Question

A man has two more nickels than dimes. He has $1.15 in all. How many coins of each kind does he have?

Answer 1

Answer:

Man has 9 Nickels and 7 dimes.

Step-by-step explanation:

Let Number of dimes be x

and Number of Nickels be y

As Per given data he has two more nickels than dimes.

$y= 2+x$

As we know that

1 = 100 cent

1 nickel is worth = 5 cents = $\frac{5}{100}$= 0.05

1 dime is worth = 10 cents = $\frac{10}{100}$ 0 .1

According to given data he has $\ \ 1.15$ total nickel and dimes

So the equation becomes,

$0.05y+0.1x=1.15$

but $y= 2+x$

$0.05(2+x)+0.1x=1.15\\0.1+0.05x+0.1x=1.15\\0.1+0.15x= 1.15\\0.15x= 1.15-0.1\\0.15x=1.05\\x=\frac{1.05}{0.15}= 7$

Substituting value of x in equation $y = 2+x$ we get

$y= 2+7 =9$

Man has 9 Nickels and 7 dimes.