Math Workout for the GMAT
Part III Content and Strategy Review
Chapter 5 Solutions Beyond Algebra
For many people, algebra is not fun. It can be timeconsuming, and it’s easy to make mistakes with algebra. If algebra isn’t your favorite way to crunch numbers, fear not. In this chapter, you’ll learn some powerful methods for avoiding algebra in a lot of questions on the Quantitative section of the GMAT. If you happen to like algebra, well, you’re in luck, too! The next chapter is about facing algebra headon. So no matter your preferred method for cracking math questions, these next two chapters will be of use. Use the system that you’re most comfortable with, be it avoiding algebra or facing algebra. If you have time, you should get comfortable with both systems, so that you can use one system to solve the problem and the other system to check your work. For now, let’s dive into strategies for algebra avoidance. You’ll need some practice to be comfortable using these techniques and to be able to recognize opportunities to use them. First, however, a question: Which of the following two problems would you rather see on the GMAT?
1.Otto has a fivedollar bill. He goes to the store and buys 3 pieces of candy that cost 50 cents each. How much change, in dollars, does Otto receive?
$0.50 

$1.50 

$2.00 

$3.00 

$3.50 
2.Oscar has an xdollar bill. He goes to the store and buys y pieces of candy that cost z cents each. How much change, in dollars, does Oscar receive?
x – yz 

xy – z 

x – 

100x – yz 

Why does the second question seem so much harder? It’s identical to the first question except that it contains variables instead of numbers. For many people, the variables make this question seem harder. If that’s how you feel, then you’re in the right place. The techniques in this chapter will help you transform algebra questions into arithmetic questions and make the world a better place.
By the way, the answer to question 2 is not (A). You’ll see how to solve it in a short while.
PLUGGING IN
As you saw with the problem about Oscar’s candy, variables can make a question considerably tougher. Wouldn’t it be wonderful if there were a way to get rid of those variables? Well, there is. It’s a method called Plugging In. Whenever you see a problem with variables in the answer choices, you should consider this approach.
Here’s the method.
1. Replace every variable in the problem with a number. Just make one up.
2. Work the problem using the numbers you plugged in. You’ll get a number for an answer. This number is your target. Circle it on your noteboard.
3. Plug your madeup numbers into the variables in the answers. Check all five and see which one matches your answer from step 2.
Try this approach with the Oscar example.
2.Oscar has an xdollar bill. He goes to the store and buys y pieces of candy that cost z cents each. How much change, in dollars, does Oscar receive?
x – yz 

xy – z 

x – 

100x – yz 

1. Make up numbers for the variables. Maybe x = 10 dollars, y = 4 pieces of candy, and z = 50 cents. You can try a different set if you like. Plugging In works with any numbers.
2. Work the problem using these numbers. If Oscar buys 4 pieces of candy for 50 cents each, he spends 4 × 50 = 200 cents, or 2 dollars. He started with 10 dollars, so he gets 10 − 2 = 8 dollars in change. It’s important to make sure that your answer is in the units (dollars or cents) specified by the problem. So the numerical answer (your target) is 8. Circle it on your noteboard.
3. Plug your madeup numbers into the answer choices. Using x = 10, y = 4, and z = 50, the answers are
As you can see, (C) is the only answer that matches your target of 8. Note that the obvious answer choice, (A), is not correct. The problem with (A) is that it doesn’t account for the difference in units (dollars versus cents) between x and z. Many algebra problems contain one or more potential pitfalls like that, and the test writers will include trap answers to snare people who make those mistakes.
Why Use Plugging In?
Some people are initially skeptical about Plugging In. “That’s not the real way to answer the question,” they might say. Or, “Why should I learn another way to solve the problem, instead of using algebra?”
Well, there are several reasons that Plugging In is superior to algebra in some cases. First, you are much more likely to make a mistake when manipulating variables than when using regular numbers. Your brain processes 5 + 3 more easily than it does x + y. That’s what makes the Oscar example tougher than the Otto example at the beginning of the chapter. In doing algebra, you have to concentrate on the variables, and that leaves you open to mistakes such as forgetting to convert units.
Second, Plugging In makes tough problems easier. Because of the adaptive nature of the test, you won’t see any problems that will be easy for you. If you’re good at algebra, the questions will just get tougher and tougher until working them algebraically is impractical. Plugging In is a great way to bring the difficulty of a question down a notch or two.
Third, Plugging In is faster. On some of the easiest problems, Plugging In may not give you much of an edge when it comes to pacing. The main advantage with easier problems is the accuracy it provides. However, on the harder problems, Plugging In is simply faster. So, not only is it more accurate on the harder problems, but it also speeds you up.
Finally, it’s always a good idea to have plenty of tools at your disposal. If one way of attacking a problem doesn’t work, you simply try another. Plugging In is a tool that works well in a variety of situations, making it a nice option when you’re uncertain how to approach a particular question.
Here’s an example of a problem that is very tough to solve algebraically, but much easier to solve via Plugging In.
3.The sum of two integers is x. If the larger integer is greater than the smaller integer by 8, what is the product of x and the smaller integer, in terms of x ?
x^{2} + 12x 

