A payment of ₹120 is made with ₹10, ₹5 and ₹2 coins. A total of 25 coins are used. Which of the following is the number of ₹10 coins used in the payment?
- (a)10
- (b)4
- (c)8
- (d)6
Answer
Why
Correct — C. Let a, b and c be the numbers of ₹10, ₹5 and ₹2 coins.
a + b + c = 25
10a + 5b + 2c = 120
Double the count equation: 2a + 2b + 2c = 50
Subtract it from the value equation: 8a + 3b = 70
3b = 70 − 8a must be a non-negative multiple of 3, and that happens only at a = 2, a = 5 and a = 8.
Of the four printed values, 8 is the one on that list: a = 8 gives b = 2 and c = 15, and 80 + 10 + 30 = ₹120 in 8 + 2 + 15 = 25 coins → option (c).
Why the others are wrong
- (a)10 — Ten ₹10 coins are already ₹100, leaving 15 coins to carry ₹20. The smallest 15 coins can make is 15 × ₹2 = ₹30, so ₹20 is unreachable.
- (b)4 — a = 4 makes 3b = 38, which is not a multiple of 3. Four ₹10 coins leave 21 coins to carry ₹80, and no split of ₹5s and ₹2s does that.
- (d)6 — a = 6 makes 3b = 22, again not a multiple of 3. Six ₹10 coins leave 19 coins to carry ₹60, and 5b + 2c = 60 with b + c = 19 forces b = 22 ⁄ 3.
Concept
Two equations and three unknowns — the algebra alone cannot close this. What closes it is the integer constraint: coin counts are whole numbers and cannot be negative.
Eliminating c between the count equation and the value equation leaves a single line, 8a + 3b = 70. Every whole-number point on it is a genuine way to pay, and there are three: (a, b) = (2, 18), (5, 10) and (8, 2).
The item is answerable because exactly one of those three values of a is printed among the options.
The stem asks which of the following is the number of ₹10 coins, not how many there are. That wording matters here: three different payments satisfy both conditions, and the options are what single one of them out.
Key facts
- Eliminating the ₹2 coins from 25 coins worth ₹120 leaves 8a + 3b = 70.
- 8a + 3b = 70 has three non-negative whole-number solutions: (2, 18), (5, 10) and (8, 2).
- a = 8 pays 8 × ₹10 + 2 × ₹5 + 15 × ₹2 = ₹120 using 25 coins.
Study next
Common traps
- Assuming the payment is unique and hunting for a single algebraic answer.
- Checking an option against the ₹120 total but forgetting the 25-coin total.
- Eliminating the ₹10 coins instead of the ₹2 coins, which leaves a messier line than 8a + 3b = 70.
SSC sets these with three denominations, one stated total value and one stated total count, so two equations have to be finished off by whole-number reasoning rather than by more algebra.
The same shape is asked at Quant Q.9 of 09 Sep 2024, 09:00 — ₹100 paid in ₹5, ₹2 and ₹1 coins with 40 coins in all.
Related PYQs
No directly related past PYQ was found.