Jose has $2.50 in nickels and dimes in his jar. If he has 11 more nickels than dimes, how many nickels does Jose have in his jar?

Question

zepdrix · Answer

If we let $\large d$ be dimes
and $\large n$ be nickles

The value of the total number of dimes would be $\large .1d$
(That's 10 cents times each dime we have).

If we do the same with the nickles $\large .05n$
(5 cents times the number of $\large n$ nickles)

If all of his nickles and dimes add up to $2.50, we can write it like this.

$\large .1d+.05n=2.50$
Some number of nickles and dimes at their given values add up to 2.50.

zepdrix · Answer

They gave us a little more information,
`he has 11 more nickels than dimes`

Which we can write this way,
$\large d+11=n$
How ever many dimes he might have, his nickles will add up to 11 more than that.

anonymous · Answer

Ok I underrstand up to here

zepdrix · Answer

Let's rewrite this equation in terms of nickles. So we'll subtract 11 from each side.$$\large d+11=n \qquad \rightarrow \qquad \color{green}{d=n-11}$$

We want to use both of these equations in order to solve for $\large n$, the number of nickles. So we'll use the green equation above, and plug it into the other equation we figured out.$$\large .1\color{green}{d}+.05n=2.50 \qquad \rightarrow \qquad .1\color{green}{(n-11)}+.05n=2.50$$

anonymous · Answer

so do i solve this and It will show the answer to the number of nickels

anonymous · Answer

the number of dimes is 13 correct?

zepdrix · Answer

Yes, with the way we setup the problem, we let $\large n$ represent the number of nickles in the jar.

So if we can solve for $\large n$, the problem is solved :)

zepdrix · Answer

dimes? ummmm yah that sounds correct :)

zepdrix · Answer

If the number of dimes is 13, we can plug that into this equation.

$\large d=n-11 \qquad \rightarrow \qquad 13=n-11$
to solve for n.

anonymous · Answer

24!!

zepdrix · Answer

yay good job \c:/

anonymous · Answer

Thanks