Question

A jar contains $\$36.25$ in dimes and quarters. There are $250$ coins in the jar. How many quarters are there in the jar?

Answer 1

Hint: We first assume that there are x number of quarter coins and y number of dime coins in the jar. Then, we get two equations in x and y, one being the sum of the number of coins as $250$ and the other one being the sum of values of the coins being $\$36.25$ .

Complete step-by-step solution:
The dime is a ten-cent coin. Now, hundred cents makes a dollar. This means one dime is equal to one-tenth of a dollar. One quarter, as the name suggests, is equal to one-fourth of a dollar as quarter means one-fourth. So, from this discussion, we have got,
$\begin{align}
  & 1\text{dime}=\dfrac{1}{10}\times \$1=\$0.1\\&1quarter=\dfrac{1}{4}\times\$1=\$0.25\\\end{align}$
We are given that the total number of coins in the jar is equal to $250$ and that the total value of all the coins added together is $\$36.25$ . Let the number of quarter coins in the jar be x and the number of dime coins in the jar be y. Then, applying the given conditions in the problem, we can develop two equations which are,
$\begin{align}
  & x+y=250....\left( 1 \right) \\
 & 0.25x+0.1y=36.25....\left( 2 \right) \\
\end{align}$
Let us find the value of y in terms of x from equation $\left( 1 \right)$ . By doing so, we convert the two variable problems into a single variable problem. This gives,
$\Rightarrow y=250-x$
Substituting this value of y in equation $\left( 2 \right)$ , we get,
$\Rightarrow 0.25x+0.1\left( 250-x \right)=36.25$
Upon simplification of the above equation, we get,
$\Rightarrow 0.25x+25-0.1x=36.25$
Subtracting $0.1x$ from $0.25x$ , the equation thus becomes,
$\Rightarrow 0.15x+25=36.25$
Subtracting $25$ on both sides of the above equation, we get,
\[\begin{align}
  & \Rightarrow 0.15x=36.25-25 \\
 & \Rightarrow 0.15x=11.25 \\
\end{align}\]
Dividing both sides of the equation by $0.15$ we get,
\[\Rightarrow x=\dfrac{11.25}{0.15}=75\]
Therefore, we can conclude that there are $75$ quarter coins in the jar.

Note: We need to know the value of one dime and one quarter for this problem and must solve the equations carefully. In the end, we must put the values of the variables in the equation and see if they satisfy or not as a way to cross-check. We can also treat the problem as a single variable one by taking the number of dime coins as $250-x$ instead of y.