Question

A man has $34$ coins in his pocket, all of which are $\mathrm{dimes}$and $\mathrm{quarters}$. If the total value of his change is $565$ $cents$. How many $\mathrm{dimes}$and how many $\mathrm{quarters}$does he have?

Answer 1

Step-1: Form a pair of linear equations:

Given that,

• Man has $34$ coins in his pocket, all of which are $\mathrm{dimes}$and $\mathrm{quarters}$
• Total value of his change is $565$ $cents$

Let the number of $\mathrm{dimes}$be $\mathrm{x}$ and the number of $\mathrm{quarters}$be $\mathrm{y}$

Then, according to the question $\mathrm{x}+\mathrm{y}=34$ $...\left(\mathrm{i}\right)$

As $1$ $\mathrm{dime}=10\mathrm{cents}$ and $1\mathrm{quarter}=25\mathrm{cents}$

So, $10\mathrm{x}+25\mathrm{y}=565$ $...\left(\mathrm{ii}\right)$

Step-2: Find the number of $\mathrm{dimes}$and $\mathrm{quarters}$:

Solve $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ to find the value of $\mathrm{x}$ and $\mathrm{y}$

From $\left(\mathrm{i}\right)$, we get; $\mathrm{x}=34-\mathrm{y}$ $...\left(\mathrm{iii}\right)$

Substitute the above value of $\mathrm{x}$ in $\left(\mathrm{ii}\right)$,

$10\left(34-\mathrm{y}\right)+25\mathrm{y}=565$

$\Rightarrow$$340-10\mathrm{y}+25\mathrm{y}=565$

$\Rightarrow$ $15\mathrm{y}=565-340$

$\Rightarrow$ $15\mathrm{y}=225$

$\Rightarrow$ $\mathrm{y}=15$

Now substitute $\mathrm{y}=15$ in $\left(\mathrm{iii}\right)$, we get;

$\mathrm{x}=34-15$

$\Rightarrow$$\mathrm{x}=19$

Hence, he has $19$ $\mathrm{dimes}$and $15$ $\mathrm{quarters}$.