R pays ₹100 to P with ₹5, ₹2 and ₹1 coins. The total number of coins used for…

Concept

When a fixed total amount is paid with a fixed total number of mixed-denomination coins, write two linear equations — one for the total value, one for the total count — and subtract them. Subtraction cancels the lowest denomination (₹1 here, whose coefficient is 1 in both equations) and leaves one relation between the higher-denomination counts. With three coin types but only two equations the system is not fully determined, so this relation gives a RANGE of integer possibilities; the offered choices then single out the intended one.

Method without testing each option (the assume-all-₹1 shortcut)

This directly answers the request for a non-trial-and-error approach: instead of plugging in 16, 17, 18, 13 one by one, derive the governing relation in one stroke.

1. Pretend all 40 coins were ₹1. That pays only 40 × ₹1 = ₹40, which is ₹60 short of ₹100.

2. Upgrading one ₹1 coin to a ₹5 coin adds ₹4 to the value (5 − 1); upgrading one ₹1 to a ₹2 adds ₹1 (2 − 1). The coin count never changes, only the value does.

3. So the ₹60 shortfall is covered entirely by these upgrades: 4 × (number of ₹5 coins) + 1 × (number of ₹2 coins) = 60, i.e. 4x + y = 60, where x = ₹5 coins and y = ₹2 coins.

Algebraic form (same relation)

1. Let x, y, z be the counts of ₹5, ₹2, ₹1 coins.

2. Value: 5x + 2y + z = 100.

3. Count: x + y + z = 40.

4. Subtract the count equation from the value equation: 4x + y = 60 — the same relation as the shortcut.

5. Hence y = 60 − 4x and z = 40 − x − y = 3x − 20.

Why the choices are still needed

Requiring every count to be a positive whole number gives y = 60 − 4x ≥ 1 ⟹ x ≤ 14 and z = 3x − 20 ≥ 1 ⟹ x ≥ 7. So x could be any of 7, 8, 9, 10, 11, 12, 13, 14 — the data alone do not fix a single value. This is why the question is multiple-choice: of the four printed values, three exceed 14 (they would force a negative number of ₹2 coins), so the only printed value inside the feasible band 7 ≤ x ≤ 14 is 13. That is the intended answer, and it is the answer recorded in the official keys.

Cross-check

With x = 13: y = 60 − 4(13) = 8 and z = 3(13) − 20 = 19. Count: 13 + 8 + 19 = 40 ✓. Value: 5(13) + 2(8) + 1(19) = 65 + 16 + 19 = 100 ✓. Both conditions hold, so the number of ₹5 coins is 13.