Let’s be honest, word problems can be confusing, even more so when they ask about percents! We know what it means to say “what is 42% of 100,” but it’s less straightforward to say “42 is 76% of what number.” Many students, whether they’re in 5th grade or 10th grade, struggle to turn percent word problems into manageable number problems.
Daunting as percent word problems may be, I have a solution: “Awesome = Peanut Butter”! This is the equation I teach all my students: A = P × B
- A = Answer (as in, the number that is a percent of something else)
- P = Percent (can be written as (x/100) or as a decimal.)
- B = Beginning value (as in, the number that you are taking a percent of)
This format will work for any percent word problem―you just need to identify which value is which! Let’s look at a couple of examples:
1. What number is 50% of 100?
- Using our handy-dandy format, let’s ask ourselves, “which value am I missing: A, P, or B?” In this case, I have the percent value and the beginning value, so I must be missing the answer value. So we write out our problem with the values that we know:
A = (50/100) × 100 - Now that we have our equation, we can cross-cancel on the right and see that A = 50.
2. 50 is what percent of 100?
- Again using our same format, let’s ask ourselves, “which value am I missing: A, P, or B?” In this case, I have the answer value and the beginning value, so I must be missing the percent value. So we write out our problem with the values that we know:
50 = (P/100) × 100 - Now that we have our equation, we can again cross-cancel on the right and see that P = 50.
3. 50 is 50% of what number?
- Once again, we are going to rely on our format: “which value am I missing: A, P, or B?” In this case, I have the answer value and the percent value, so I am missing the beginning value. We write out our problem with the values that we know:
50 = (50/100) × B - Now that we have our equation, we can simplify the fraction on the right to (1/2) and multiply across.
50 = (1/2) × B
50 = (B/2) - Finally, we’ll multiply both sides by 2 to remove the fraction.
50 × 2 = (B/2) × 2
100 = B
Now that we have seen our equation in action, let’s apply it to a real word problem.
- Mason was able to sell 35% of his vegetables before noon. If Mason had 200 kg of vegetables in the morning, how many grams of vegetables was he able to sell by noon?
- Which value are we missing? We know that Mason had 200 vegetables in the Beginning, and we know that he sold 35 Percent. We are looking for the Answer value.
- A = (35/100) × 200
- A = (35/1) × 2
- A = 35 × 2
- A = 70
- Therefore we know that Mason sold 70 kg of vegetables.
So when next we come across one of the dreaded percent word problems, never fear! Remember that “Awesome = Peanut Butter” and solve away! Ready for your student to approach word problems and other topics with confidence? Reach out to a Director today to find a 1:1 math tutor who can instill that confidence!
