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Re: A purse contains 5 -cent coins and 10 -cent coins worth a total of
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17 Apr 2026, 05:21
Explanation
Let
x = number of 5‐cent coins
y = number of 10‐cent coins
\(5x+10y=175\) (since 1.75=175 cents)
The 5‐cent coins are replaced with 10‐cent coins: instead of x 5‐cent coins, we now have x 10‐cent coins.
The 10‐cent coins are replaced with 5‐cent coins: instead of y 10‐cent coins, we now have y 5‐cent coins.
So total value in cents after replacement:
\(10x+5y=215\) (since 2.15=215 cents)
\(5x+10y=175 (1)\)
\(10x+5y=215 (2)\)
Multiply (1) by 2:
\(10x+20y=350\)
Subtract (2) from that:
\((10x+20y)−(10x+5y)=350−215\)
\(15y=135\)
\(y=9\)
Substitute y=9 into (1):
\(5x+10(9)=175\)
\(5x+90=175\)
\(5x=85\)
\(x=17\)
\(x+y=17+9=26\)
Answer: A