Ratio and Average — Three Numbers Three numbers are in the ratio 4 : 5 : 6. Their arithmetic average is 25. Based on this information, determine the largest of the three numbers.

Introduction / Context:This problem uses the standard technique for ratio sets: represent the three numbers as multiples of a common factor and then use the average to determine that factor. It tests fluency with averages, ratios, and quick algebra—skills that are common in competitive aptitude exams.

Given Data / Assumptions:

• The numbers are proportional to 4 : 5 : 6.
• Let the numbers be 4k, 5k, and 6k for some positive k.
• Their average (arithmetic mean) is 25.
• We are asked to find the largest number among the three.

Concept / Approach:Average of three quantities equals (sum of the three)/3. When each number is written as a multiple of k, the sum becomes (4k + 5k + 6k) = 15k. Hence the average is (15k) / 3 = 5k. Equating 5k to the given average provides k directly, and then each term follows by simple multiplication.

Step-by-Step Solution:Let the numbers be 4k, 5k, 6k.Average = (4k + 5k + 6k)/3 = 15k/3 = 5k.Set 5k = 25 → k = 5.Largest number = 6k = 6 * 5 = 30.

Verification / Alternative check:Numbers are 20, 25, 30. Their average is (20 + 25 + 30)/3 = 75/3 = 25, confirming consistency with the given average.

Why Other Options Are Wrong:42 and 36 correspond to incorrect k values; 32 and 28 also result from misreading the ratio or mixing average with sum. Only 30 aligns with k = 5.

Common Pitfalls:Using 4 + 5 + 6 = 15 directly as the largest term; dividing 25 by 3 instead of equating it to 5k; forgetting that the largest corresponds to the largest ratio term (6k).

Final Answer:30