2x^{2} − 8 

x^{2} − 8x 

First, try to solve this problem algebraically. Go ahead, take your time.…
Tough, wasn’t it? Now work this problem with Plugging In. Pick numbers for the two integers. You do need to follow the restriction that one is greater than the other by 8. Suppose you choose 2 and 10. You’re told that x is the sum of the integers, so x = 2 + 10 = 12. The question asks for the product of x and the smaller integer, so your target is 12 × 2 = 24.
Finally, plug x = 12 into the answer choices, to see which one matches your target of 24.
Choice (A) becomes 12^{2} + 12(12) = 288.
Choice (B) becomes 2(12^{2}) − 8 = 280.
Choice (C) becomes − 4(12) = 24.
Choice (D) becomes 12^{2} − 8(12) = 48.
Choice (E) becomes + 4(12) − 8 = 112.
Answer choice (C) matches your target, so it is the correct answer.
As you can see, Plugging In was much easier than working the algebra for that problem. You should practice this method, so that you can use it effectively on tough algebra problems.
Hidden Variables
You’ve seen how Plugging In works when the answer choices contain variables, but sometimes the variable in the answers isn’t quite so obvious. In fact, it may be invisible. If the answers contain fractions or percents, check the question. If it asks for a fraction or percent of some unknown amount, that unknown amount is the invisible variable. Plug in a number for that variable and use the steps you just learned. Here’s an example.
3.Marty spends of his weekly allowance on baseball cards. He spends of the rest on bubble gum. He spends of his allowance on soda pop. If Marty has no other expenses and saves the rest of his allowance in a piggy bank, what fraction of his allowance does he save in the piggy bank?
The question asks for a fraction of Marty’s allowance, so Marty’s allowance is the invisible variable in the answer choices. First, make up a number, say $60, for Marty’s allowance. Second, work the problem using that number. Marty spends × 60 = 20 on baseball cards. That leaves 60 − 20 = 40 to spend. He spends of the rest or × 40 = 10 on bubble gum. That leaves 40 − 10 = 30 dollars. He spends × 60 = 10 on soda pop, leaving 30 − 10 = 20 dollars for his piggy bank. Third, you need to convert that number to match the answers. He saves 20 dollars out of the original 60 or = . The answer is (C).
Choosing Good Numbers
When you use the Plugging In method, you should exercise some care in selecting the numbers you plug in. Although the method will work with any numbers, you’ll make things easier if you follow two general guidelines.
First, plug in numbers that will work nicely with the calculations necessary to solve the problem. In the previous example, you start with Marty’s allowance and multiply by the fractions , , and . Although you could plug in an ugly number, such as $11.17, for his allowance, that would make the calculations much tougher than necessary. Instead, you should choose a number that’s easily divisible by 3, 4, and 6, such as $60.
Second, avoid numbers that make more than one answer match your target. Suppose you plug in x = 4 for a certain question, and your target works out to be 16. If 4x and x^{2} are both among the answer choices, you have a problem: Both match your target. How can you tell which of them is the real answer?
If you do end up with more than one answer matching your target, simply rework the problem by plugging in a different number. Then, check only the answers that seemed correct the first time.
Of course, it would be much better to prevent that problem in the first place. To do that, avoid plugging in numbers that appear in the problem or the answer choices. You should also avoid 0 and 1, because they are numbers that do weird things.
Here is a summary of guidelines to follow when selecting numbers to plug in.
Choose numbers that work nicely.
· For problems involving fractions, use a multiple of the denominators.
· For problems involving different units, use a multiple of the conversion number (for example, 120 minutes for a minutestohours conversion).
· For problems involving percents, use 100 or a multiple of 100.
Avoid numbers that cause problems.
· Don’t use 0 or 1.
· Don’t use numbers you see in the question or answer choices.
· Don’t use the same value for several variables.
· Don’t use conversion numbers (although multiples of them are okay).
The next example shows how choosing a bad number can create extra work for you or possibly even lead to a wrong answer.
4.A machine can produce p fan blades in an hour. How many fan blades can this machine produce in q minutes?
pq 

60pq 

1. Plug in some numbers for the variables. Suppose p = 10 fan blades in an hour and q = 60 minutes.
2. Find a numerical answer. The machine produces 10 fan blades in an hour, so it produces 10 fan blades in 60 minutes, because 60 minutes is one hour. So the answer is 10. Circle it on your noteboard.
3. Plug your numbers into the answer choices to see which one matches.
For (A), you get pq = (10)(60) = 600. The answer should be 10, so eliminate (A).
For (B), you get = = 10. That seems right. However, you remember that you’re supposed to check all five choices, so you keep going.
For (C), you get 60pq = (60)(10)(60) = 36,000. That’s certainly not 10, so eliminate (C).
For (D), you get = = = . That’s not 10, so eliminate (D).
For choice (E), you get = = 10.
Wait a second, both (B) and (E) can’t be right! Now you need to try some different numbers and check (B) and (E) again.
Where did this all go wrong? The problem started with setting q = 60. Because 60 shows up in four of the answer choices, it’s possible that q = 60 will make some answers equal in value. In general, you should never plug in a unitconversion number. To avoid that problem, use a different value for q. You want to keep the calculations simple, so use a number that plays nicely with 60, such as 30 or 120. Start with these numbers so that you don’t have to work the problem twice.
Try the problem with a different number for q, such as q = 120 minutes. Keep p = 10 fan blades per hour. In 120 minutes, the machine can produce 20 fan blades, because 120 minutes is 2 hours. So the answer is 20 fan blades. When you plug p = 10 and q = 120 into the answer choices, you get
pq = (10)(120) = 1,200 

= = 5 

60pq = (60)(10)(120) = 72,000 

= = 

= = 20 
So the correct answer turns out to be (E).
There are also a couple of guidelines you should follow when using Plugging In on a problem with several variables. First, use a different number for each variable. Failure to do so can result in several answers all seeming to work. This is essentially the same problem as using a number that appears in the answer choices.
Second, if the problem contains several variables all related in an equation, you’re going to make up numbers for all the variables except one. Solve the equation to find the value for that last variable.
Here’s an example:
5.If a = , what is the value of b in terms of a and c ?
ac − 5 

c(a − 5) 

(a − 5)(c − 5) 

5ac 
1. Plug in numbers for the variables. There is an equation, so you’ll make up numbers for only two of the three variables. If you start with c = 3, you probably want the top of the fraction to be a multiple of 3, such as 12. Let b = 7 so that the top is 12. Now, solve the equation to find the value of a, which is a = = = 4.
2. Find a numerical answer. In this case, the question just asks for the value of b, which is 7. Circle it on your noteboard.
3. Plug your numbers into the answer choices. You get
= = 

ac − 5 = (4)(3) − 5 = 12 − 5 = 7 

c(a − 5) = 3(4 − 5) = 3(−1) = −3 

(a − 5)(c − 5) = (4 − 5)(3 − 5) = (−1)(−2) = 2 

5ac = (5)(4)(3) = 60 
As you can see, the only answer that matches is (B).
DRILL 1
Answers can be found on this page.
1.Alice is twice as old as Brian and Cathy is 6 years younger than Brian. If Alice is a years old, how old is Cathy in terms of a ?
a + 6 

a − 6 

2a − 6 
2.All the widgets manufactured by Company X are stored in two warehouses, A and B. Warehouse A contains three times as many widgets as does warehouse B. If of the widgets in warehouse A and of the widgets in warehouse B are defective, what fraction of all the widgets are NOT defective?
3.Of all the players in a professional baseball league, are foreignborn, including of the pitchers. If of the players are pitchers, what percentage of the players who are not pitchers are foreignborn?
100% 

75% 

66% 

50% 

25% 
Plugging In The Answers
Plugging In The Answers is another variation of Plugging In; it also turns algebra problems into arithmetic problems. You use Plugging In when the answer choices contain variables—visible or invisible. Use Plugging In The Answers when the answer choices are numbers. Alternatively, if you feel the urge to write an algebraic equation, you should probably just Plug In The Answers. Essentially, you try the answer choices to see which one fits with the information in the question. Here’s the stepbystep method.
1. Identify the specific number for which the question asks. In other words, what value do the answer choices represent?
2. Try answer (C). Work through the problem using that number. If it works, you’re done. Otherwise,
3. Try a different answer choice. If answer (C) was too big, try a smaller answer. If (C) was too small, try a bigger number. Repeat until you find the answer that fits the information in the question.
Try this example.
1.Oliver has twice as many marbles as Ted. Ted has twice as many marbles as Merrill. If the total number of marbles among the three is 28, how many marbles does Ted have?
4 

8 

12 

14 

16 
1. Identify the specific number asked for in the question. The question is “how many marbles does Ted have?” so the answer choices represent Ted’s marbles.
2. Using answer (C), work through the question. If Ted has 12 marbles, then Oliver has twice that, or 2 × 12 = 24 marbles. Merrill has half as many marbles as Ted, or = 6 marbles. The total is 12 + 24 + 6 = 42 marbles. The total is supposed to be 28, so eliminate (C).
3. Try a different answer. The total from (C), 42, was too big, so you should look for a smaller number. Try (B). If Ted has 8 marbles, then Oliver has 2 × 8 = 16 marbles. Merrill has = 4 marbles. The total is 8 + 4 + 16 = 28. That matches the number in the question, so (B) is the correct answer.
Plugging In The Answers works when the question asks for a single number, such as “What is x?” You cannot use it when the question asks for multiple numbers, such as “What is x + y?” That’s because you won’t know how much of the answer is x and how much is y.
Plugging In The Answers offers the same advantages that regular Plugging In does. First, it helps you avoid many of the careless mistakes that people make when doing algebra. Numbers are simply easier to work with than variables. Second, it’s a great tool for tackling really tough problems that are impossibly difficult to solve algebraically.
It may take a little practice to get the hang of Plugging In and Plugging In The Answers, but they are invaluable for simplifying algebra questions, particularly word problems. Take the time to get familiar with these techniques.
DRILL 2
Answers can be found on this page.
1.Rob is twice as old as Jodie is now. In 10 years, Rob will be 20 years older than Jodie is at that time. How old is Jodie now?
10 

