3.1 PS X-Ray
3.3 PS Step Method
3.4 Practice Problems
Later in the chapter we’ll present a step method that you should employ on every Problem Solving question you face. However, since different problems call for different approaches, one of the steps, “Plan the Attack,” is open-ended and calls for you to choose the most effective approach to the problem at hand. So before we get to the step method itself, we’ll first demonstrate a standard approach, as well as a few alternative approaches that may come in handy in particular situations. Each approach discussed in this section represents a different way to use the math concepts you reviewed in the previous chapter. We’ll cover the following:
  • Standard Applications of Math Concepts
  • Alternative Approaches for Special Cases
Standard Applications of Math Concepts
There’s no need to make things more complicated than they need to be; some questions require nothing more than straightforward applications of the concepts you learned in chapter 2. This doesn’t necessarily mean that such questions will be easy, since some of the concepts themselves can be complex, and the test makers occasionally complicate matters by sprinkling traps among the choices. Easier questions often require the application of a single concept, while harder questions may involve multiple concepts. Some may even require you to draw your own diagram when none is given. Regardless of the difficulty level, the standard application approach is the same: Scope out the situation, decide on what concept or concepts are being tested, and then use what you know about those concepts to answer the question before looking at the choices. If you’ve done your work well, the answer you get will be among the choices on the screen, and you’ll click it and move on.
Let’s look at a few examples spanning various difficulty levels. We’ll take a look at single-concept questions based around one particular math concept, and multiple-concept questions that require you to make use of numerous bits of math knowledge to arrive at the answer. As you’ll see, we’ve bolded all our Math 101 concepts as they come up to make it easier for you to navigate through our explanations.
Single-Concept Questions
Here’s an example of the most basic kind of Problem Solving question you’ll see:
What is the value of x if 3x – 27 = 33?
(A) 2
(B) 11
(C) 20
(D) 27
(E) 35
The math concept in play here is equations with one variable, something you likely remember from junior high school. There’s nothing to do here but apply the concept: First isolate the variable by adding 27 to both sides to get 3x = 60, and then divide both sides by 3 to get x = 20, choice C. No doubt the test makers include 2 among the answer choices to trap people who accidentally subtracted 27 from both sides, yielding 3x = 6 and x = 2. 11, choice B, is what you get if you divide one number in the problem (33) by another (3), and 27, choice D, appears in the problem itself. Assuming you didn’t fall for any of these traps, there’s not much to it: Just apply a single, fairly basic concept directly to the problem to pick up the point.
Not all single-concept questions are necessarily so straightforward, however, especially as you get on in the section. Try this one on for size:
At a local golf club, 75 members attend weekday lessons, 12 members attend weekend lessons, and 4 members attend both weekday and weekend lessons. If 10 members of the club do not attend any lessons, how many members are in the club?
(A) 65
(B) 75
(C) 82
(D) 93
(E) 101
There’s only one concept in play here, but if you don’t know it, you’re in for a very tough time. You need the formula for group problems with two groups: group 1 + group 2 – both + neither = total. This is a formula you probably didn’t learn in junior high or high school, or most likely forgot even if you did. You probably won’t see a question like this early in the section, but if you’re doing well, the CAT’s going to challenge you and start spitting out questions from the harder end of the question pool. In any case, it’s really only testing whether you’ve done your homework and memorized the formula.
If you did, then you’d be in great shape, since the math itself is not particularly difficult: If we let group 1 be the 75 members who attend weekday lessons and let group 2 be the 12 members who attend weekend lessons, we get: 75 + 12 – 4 + 10 = total. Solving for total gives 93, choice D. Notice how choice B, 75, is a number contained in the problem, while choice E, 101, is what you get if you mistakenly add 4 instead of subtract it.
Multiple-Concept Questions
Some Problem Solving questions require you to pull together two or more choice tidbits from your arsenal of essential math concepts. One of the most common examples of a multiple-concept question involves geometric formulas that generate equations that need to be solved arithmetically and/or algebraically. Here’s an example:
If AB = BC and x = 60, what is the length of CE in rectangle ACDE?
(A) 4
(D) 8
(E) 12
This is a bit more involved than a typical single-concept question because there are a number of geometry concepts you need to know and some genuine opportunities to slip up on the arithmetic end too. It’s a mish-mash problem to boot, involving three triangles and a rectangle, so if you don’t know the special and exciting properties of these geometric figures, you’re pretty much sunk right there. If you do, then you should be able to at least formulate the correct equation for line EC, but then you still have to crunch the numbers to solve it. Let’s see what a solid effort on this question might look like.
