R pays ₹100 to P with ₹5, ₹2 and ₹1 coins. The total number of coins used for paying are 40. What is the number of coins of denomination ₹5 in the payment?
Question:
R pays ₹100 to P with ₹5, ₹2 and ₹1 coins. The total number of coins used for paying are 40. What is the number of coins of denomination ₹5 in the payment?
R pays ₹100 to P with ₹5, ₹2 and ₹1 coins. The total number of coins used for paying are 40. What is the number of coins of denomination ₹5 in the payment?
Options
Answer: (4) 13
Explanation:
Let x = number of ₹5 coins y = number of ₹2 coins z = number of ₹1 coins Total coins: x + y + z = 40 Total value: 5x + 2y + z = 100 Subtract the first equation from the second: (5x + 2y + z) − (x + y + z) = 100 − 40 4x + y = 60 So y = 60 − 4x Substitute in x + y + z = 40: x + (60 − 4x) + z = 40 −3x + 60 + z = 40 z = 3x − 20 For y and z to be non-negative: y ≥ 0 ⇒ 60 − 4x ≥ 0 ⇒ x ≤ 15 z ≥ 0 ⇒ 3x − 20 ≥ 0 ⇒ x ≥ 7 Check answer options: 13, 16, 17, 18 Only x = 13 satisfies conditions: y = 60 − 52 = 8 z = 39 − 20 = 19 Thus number of ₹5 coins = 13. Final Answer: 13
Explanation:
Let x = number of ₹5 coins y = number of ₹2 coins z = number of ₹1 coins Total coins: x + y + z = 40 Total value: 5x + 2y + z = 100 Subtract the first equation from the second: (5x + 2y + z) − (x + y + z) = 100 − 40 4x + y = 60 So y = 60 − 4x Substitute in x + y + z = 40: x + (60 − 4x) + z = 40 −3x + 60 + z = 40 z = 3x − 20 For y and z to be non-negative: y ≥ 0 ⇒ 60 − 4x ≥ 0 ⇒ x ≤ 15 z ≥ 0 ⇒ 3x − 20 ≥ 0 ⇒ x ≥ 7 Check answer options: 13, 16, 17, 18 Only x = 13 satisfies conditions: y = 60 − 52 = 8 z = 39 − 20 = 19 Thus number of ₹5 coins = 13. Final Answer: 13
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