# The number of coins of each type.

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.6, Problem 34E
To determine

## To find: The number of coins of each type.

Expert Solution

Mary has 5 quarters, 10 dimes and 15 nickels.

### Explanation of Solution

Given:

Mary has total amount \$3.00.

Dimes are twice than quarters and nickels are five more than dimes.

Calculation:

Let the coins of quarters be x.

Tabulate the given information into the language of algebra.

 In words In Algebra Coins of quarters x Coins of dimes 2x Coins of nickels 2x+5

Model the equation for the above information.

Coinsofquarters+Coinsofnickels+Coinsofdimes=Totalcoins0.25x+0.10×2x+0.05(2x+5)=3

Simplify the above equation for x,

0.25x+0.20x+0.10x+0.25=30.55x=2.75x=5

Coins of dimes,

dimes=2x=2×5=10

Coins of nickels,

nickles=2x+5=2×5+5=15

Thus, Mary has 5 quarters, 10 dimes and 15 nickels.

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