Amol bought 6 pens, 8 pencils and 4 erasers. Rajan bought 7 pens, 8 erasers…

Concept: When two purchases are linked through a known ratio of their totals (e.g. one total is a fixed multiple of the other), write each purchase as a linear equation in the unit costs. Scale one equation so the coefficients of the variables you want to eliminate match the other equation, then subtract — this isolates the one unknown you need, as a fraction of the total.

Application:

1. Let the cost of one pen, one pencil and one eraser be p, q and r respectively, and let Amol's total payment be P.

2. Amol's purchase gives the equation: 6p + 8q + 4r = P ...(i)

3. Rajan paid an amount that was "half more" than Amol's, i.e. 1.5P, so: 7p + 16q + 8r = 1.5P ...(ii)

4. Multiply equation (i) by 2 so the pencil coefficient (8→16) and the eraser coefficient (4→8) both match equation (ii): 12p + 16q + 8r = 2P ...(iii)

5. Subtract (ii) from (iii): the pencil and eraser terms cancel, leaving 5p = 0.5P, so p = P/10.

6. Amol's spend on pens is 6p = 6 × P/10 = 3P/5 = 0.6P — that is, 60% of the total amount Amol paid.

Cross-check:

Take P = 100, so p = 10 and 6p = 60 (60% of 100). Equation (i) then needs 8q + 4r = 40; for instance q = 4, r = 2 works (8 × 4 + 4 × 2 = 40). Checking equation (ii): 7(10) + 16(4) + 8(2) = 70 + 64 + 16 = 150 = 1.5 × 100 — consistent, confirming the pen's cost and the 60% share.