SOLUTION: There are twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes, he would have 90 cents more than what he has now. Has many dimes an
Question 1104494: There are twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes, he would have 90 cents more than what he has now. Has many dimes and quarters does he have? Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52887) (Show Source): You can put this solution on YOUR website! .
D + Q = 20.
25D + 10Q = 10D + 25Q + 90
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D + Q = 20,
15D - 15Q = 90
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D + Q = 20
D - Q = 6
----------------Add the equations
2D = 26 ====> D = 13.
Answer. 13 dimes and (20-13) = 7 quarters.
Each time a quarter and a dime are switched, the total value of the coins changes by 15 cents.
If switching all the dimes for quarters and all the quarters for dimes changes the total value by 90 cents, then the difference between the numbers of dimes and quarters must be 90/15 = 6.
Since there are 20 coins in all, and since the difference between the numbers of dimes and quarters is 6, there are 13 of one coin and 7 of the other.
Finally, since the value of the coins would increase by 90 cents, he must have more of the smaller value coin -- so he has more dimes than quarters.
Well.... At least, that is the way I would solve it. For me, it seems the logical reasoning path is easier than the formal algebraic solution, which might look something like this....:
Let d be the number of dimes; then (20-d) is the number of quarters.
The value of the coins in cents is .
The value of the coins if the dimes and quarters were switched is .
The value with the coins switched is 90 cents more than the actual value: