Question:
A change-making machine contains one-rupee, two-rupee and five-rupee coins. The total number of coins is 300. The amount is Rs. 960. If the numbers of one-rupee coins and two-rupee coins are interchanged, the value comes down by Rs. 40. The total number of five-rupee coins is
A change-making machine contains one-rupee, two-rupee and five-rupee coins. The total number of coins is 300. The amount is Rs. 960. If the numbers of one-rupee coins and two-rupee coins are interchanged, the value comes down by Rs. 40. The total number of five-rupee coins is
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Let the numbers of one-rupee and two-rupee coins be x and y. The interchange changes the value by exactly the difference \(y - x\), which gives one equation quickly.
Updated On: Nov 4, 2026
- 140
- 100
- 60
- 150
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The Correct Option is A
Solution and Explanation
Step 1: Set up the unknowns:
Let the numbers of one-rupee, two-rupee and five-rupee coins be \(x, y, z\).
Number of coins: \(x + y + z = 300\).
Value: \(x + 2y + 5z = 960\).
Step 2: Use the interchange:
After the interchange there are \(y\) one-rupee coins and \(x\) two-rupee coins, so the new value is \(y + 2x + 5z = 960 - 40 = 920\).
Subtracting from the original value: \((x + 2y) - (2x + y) = 40\), so \(y - x = 40\).
Step 3: Solve:
Subtract the coin-count equation from the value equation: \(y + 4z = 660\).
Put \(y = x + 40\) in \(x + y + z = 300\): \(2x + z = 260\), so \(z = 260 - 2x\).
Then \(x + 40 + 4(260 - 2x) = 660\), so \(1080 - 7x = 660\) and \(x = 60\).
Hence \(y = 100\) and \(z = 260 - 120 = 140\).
Step 4: Check:
\(60 + 100 + 140 = 300\); \(60 + 200 + 700 = 960\); after the interchange \(100 + 120 + 700 = 920\), a drop of 40. All conditions hold.
Final Answer:
The correct option is (1): 140.
Let the numbers of one-rupee, two-rupee and five-rupee coins be \(x, y, z\).
Number of coins: \(x + y + z = 300\).
Value: \(x + 2y + 5z = 960\).
Step 2: Use the interchange:
After the interchange there are \(y\) one-rupee coins and \(x\) two-rupee coins, so the new value is \(y + 2x + 5z = 960 - 40 = 920\).
Subtracting from the original value: \((x + 2y) - (2x + y) = 40\), so \(y - x = 40\).
Step 3: Solve:
Subtract the coin-count equation from the value equation: \(y + 4z = 660\).
Put \(y = x + 40\) in \(x + y + z = 300\): \(2x + z = 260\), so \(z = 260 - 2x\).
Then \(x + 40 + 4(260 - 2x) = 660\), so \(1080 - 7x = 660\) and \(x = 60\).
Hence \(y = 100\) and \(z = 260 - 120 = 140\).
Step 4: Check:
\(60 + 100 + 140 = 300\); \(60 + 200 + 700 = 960\); after the interchange \(100 + 120 + 700 = 920\), a drop of 40. All conditions hold.
Final Answer:
The correct option is (1): 140.
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