Word problems on the ACT Math Test often do not require particularly difficult algebra to solve. Most of the time, solving simple linear equations is all that you need to do! Here’s some quick tips to help you rock Word Problems on the ACT!

Start by writing down what the question is really asking. What do the answer choices represent? The number of marine animals in an aquarium? The cost per person for a haircut? This is usually in the last sentence of the word problem.

Assign variables to unknown quantities. Pick variables that are logical and clear. Let’s say a question is asking about the price of a child’s movie ticket versus an adult’s movie ticket. Choose “c” to represent the cost of one child’s ticket, and choose “a” to represent to cost of one adult’s ticket.

Pull out the key phrases and ‘translate’ them from English to Math. Look for these common words and phrases. Remember to pay special attention to subtraction and division since what comes first in the sentence does not necessarily come first in the expression (i.e. “4 less than x” = x – 4).

If you are having a hard time setting up the equations, try to work backwards using the answer choices. If the answer choices represent the cost of the child’s movie ticket, start with choice (C) and ask yourself, what if C is correct? What else has to be true? The right answer will agree with the given numbers and the relationship between the variables that is described in the question stem. This is also an excellent way to check your work! If you think you’ve found the correct answer, plug it in to make sure!

Try a practice Word Problem on your own!

Stephanie has built a new pool that is 5 ft x 5 ft x 4ft to replace her old one. The old pool is 5 feet long, 4 feet wide, and 2 feet high. It is currently half full. If Stephanie pours all of the water currently in the old pool into the new one, what fraction of the new pool will be filled?

(A) 1/8

(B) 1/6

(C) 1/5

(D) 1/4

(E) 1/3

Volume of a rectangular solid = length x width x height. Using this formula, we find that the volume of the old pool is .The pool is half full which mean it currently holds of water. The volume of the second pool is . To find the fraction of the water to the total volume, divide the amount of water added to the pool by its total capacity. 20/100 = 1/5, or (C).