SOLUTION: There are 52 coins in a purse. Some of them are quarters, and some are dimes.
The total value of all coins is $7.45. how many were quarters and how many were dimes?
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Question 745011: There are 52 coins in a purse. Some of them are quarters, and some are dimes.
The total value of all coins is $7.45. how many were quarters and how many were dimes?
Found 2 solutions by stanbon, algebrahouse.com:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
There are 52 coins in a purse. Some of them are quarters, and some are dimes.
The total value of all coins is $7.45. how many were quarters and how many were dimes?
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Quantity Eq: d + q = 52 coins
Value Eq::: 10d+25q = 745 cents
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Modify for elimination:
10d + 10q = 520
10d + 25q = 745
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Subtract and solve for "q":
15q = 225
q = 15 (# of quarters)
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Solve for "d":
d + q = 52
d + 15 = 52
d = 37 (# of dimes)
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Cheers,
Stan H.
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Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
There are 52 coins in a purse. Some of them are quarters, and some are dimes.
The total value of all coins is $7.45. how many were quarters and how many were dimes?
x = number of quarters
52 - x = number of dimes {there are 52 coins total}
0.25x + 0.1(52 - x) = 7.45 {number of coins times value of coin equals total value}
0.25x + 5.2 - 0.1x = 7.45 {used distributive property}
0.15x + 5.2 = 7.45 {combined like terms}
0.15x = 2.25 {subtracted 5.2 from each side}
x = 15 {divided each side by 0.15}
52 - x = 37 {substituted 15, in for x, into 52 - x}
15 quarters and 37 dimes in the purse
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