Two whole numbers have sum 64. Which of the following ratios cannot be their ratio?

Introduction / Context:If two whole numbers are in ratio a : b, then the actual numbers are ak and bk for some positive integer k, and their sum is (a + b)k. This sum must equal 64; hence 64 must be divisible by (a + b). We use this divisibility condition to find the impossible ratio.

Given Data / Assumptions:

• a and b are positive integers representing the ratio.
• Sum (a + b)k = 64 for some integer k.
• Both numbers are whole (nonnegative integers).

Concept / Approach:Compute (a + b) for each option, then check if 64/(a + b) is an integer. If not, that ratio cannot produce whole-number addends summing to 64.

Step-by-Step Solution:5 : 3 ⇒ a + b = 8 ⇒ 64/8 = 8 (OK)7 : 1 ⇒ a + b = 8 ⇒ 64/8 = 8 (OK)3 : 4 ⇒ a + b = 7 ⇒ 64/7 not an integer (Not possible)9 : 7 ⇒ a + b = 16 ⇒ 64/16 = 4 (OK)

Verification / Alternative check:Construct numbers for possible cases (e.g., 5k and 3k with k = 8 give 40 and 24) to confirm. For 3:4, you cannot reach sum 64 with an integer k.

Why Other Options Are Wrong:They are actually possible; only 3 : 4 fails the divisibility test.

Common Pitfalls:Confusing the sum 64 with a single term; forgetting k must be an integer to keep numbers whole.

Final Answer:3 : 4