20 

30 

40 

50 
2.John and Mark each own a collection of baseball cards. The two collections combined contain 120 cards. If John were to trade 5 cards to Mark and receive 2 of Mark’s cards in return, John would have 22 more cards than Mark does. How many cards does Mark possess before the proposed trade?
85 

74 

52 

46 

35 
Must Be Problems
Some questions ask “what must be true?” or some variation on that. They supposedly test your knowledge of the properties of various types of numbers (for example, odd, even, positive, negative, and so on). However, you can easily solve these questions with the Plugging In technique. The difference is that you need to plug in more than once.
Try numbers from both sides of whichever issue the question is testing. For example, if the question is testing odd versus even, plug in both an odd number and an even number. After you plug in an odd number, eliminate any answers that aren’t true. Then plug in an even number and eliminate any answers that aren’t true. In most cases, you’ll be left with one answer at this point. If not, continue plugging in different numbers until you’re left with one answer.
If you need to plug in more numbers, try 1 and 0. Earlier, you avoided those numbers because they did weird things. Now, however, you want the answers to do weird things so that you can eliminate them.
Look at this next example.
1.If the product xy is negative, which of the following must be true?
x < y 

x < 0 

y < 0 

< 0 

x + y < 0 
The question is testing positive versus negative, so you’ll want to try both kinds of numbers. You are constrained by the fact that xy is negative. The numbers you plug in must fit that condition. You might start with x = 2 and y = −3. (A) is false, so you can eliminate it. (B) is also false, so eliminate it. (C) is true in this case, so keep it. (D) is true in this case, so keep it. (E) is true in this case, so keep it.
Next, try a different set of numbers. Try reversing which variables are positive and negative. Let x = −2 and y = 3. Now you need to check only (C), (D), and (E) because you’ve already eliminated the others. (C) is false, so eliminate it. (D) is true in this case, so keep it. (E) is false, so eliminate it. The correct answer must be (D).
DRILL 3
Answers can be found on this page.
1.If x is an integer, which of the following must be odd?
3x 

2x 

3x + 1 

4x 

4x + 1 
2.If p, q, and r are nonzero numbers and p = q – r, which of the following must equal 0 ?
p – r 

− 1 

p – (q + r) 

− 1 
Comprehensive Solutions Beyond Algebra Drill
Answers can be found on this page.
Remember!
For Data Sufficiency problems in this book, we do not supply the answer choices. The five possible answer choices are the same every time.
1.If Joe Bob was 25 years old 5 years ago, how old was he x years ago?
x − 30 

x − 20 

30 – x 

20 – x 

20 + x 
2.Increasing the original length of a racetrack by 15% and then increasing the new length by 10% is equivalent to increasing the original length by
30.0% 

27.5% 

26.5% 

25.0% 

12.5% 
3.At a certain bakery, of the cookies sold in one week were chocolate chip and of the remaining cookies sold were oatmeal raisin. If x of the cookies sold were oatmeal raisin, how many were chocolate chip?
x 

x 

x 

x 

x 
4.If 50% of the money in a certain portfolio was invested in stocks, 20% in bonds, 15% in real estate, and the remaining $37,500 in a money market fund, what was the total amount invested in the portfolio?
$100,000 

$125,000 

$175,000 

$250,000 

$375,000 
5.Of all the pies baked in a certain bakery, are apple pies, are cherry pies, are pecan pies, and the rest are coconut cream pies. If the combined number of pecan pies and coconut cream pies is 40, how many pies total did the bakery bake?
56 

84 

91 

105 

112 
6.Bill buys two types of soda. He buys m bottles of Brand A at $0.50 each. He buys n bottles of Brand B at $0.60 each. What is Bill’s average cost in cents for a bottle of soda, in terms of m and n ?
7.If a, b, and c are nonzero integers, which of the following must be an integer?
+ 1 

abc − 1 

− 1 

+ 1 

+ + 
8.In a certain school, of the students are boys and the rest are girls. Of the boys, play soccer. If the number of girls who play soccer equals the number of boys who play soccer, what fraction of the girls play soccer?
9.Jason has a handful of dimes and quarters. There are a total of 22 coins. If the total value of the coins is $3.25, how many dimes does Jason have?
7 

