Question 1101094: Carol has some dimes and quarters if she has 25 coins worth a total of $3.70 how many of each type of coin does she have? Found 2 solutions by algebrahouse.com, greenestamps:Answer by algebrahouse.com(1659) (Show Source):
10x + 25(25 - x) = 370 {value of coin times number of coins equals total value}
10x + 625 - 25x = 370 {used distributive property}
-15x + 625 = 370 {combined like terms}
-15x = -255 {subtracted 625 from each side}
x = 17 {divided each side by -15}
25 - x = 8 {substituted 17, in for x, into 25 - x}
You can put this solution on YOUR website! Let me offer a couple of alternative methods for solving this kind of problem. The other tutor uses a single variable and substitution, which is fine. But I personally prefer either of these other methods.
Look at all three methods of solution and try each one, to find which one "works" best for you.
1. using two variables and elimination
let d = the number of dimes
and q = the number of quarters
Then
(a) (there are 25 coins total)
(b) (the total value of the coins is $3.70, or 370 cents)
Multiply the first equation by 10 (to get "10d"), then subtract the resulting equation from the second equation; this eliminates variable d, leaving an equation you can solve for q:
Then substitue q=8 in either original equation to find d:
2. using logical analysis -- which ends up doing basically the same calculations as the elimination method above.
If all 25 coins were dimes, the total value would be 250 cents; that is 120 cents less than the actual total of 370 cents.
Each quarter is worth 15 cents more than each dime; so each time we trade in a dime for a quarter, we keep the same number of coins but increase the total by 15 cents.
To increase the total by 120 cents to get the actual total of 370 cents, the number of quarters has to be 120/15 = 8.
And then the number of dimes is 25-8 = 17.
