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Word Problems About Money Help

By — McGraw-Hill Professional
Updated on Sep 27, 2011

Introduction to Money Word Problems

Coin problems are also common algebra applications. Usually the total number of coins is given as well as the total dollar value. The question is normally “How many of each coin is there?”

How to Solve Money Word Problems

Let x represent the number of one specific coin and put the number of other coins in terms of x . The steps involved are:

  1. Let x represent the number of a specific coin;
  2. Write the number of other coins in terms of x (this skill was developed in the “age” problems);
  3. Multiply the value of the coin by its number; this gives the total amount of money represented by each coin;
  4. Add all of the terms obtained in Step 3 and set equal to the total money value;
  5. Solve for x ;
  6. Answer the question. Don’t forget this step! It is easy to feel like you are done when you have solved for x , but sometimes the answer to the question requires one more step.

Examples

As in all word problems, units of measure must be consistent. In the following problems, this means that all money will need to be in terms of dollars or in terms of cents. In the examples that follow, dollars will be used.

Example 1:

Terri has $13.45 in dimes and quarters. If there are 70 coins in all, how many of each coin does she have?

Let x represent the number of dimes. Because the number of dimes and quarters is 70, 70 − x represents the number of quarters. Terri has x dimes, so she has $0.10 x in dimes. She has 70 − x quarters, so she has $0.25(70 − x ) in quarters. These two amounts must sum to $13.45.

Increasing/Decreasing by a Percent Examples

Terri has 27 dimes and 70 − x = 70 – 27 = 43 quarters.

Example 2:

Bobbie has $1.54 in quarters, dimes, nickels, and pennies. He has twice as many dimes as quarters and three times as many nickels as dimes. The number of pennies is the same as the number of dimes. How many of each coin does he have?

Nickels are being compared to dimes, and dimes are being compared to quarters, so we will let x represent the number of quarters. Bobbie has twice as many dimes as quarters, so 2 x is the number of dimes he has. He has three times as many nickels as dimes, namely triple 2 x : 3(2 x ) = 6 x . He has the same number of pennies as dimes, so he has 2 x pennies.

How much of the total $1.54 does Bobbie have in each coin? He has x quarters, each worth $0.25, so he has a total of 0.25 x (dollars) in quarters. He has 2 x dimes, each worth $0.10; this gives him 0.10(2 x ) = 0.20 x (dollars) in dimes. Bobbie has 6 x nickels, each worth $0.05. The total amount of money in nickels, then, is 0.05(6 x ) = 0.30 x (dollars). Finally, he has 2 x pennies, each worth $0.01. The pennies count as 0.01(2 x ) = 0.02 x (dollars).

The total amount of money is $1.54, so

Increasing/Decreasing by a Percent Examples

Bobbie has 2 quarters; 2 x = 2(2) = 4 dimes; 6 x = 6(2) = 12 nickels; and 2 x = 2(2) = 4 pennies.

Find practice problems and solutions at Word Problems About Money Practice Problems - Set 1.

Money Word Problems - One Quantity with Different Investments and Interest Rates

Some money problems involve one quantity divided into two investments paying different interest rates. Such questions are phrased “How much was invested at ___%?” or “How much was invested at each rate?”

Example

A woman had $10,000 to invest. She deposited her money into two accounts—one paying 6% interest and the other Increasing/Decreasing by a Percent Examples interest. If at the end of the year the total interest earned was $682.50, how much was originally deposited in each account?

You could either let x represent the amount deposited at 6% or at Increasing/Decreasing by a Percent Examples . Here, we will let x represent the amount deposited into the 6% account. Because the two amounts must sum to 10,000, 10,000 – x is the amount deposited at Increasing/Decreasing by a Percent Examples . The amount of interest earned at 6% is 0.06 x , and the amount of interest earned at Increasing/Decreasing by a Percent Examples is 0.075(10,000 – x ). The total amount of interest is $682.50, so 0.06 x + 0.075(10,000 – x ) = 682.50.

Increasing/Decreasing by a Percent Examples

The woman deposited $4500 in the 6% account and 10,000 – x = 10,000 – 4500 = $5500 in the Increasing/Decreasing by a Percent Examples account.

Find practice problems and solutions at Word Problems About Money Practice Problems - Set 2.

More practice problems for this concept can be found at: Algebra Word Problems Practice Test.

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