In a school, there were 1582 students and the ratio of the number of the boys…

Concept: When a total is shared in the ratio a : b, the whole is made of (a + b) equal parts, so one part equals the total divided by (a + b), and the two shares are a parts and b parts. A ratio fixes only relative size, never absolute size, so as soon as either quantity is increased or decreased the ratio has to be rebuilt from the new absolute counts: a required ratio p : q on the new counts becomes the cross-multiplied equation q × (new first count) = p × (new second count).

Application:

1. Total students = 1582 and boys : girls = 4 : 3, so the whole consists of 4 + 3 = 7 equal parts.

2. One part = 1582 ÷ 7 = 226.

3. Boys at the start = 4 × 226 = 904 and girls at the start = 3 × 226 = 678, and these add back to 904 + 678 = 1582.

4. After 30 girls join, the girls number 678 + 30 = 708.

5. Let x be the number of boys who left, so the boys number 904 − x.

6. The new ratio requires (904 − x) : 708 = 7 : 6, which cross-multiplies to 6 × (904 − x) = 7 × 708.

7. 5424 − 6x = 4956, so 6x = 5424 − 4956 = 468.

8. x = 468 ÷ 6 = 78.

Cross-check:

• Boys remaining = 904 − 78 = 826 and girls = 708; dividing both by 118 gives 826 : 708 = 7 : 6, which is exactly the ratio the question states.

• Second method: the 708 girls occupy 6 parts of the new ratio, so one new part = 708 ÷ 6 = 118 and the boys must occupy 7 × 118 = 826; the drop from 904 to 826 is 78.

Hence the number of boys who left the school is 78.