First, you’re best off redrawing the diagram on your scratch paper, since you wouldn’t want to keep all the information you’re going to add to it in your head. Since AB and BC are equal, ∠BAC and ∠BCA must be equal since the angles in a triangle opposite from equal sides are equal (concept 1). Since the third angle labeled x equals 60°, ∠BAC and ∠BCA together must total 120° because the three angles of a triangle add up to 180° (concept 2). Since we determined that ∠BAC and ∠BCA are equal, they both must be 60°. Notice anything now? A triangle with three equal angles is an equilateral triangle (concept 3). Since all three sides in an equilateral triangle are equal (concept 4), AB = BC = AC = 4. Since ACDE is a rectangle, and opposite sides of a rectangle are equal (concept 5), AC = ED = 4. By now your sketch should look like this:
Now that we have two sides of right triangle ECD, we have everything we need to figure out the length of EC, thanks to the Pythagorean theorem: x 2 + y2 = z2 where x and y are the sides and z is the hypotenuse (concept 6). Substituting 4 and 8 as the sides and EC as the hypotenuse gives us:
(EC)2 = 42 + 82
For convenience, we’ll denote all of the ensuing arithmetic, including simplifying the radical, as concept 7:
Voila!—choice B. Check out the traps: 4 (A) is a number calculated along the way; 8 (D) is a number given in the problem; and 12 (E) is what you get if you add the two known sides of triangle ECD together.
Notice that no fewer than seven math concepts made their way into this problem—none of them particularly earth-shattering or treacherous, mind you, but still adding up to a medium-level challenge with plenty of potential pitfalls.
Alternative Approaches for Special Cases
The standard “do question, look for answer” approach is all well and good in many cases, but some questions call out for alternative approaches. When the question contains variables in the answer choices, making up numbers and substituting them into the problem is often very effective. Conversely, when the answer choices contain actual numbers, you may benefit from simply plugging them into the given situation to see which one works, instead of hacking through some difficult arithmetic or algebra. Let’s take a look at each of these strategies, one by one.
Making Up Numbers
Which of the following problems would you rather be faced with on test day?
  • Question 1: If x apples cost y cents, how much will z apples cost in dollars?
  • Question 2: If 5 apples cost 50 cents, how much will 10 apples cost in dollars?
If you’re like most people, question 2 looks much easier, and you probably wouldn’t have much trouble solving it: If you double the number of apples, you double the number of cents. One hundred cents equals one dollar. Done.
The difference between question 1 and question 2 is simple. We replaced the variables in question 1 with some made-up numbers, thus creating the easier question 2. So, if you see x, y, m, n, or any other variables in both the question and the answer choices, see if you can avoid using complicated algebra by making up numbers and inserting them into the problem. You don’t want to just make up any old numbers, however—you want numbers that will simplify the problem. Use the following guidelines:
  • Pick easy numbers. Although you could choose 582.97 as a value, you definitely wouldn’t be making the problem any easier. Stick to relatively small, whole numbers whenever possible.
  • Avoid 0, 1, and any numbers used in the problem. The numbers 0 and 1 have unique properties that may skew the results when used for this technique, so don’t substitute either of those into the problem. (We’ll give you the exact opposite advice in the Quantitative Comparisons chapter, since in those questions the special properties of 0 and 1 come in handy.) Also, since the test makers sometimes use numbers from the question to construct distractors, you may get yourself into trouble by selecting those as well. If, for example, the problem contains the expression 3a + 5, don’t use 3 or 5. You shouldn’t have any trouble avoiding the few numbers used in the question itself, just for good measure.
  • Choose different numbers for different variables. For example, if the problem contains the variables m and n, you wouldn’t want to choose 2 for both. Instead, you might choose 2 for m and 3 for n.
  • Pay attention to units. If a problem involves a change in units (such as minutes to hours, pennies to dollars, feet to yards, and so on), choose a number that works well for both units. For example, 120 would be a good choice for a vari- able representing minutes, since 120 minutes is easily converted into 2 hours.
  • Obey the rules of the problem. Occasionally, the problem may include specific requirements for variables. For example, if the problem says that x must be negative, you can’t make up a positive value for x.
  • Save dependent variables for last. If the value of one variable is determined by the value of one or more other variables, make up numbers for those other variables first. That will automatically determine the value of the variable that depends on the value of the others. For example, if the problem states that a = b + c, a is dependent on b and c. Choose values for b and c first, and the value of a will then simply emerge as the sum of b and c.
Once you’ve selected your values, an actual number will emerge when you work the problem out with the numbers you’ve selected. All you need to do then is check which answer choice contains an expression that yields the same value when you make the same substitutions. This will make more sense in the context of an example, so let’s apply the strategy to the following question.
A gear makes r rotations in m minutes. If it rotates at a constant speed, how many rotations will the gear make in h hours?
Sure, you could crunch through this algebraically, and if that floats your boat, great. However, if you’re among those who get a headache from just looking at questions like this, making up numbers may be just the way to go. Here’s how.
