Question

a coin bank that accepts only nickels and dimes contains $9.15. There are 3 more than twice as many nickels there are dimes. How many of each coin are in the bank.

Answer 1

45 dimes and 93 nickels were in bank

Answer:

solution

Let "n" be the number of nickels

Let "d" be the number of dimes

We know that,

value of 1 nickel = $ 0.05

value of 1 dime = $ 0.10

Given that There are three more than twice as many nickels as there are dimes

Number of nickels = 3 + 2(number of dimes)

n = 3 + 2d ---- eqn 1

Also given that coin bank that excepts only nickels and dimes contains $9.15

number of nickels x value of 1 nickel + number of dimes x value of 1 dime = 9.15

n \times 0.05 + d \times 0.10 = 9.15n×0.05+d×0.10=9.15

0.05n + 0.10d = 9.15 ---- eqn 2

Let us solve eqn 1 and eqn 2

Substitute eqn 1 in eqn 2

0.05(3 + 2d) + 0.10d = 9.15

0.15 + 0.1d + 0.10d = 9.15

0.2d = 9

d = 45

From eqn 1

n = 3 + 2(45)

n = 3 + 90 = 93

n = 93

Thus 45 dimes and 93 nickels were in bank