How do I solve this word problem? David has nickles, dimes, and quarters worth $15.75. He has 10 more dimes than nickles and twice as many quarters as dimes. How many of each does he have?

Question

anonymous · Answer

Think of it this way....q represents how much in quaters...d for dimes and n for nickles.
There are three equations here. The first one says the total number of money here is 15.75 so... q + d + n = 15.75
Then  ten more dimes than nickles is 10d = n 
And twice as many quarters than dimes is
2q = d

anonymous · Answer

So do I have to solve for each of them separately? Thanks, by the way!

whpalmer4 · Answer

I would do it slightly differently:  I would let $q$ be the number of quarters, $d$ the number of dimes, $n$ the number of nickels.  That means that the value of the coins is given by $$25q+10d+5n$$ and as we know the total value is $15.75 (or 1575 cents), our first equation would be $$25q+10d+5n=1575$$

Now, we also know that there are 10 more dimes than nickels, so $d= 10+n$ is another equation.  Finally, twice as many quarters as dimes is $q = 2d$, which is our third equation.

We have 3 equations in 3 unknowns, which is sufficient to determine the answer.  We can make it into 1 equation in 1 unknown by doing substitutions with the coin relationships:

$$25q+10d+5n = 1575$$Use $q=2d$:
$$25(2d)+10d + 5n = 1575$$Now use $d=10+n$
$$25(2(10+n))+10(10+n)+5n=1575$$
Expand that out and solve for $n$.  Once you have that, use the two formulas relating the number of dimes and quarters and nickels to find the values of $d$ and $q$.

anonymous · Answer

Thanks very Much :)

whpalmer4 · Answer

To check your answer, plug d, q and n into 25q + 10d + 5n and make sure the result is 1575.

kenljw · Answer

25(2(10+n))+10(10+n)+5n=1575  factor
(10+n)((25 2+10))+5n=1575
(n+10)12+n=315  divide by 5
13n +120=315
n=15