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Formulas and Mixtures Word Problems Study Guide

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Updated on Aug 24, 2011

Find practice problems and solutions for these concepts at Formulas and Mixtures Word Problems Practice Problems.

Some word problems can be solved using a specific formula. We must use the context and keywords of the problem to figure out which formula to use, and then substitute the values given in the problem into the formula to solve for the missing value. We'll look at some simple formulas, and then a more complex type of a word problem—a mixture problem.

Formulas: Subject Review

The formula for distance is D = rt, where D, distance, is equal to the product of r, rate, and t, time.

If an object travels at 30 miles per hour for three hours, it will travel 30 × 3 = 90 miles.

Sometimes we need to rearrange a formula to solve for a certain value. For instance, D = rt, but if we are given distance and rate and are looking for time, we must rewrite the formula to solve for time. Dividing both sides of the equation by r gives us t = . To find time given distance and rate, we must divide the distance by the rate.

In the same way, I = prt, but if we are looking for the principal given the interest, rate, and time, we must divide both sides of the formula by r and t to get p alone on one side of the equation. Principal is equal to . The formula for interest is I = prt, where I, interest, is equal to the product of p, principal, r, rate, and t, time. The rate is given as a percent and should be converted to a decimal.

A principal of $500, at a rate of 4% per year for six years, will make ($500)(.04)(6) = $120.

Given a temperature in Fahrenheit, F, we can convert to Celsius, C, or given a temperature in Celsius, we can convert to Fahrenheit: C = (F – 32) and F = C + 32.

If the temperature is 86° Fahrenheit, then the temperature in Celsius is (86 – 32) = (54) = 30°. If the temperature is 20° Celsius, then the temperature in Fahrenheit is (20) + 32 = 36 + 32 = 68°.

The formula for velocity is V = at, where V, velocity, is equal to the product of a, acceleration, and t, time.

If an object at rest accelerates at 5 meters per second for ten seconds, then its velocity is (5)(10) = 50 meters per second.

There are many formulas for converting one unit of measure to another. Here are just a few:

  • To convert from grams, meters, or liters to milligrams, millimeters, or milliliters, multiply by 1,000.
  • To convert from milligrams, millimeters, or milliliters to grams, meters, or liters divide by 1,000.
  • To convert from inches to feet, divide the number of inches by 12.
  • To convert from feet to inches, multiply the number of feet by 12.
  • To convert from feet to yards, divide the number of feet by 3.
  • To convert from yards to feet, multiply the number of yards by 3.
  • To convert from ounces to pounds, divide the number of ounces by 16.
  • To convert from pounds to ounces, multiply the number of pounds by 16.

Example

David's model car travels at 2 feet per second. If the car travels for ten seconds, how far will it travel?

Read the entire word problem.

We are given the rate at which a model car travels and the time in which it travels.

Identify the question being asked.

We are looking for the distance it travels.

Underline the keywords.

The keyword per can signal multiplication or division.

Cross out extra information and translate words into numbers.

There is no extra information in this problem.

List the possible operations.

To find the distance an object travels, we must multiply the rate by the time. The keyword per signals multiplication, since the rate gives us the distance traveled in one second, and we are looking for the distance traveled in ten seconds.

Write number sentences for each operation.

Our number sentence is the distance formula, with 2 substituted for r and 10 substituted for t: D = (2)(10)

Solve the number sentences and decide which answer is reasonable.

20 feet = (2)(10)

Check your work.

Since D = rt, we can check our distance answer by dividing the distance by the rate to see if that quotient equals the given time, 10 seconds: = 10 seconds. We also could have divided distance by the time to see if that quotient equals the given rate, 2 feet per second: = 2 feet per second.

At the same time we study a word problem for keywords, we should also search for words that tell u to use a specific formula. The word per can mean "rate," and the word travel often means that we are working with a distance. The words Celsius and Fahrenheit may indicate that we need to convert from one unit to the other.

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