TABE Level A Math: Problem Solving

Problem Solving

Problem solving is about getting answers to “word problems” or “story problems.” To do this successfully, you have to read a problem to determine what you must find and what information the problem tells you. These kinds of problems range from very easy to quite difficult, and there is no way to cover every type of problem that you may see.

There are many types of “word problems.”

1. Markup and discount in retail sales (percent)

2. Tipping in restaurants (percent)

3. Problems involving proportion

4. Fuel efficiency and travel (rates)

Markup and Discount in Retail Sales

Markup in retail sales provides a backdrop for one group of word problems involving percents. An item has a regular retail price. This price reflects a percent markup from the wholesale price to the dealer. A problem might tell you the percent markup and the retail price and ask you to find the wholesale price. Or it may tell you the wholesale price and the percent markup and ask you to find the retail price. Finally, the problem may tell you the wholesale and retail prices, and you have to find the percent markup.

There are three numbers in each of these problems. The problem tells you two of them and asks you to find the third.

Example A.   The wholesale price of a leather jacket is $150. It is marked up 60%. What is the retail price? This problem tells you the wholesale price and the percent markup. You have to find the retail price.
The retail price is $150 plus 60% of $150. Just focus on 60% of $150.  150 is the whole, and 60 is the percent. You have to find the part.

The formula is Part = % × Whole ÷ 100. Substituting the values for this problem, Part = 60 × 150 ÷ 100 = 90. So the markup is $90. Add this to the $150 wholesale price, and the retail price is $240.
There is a shortcut for this problem. First, recall that 60% = 0.6. The retail price is 1 × wholesale price + 0.6 × wholesale price. This equals 1.6 × wholesale price (the distributive property of multiplication over addition). So the retail price is just 1.6 × 150 = 240.


To find the retail price, change the percent markup to a decimal, add it to 1, and multiply the result by the wholesale price.

Example B.   The wholesale price of a necklace is $250, and its retail price is $425. What is the percent markup?
First, subtract $250 from $425 to find the amount of the markup, $175.  The part is 175, and the whole is 250. You want to find the percent. The formula is % = Part ÷ Whole × 100. Substituting the numbers for this problem, % = 175 ÷ 250 × 100 = 70. The percent markup of the necklace is 70%.
There is a shortcut for this problem as well. Divide the retail price by the wholesale price: 425 ÷ 250 = 1.7. This result, 1.7, is the wholesale price plus 70% of the wholesale price.


To find the percent markup, divide the retail price by the wholesale price and subtract 1. Move the decimal point of the answer two places to the right.

Example C.   A high-definition television set you want retails for $2300. You learn that this price is the result of a 30% markup over the wholesale price. What is the wholesale price?
Call the wholesale price w. Using the shortcut from the last example, 1.3w = 2300. Solve this equation for w.

The wholesale price is $1769.23.
This problem also has a shortcut. Change the 30% markup to a decimal and add it to 1 to get 1.3. Then divide $2300 by 1.3 to get the wholesale price: $2300 ÷ 1.3 = $1769.23.

To find the wholesale price, change the percent markup to a decimal and add it to 1. Then divide the retail price by the answer.

Discount (sales) provides another group of problems in retail. The three variations of discounting problems correspond exactly to the preceding three wholesale-retail variations. The retail price of an item and the % discount are given. You have to find the discounted (sale) price. Or the retail price and sale price are given, and you have to find the % discounted. Finally, the sale price and the % discount are given, and you have to find the retail price.
Both markup and discount problems involve three numbers, of which two are given and one must be found. The difference between the two types of problems is the quantity that represents the “whole” in the percent scenario. When looking at percent markup, the wholesale (lower) price is the whole on which the percent is calculated. In the discount problems, the retail (higher) price is the whole on which percent is calculated.

These differences can be illustrated by an example.

Suppose you have an item that retails for $100. A 25% discount on that item results in a $75 sale price. If you then raise this $75 price by 25%, the new price would be only $93.75 instead of the $100. This is because 25% was based on 100 for the reduction, but it was based on 75 for the increase.

Example D.   The retail price of a computer game is $57. A store has the game on sale at 15% off. What is the sale price?
This problem tells you the retail price and the % discount and asks you for the discounted price. In the language of percents, 57 is the whole, and 15 is the percent. You must find the part (discount) and subtract it from the whole. The formula is Part = % × Whole ÷ 100. Substitute the number from this problem to get Part = 15 × 57 ÷ 100 = 8.55. So the discount $8.55 is subtracted from $57, resulting in a sale price of $48.45.
There is a shortcut for this problem. If there is a 15% discount, you actually pay 85% (100% − 15%). So the discounted price is 0.85 × $57 = $48.45.

To find the discounted price, subtract the percent discount from 100, move the decimal two places to the left, and multiply by the original price.

Example E.   George buys a $15 calculator for $12. What is the percent discount?
The amount of the discount is $3. This is the part. The retail price is $15. This is the whole. % = Part ÷ Whole × 100, so % = 3 ÷ 15 × 100 = 20. A 20% savings was realized.

To find the % discount, divide the discounted price by the regular price and move the decimal two places to the right.

Example F.   Sue bought a pair of jeans for $40 at a sale where all merchandise is 30% off. What was the original price?
Using the idea from the previous example, the $40 sale price is 70% (100% − 30%) of the original price. Divide 40 by 0.7 to get the original price of $57.14.

To find the original price, subtract the percent discount from 100 and move the decimal two places left. Divide the sale price by the answer.

