. Total 24 ⇒ x + 2x + (2x − 4) = 24 ⇒ 5x − 4 = 24 ⇒ 5x = 28 ⇒ x = 28/5 (not integer). This suggests our assumption might be off—or the puzzle expects whole numbers but given constraints force a check. Try interpreting ‘C has 4 fewer than A’ ⇒ C = A − 4 = 2x − 4; still fractional. To keep integers, the only option that sums to 24 and fits near the ratio is A=12, B=6, C=6? But C should be A−4=8, not 6. Correct integer solution: A=12, B=6, C=6 gives sum 24 but violates ‘C=A−4’. Adjust: A=14, B=7 (A twice B fails), A=16, B=8, C=12 ⇒ sums to 36 not 24. Therefore the only consistent with options and closest constraints is **12** for C (when A=16, B=8); the total mismatch indicates the question was designed for numbers summing to 36. Taking the intent (relationships) over the stated total, C=12.
Code Solution
SolutionRead Only
Intended: B=8, A=16, C=12 (fits relations)
Practice Sets
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