Out of three numbers, the first is twice the second and is half of the third.…

Concept: When three quantities are linked by given multiplicative relationships to one another, express every quantity in terms of a SINGLE variable using each relationship separately, then use the total/average condition to solve for that variable — only after that compute whatever combination (sum, difference, ratio) the question actually asks for.

1. Let the second number be x.

2. Since the first number is twice the second, first = 2x.

3. Since the first number is stated to be half of the third, the third must be twice the first: third = 2 × (2x) = 4x.

4. The average of the three numbers is 56, so their sum = 56 × 3 = 168.

5. Sum in terms of x: 2x + x + 4x = 7x = 168, so x = 24.

6. First = 2 × 24 = 48 and Third = 4 × 24 = 96.

7. Required difference = Third − First = 96 − 48 = 48.

Cross-check: Taking the first number itself as a, the same relationships give second = a/2 and third = 2a. The sum a + a/2 + 2a = 3.5a = 168 gives a = 48, second = 24 and third = 96 — the identical set of numbers, confirming the difference independently.

Hence, the difference between the first and third numbers is 48.