Thomas has a collection of 25 coins some are dimes and some are quarters. If the total value of all the coins is $5.05, how many of each kind of coin are there?

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Answer 1 · 2016-11-24T12:11:30

Thomas has 8 dimes and 17 quarters

To start, let's call the number of dimes Thomas has $d$ $d$ and the number of quarters he has $q$ $q$ .

Then, because we know he has 25 coins we can write:

$d + q = 25$ $d + q = 25$

We also know the combination of dimes and quarters add up to $$ 5.05$ $$5.05$ so we can also write:

$0.10 d + 0.25 q = 5.05$ $0.10d + 0.25q = 5.05$

Solving the first equation for $q$ $q$ gives:

$d + q - d = 25 - d$ $d + q - d = 25 - d$

$q = 25 - d$ $q = 25 - d$

We can now substitute $25 - d$ $25 - d$ for $q$ $q$ in the second equation and solve for $d$ $d$ :

$0.10 d + 0.25 (25 - d) = 5.05$ $0.10d + 0.25(25 - d) = 5.05$

$0.10 d + 6.25 - 0.25 d = 5.05$ $0.10d + 6.25 - 0.25d = 5.05$

$6.25 - 0.15 d = 5.05$ $6.25 - 0.15d = 5.05$

$6.25 - 0.15 d + 0.15 d - 5.05 = 5.05 + 0.15 d - 5.05$ $6.25 - 0.15d + 0.15d - 5.05 = 5.05 + 0.15d - 5.05$

$1.20 = 0.15 d$ $1.20 = 0.15d$

$\frac{1.20}{0.15} = \frac{0.15 d}{0.15}$ $1.20/0.15 = (0.15d)/0.15$

$d = 8$ $d = 8$

We can now substitute $8$ $8$ for $d$ $d$ in the solution of the first equation and calculate $q$ $q$ .

$q = 25 - 8$ $q = 25 - 8$

$q = 17$ $q = 17$