I bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and…

When two purchases share some item quantities but differ in others, you can scale one total so the shared items' quantities line up with the other purchase; subtracting then cancels those shared terms and isolates the price of the differing item as a fraction of the original total — without needing to know each individual price.

Applying this to the given amounts:

1. Let the price of one pen, one pencil, and one eraser be P, L, and E. Your total spend is x = 5P + 7L + 4E.

2. Double your total: 2x = 10P + 14L + 8E. This is chosen because the pencil and eraser quantities (14 pencils, 8 erasers) now match Rajan's purchase exactly.

3. Rajan's total is 6P + 14L + 8E, and the problem states this equals one-and-a-half times your total, i.e. 1.5x.

4. Subtract Rajan's total from your doubled total: the 14 pencils and 8 erasers cancel on both sides, leaving 2x − 1.5x = (10P − 6P) = 4P, i.e. 0.5x = 4P.

5. So x = 8P, meaning the price of one pen is exactly one-eighth of your total spend: P = x/8.

6. Your spend on pens alone is 5P = 5(x/8) = 5x/8 of your total.

7. As a percentage: (5/8) × 100 = 62.5%.

Cross-check: because the elimination step matched the pencil and eraser coefficients exactly (14 and 8 on both sides), the pencil and eraser prices L and E cancel out completely — the ratio P/x = 1/8 holds no matter what L and E individually are, as long as 5P + 7L + 4E = x. For example, P = 1, and any L, E with 7L + 4E = 3 (say L = 3/7, E = 0) gives x = 8, and Rajan's total 6(1) + 14(3/7) + 8(0) = 6 + 6 = 12 = 1.5 × 8, confirming the relationship independent of the exact pencil/eraser split.