If the price of 3 pens, 2 pencils and 4 erasers is ₹ 92 and the price of 8…

When two equations connect the same three unknowns and one equation is a whole-number multiple of a combination that also appears in the other, there is no need to solve for every unknown individually. Simplify the second equation until it isolates that shared combination as a single number, then substitute that number straight into the first equation to solve directly for the remaining unknown.

1. Let the price of one pen, one pencil and one eraser be ₹p, ₹q and ₹r respectively. The given information becomes two equations: 3p + 2q + 4r = 92 and 8q + 16r = 68.

2. Divide every term of the second equation by 4: 2q + 4r = 17. This is exactly the (2q + 4r) block already sitting inside the first equation.

3. Substitute 2q + 4r = 17 into the first equation: 3p + 17 = 92.

4. Solve for p: 3p = 92 − 17 = 75, so p = 75 ÷ 3 = 25.

5. The price of 24 pens is 24 × p = 24 × 25 = ₹600.

Independent check: pick any pair of q, r that satisfies 2q + 4r = 17, say q = 1.5 and r = 3.5, and plug them into the original first equation: 3(25) + 2(1.5) + 4(3.5) = 75 + 3 + 14 = 92, which matches the given total — confirming the value of p used above.