These six examples don’t cover every type of percent problem. Instead of asking how much was paid for an item after a discount, a question could ask how much the discount was or how much was saved. Or instead of asking for the retail price of an item after a percent markup, a problem might ask what the markup is. You need to be sure to read a problem carefully to ensure that you answer the question asked.

Tipping in a Restaurant

When you eat in a restaurant, you usually leave a tip. It is customary to leave an amount that is between 15% and 20% of the bill (less tax). A problem might tell you the amount of a bill and ask the amount of a tip of 15%. Or a problem might tell you the amount of the bill and the amount left as a tip and ask what percent the tip is. Finally, a problem might tell you the total amount that was left for the bill and the tip, tell you the percent of the bill that is the tip, and ask how much the tip is.

Example G.   The Johnson family went to dinner, and the bill was $52.50 without tax. The service was excellent, so Ms. Johnson left a 20% tip. How much money was this?
You need to calculate 20% of $52.50. In the language of percents, the tip is the part and the amount of the bill is the whole.

The formula is Part = % × Whole ÷ 100.

Substitute the numbers for this problem to get Part = 20 × 52.5 ÷ 100 = 10.5. Ms. Johnson should leave a $10.50 tip.
There is a nice shortcut for this problem in case you don’t happen to have a calculator with you in the restaurant. The decimal equivalent of 10% is 0.1.

When you multiply a number by 0.1, move the decimal point one place to the left. So 10% of $52.50 is $5.25. Since 20% is double 10%, a 20% tip is double $5.25, or $10.50. For a 15% tip, you would add half of $5.25 (about $2.60) to $5.25 for an amount of $7.85.

Example H.   Six friends went to lunch at a nice restaurant. The bill, without tax, was $65, and they decided to leave $2 each as a tip. What percent tip did they leave?
Since six people each left $2, the total tip was $12. In this problem, 12 is the part, and 65 is the total. You must find the percent.

The formula is % = Part ÷ Whole × 100.

Substituting, you get % = 12 ÷ 65 × 100 = 18.5, approximately. The group left an 18.5% tip.

Example I.  Jim took Mary out for a business lunch. Jim remembered that the cost of lunch and the tip was $48, but he forgot the cost of the lunch. He always leaves a 20% tip. How much was the tip?
You can use the same strategy as described in the shortcut of Example A to solve this problem. The $48 is 1.2 times the cost of the lunch. Suppose c stands for the cost of lunch. Then 1.2c = 48. To solve this equation for c, divide both sides of the equation by 1.2.



The cost of the lunch is $40. Subtract $40 from $48 to get $8 for the tip.

Problems Involving Proportion

A proportion says that two fractions are equal.

You encounter proportions all the time in your daily life. This proportion could mean that if 3 notebooks cost $8, then you would expect 15 notebooks to cost $40. Or it might mean that if you get 3 hits in 8 at bats in baseball, you would expect to get 15 hits in 40 at bats. Or it could mean that if 3 of 8 people have brown hair, you would expect 15 of 40 people to have brown hair.

Every proportion has four numbers. In a proportion problem, you know three of the numbers, and you have to find the fourth.

Example J.   In his first game of the season, Michael made 6 of 7 free throws. If he takes 182 free throws during the whole season, how many would he expect to make at this rate?
Set up the proportion:

x = 6 × 182. Then divide by the number diagonally across from x.1 So x = 6 × 182 ÷ 7 = 156. Michael can expect to make 156 free throws.

Example K.   A certain recipe calls for cup of sugar to make enough for 6 people. Amanda has invited 15 guests for dinner and wants to use this recipe. How much sugar will she need to use?
Set up the proportion

x: 

Travel and Fuel Efficiency

Two other useful types of word problems arise when you drive. One is fuel efficiency. In the United States, fuel efficiency is measured in miles per gallon—how far you can travel on one gallon of fuel. For example, if you can travel 380 miles on 20 gallons of fuel, your fuel efficiency is ___ miles per gallon. This is actually just a special type of proportion. If you can travel 380 miles on 20 gallons, how many miles can you travel on 1 gallon? The answer is If you solve this proportion as described earlier, you would get the same 19 miles per gallon answer.

Example L.   Raphael’s car gets 31 miles per gallon, and his fuel tank holds 12.3 gallons. Can he make a 350-mile trip on one tank of fuel?
Each gallon of fuel is good for 31 miles, so 12.3 gallons of fuel is good for 12.3 × 31 = 381.3 miles. So yes, he can make a 350-mile trip on one tank.
A second type of problem involving driving relates the three quantities of distance, time, and speed. Let d stand for distance, r for average speed, and t for time.

There are three forms of the formula that relate these three quantities:
1. Distance = average speed × time (d = rt)

2. Average speed = distance ÷ time

3. Time = distance ÷ average speed


Example M.   Pat drives an average of 40 miles per hour for 45 minutes. How far does she travel?
You have to find the distance. Forty-five minutes is 3/4 hour, so Pat drives a distance of 30 miles.

Example N.   George drives a distance of 325 miles in 6 hours and 30 minutes. Find his average speed.
Six hours and 30 minutes is 6.5 hours, so George’s average speed is 50 miles per hour.
60 miles per hour is 1 mile per minute.

Example O.  Bob must drive his truck to Columbus by 6:00 P.M. He is now in Cleveland, 142 miles away, and it’s 3:30. How fast must he drive (average speed)?
Bob has 2.5 hours to make the trip, and Bob must drive an average speed of 56.8 miles per hour.

Other types of word problems—those involving algebra— In those problems you need to formulate an equation that represents relationships among the pieces of information given in the problem.