The table given below shows the total number of wooden and mechanical pencils…

Concept

Each shop gives two facts: the TOTAL pencils (wooden + mechanical) and the TOTAL price. With unit prices fixed (wooden = Rs 5, mechanical = Rs 20), each shop is a pair of linear equations in the unknown counts. Let w = wooden and m = mechanical for a shop:

1. Count equation: w + m = (total pencils).

2. Price equation: 5w + 20m = (total price).

Dividing the price equation by 5 gives w + 4m = (price ÷ 5). Subtracting the count equation eliminates w and isolates the mechanical count: 3m = (price ÷ 5) − (total pencils). Then w = total − m. Finally, average of k chosen quantities = (their sum) ÷ k.

Application

Shop A — wooden count:

1. w + m = 38 and 5w + 20m = 490, so w + 4m = 98.

2. Subtract: 3m = 98 − 38 = 60, so m = 20 and w = 38 − 20 = 18. Wooden in A = 18.

Shop B — mechanical count:

1. w + m = 47 and 5w + 20m = 415, so w + 4m = 83.

2. Subtract: 3m = 83 − 47 = 36, so m = 12. Mechanical in B = 12.

Shop E — mechanical count:

1. w + m = 40 and 5w + 20m = 425, so w + 4m = 85.

2. Subtract: 3m = 85 − 40 = 45, so m = 15. Mechanical in E = 15.

Required average = (18 + 12 + 15) ÷ 3 = 45 ÷ 3 = 15.

Cross-check

Verify each split satisfies its price equation: A → 5(18) + 20(20) = 90 + 400 = 490 ✓; B → 5(35) + 20(12) = 175 + 240 = 415 ✓; E → 5(25) + 20(15) = 125 + 300 = 425 ✓. The three required quantities (18, 12, 15) sum to 45, and 45 ÷ 3 = 15.