Three numbers are in the ratio 2 : 3 : 5. If the sum of the first and the…

Concept — When quantities are in a fixed ratio a : b : c, all three terms share one common multiplier x, so they can be written as ax, bx and cx. Any statement linking the terms then turns into a single linear equation in x, and the total of the terms is (a + b + c)x. Because every term carries the same x, any relation of the form “some terms together are k more than another term” collapses to (difference of the coefficients) × x = k, so x = k ÷ (that coefficient difference), and the quantity asked for follows as a multiple of x — the individual numbers never have to be found separately.

Application — Turn the given condition into an equation in x.

1. Let the three numbers be 2x, 3x and 5x.

2. Sum of the first and the third number = 2x + 5x = 7x.

3. “156 more than the second number” means 7x = 3x + 156.

4. Subtract 3x from both sides: 7x − 3x = 156, so 4x = 156.

5. Divide by 4: x = 156 ÷ 4 = 39.

6. Sum of all three numbers = 2x + 3x + 5x = 10x = 10 × 39 = 390.

Cross-check — With x = 39 the three numbers are 78, 117 and 195. Here 78 + 195 = 273 and 273 − 117 = 156, exactly the condition given, and 78 + 117 + 195 = 390.

Shortcut — The condition fixes 4x = 156 while the quantity asked for is 10x, so the total is (10 ÷ 4) × 156 = 2.5 × 156 = 390.

Generalisation — For a ratio a : b : c in which the first and the third term together exceed the second by k, the same three moves give x = k ÷ (a + c − b) and a total of (a + b + c) × k ÷ (a + c − b). Here a + c − b = 2 + 5 − 3 = 4 and a + b + c = 10, so the total is 10 × 156 ÷ 4 = 390. Changing which terms are combined, or asking for one of the numbers instead of the total, only changes which coefficients appear; the method itself does not change.