All ACT Math Resources
Example Questions
Example Question #1 : Prisms
The length of a box is 3 times the width. Which of the following gives the length (L inches) in terms of the width (W inches) of the box?
L = 3W
L = 3/W
L = ½ (3W)
L = W + 3
L = 3W
When reading word problems, there are certain clues that help interpret what is going on. The word “is” generally means “=” and the word “times” means it will be multiplied by something. Therefore, “the length of a box is 3 times the width” gives you the answer: L = 3 x W, or L = 3W.
Example Question #82 : Solid Geometry
The width of a box, in inches, is 5 inches less than three times its length. Which of the following equations gives the width, W inches, in terms of the length, L inches, of the box?
W=3-5L
W=5-3L
W=3L-5
W=5L-3
W=3L-5
We notice the width is “5 inches less than three times its width,” so we express W as being three times its width (3L) and 5 inches less than that is 3L minus 5. In this case, W is the dependent and L is the independent variable.
W = 3L - 5
Example Question #972 : Act Math
Sturgis is in charge of designing a new exhibit in the shape of a rectangular prism for a local aquarium. The exhibit will hold alligator snapping turtles and needs to have a volume of . Sturgis knows that the exhibit will be long and go back into the wall.
What will the height of the new exhibit be?
This sounds like a geometry problem, so start by drawing a picture so that you know exactly what you are dealing with.
Because we are dealing with rectangular prisms and volume, we will need the following formula:
Or
We are solving for height, so you can begin by rearranging the equation to get by itself:
Then, plug in our knowns (, and )
Here is the problem worked out with a corresponding picture:
Example Question #1 : How To Find The Length Of An Edge Of A Prism
Sturgis is in charge of designing a new exhibit in the shape of a rectangular prism for a local aquarium. The exhibit will hold alligator snapping turtles and needs to have a volume of . Sturgis knows that the exhibit will be long and go back into the wall.
If three-quarters of the exhibit's volume will be water, how high up the wall will the water come?
Cannot be determined with the information provided
The trickiest part of this question is the wording. This problem is asking for the height of the water in the exhibit if the exhibit is three-quarters full. We can find this at least two different ways.
1) The longer way requires that we begin by finding three quarters of the total volume:
Now we go back to our volume equation, and since we are again looking for height, we want it solved for :
Becomes
2) The easier way requires that we recognize a key detail. If we take three-quarters of the volume without changing our length or width, our new height will just be three-quarters of the total height. We can solve for the total height of the exhibit by using the volume equation and rearranging it to solve for :
At this point, we can substitute in our given values and solve for :
So, the total height of the exhibit is . We can now easily solve for three-quarters of the total height: