# Money Word Problems

First, we will look at a money word problem involving calculating Simple Interest. Simple Interest word problems are based on the formula for Simple Interest and the formula for Amount. Then, we will look at a money word problem that involves coins and dollar bills.

Related Topics:

Compound Interest Word Problems
Other Algebra Word Problems

### Simple Interest Word Problems

Formula for Simple Interest

i = prt

i represents the interest earned
p represents the principal which is the number of dollars invested
t represents the time the money is invested which is generally stated in years or fractions of a year.

Formula for Amount

A = p + i

A represents what your investment is worth if you consider the total amount of the original investment (p) and the interest earned (i)

Example:

James needs interest income of \$5,000. How much money must he invest for one year at 7%? (Give your answer to the nearest dollar)

Solution:

5,000 = p(0.07)(1)

p = 71,428.57

He must invest \$71,429

### Simple Interest Word Problems

Example:
Pam invested \$5000. She earned 14% on part of her investment and 6% on the rest. If she earned a total of \$396 in interest for the year, how much did she invest at each rate?
Note that this problem requires a chart to organize the information. The chart is based on the interest formula, which states that the amount invested times the rate of interest = interest earned. The chart is then used to set up the equation.

How to solve simple interest word problems?
Example:
Suppose \$7,000 is divided into two bank accounts. One account pays 10% simple interest per year and the other pays 5%. After three years there is a total of \$1451.25 in interest between the two accounts. How much was invested into each account (rounded to the nearest cent)?

### Dollar and Coin Word Problems

Example:

Paul has \$31.15 from paper route collections. He has 5 more nickels than quarters and 7 fewer dimes than quarters. How many of each coin does Paul have?

Solution:

Let x be the number of quarters
x + 5 be the number of nickels
x – 7 be the number of dimes

25x + 5(x + 5) + 10(x – 7) = 3,115
25x + 5x + 25 + 10x – 70 = 3,115
40x = 3,160
x = 79

Paul has 79 quarters, 84 nickels and 72 dimes.

### Algebra Money Word Problems

Algebra Money Word Problems with two variables (x and y). How to solve equations with two variables?
Example:
David has only \$5 bills and \$10 bills in his wallet. If he has 5 bills totaling \$35, how many of each does he have?

### Coin Word Problem

Example:
Martin has a total of 19 nickels and dimes worth \$1.65. How many of each type of coin does he have? Note that this problem requires a chart to organize the information. The chart is based on the total value formula, which states that the number of coins times the value of each coin = the total value. The chart is then used to set up the equation.

How to solve word problems involving coins and money?
Example:
You have three times as many quarters as dimes and the total amount of money is \$6.80. How many quarters and dimes do you have? How to solve algebra word problem involving money?
Example:
You have 6 times as many quarters as dimes and the total amount of money ia \$8.00. How many quarters and dimes do you have? How to solve a coin word problem involving pennies and nickels?
Example:
A pile of 16 coins consists of pennies and nickels. The total amount of money is 36 cents. How many nickels and pennies do you have? How to solve a word problem involving stamps?
Example:
You bought 16 stamps consisting of 37-cent stamps and 23-cent stamps. If the total cost of the stamps is \$4.10, find the number and types of stamps purchased.

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