8 

11 

12 

15 
10.If both a and b are nonzero integers, which of the following must be positive?
I. a^{2} + b^{2}
II. a^{2} – b^{2}
III.(a – b)^{2}
I only 

II only 

III only 

I and II 

I, II, and III 
11.If y ≠ −7, then =
y^{2} − 5y + 1 

y^{2} − 2y − 1 

y^{2} + 5y − 15 

2y^{2} − 3y − 1 

2y^{2} − 5y + 1 
12.Anjeanette inserts light bulbs in a row of light sockets in the repeating pattern red bulb, white bulb, white bulb, red bulb, white bulb, and white bulb. If she inserts a total of n red bulbs in the row, including the first and last bulbs, how many white bulbs did she insert in the row?
2n − 2 

2n − 1 

2n + 1 

3n − 2 

3n + 1 
13.Alex paid a $12 fee to receive a 10% discount off of the list price on all books he bought in a sixmonth period. Victor paid a $15 fee to receive a 15% discount off of the list price on all books he bought in the same period. If each bought books with a total list price of b during that period and each paid the same net amount, including the discount and the fee, what is b ?
$54 

$60 

$66 

$100 

$120 
14.If n > 6 and 3m + 5n = 0, then which of the following must be true?
m > 10 

m < −10 

m = 10 

n < 10 

n > 10 
15.If x is a multiple of 5, y is a factor of 16, and z is a factor of 40, which of the following must be true?
I. xy is a multiple of z.
II. xy is not a factor of z.
III.xy is a factor of z.
None of the above 

I only 

II only 

III only 

I and II only 
16.If p and q are consecutive even integers and p < q, which of the following must be divisible by 3 ?
p^{2} + pq 

pq^{2} + pq 

p^{2}q – pq 

p^{2}q^{2} 

pq^{2} – pq 
17.A tank holds x gallons of a saltwater mixture that is 20% salt by volume. One fourth of the water is evaporated, leaving all of the salt. When 10 gallons of water and 20 gallons of salt are added, the resulting mixture is 33% salt by volume. What is the value of x ?
37.5 

75 

100 

150 

175 
18.Emma’s piggy bank contains 12 cents more than Robert’s piggy bank does, and x is the sum, in cents, of the money in their piggy banks. If half of the money in Emma’s piggy bank is moved to Robert’s and then x cents is added to each piggy bank, how much money, in cents, will Robert’s piggy bank then contain?
x − 3 

x + 3 

x − 3 

x + 3 

x + 6 
19.If u and v are distinct prime numbers, which of the following CANNOT be a prime number?
uv + 3v − 2 