The variables in this problem are r, m, and h. We can make up whatever values we’d like for these, as long as the values we choose make it easy to work the problem. For r, the number of rotations, let’s choose something small, like 3. For m, the number of minutes, we should choose a value that will make it easy to convert to hours: 120 works well, since 120 minutes is the same as 2 hours. Finally, for h, we should choose something small again. Remember that we need to choose a different value for each variable, so let’s use 4. Now that we have our numbers, simply plug them into the situation:

A gear makes 3 rotations in 120 minutes. If it rotates at a constant speed, how many rotations will the gear make in 4 hours?

Okay, much better—that’s something we can sink our teeth into. 120 minutes is the same as 2 hours, during which time the gear rotates 3 times. If it rotates 3 times in 2 hours, how many times will it rotate in 4 hours? That’s just twice as much time, so it will make twice as many rotations: 2 × 3 = 6, and so 6 is what we get when we substitute our values into the problem. Now we have to find the answer choice that’s equal to 6 when the same values are substituted for its variables. Just work your way down the list, using 3 for r, 120 for m, and 4 for h:
A: That’s not 6, the answer we seek, so move on.
B: Yup—this is exactly what we’re looking for, so B is correct. If you’re sure of your work, there’s no need to even continue with the choices; you’d just click B and move on to the next question. For practice, though, let’s see how the other three pan out:
C: Way too big.
D: Way too small.
E: Four times bigger than what we’re after. Just what we thought: B is the only choice that matches the number we derived from our made-up numbers, so B gets the point.
You’ll get more practice with this strategy as we go forward. Let’s now move on to our other specialty technique, an exercise in role reversal that we call . . .
Working Backward
When the question includes an equation (or a word problem that can be translated into an equation), and the answer choices contain relatively simple numbers, then it may be possible to plug the choices into the equation to see which one works. Working backward from the choices in this manner may help you avoid setting up or solving complicated equations and can save you time as well because of a neat wrinkle of this technique: Since the choices in math questions are usually written in either ascending or descending order, you can start with the middle choice, choice C, and either get the answer immediately or at least eliminate three choices for the price of one. Here’s how.
Let’s say the answer choices are in ascending order. If you start by plugging in C, then even if that choice doesn’t work, you can use the outcome to determine whether you need to plug in a smaller or larger number. If you need a smaller number, then D and E are out of the question, and you can go right to test choice A or B. If instead you need a larger number, chop A and B and try D or E. Notice another nice feature: When you plug in for the second time, that choice will either work or leave only one choice standing. If you follow this alternative approach, you shouldn’t ever have to check more than two choices.
As always, math strategies make the most sense in the context of examples, so we’ll demonstrate using the following question.
A classroom contains 31 chairs, each of which has either a cushion or a hard back. If the room has five more cushion chairs than hard-backed chairs, how many cushion chairs does it contain?
(A) 10
(B) 13
(C) 16
(D) 18
(E) 21
Now if you happen to be an algebra whiz, you’d go ahead and use the information to set up a pair of simultaneous equations to solve the problem. However, you may find it easier to work backward instead. Since the choices are in ascending order, we’ll start with the middle one and pretend it’s correct. If it really is correct, then plugging it into the problem’s scenario will cause all the numbers to work out, so let’s see if it does.
The question is looking for the number of cushion chairs, which for the moment we’re assuming to be 16. We can bounce that number off the information in the beginning of the second sentence (5 more cushion chairs than hard chairs) to determine that with 16 cushion chairs, there would have to be 11 hard chairs. Now all we have to do is check whether this scenario matches the information in the first sentence. Would that give us 31 chairs total? Nope: 16 + 11 = 27, so the numbers don’t jibe. That tells us three things: Choice C isn’t correct, choice B isn’t correct, and choice A isn’t correct. We can knock out A and B along with C because they’re both smaller than C, and if the number in C isn’t big enough to get us to our required 31 chairs, A and B ain’t gonna cut it either.
Now let’s try D—if it works, it’s correct, and if it doesn’t, we can select E without even trying it out: 18 cushion chairs means 18 – 5 = 13 hard chairs and 18 + 13 = 31 chairs total. That matches the information in the question, so D is correct. Note that you could have worked the numbers the other way: If there are 31 chairs total, and we assume there are 18 cushion chairs, then there would have to be 13 hard chairs. That matches the information in the second sentence that requires 5 more cushion chairs than hard ones. Either way you slice it, the number 18 fits the bill when plugged back into the situation, and we didn’t have to bother with creating and solving simultaneous equations.
Use your judgment as to when to work backward. If there are numbers in the answer choices, then consider it, but don’t do it if the numbers are unwieldy, such as complex fractions. One of the skills that the best math test takers possess is the ability to determine the most effective way to work through the problems. We’ve shown you standard applications and a few powerful alternatives. Now let’s assimilate what you’ve learned into the general step method you’ll use for all Problem Solving questions.
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