An amount of ₹120 is paid using ₹10, ₹5 and ₹2 coins. If the number of coins of ₹10 and ₹2 are interchanged, then the amount becomes ₹304. If the number of ₹5 and ₹2 coins are interchanged, then the amount is ₹165. How many ₹2 coins are used in the payment of ₹120?
25
Let a, b and c be the numbers of ₹10, ₹5 and ₹2 coins respectively.
Original payment: \(10a + 5b + 2c = 120\).
Interchanging the ₹10 and ₹2 counts: \(2a + 5b + 10c = 304\).
Interchanging the ₹5 and ₹2 counts: \(10a + 2b + 5c = 165\).
Subtract the first equation from the second: \(-8a + 8c = 184\), so \(c - a = 23\).
Subtract the third equation from the first: \(3b - 3c = -45\), so \(c - b = 15\).
Using \(a = c - 23\) and \(b = c - 15\) in \(10a + 5b + 2c = 120\) gives \(10(c-23) + 5(c-15) + 2c = 120\).
Expand: \(17c - 305 = 120\), so \(17c = 425\) and \(c = 25\).
Hence, the number of ₹2 coins used is 25.
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Important Questions from Linear Equation in 2 or more Variables
If \(3x+6y+9z = \dfrac{20}{3}, 6x+9y + 3z = \dfrac{17}{3}\) and \(18x+ 27y - z = \dfrac{113}{9}\) , then what is the value of \(75x+113y \ ?\)
If the system of equations 2x - 3y - 3 and -4x + qy - p/2 is inconsistent which of the following cannot be the value of p ?
If \(4x + \dfrac{6}{y}= 15 \) and \(6x - \dfrac{8}{y} = 14\) , then the value of p in y = px - 2 is :
Simplify: 7x + 3x(x – 4) = ?
A. 10x + 12
B. 10x – 12
C. 3x2 + 5x
D. 3x2 – 5xIf 5x + y = 17 and xy = 6, then what is the value of 125x3 + y3 ?