uv – u + 2v − 2 

uv + 2u – v − 2 

uv + 3u − 2v − 6 

uv + u + v + 1 
Challenge!
Take a crack at this highlevel GMAT question.
20.If a = x – b and b = a – y, what is the value of ab ?
ANSWERS AND EXPLANATIONS
Drill 1
1.CFirst, make up a number for a, such as a = 20. Second, solve for a numerical answer. If Alice is 20, then Brian is 10. If Brian is 10, then Cathy is 10 − 6 = 4. So the answer is 4. Third, plug a = 20 into the answers and find the one that equals 4. The only one that matches is (C).
2.CFirst, make up some numbers for the number of widgets in A and B. Suppose A contains 300 widgets and B contains 100 widgets. Second, find a numerical answer. Warehouse A contains × 300 = 20 defective widgets. Warehouse B contains × 100 = 5 defective widgets. The total number of widgets is 100 + 300 = 400 and the number of defective widgets is 20 + 5 = 25. So the number of nondefective widgets is 400 − 25 = 375. The fraction of widgets that are not defective is = . Choose (C).
3.AFirst, plug in a number for the total number of players. The number should be compatible with , , and , so try 240 players. Second, solve for a numerical answer. You may want to use a grid like the ones you learned about previously. See the example below. If there are 240 players total, then there are × 240 = 120 foreignborn players and 240 − 120 = 120 nativeborn players. There are × 240 = 180 pitchers and 240 − 180 = 60 nonpitchers. Of the pitchers, × 180 = 60 are foreignborn, leaving 180 − 60 = 120 nativeborn pitchers. Of the 120 foreignborn players, 60 are pitchers, leaving 120 − 60 = 60 foreignborn nonpitchers. So of the 60 nonpitchers, 60 are foreignborn. That’s 100%. Choose (A).
Drill 2
1.BStart with (C). If Jodie is 30 years old now, then Rob is 2 × 30 = 60 years old. In 10 years, Rob will be 60 + 10 = 70 and Jodie will be 30 + 10 = 40 years old. However, Rob will be 70 − 40 = 30 years older than Jodie, not the stated 20 years. Try a different answer. In (B), if Jodie is 20 years old, then Rob is 2 × 20 = 40. In 10 years, Rob will be 40 + 10 = 50 and Jodie will be 20 + 10 = 30. Rob will be 50 − 30 = 20 years older, so (B) is correct.
2.DFirst, figure out what the answer represents. In this case, it is the number of cards Mark owns at the start. Second, use (C) and work through the problem. If Mark owns 52 cards, then John owns 120 − 52 = 68 cards. John then trades 5 cards to Mark and gets 2 back. That’s a net of 3 cards to Mark. So John now has 68 − 3 = 65 and Mark has 52 + 3 = 55. That’s a difference of 10 cards, not 22 cards. Eliminate (C). Third, try another answer. In (B), Mark starts with 74 cards, so John has 120 − 74 = 46 cards. After the trade, John has 46 − 3 = 43 cards and Mark has 74 + 3 = 77 cards. John has fewer cards than Mark, so that’s definitely the wrong direction. Eliminate (B) and (A) and try (D). Mark has 46 cards, so John has 120 − 46 = 74 cards. After the trade, John has 74 − 3 = 71 cards and Mark has 46 + 3 = 49 cards. That’s a difference of 71 − 49 = 22 cards, which matches the information in the question. Choose (D).
Drill 3
1.EThe issue is odd versus even, so plug in numbers of both types. Let x = 2, an even number. (A) is 6, an even number, so eliminate it. (B) is 4, an even number, so eliminate it. (C) is 7, an odd number, so keep it. (D) is 8, an even number, so eliminate it. (E) is 9, an odd number, so keep it. Now try an odd number, such as x = 3. (C) is 10, an even number, so eliminate it. (E) is 13, an odd number, so (E) is the correct answer.
2.EThis question doesn’t really involve an odd/even or positive/negative issue, so you’ll just need to plug in different sets of numbers until you’re left with one answer. Try p = 3, q = 6, and r = 3. Answer (A) is 3 − 3 = 0. Keep it. Answer (B) is = . Eliminate it. (C) is − 1 = 3 − 1 = 2. Eliminate it. (D) is 3 – (6 + 3) = − 6. Eliminate it. (E) is − 1 = 1 − 1 = 0. Keep it. Try a different set of numbers, such as p = 1, q = 3, and r = 2. Now (A) is 1 − 2 = −1. Eliminate it. (E) is − 1 = 1 − 1 = 0. The correct answer must be (E).
Comprehensive Solutions Beyond Algebra Drill
1.CLet x = 10. If Joe Bob was 25 years old 5 years ago, he must be 30 years old now. So 10 years ago, he was 20 years old. Plug x = 10 into the answers and see which one matches. The answer is (C).
2.CPlug in a number for the original length, say 100 yards. Increasing the length by 15% adds 15 yards for a new total of 100 + 15 = 115 yards. Increasing this new length by 10% adds 0.10 × 115 = 11.5 yards, for a total length of 115 + 11.5 = 126.5 yards. That’s an overall increase of 26.5 yards or = 26.5%. Choose (C).
3.ESuppose that the bakery sold a total of 20 cookies. Then there are × 20 = 5 chocolate chip cookies, leaving 20 − 5 = 15 cookies. Of these remaining cookies, × 15 = 3 are oatmeal raisin, so x = 3. The numerical answer for the number of chocolate chip cookies is 5. Plug x = 3 into the answers and see which one matches. The answer is (E).
4.DStart with (C). If there is $175,000 in the portfolio, there is 0.5 × $175,000 = $87,500 in stocks, 0.2 × $175,000 = $35,000 in bonds, 0.15 × $175,000 = $26,250 in real estate, and $175,000 – $87,500 – $35,000 – $26,250 = $26,250 left for the money market fund. That doesn’t match the $37,500 stated in the question, so eliminate (C). You need a bigger number, so try (D). If the total amount is $250,000, then there is 0.5 × $250,000 = $125,000 in stocks, 0.2 × $250,000 = $50,000 in bonds, 0.15 × $250,000 = $37,500 in real estate, and $250,000 – $125,000 – $50,000 – $37,500 = $37,500 left for the money market fund. That matches the information in the question, so choose (D).
5.ENormally, you’d start by checking (C). However, both (C) and (D) are odd. Because apple pies are of the total, you can’t have an odd total. Eliminate (C) and (D). Now check the middle answer of what’s left, (B). If there are 84 pies total, there are × 84 = 42 apple pies, × 84 = 12 cherry pies, and × 84 = 21 pecan pies. That leaves 84 − 42 − 12 − 21 = 9 coconut cream pies. The total of pecan and coconut cream pies is 21 + 9 = 30 pies. That’s too few. Eliminate (B) and try a bigger answer. In (E), there are 112 pies total. So there are × 112 = 56 apple pies, × 112 = 16 cherry pies, and × 112 = 28 pecan pies. That leaves 112 − 56 − 16 − 28 = 12 coconut cream pies. The total of pecan and coconut cream pies is 28 + 12 = 40. That matches the information in the question, so choose (E).
6.DPlug in numbers for m and n. Let m = 10 bottles at 50 cents each for a cost of 10 × 50 = 500 cents. Let n = 15 bottles at 60 cents each for a cost of 15 × 60 = 900 cents. Bill’s total cost is 500 + 900 = 1,400 cents for 10 + 15 = 25 bottles of soda. So his average cost is = 56 cents per bottle. Plug m = 10 and n = 15 into the answers to see which one equals 56. Remember that you need an answer in cents, not in dollars. (A) is 0.56, which is the trap answer in dollars. (B) is . (C) is . (D) is 56. (E) is , which is about 13. Choose (D).
7.BPlug in some numbers, such as a = 2, b = 3, and c = 4. (A) is + 1 = + 1 = . That’s not an integer, so eliminate it. (B) is (2)(3)(4) − 1 = 24 − 1 = 23. That is an integer, so keep it. (C) is − 1 = − 1 = = . That’s not an integer, so eliminate it. (D) is + 1 = + 1 = . That’s not an integer, so eliminate it. (E) is + + = . That’s not an integer, so eliminate it. Choose (B).
8.EMake up a number for the total number of students, say 200. There are × 200 = 120 boys and 200 − 120 = 80 girls. There are × 120 = 30 soccer players among the boys, so there are also 30 soccer players among the girls. The fraction of girls that play soccer is = . Choose (E).
9.EStart Plugging In The Answers with (C). He has 11 dimes worth 11 × 0.10 = $1.10. He has 22 − 11 = 11 quarters worth 11 × 0.25 = $2.75. That’s a total value of 1.10 + 2.75 = $3.85. That doesn’t match the $3.25 from the question. Eliminate (C). You want less money, so you need fewer quarters and more dimes. Try (D). Jason has 12 dimes worth 12 × 0.10 = $1.20. He has 22 − 12 = 10 quarters worth 10 × 0.25 = $2.50. That’s a total value of 1.20 + 2.50 = $3.70, which is still too big. Try (E). Jason has 15 dimes worth 15 × 0.10 = $1.50. He has 22 − 15 = 7 quarters worth 7 × 0.25 = $1.75. That’s a total value of 1.50 + 1.75 = $3.25, which matches the information in the question. Choose (E).
10.APlug in numbers for a and b, such as a = 2 and b = 3. I is 4 + 9 = 13. That’s positive, so keep it. II is 4 − 9 = −5. That’s negative, so you can eliminate it. That means you eliminate (B), (D), and (E). III is (−1)^{2} = 1. That’s positive, so keep it. Try strange numbers such as a = 2 and b= 2. I is 4 + 4 = 8. That’s positive, so keep it. III is (0)^{2} = 0. That’s not positive, so eliminate it and (C). The answer must be (A).
11.BSince there are variables in the answer choices, you should try Plugging In. If y = 3, for example, the original expression becomes = = . Now, plug y = 3 into each answer and see which one equals 2. Answer (B) becomes 3^{2} − 2(3) − 1 = 9 − 6 − 1 = 2. None of the other answers equals 2, so choose (B).
12.AThe repeating pattern is 3 bulbs (red, white, white), and you want to start and end with a red bulb. Since there are variables in the answer choices, try Plugging In. Suppose the total is 7 bulbs, so that the sequence is red, white, white, red, white, white, red. There are 3 red bulbs, so n= 3. There are 4 white bulbs, so your target is 4. Plug n = 3 into the answer choices to see which one equals 4. (A) equals 4; (B) equals 5; (C) equals 7; (D) equals 7; and (E) equals 10. Choose (A).
13.BWith numerical answer choices, you should consider Plugging In The Answers. Start with (C), so each person bought books totaling $66 in list price. Alex received a 10% discount, which is 0.1 × 66 = $6.60 and paid a $12 fee, so his total is 66 − 6.60 + 12 = $71.40. Victor received a 15% discount, which is 0.15 × 66 = $9.90 and paid a $15 fee, so his total is 66 − 9.90 + 15 = $71.10. Close, but not the same. Eliminate (C) and try (B), in which each person buys books totaling $60 in list price. Alex’s 10% discount is 0.1 × 60 = $6.00, so he pays 60 − 6.00 + 12 = $66.00. Victor’s 15% discount is 0.15 × 60 = $9.00, so he pays 60 − 9.00 + 15 = $66.00. The two people paid the same price, as the question stipulated, so (B) is correct.
14.BIt’s a “must be” problem, so plug in a set of numbers. Let n = 15, so that 3m + 75 = 0. That means m = −25. You can eliminate (A), (C), and (D). Now try another number, such as n = 9. That gives the equation 3m + 45 = 0, so m = −15. Although (B) is still true, (E) isn’t. Cross off (E) and choose (B).
15.AIt’s a “must be” question, so plug in one or more sets of numbers. Suppose x = 10, y = 2, and z = 8. Be sure the numbers you choose are consistent with the conditions described in the problem. Now, see which answers are true. Roman numeral I says that 20 is a multiple of 8. That’s not true, so eliminate (B) and (E). II says that 20 is not a factor of 8. That’s true, so don’t eliminate anything. III says that 20 is a factor of 8. That’s not true, so eliminate (D). The question is whether II is always true or only sometimes, depending on the numbers you choose. Try to find a set of numbers that disproves II. Suppose x = 10, y = 2, and z = 20. Now, II says that 20 is not a factor of 20, which is a false statement. Eliminate (C) and choose (A).
16.EIt’s a “must be” problem, so plug in one or more sets of numbers. Suppose p = 2 and q = 4. Answer (A) becomes 2^{2} + (2)(4) = 4 + 8 = 12. That’s divisible by 3, so keep it. Answer (B) becomes (2)(4^{2}) + (2)(4) = 32 + 8 = 40. That’s not divisible by 3, so eliminate (B). Answer (C) becomes (2^{2})(4) − (2)(4) = 16 − 8 = 8. That’s not divisible by 3, so eliminate (C). Answer (D) becomes (2^{2})(4^{2}) = (4)(16) = 64. That’s not divisible by 3, so eliminate (D). Answer (E) becomes (2)(4^{2}) − (2)(4) = 32 − 8 = 24. That is divisible by 3, so keep it. Try a new set of numbers, such as p = 4 and q = 6. Now, (A) becomes 4^{2} + (4)(6) = 16 + 24 = 40. That’s not divisible by 3, so eliminate (A). Answer (E) becomes (4)(6^{2}) − (4)(6) = 144 − 24 = 120. That is still divisible by 3. Choose (E).
17.DWith numerical answers to a story problem, try Plugging In The Answers. Start with (C), which is 100 gallons of mixture. It’s 20% salt, so there are 20 gallons of salt and 80 gallons of water. Take away one fourth of the water, or 20 gallons, leaving 60 gallons of water and 20 gallons of salt. Finally, add 10 gallons of water and 20 gallons of salt to get 40 gallons of salt and 70 gallons of water, or 40 + 70 = 110 gallons of the mixture. The salt percentage is = 36.36...%, not 33%, so eliminate (C) and try a different answer. With (D), you start with 150 gallons, including 0.2 × 150 = 30 gallons of salt and 120 gallons of water. Evaporating one fourth of the water leaves 30 gallons of salt and 90 gallons of water. Adding new salt and water gives you 30 + 20 = 50 gallons of salt and 90 + 10 = 100 gallons of water, or 50 + 100 = 150 gallons total. The salt percentage is = = 33%, so (D) is correct.
18.CIt’s easier to plug in the numbers for Emma and Robert and then find x than vice versa. Suppose Robert has 10 cents and Emma has 10 + 12 = 22 cents. That makes x = 10 + 22 = 32. Taking half of Emma’s money, 11 cents, and giving it to Robert gives him 10 + 11 = 21 cents. Now, x = 32 cents is given to each child, so Robert has 21 + 32 = 53 cents. That’s your target. Plug x = 32 into the answers to see which one equals 53. Answer choice (C) becomes (32) − 3 = 56 − 3 = 53. None of the other answers equal 53, so choose (C).
19.EThis is similar to a “must be” problem. Plug in several sets of numbers and eliminate answer choices that are prime. Suppose u = 2 and v = 3. Choice (A) becomes (2)(3) + 3(3) − 2 = 13; that’s prime, so eliminate it. Choice (B) becomes (2)(3) − 2 + 2(3) − 2 = 8; that’s not prime, so keep it. Choice (C) becomes (2)(3) + 2(2) − 3 − 2 = 5; that’s prime, so eliminate it. Choice (D) becomes (2)(3) + 3(2) − 2(3) − 6 = 0; that’s not prime, so keep it. Choice (E) becomes (2)(3) + 2 + 3 + 1 = 12; that’s not prime, so keep it. Try switching the numbers, so that u = 3 and v= 2. (B) becomes (3)(2) − 3 + 2(2) − 2 = 5; that’s prime, so eliminate it. (D) becomes (3)(2) + 3(3) − 2(2) − 6 = 5; that’s prime, so eliminate it. (E) becomes (3)(2) + 3 + 2 + 1 = 12; that’s not prime, so keep it. You’ve eliminated (A), (B), (C), and (D), so choose (E).
20.DIf you attempted the typical algebraic solution for this question, you are likely surprised by this answer. Instead of spending time trying to manipulate the equations to yield the correct answer, you should Plug In! Pick a number for x and b. Make x = 5 and b = 2. Therefore, a = 3, and y = 1. The target for this question is ab, or (3)(2) = 6. Now, plug x − 5 and y − 1 into the answers until one works. Only (D) yields 6. Choose (D).