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KiRa [710]
3 years ago
9

A coin bank that excepts only nickels and dimes contains $9.15. There are three more than twice as many nickels as there are dim

es. How many of each coin are in the bank?
Mathematics
1 answer:
valentinak56 [21]3 years ago
3 0

45 dimes and 93 nickels were in bank

<em><u>Solution:</u></em>

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

<em><u>Given that There are three more than twice as many nickels as there are dimes</u></em>

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

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

<em><u>Also given that coin bank that excepts only nickels and dimes contains $9.15</u></em>

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 = 9.15

0.05n + 0.10d = 9.15  ---- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

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

<h3>d = 45</h3>

From eqn 1

n = 3 + 2(45)

n = 3 + 90 = 93

<h3>n = 93</h3>

Thus 45 dimes and 93 nickels were in bank

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