Acoin bank that excepts only nickels and dimes contains $9.15. there are three more than twice as many nickels as there are dimes. how many of each coin are in the bank?

45 dimes and 93 nickels were in bank

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

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

From eqn 1

n = 3 + 2(45)

n = 3 + 90 = 93

Thus 45 dimes and 93 nickels were in bank

i think the correct option is the second option.

hreryryrranswer:

step-by-step explanation: