﻿ ﻿THE SAT MATH TEST: PROBLEM SOLVING AND DATA ANALYSIS - McGraw-Hill Education SAT 2017 Edition (Mcgraw Hill's Sat) (2016)

## McGraw-Hill Education SAT 2017 Edition (Mcgraw Hill's Sat) (2016)

### THE SAT MATH TEST: PROBLEM SOLVING AND DATA ANALYSIS

1. Working with Data
2. Working with Rates, Ratios, Percentages, and Proportions
3. Working with Tables of Data
4. Working with Graphs of Data

The SAT Math: Problem Solving and Data Analysis

Why is problem-solving and data analysis important on the SAT Math test?

About 26% (15 out of 58 points) of the SAT Math questions fall under the category of Problem Solving and Data Analysis. Questions in this category test your ability to

create a representation of a problem, consider the units involved, attend to the meaning of quantities, and [apply reasoning about] ratios, rates, and proportional relationships .

They also assess your skill in

interpreting and synthesizing data, [as well as identifying] quantitative measures of center, the overall pattern, and any striking deviations from the overall pattern in different data sets .

The specific topics include

• using rates, ratios, and proportional relationships to solve problems
• evaluating and analyzing data gathering methods
• calculate and use statistics of “central tendency” like mean, median, and mode
• basic measures of data “spread” such as standard deviation, range, and confidence intervals
• solving problems concerning percentages and percent change
• analyzing scatterplots, pie graphs, tables, histograms, and other graphs
• exploring linear, quadratic, and exponential relationships in data

How are these skills used?

Analyzing and drawing inferences from data are core skills not only in mathematics and the physical sciences, but also in social sciences such as psychology, sociology, and economics. Since these subjects constitute a substantial portion of any liberal arts curriculum, colleges consider these to be essential college preparatory skills.

Sound intimidating? It”s not.

If you take the time to master the four core skills presented in these 16 lessons, you will gain the knowledge and practice you need to master SAT Math problem-solving and data analysis questions.

Skill 1: Working with Data

Lesson 1: Working with averages (arithmetic means)

The average (arithmetic mean) of four numbers is 15. If one of the numbers is 18, what is the average (arithmetic mean) of the remaining three numbers?

The average (arithmetic mean) of any set of numbers is calculated with the formula

But it is helpful to notice that this form can take two other forms:

(Medium ) In this problem, we are given the average of the set and the number of numbers in the set. So we can use the formula sum = average × # of numbers to find the sum of these numbers: 15 × 4 = 60. If one of these numbers is 18, then the sum of the remaining three numbers is 60 − 18 = 42. Using the first formula above gives us an average of 42/3 = 14.

Ms. Aguila”s class, which has 20 students, scored an average of 90% on a test. Mr. Bowle”s class, which has 30 students, scored an average of 80% on the same test. What was the combined average score for the two classes? (Disregard the % symbol when gridding. For instance, enter 74% as 74.)

(Medium ) Can we just take the average of the scores for the two classes, and say the overall average is (90 + 80)/2 = 85? No, because there are more students in the second class, so we can”t “weigh” the two classes equally. Using the formula above we can calculate the sum of all of the scores in both classes. In Ms. Aguila”s class, the sum of the scores is 90 × 20 = 1,800, and the sum of the scores in Mr. Bowe”s class is 80 × 30 = 2,400. Therefore, the sum of all of the scores in the two classes combined is 1,800 + 2,400 = 4,200. Since there are 50 students altogether in the two classes, the combined average is 4,200/50 = 84.

Lesson 2: Working with medians and modes

The median of 1, 6, 8, and k is 5. What is the average (arithmetic mean) of these four numbers?

The median of any set of numbers is the number that divides the ordered set into two equal sets. In other words, half of the numbers should be less than or equal to the median, and half the numbers should be greater than or equal to the median. To find a median,

1. Put the numbers in increasing (or decreasing) order.
2. If there are an odd number of numbers, the median is the middle number.
3. If there are an even number of numbers, the median is the average of thetwo middle numbers.

(Medium ) Using this definition we can find the value of k . The tricky part is step 1, since we don”t know where k should be when we put the numbers in order. Clearly, however, there are only four possibilities to consider.

If k is the least of these numbers, then the correct ordering is k , 1, 6, 8. Since there are an even number of numbers, the median is the average of the middle two: (1 + 6)/2 = 3.5. But this contradicts the given fact that the median is 5, so that doesn”t work. Putting k in the next slot gives us an order of 1, k , 6, 8. In this case, the median would be (k + 6)/2.

Multiply by 2:

k + 6 = 10

Subtract 6:

k = 4

Notice that this confirms our assumption that k is between 1 and 6, so k must equal 4. Now we must find the average of these four numbers: (1 + 4 + 6 + 8)/2 = 19/2 = 9.5.

The table above shows the results of 50 rolls of a die, with two missing values labeled a and b . If the mode of these 50 rolls is 2, what is the greatest possible average (arithmetic mean) value of these rolls?

The mode of a set of numbers is the number that appears the most frequently . This means that not every set of numbers has a mode. For instance, in the set 1, 1, 2, 3, 4, the mode is 1, but the set 1, 2, 3, 4 does not have a mode, because every numbers occurs once.

(Hard ) This data set has 50 numbers, each representing a roll of a die. If the mode is 2, then 2 is the most frequent roll. Since the table above shows that the highest known frequency is 10 (for a roll of 1), then a (the number of times a 2 was rolled) must be at least 11. We also know that the total number of rolls is 50, so 10 + a + b + 7 + 9 + 9 = 50, and therefore a + b = 15. The question asks us to find the greatest possible average of these rolls, so we want to maximize the sum of all of the rolls. This means that we want b (the number of times a 3 was rolled) to be as great as possible. Since b = 15 − a , then the greatest b can be is 15 − 11 = 4. Therefore a = 11 and b = 4. Now we have to find the average of these 50 numbers: [(1)(10) + (2)(11) + (3)(4) + (4)(7) + (5)(9) + (6)(9)]/50 = (10 + 22 + 12 + 28 + 45 + 54)/50 = 3.42.

The SAT Math test may occasionally ask you about the “spread” of a set of data. You will NOT have to calculate technical statistics like variance , standard deviation , or margin of error , but you might be asked to answer more basic questions about the “spread” of a set of data, as in the questions below.

The “range” of a set of data is defined as the absolute difference between the least value and the greatest value in the set. If five positive integers have an average (arithmetic mean) of 10, what is the greatest possible “range” of this set?

(Medium-hard ) If five numbers have an average of 10, then their sum must be 5 × 10 = 50. If we want the greatest possible “range,” then we must maximize one of these numbers by minimizing the sum of the other numbers. Since the smallest positive integer is 1, we can minimize the sum of the other four numbers by setting them all equal to 1. This gives us 1 + 1 + 1 + 1 + x = 50, so x = 46. This gives us a maximum “range” of 46 − 1 = 45.

The “absolute deviation” of a number in a set is the absolute difference between that number and the average (arithmetic mean) of the set. The “average absolute deviation” of a set is the average (arithmetic mean) of all of the absolute deviations in the set. Which of the following sets has the greatest “average absolute deviation?”

1. A) 2, 2, 2, 2
2. B) 2, 3, 4, 5
3. C) 4, 4, 5, 5
4. D) 4, 4, 4, 5

(Medium ) When a question introduces a new term, read its definition carefully —several times, if necessary. This question gives us two new terms. To understand these terms, let”s apply them to a simple set of numbers, such as the set in choice (A). The average of this set is 2; therefore, the “absolute deviation” of each of these numbers is its “absolute difference” from 2, which is 0 for each number. The “average absolute deviation” of the entire set is defined as the average of these “absolute deviations,” which is, of course, (0 + 0 + 0 + 0)/4 = 0.

Notice that the “average absolute deviation” of a number set is a measure of “spread.” Since the numbers in (A) are bunched up as tightly as possible, their “average absolute deviation” is 0. Now let”s look at the remaining choices. Which seems to have the greatest “spread?” Once you”ve made your guess, do the calculations and see if you”re right. For confirmation, you should find that the “average absolute deviations” are (A) 0, (B) 1, (C) 0.5, and (D) 0.375. Therefore, the correct answer is (B).

Lesson 4: Variations and drawing inferences from data

The variables x and y vary directly if they have a constant ratio , that is,

(where k is a constant)

The variables x and y vary inversely if they have a constant product , that is,

(where k is a constant)

Given the ordered pairs in the table above, which of the following could be true?

1. A) yvaries directly asx
2. B) yvaries inversely asx
3. C) yvaries directly as the square ofx
4. D) yvaries inversely as the square ofx

(Medium-hard ) Using the definitions above, we can see whether y and x vary directly or inversely. Do they have a constant ratio ? No: 5/1 ≠ 20/2. Therefore, they do not vary directly and (A) is incorrect. Do they have a constant product ? No: 1 × 5 ≠ 2 × 20. Therefore, they do not vary inversely, and (B) is incorrect. To check (C), we must ask: do y and x 2 have a constant ratio? Yes: 5/(1)2 = 20/(2)2 = 5, therefore the correct answer is (C).

If y varies directly as x , then the graph of their relation in the xy -plane is a line through the origin :

If y varies inversely as x , then the graph of their relation in the xy -plane is a hyperbola that approaches, but does not touch, thex -and y -axes :

The variables x and y are related by an equation of the form y = Ax n where A and n are both positive real numbers. Based on the data in the table above, what is the value of y when x = 3?

1. A) 40.5
2. B) 46.0
3. C) 54.0
4. D) 64.0

(Medium ) In order to find the value of y when x = 3, we must find the specific equation relating x and y . We can find the values of A and n in the equation y = Ax n by first plugging in the values from the table. Plugging in the first ordered pair gives us 12 = A (2) n and plugging in the second ordered pair gives us 96 = A (4) n .

(Notice that this means that y varies directly as x 3 .)

Substitute n = 3 into either

Therefore, the equation that relates x and y is y = 1.5x 3 . Finally, we find y when x = 3 by substituting into the equation: y = 1.5(3)3 = 1.5(27) = 40.5, so the correct answer is (A).

Exercise Set 1 (No Calculator)

1

The “range” of a set of data is defined as the absolute difference between the least value and the greatest value in the set. Four positive integers have an average (arithmetic mean) of 7.5.

1. What is the greatest possible range of this set?
2. What is the least possible range of this set?

2

If the median of 2, 4, 6, and b is 4.2, what is the average (arithmetic mean) of these four numbers?

3

The average (arithmetic mean) of 2, 5, 8 and k is 0. What is the median of these numbers?

4

A set of numbers has a sum of 48 and an average of 6. How many numbers are in the set?

5

If the average (arithmetic mean) of 4 and x is equal to the average (arithmetic mean) of 2, 8, and x , what is the value of x ?

6

The median of a set of 22 consecutive even integers is 25. What is the largest number in the set?

7

If p varies inversely as q and p = 4 when q = 6, the which of the following is another solution for p and q?

1. A) p= 8 andq = 12
2. B) p= 8 andq = 10
3. C) p= 12 andq = 1
4. D) p= 12 andq = 2

8

A set of n numbers has an average (arithmetic mean) of 3k and a sum of 12m , where k and m are both positive. Which of the following is equivalent to n?

1. A)
2. B)
3. C)
4. D)

9

If y varies inversely as the square of x , then when x is multiplied by 4, y will be

1. A) divided by 16
2. B) divided by 2
3. C) multiplied by 2
4. D) multiplied by 16

10

Let f (x , y ) = Ax 2 y 3 where A is a constant. If f (a, b ) = 10, what is the value of f (2a , 2b )?

1. A) 100
2. B) 260
3. C) 320
4. D) 500

11

A set of four integers has a mode of 7 and a median of 4. What is the greatest possible average (arithmetic mean) of this set?

1. A) 3.50
2. B) 3.75
3. C) 4.00
4. D) 4.25

Exercise Set 1 (Calculator)

12

Four positive integers have a mode of 4 and a median of 3. What is their sum?

13

Five different integers have an average (arithmetic mean) of 10. If none is less than 5, what is the greatest possible value of one of these integers?

14

If b varies inversely as a , and b = 0.5 when a = 32, then for how many ordered pairs (a , b ) are a and b both positive integers?

15

The median of 11 consecutive integers is 28. What is the least of these integers?

16

If y = Ax 3 and y = 108 when x = 3, then for what value of x does y = 62.5?

17

A set of four positive integers has a median of 2 and a mode of 2. If the average (arithmetic mean) of this set is 3, what is the largest possible number in the set?

18

If y varies inversely as x and the graph of their relation in the xy -plane passes through the point (2, 15), what is the value of y when x = 4?

19

A six-sided die was rolled 30 times and the results tabulated above. What is the difference between the average (arithmetic mean) of the rolls and the median of the rolls?

1. A) 0.1
2. B) 0.2
3. C) 0.3
4. D) 0.4

20

If y varies inversely as the square of x , and y = 4 when x = 2, then what is the value of y when x = 3?

1. A)
2. B)
3. C) 3
4. D) 9

21

At a fixed temperature, the volume of a sample of gas varies inversely as the pressure of the gas. If the pressure of a sample of gas at a fixed temperature is increased by 50%, by what percent is the volume decreased?

1. A) 25%
2. B)
3. C) 50%
4. D)

22

If the graph of y = f (x ) in the xy -plane contains the points (4, 3) and (16, 6), which of the following could be true?

1. A) yvaries directly as the square ofx
2. B) yvaries inversely as the square ofx
3. C) yvaries directly as the square root ofx
4. D) yvaries inversely as the square root ofx

No Calculator

1a . 26 If the average of 4 numbers is 7.5, they must have a sum of 4 × 7.5 = 30. To maximize the range, we must maximize one of the numbers by minimizing the other 3 by setting them all equal to 1 (the smallest positive integer). The numbers therefore are 1, 1, 1, and 27, and the range is 27 − 1 = 26.

1b . 1 To minimize the range, we “cluster” the numbers as closely together as possible. The tightest cluster of integers with a sum of 30 is 7, 7, 8, and 8, which gives a range of 8 − 7 = 1.

2 . 4.1 If the set contains four numbers, its median is the average of the middle two numbers, so the middle two numbers must have a sum of (2)(4.2) = 8.4. Thus the four numbers must be 2, 4, 4.4, and 6. (Notice that the question did not say that all numbers were integers.) The average of these is 16.4/4 = 4.1.

3 . 3.5 If the average of these numbers is 0, their sum must be (4)(0) = 0, and therefore k = −15 and the numbers, in increasing order, are −15, 2, 5, and 8. The median is (2 + 5)/2 = 3.5.

4 . 8 6 = 48/n , so n = 8.

5 . 8

Cross-multiply:

12 + 3x = 20 + 2x

Subtract 2x and 12:

x = 8

6 . 46 The median divides the set into two equal parts, so 11 of these numbers must be less than 25 and 11 must be greater than 25. Since they are consecutive even integers, the 11 numbers above the median must be 26, 28, 30, 32, . . . 46.

7 . D If p and q vary inversely, their product is a constant. 4 × 6 = 24, and the only other pair with a product equal to 24 is (D) 12 and 2.

8 . A n = sum/average = 12m /3k = 4m /k

9 . A The equation relating x and y is y = k /x 2 . If x = 1, then y = k . If x is multiplied by 4, then x = 4 and y = k/ 16, so y has been divided by 16.

10 . C f (a , b ) = Aa 2 b 3 = 10. f (2a , 2b ) = A (2a )2 (2b )3 = 32(Aa 2 b 3 ) = 32(10) = 320.

11 . B If this set has a mode of 7, then at least two of the numbers are 7. If the median is 4, then the two middle numbers must have a sum of (2)(4) = 8. Therefore the two middle numbers are 1 and 7, and the sequence must be n , 1, 7, 7. To maximize the average, we must maximize n , but n can”t be 1, because then the set would not have a mode of 7. It must be the next lower integer, 0, and the average is (0 + 1 + 7 + 7)/4 = 3.75.

Calculator

12 . 11 The only four numbers that satisfy these conditions are 1, 2, 4, and 4.

13 . 24 If the average of five numbers is 10, their sum is 5 × 10 = 50. To maximize one, we must minimize the sum of the other four. If none is less than five, and all are different integers, they are 5, 6, 7, 8, and 24.

14 . 5 If the variables vary inversely, their product is constant. (0.5)(32) = 16. The only pairs of positive integers with a product of 16 are (1, 16), (2, 8), (4, 4), (8, 2), and (16, 1).

15 . 23 If the middle number is 28, there are five numbers less than 28, and five greater. Since they are consecutive integers, the least is 28 − 5 = 23.

16 . 2.5 Since 108 = A (3)3 , A = 4, so if 62.5 = 4x 3 , x = 2.5.

17 . 7 At least two of the integers must be 2 and none can be less than 1. If the sum must be 4 × 3 = 12, the set including the largest possible number is 1, 2, 2, and 7.

18 . 7.5 The product of x and y is 2 × 15 = 30, so y = 30/4 = 7.5.

19 . C Average = (1 × 4 + 2 × 5 + 3 × 4 + 4 × 6 + 5 × 5 + 6 × 6)/30 = 3.7. Median = average of 15th and 16th roll: (4 + 4)/2 = 4. 4 − 3.7 = 0.3.

20 . A y and x 2 must have a constant product of 4 × 22 = 16. Therefore, y = 16/9.

21 . B Pick values for the original pressure and volume, such as 2 and 3. If they vary inversely, their product is the constant 2 × 3 = 6. If the pressure is increased by 50%, it becomes (1.5)(2) = 3, and so the volume becomes 6/3 = 2, a change of −33 1/3%.

22 . C For both ordered pairs, is a constant: , so y is directly proportional to the square root of x .

Skill 2: Working with Rates, Ratios, Percentages, and Proportions

Lesson 5: Rates and unit rates

On a sunny day, a 50 square meter section of solar panel array can generate an average of 1 kilowatt-hour of energy per hour over a 10-hour period. If an average household consumes 30 kilowatt-hours of energy per day, how large an array would be required to power 1,000 households on sunny days?

1. A) 1,500 square meters
2. B) 15,000 square meters
3. C) 150,000 square meters
4. D) 15,000,000 square meters

(Medium ) This is clearly a “rate problem,” because it includes two “per” quantities. When working with rates, keep two important ideas in mind:

The units for any rate can be translated to give the formula for the rate. For instance, if a word problem includes the fact that “a rocket burns fuel at a rate of 15 kilograms per second,” this fact can be translated into a formula as long as we remember that per means divided by :

Any “rate fact” in a problem can be interpreted as a “conversion factor.” For instance, if “a rocket burns fuel at a rate of 15 kilograms per second,” then in the context of that problem, one second of burning is equivalent to 15 kilograms of fuel being burned. Therefore, as we discussed in Chapter 7 , Lesson 4, we are entitled to use either of the following conversion factors in this problem:

Just as we did in Chapter 7 , Lesson 4, we can solve this problem by just noticing that it is essentially a conversion problem . The question asks “how large an array (in square meters) would be required to power 1,000 households on sunny days?” So we can treat the problem as a conversion from a particular number of households to a particular number of square meters of solar panels :

Note very carefully how (1) all of the units on the left side of the equation cancel except for “square meters” (which is what we want), and (2) each conversion factor represents an explicit fact mentioned in the problem.

Many rate problems can be easily managed with the “rate pie”:

This is a simple graphical device to organize information in a rate problem. It is simply a way of expressing all three forms of the “rate equation” at once: distance = rate × time ; rate = distance/time ; and time = distance/rate . For example, if a word problem states that “Maria completes an x -mile bicycle race at an average speed of z miles per hour,” your “rate pie” should look like this:

First, we plug the given values in: x miles goes in for distance, and z miles per hour goes in for rate. Then, as soon as two of the spaces are filled, we simply perform the operation between them (in this case division) and put the result in the final space. In this case, the time Maria took to complete the race was x/z hours.

A water pump for a dredging project can remove 180 gallons of water per minute, but can work only for 3 consecutive hours, at which time it requires 20 minutes of maintenance before it can be brought back online. While it is offline, a smaller pump is used in its place, which can pump 80 gallons per minute. Using this system, what is the least amount of time it would take to pump 35,800 gallons of water?

1. A) 3 hour 10 minutes
2. B) 3 hours 15 minutes
3. C) 3 hours 25 minutes
4. D) 3 hours 30 minutes

(Hard ) If we want to pump out the water as quickly as possible, we want to use the stronger pump for the maximum three hours. To find the total amount of water pumped in that time, we do the conversion:

So after 3 hours, there are still 35,800 − 32,400 = 3,400 gallons left to pump. At that point, the smaller pump must be used for a minimum of 20 minutes, which can pump

which still leaves 3,400 − 1,600 = 1,800 gallons left. Notice that we have already taken 3 hours and 20 minutes, and as yet have not finished pumping. This means that choices (A) and (B) are certainly incorrect. So how long will it take to pump the remaining 1,800 gallons? Now that we can bring the stronger pump online, it will only take 1,800 gallons × (1 minute/180 gallons) = 10 more minutes; therefore, the correct answer is (D).

Although you don”t need to construct a graph of this situation to solve the problem, graphing helps show the overall picture:

Notice that the line has a slope of 180 for the first 180 minutes, then 80 for the next 20 minutes, and then 180 for the next 180 minutes, and crosses the line y = 35,800 at 210 minutes.

In the graph of any linear function, y in terms of x , the slope of the line is equivalent to the unit rate of the function, that is, the rate at which y increases or decreases for every unit increase in x .

Lesson 6: Ratios: part-to-part and part-to-whole

A marathon offers \$5,000 in prize money to the top three finishers. If the first-, second-, and third-place prizes are distributed in a ratio of 5:4:1, how much money, in dollars, does the second-place finisher receive?

(Easy ) When given a “part-to-part” ratio, such as 5:4:1 (which is of course, really a part-to-part-to-part ratio), it often helps to add up the parts and consider the whole. This prize is divided into 5 + 4 + 1 = 10 equal parts, so the winner gets 5/10 of the prize money, the second-place finisher gets 4/10 of the prize money, and the third-place finisher gets 1/10 of the prize money. The second-place finisher therefore takes home (4/10) × \$5,000 = \$2,000.

If you are given a part-to-part ratio, it is often helpful to add up the parts and then divide each part by the sum. For instance, if a paint mixture is a 2:5 combination of red and yellow, respectively, the “whole” is 2 + 5 = 7, which means that the mixture is 2/7 red and 5/7 yellow.

Bronze is an alloy (a metallic mixture) consisting of copper and tin. If 50 kg of a bronze alloy of 20% tin and 80% copper is mixed with 70 kg of a bronze alloy of 5% tin and 95% copper, what fraction, by weight, of the combined bronze alloy is tin?

1. A) 5/48
2. B) 9/80
3. C) 1/8
4. D) 1/4

(Medium ) The combined alloy will weigh 50 kg + 70 kg = 120 kg. The total weight of the tin comes from the two separate alloys: (0.20)(50) + (0.05)(70) = 10 + 3.5 = 13.5 kg. Therefore the fraction of the combined alloy that is tin is 13.5/120, which simplifies to 9/80.

Exercise Set 2 (Calculator)

1

If a train travels at a constant rate of 50 miles per hour, how many minutes will it take to travel 90 miles?

2

Two cars leave the same point simultaneously, going in the same direction along a straight, flat road, one at 35 miles per hour and the other at 50 miles per hour. After how many minutes will the cars be 5 miles apart?

3

If a \$6,000 contribution is divided among charities A , B , and C in a ratio of 8:5:2, respectively, how much more, in dollars, does charity A receive than charity C ?

4

If a car traveling at 60 mph is chasing a car travelling at 50 mph and is ¼ mile behind, how many minutes will it take the first car to catch the second?

5

A truck”s gas tank can hold 18 gallons. If the tank is 2/3 full and the truck travels for 4 hours at 60 miles per hour until it runs out of gas, what is the efficiency of the truck, in miles per gallon?

6

A motorcycle has a fuel efficiency of 60 miles per gallon when it is cruising at a speed of 50 miles per hour. How many hours can it travel at 50 miles per hour on a full tank of gas, if its tank can hold 10 gallons?

7

If the ratio of a to b is 3 to 4, and the ratio of a to c is 5 to 2, what is the ratio of b to c ?

1. A) 3 to 10
2. B) 3 to 5
3. C) 5 to 3
4. D) 10 to 3

8

A paint mixture consists of a 3:2:11 ratio of red, violet, and white, respectively. How many ounces of violet are needed to make 256 ounces of this mixture?

1. A) 32
2. B) 36
3. C) 46
4. D) 48

9

A pool that holds 20,000 gallons is ¼ full. A pump can deliver g gallons of water every m minutes. If the pumping company charges d dollars per minute, how much will it cost, in dollars, to fill the pool?

1. A)
2. B)
3. C)
4. D)

10

Yael travels to work at an average speed of 40 miles per hour and returns home by the same route at 24 miles per hour. If the total time for the round trip is 2 hours, how many miles is her trip to work?

1. A) 25
2. B) 30
3. C) 45
4. D) 60

11

A hare runs at a constant rate of a miles per hour, and a tortoise runs at a constant rate of b miles per hour, where 0 < b < a . How many more hours will it take the tortoise to finish a race of d miles than the hare?

1. A)
2. B)
3. C)
4. D)

12

Janice can edit 700 words per minute and Edward can edit 500 words per minute. If each page of text contains 800 words, how many pages can they edit, working together, in 20 minutes?

13

If a printer can print 5 pages in 20 seconds, how many pages can it print in 5 minutes?

14

Traveling at 40 miles per hour, Diego can complete his daily commute in 45 minutes. How many minutes would he save if he traveled at 50 miles per hour?

15

If and , what is

16

If a cyclist races at 30 miles per hour for 1/2 of the distance of a race, and 45 miles per hour for the final 1/2 of the distance, what is her average speed, in miles per hour, for the entire race?

17

Anne can paint a room in 2 hours, and Barbara can paint the same room in 3 hours. If they each work the same rate when they work together as they do alone, how many hours should it take them to paint the same room if they work together?

18

What is the average speed, in miles per hour, of a sprinter who runs ¼ mile in 45 seconds? (1 hour = 60 seconds)

1. A) 11.25
2. B) 13.5
3. C) 20
4. D) 22

19

A car travels d miles in t hours and arrives at its destination 3 hours late. At what average speed, in miles per hour, should the car have gone in order to arrive on time?

1. A)
2. B)
3. C)
4. D)

20

In three separate 1-mile races, Ellen finished with times of x minutes, y minutes, and z minutes, respectively. What was her average speed, in miles per hour , for all three races?

1. A)
2. B)
3. C)
4. D)

21

Sylvia drove 315 miles and arrived at her destination in 9 hours. If she had driven 10 miles per hour faster, how many hours would she have saved on the trip?

1. A) 1.75 hours
2. B) 2.00 hours
3. C) 2.25 hours
4. D) 2.50 hours

1 . 108 time = distance/rate = 90 miles/50 mph = 1.8 hours = 1.8 hour × 60 min/hour = 108 minutes.

2 . 20 The fast car is moving ahead of the slow car at a rate of 50 − 35 = 15 mph, and so it will be 5 miles ahead after 5 ÷ 15 = 1/3 hour = 20 minutes.

3 . 2,400 Since 8 + 5 + 2 = 15, charity A receives 8/15 of the contribution, and charity C receives 2/15. The difference is 6/15, or 2/5, of the total, which is (2/5)(\$6,000) = \$2,400.

4 . 1.5 Since the faster car is catching up to the slower car at 60 − 50 = 10 mph, it will take (1/4 mile)/(10 mph) = 1/40 hours = 60/40 minutes = 1.5 minutes.

5 . 20 The tank contains (2/3)(18) = 12 gallons, and travels (4 hours)(60 mph) = 240 miles, so its efficiency is 240/12 = 20 miles per gallon.

6 . 12 With 10 gallons of gas and an efficiency of 60 miles per gallon, the car can travel 10 × 60 = 600 miles. At 50 miles an hour this would take 600/50 = 12 hours.

7 . D

8 . A According to the ratio, the mixture is 2/(3 + 2 + 11) = 2/16 = 1/8 violet. Therefore 256 ounces of the mixture would contain (1/8)(256) = 32 ounces of violet paint.

9 . C If the pool is ¼ full, it requires (3/4)(20,000) = 15,000 more gallons.

10 . B Let x = the distance, in miles, from home to work. Since time = distance/rate , it takes Yael x /40 hours to get to work and x /24 hours to get home.

Simplify:

Multiply by 15:

x = 30 miles

11 . B The tortoise would take d /b hours to complete the race, and the hare would take d /a hours to complete the race, so the tortoise would take hours longer.

12 . 30 Together they can edit 700 + 500 = 1,200 words per minute, so in 20 minutes they can edit

13 . 75 If the printer can print 5 pages in 20 seconds, it can print 15 pages in 1 minute, and therefore 15 × 5 = 75 pages in 5 minutes.

14 . 9 Since 45 minutes is ¾ hour, Diego”s daily commute is 40 × ¾ = 30 miles. If he traveled at 50 mph it would take him 30/50 = 3/5 hours = 36 minutes, so he would save 45 − 36 = 9 minutes.

15 . 3/10 or 0.3

Simplify:

Multiply by ¾:

16 . 36 Pick a convenient length for the race, such as 180 miles (which is a multiple of both 30 and 45). The first half of the race would therefore be 90 miles, which would take 90 miles ÷ 30 mph = 3 hours, and the second half would take 90 miles ÷ 45 mph = 2 hours. Therefore, the entire race would take 3 + 2 = 5 hours, and the cyclist”s average speed would therefore be 180 miles ÷ 5 hours = 36 miles per hour.

17 . 1.2 or 6/5 Anne”s rate is 1/2 room per hour, and Barbara”s rate is 1/3 room per hour, so together their rate is 1/3 + 1/2 = 5/6 room per hour. Therefore, painting one room should take (1 room)/(5/6 room per hour) = 6/5 hours.

18 . C

19 . B In order to arrive on time, it would have to travel the d miles in t − 3 hours, which would require a speed of d /(t − 3) mph.

20 . D

21 . B Sylvia traveled at 315/9 = 35 miles per hour. If she had traveled at 35 + 10 = 45 miles per hour, she would have arrived in 315/45 = 7 hours, thereby saving 2 hours.

Lesson 7: Interpreting percent problems

What number is 5 percent of 36?

When interpreting word problems, remember that statements about quantities can usually be translated into equations or inequalities . Here”s a simple translation key:

(Easy ) Notice that this enables us to translate the question into an equation, which can be solved to get the answer:

28 is what percent of 70?

Again, let”s use the glossary to translate and then solve:

Simplify:

Divide by 0.7:

40 = x

What number is 120% greater than 50?

To increase a number by x % , just multiply it by . To decrease a number by x % , just multiply it by . For instance, to increase a number by 20%, just multiply by 1.20 (because the final quantity is 120% of the original quantity, and to decrease a number by 20%, just multiply by 0.80 (because the final quantity is 80% of the original quantity).

(Easy ) If we increase a number by 120%, the resulting number is 100% + 120% = 220% of the original number. Therefore, the number that is 120% greater than 50 is 2.20 × 50 = 110.

Lesson 8: Percent change

A shirt has a marked retail price of \$80, but is on sale at a 20% discount. If a customer has a coupon for 10% off of the sale price, and if the sales tax is 5%, what is the final price of this shirt, including all discounts and tax?

1. A) \$58.80
2. B) \$60.00
3. C) \$60.48
4. D) \$61.60

(Medium ) To find the final price, we must perform three changes: decrease by 20%, decrease by 10%, and increase by 5%. This gives us (1.05)(0.90)(0.80)(\$80) = \$60.48, so the answer is (C). Notice that, since multiplication is commutative , it doesn”t matter in what order we perform the three changes; the result will still be the same.

If a population of bacteria increases from 100 to 250, what is the percent increase in this population?

1. A) 60%
2. B) 67%
3. C) 150%
4. D) 250%

To find the percent change in a quantity, just use the formula

Notice that any “percent change” is a “percent of the initial amount,” which explains why the initial amount s the value in the denominator.

(Easy ) If we know this formula, this question is straightforward: the percent change is (250 − 100)/100 × 100% = 150%, choice (C). If you mistakenly use 250 as the denominator, you would get an answer of (A) 60%, which is incorrect.

Lesson 9: Working with proportions and scales

On a scale blueprint, the drawing of a rectangular patio has dimensions 5 cm by 7.5 cm. If the longer side of the actual patio measures 21 feet, what is the area, in square feet, of the actual patio?

1. A) 157.5 square feet
2. B) 294.0 square feet
3. C) 356.5 square feet
4. D) 442.0 square feet

(Medium ) In a scale drawing, all lengths are proportional to the corresponding lengths in real life. That is, the lengths in the drawing and the corresponding lengths in real life should all be in the same ratio . We can set up a proportion here to find the shorter side of the patio, x .

When working with proportions, remember the two laws of proportions .

The Law of Cross-Multiplication

If two ratios are equal, then their “cross-products” must also be equal.

The Law of Cross-Swapping

If two ratios are equal, then their “cross-swapped” ratios must also be equal.

Cross-multiply:

7.5x = 105

Divide by 7.5:

x = 14

Therefore, the patio has dimensions 21 feet by 14 feet, and so it has an area of (21)(14) = 294 square feet. The correct answer is (B).

Exercise Set 3 (Calculator)

1

What number is 150% of 30?

2

If the areas of two circles are in the ratio of 4:9, the circumference of the larger circle is how many times the circumference of the smaller circle?

3

What number is 30% less than 70?

4

What number is the same percent of 36 as 5 is of 24?

5

David”s motorcycle uses 2/5 of a gallon of gasoline to travel 8 miles. At this rate, how many miles can it travel on 5 gallons of gasoline?

6

The retail price of a shirt is \$60, but it is on sale at a 20% discount and you have an additional 20% off coupon. If there is also a 5% sales tax, is the final cost of the shirt?

1. A) \$34.20
2. B) \$36.48
3. C) \$37.80
4. D) \$40.32

7

If the price of a house increased from \$40,000 to \$120,000, what is the percent increase in price?

1. A) 67%
2. B) 80%
3. C) 200%
4. D) 300%

8

At a student meeting, the ratio of athletes to non athletes is 3:2, and among the athletes the ratio of males to females is 3:5. What percent of the students at this meeting are female athletes?

1. A) 22.5%
2. B) 25%
3. C) 27.5%
4. D) 37.5%

9

To make a certain purple dye, red dye and blue dye are mixed in a ratio of 3:4. To make a certain orange dye, red dye and yellow dye are mixed in a ratio of 3:2. If equal amounts of the purple and orange dye are mixed, what fraction of the new mixture is red dye?

1. A)
2. B)
3. C)
4. D)

10

If the price of a stock declined by 30% in one year and increased by 80% the next year, by what percent did the price increase over the two-year period?

1. A) 24%
2. B) 26%
3. C) 50%
4. D) 500

11

A farmer has an annual budget of \$1,200 for barley seed, with which he can plant 30 acres of barley. If next year the cost per pound of the seed is projected to decrease by 20%, how many acres will he be able to afford to plant next year on the same budget?

1. A) 24
2. B) 25
3. C) 36
4. D) 37.5

12

If x is % of 90, what is the value of – x ?

13

If n is 300% less than , what is the value of |n |?

14

The cost of a pack of batteries, after a 5% sales tax, is \$8.40. What was the price before tax, in dollars?

15

If the price of a sweater is marked down from \$80 to \$68, what is the percent discount? (Ignore the % symbol when gridding.)

16

Three numbers, a , b , and c , are all positive. If b is 30% greater than a , and c is 40% greater than b , what is the value of

17

If the width of a rectangle decreases by 20%, by what percent must the length increase in order for the total area of the rectangle to double? (Ignore the % symbol when gridding.)

18

Two middle school classes take a vote on the destination for a class trip. Class A has 25 students, 56% of whom voted to go to St. Louis. Class B has n students, 60% of whom voted to go to St. Louis. If 57.5% of the two classes combined voted to go to St. Louis, what is the value of n ?

19

If 12 ounces of a 30% salt solution are mixed with 24 ounces of a 60% salt solution, what is the percent concentration of salt in the mixture?

1. A) 45%
2. B) 48%
3. C) 50%
4. D) 54%

20

If the length of a rectangle is doubled but its width is decreased by 10%, by what percent does its area increase?

1. A) 80%
2. B) 90%
3. C) 180%
4. D) 190%

21

The freshman class at Hillside High School has 45 more girls than boys. If the class has n boys, what percent of the freshman class are girls?

1. A)
2. B)
3. C)
4. D)

22

If the population of town B is 50% greater than the population of town A, and the population of town C is 20% greater than the population of town A, then what percent greater is the population of town B than the population of town C?

1. A) 20%
2. B) 25%
3. C) 30%
4. D) 40

1 . 45 1.50 × 30 = 45

2 . 1.5 Imagine that the areas are 4π and 9π. Since the area of a circle is πr 2 , their radii are 2 and 3, and their circumferences are 2(2)π = 4π and 2(3)π = 6π, and 6π ÷ 4π = 1.5.

3 . 49 70 − 0.30(70) = 0.70(70) = 49.

4 . 7.5

Cross multiply:

24x = 180

Divide by 24:

x = 7.5

5 . 100

Cross multiply:

Multiply by 5/2:

x = 100

6 . D 1.05 × 0.80 × 0.80 × \$60 = \$40.32

7 . C (120,000 − 40,000)/40,000 × 100% = 200%

8 . D The fraction of students who are athletes is 3/(2 + 3) = 3/5, and the fraction of these who are females is 5/(3 + 5) = 5/8. Therefore the portion who are female athletes is 3/5 × 5/8 = 3/8 = 37.5%.

9 . C The purple dye is 3/(3 + 4) = 3/7 red, and the orange dye is 3/(3 + 2) = 3/5 red. Therefore, a half-purple, half-orange dye is (1/2)(3/7) + (1/2)(3/5) = 3/14 + 3/10 = 18/35 red.

10 . B If the price of the stock were originally, say, \$100, then after this two-year period its price would be (0.70)(1.80)(\$100) = \$126, which is a 26% increase.

11 . D The quantity of barley seed is proportional to the acreage it can cover. The cost of seed for each acre of barley was originally \$1,200/30 = \$40 per acre. The next year, after the 20% decrease, the price would be (0.80)(\$40) = \$32 per acre. With the same budget, the farmer can therefore plant 1,200/32 = 37.5 acres of barley.

12 . 1/15 or 0.067 or 0.066

13 . 5

14 . 8.00 Let x be the price before tax:

1.05x = \$8.40

Divide by 1.05:

x = \$8.00

15 . 15 (68 − 80)/80 = −0.15

16 . 1.82 b = 1.30a and c = 1.40b , so c = 1.40(1.30a ) = 1.82a . Therefore c/a = 1.82a /a = 1.82.

17 . 150 For convenience, pick the dimensions of the rectangle to be 10 and 10. (This is of course a square, but remember that a square is a rectangle!) This means that the original area is 10 × 10 = 100. If the width decreases by 20%, the new width is (0.80)(10) = 8. Let the new length be x . Since the new rectangle has double the area, 8x = 200, and so x = 25. This is an increase of (25 − 10)/10 × 100% = 150%.

18 . 15 The total number of “St. Louis votes” can be expressed in two ways, so we can set up an equation to solve for n :

(0.56)(25) + (0.60)n = 0.575(25 + n )

Simplify:

14 + 0.6n = 14.375 + 0.575n

Subtract 14 and .575n :

0.025n = 0.375

Divide by .025:

n = 15

19 . C The total amount of salt in the mixture is (.30)(12) + (.60)(24) = 18, and the total weight of the mixture is 12 + 24 = 36 ounces, so the percent salt is 18/36 = 50%.

20 . A If the original dimensions are w and l , the original area is wl . If the length is doubled and the width decreased by 10%, the new area is (0.9l )(2w ) = 1.8wl , which is an increase of 80%.

21 . C The number of girls in the class is n + 45, and the total number of students is n + n + 45, so the percent of girls is .

22 . B B is 50% greater than A :

B = 1.5A

C is 20% greater than A :

C = 1.2A

Divide by 1.2:

Substitute:

Simplify:

B = 1.25C

Skill 3: Working with Tables of Data

Lesson 10: Using tables as problem-solving tools

A table can be useful for organizing information that falls into categories. Even if a problem does not include a table, ask yourself: does the information in this problem fall into non-overlapping categories? If so, consider setting up a table with the categories as row or column labels.

A committee determines that it will meet on the Thursday after the third Monday of every month. What is the latest date of the month on which this meeting could fall?

1. A) the 17th
2. B) the 18nd
3. C) the 24rd
4. D) the 25th

(Medium ) A calendar, which of course is a kind of table, can be handy here. But how do we fill in the numbers? Since we want the latest date of the month possible, we need to find the latest date on which the third Monday could fall. A little trial-and-error should reveal that the latest the first Monday could fall is the 7th.

Since there are seven days per week, the second Monday must be the 14th and the third must be the 21st. Therefore the meeting will be three days later, on Thursday the 24th. We don”t have to complete the entire calendar page to solve the problem.

Sam is considering buying a car, and has two models to choose from. Model N has a sticker price that is 25% higher than that of Model P, but he will receive \$1,500 in trade-in credit from the Model N dealer and only \$1,000 in trade-in credit from the Model P dealer. Even after the trade-in credits are applied to both cars, Model N is still \$2,000 more expensive than Model P (before taxes and fees). What is the sticker price of Model N?

1. A) \$10,000
2. B) \$12,000
3. C) \$12,500
4. D) \$13,000

(Medium ) Again, this problem does not give us a table, but the fact that we must keep track of two car models and two prices per car model (sticker price and post trade-in price) suggests that a table might be useful.

Since Model N has a sticker price that is 25% higher than that of Model P, if the sticker price of Model P is x , then the sticker price for Model N is 1.25x . The respective costs after trade-in, then, are 1.25x − 1,500 and x − 1,000. Now we must solve for x by setting up an equation, which comes from the fact that, even after the trade-in, Model N is still \$2,000 more expensive. Therefore 1.25x − 1500 = (x − 1,000) + 2,000.

Simplify:

1.25x − 1,500 = x + 1,000

Add 1,500 and subtract x :

0.25x = 2,500

Divide by 0.25:

x = 10,000

Now, keep in mind that the question is asking for the sticker price of Model N, not Model P. Therefore the correct answer is (1.25)(10,000) = 12,500.

Lesson 11: Finding conditional probabilities and population fractions

According to the results of the poll indicated in the table above, what percent of those who expressed an opinion on the proposal were female?

1. A) 52%
2. B) 54%
3. C) 68%
4. D) 81%

(Medium ) This question asks us to find a population fraction , that is, a part-to-whole fraction. This question is almost identical to asking a conditional probability question, namely: if a poll subject who expressed an opinion on Proposal 547 is chosen at random, what is the probability that the person is female?

The probability of an event is a part-to-whole ratio, and therefore can have only a value from 0 to 1. An event with probability 0 is impossible, and an event with probability 1 is certain.

The table indicates that the total number of respondents who expressed an opinion is 218 + 82 = 300. Of these, 120 + 42 = 162 are women, and therefore the percent of these that are women is 162/300 × 100% = 54%. Alternately, we could say that if a respondent expressing an opinion were chosen at random, that person would have a 0.54 probability of being female.

Lesson 12: Analyzing variable relations with tables

In Lesson 4 we used tables to identify direct and inverse variations between variables, but ordered pairs of variables can have many other relationships as well.

Tables—along with graphs and equations—are important tools for analyzing functional relationships between variables. When given an equation expressing y in terms of x , you should know how to generate a table of ordered pairs and use that table to graph the relationship. Alternately, you should learn to analyze the properties of a function from that table of values.

Even if our equation is complicated, like

we can just pick values for x , like −3, −2, −1, 0, 1, 2, and 3, use these to calculate the corresponding values of y (or f (x )) and create a table of ordered pairs

which we can then use to plot points on a graph.

If the variables x and y in the table above have a linear relationship, what is the value of b?

1. A) 19
2. B) 20
3. C) 21
4. D) 22

(Medium ) One way to approach this question is to use the fact that, in a linear relation, the y increases in proportion to the change in x . This rate of increase is the slope of the line. Taking the first and third ordered pairs, it seems that when x increases by 10 − 2 = 8, y in turn increases by 34 − 10 = 24. Therefore, the slope of the line is 24/8 = 3. That is, every time the x coordinate increases by 1, the y coordinate increases by 3. Now looking at the first and second ordered pairs, since x is increasing by 5 − 2 = 3, y must increase by 3(3) = 9, and so b is equal to 10 + 9 = 19, and the correct answer is (A).

Another way to look at it is that we have solved the proportion :

Simplify:

Cross multiply:

72 = 8b − 80

152 = 8b

Divide by 8:

19 = b

Exercise Set 4 (Calculator)

1

Which of the following equations best expresses the population, P , of bacteria culture A, as a function of t , in minutes?

1. A) P= 200t+ 520
2. B) P= 520(1.4)t
3. C) P= 2,000(t− 0.5)2
4. D) P= 520t+ 200

2

Which of the following equations best expresses the population, P , of bacteria culture B, as a function of t , in minutes?

1. A) P= 100t+ 500
2. B) P= 500(1.2)t
3. C) P= 2,000(t− 0.5)2
4. D) P= 500t+ 100

3

After 2 minutes, the population of culture A is what percent greater than the population of culture B?

1. A) 16.7%
2. B) 20.0%
3. C) 27.8%
4. D) 127.8%

4

If culture A continues to grow at a constant rate, at what time should its population reach 2,000?

1. A) 7 minutes 4 seconds
2. B) 7 minutes 24 seconds
3. C) 7 minutes 40 seconds
4. D) 8 minutes 20 seconds

5

By what percent did the population of culture B increase over the first 3 minutes?

1. A) 36.4%
2. B) 42.1%
3. C) 72.8%
4. D) 172.8%

Questions 6–8 refer to the following information

6

According to the table above, how much is the price of one senior ticket?

1. A) \$4.00
2. B) \$6.00
3. C) \$12.00
4. D) \$16.00

7

How much more is the cost of one adult ticket than the cost of one student ticket?

1. A) \$0.50
2. B) \$1.50
3. C) \$2.50
4. D) \$3.00

8

Which is closest to the average (arithmetic mean) price of the 250 tickets sold?

1. A) \$5.54
2. B) \$5.59
3. C) \$5.68
4. D) \$5.72

9

What is the median price of the 250 tickets sold?

1. A) \$5.00
2. B) \$5.50
3. C) \$5.75
4. D) \$6.00

10

If a meeting must take place on the third Tuesday of the month, what is the earliest date of the month on which it could take place?

1. A) the 14th
2. B) the 15th
3. C) the 22nd
4. D) the 27th

11

What is the latest date of the month on which the meeting could take place?

1. A) the 13th
2. B) the 14th
3. C) the 21st
4. D) the 26th

Questions 12–21 refer to the following information

12

From 1970 to 1990, the percent increase in the U.S. consumption of nuclear energy was closest to

1. A) 96%
2. B) 240%
3. C) 2,400%
4. D) 3,400%

13

In a pie graph representing total U.S. energy consumption in 2010, the sector representing non nuclear renewables would have a central angle measuring approximately

1. A) 8°
2. B) 12°
3. C) 24°
4. D) 30°

14

Nuclear energy and renewable energy are often grouped together in the category “non-greenhouse” energy. In 1970, approximately what percent of non-greenhouse energy was nuclear?

1. A) 0.4%
2. B) 5.6%
3. C) 5.9%
4. D) 6.4%

15

In 2010 what percent of non-greenhouse energy consumption was nuclear?

1. A) 9%
2. B) 29%
3. C) 49%
4. D) 51%

16

In the four years shown, what percent of the total energy consumed was due to non-nuclear renewables?

1. A) 5.6%
2. B) 6.8%
3. C) 7.4%
4. D) 7.9%

17

What was the percent increase in fossil fuel energy consumption between 1950 and 2010?

1. A) 28%
2. B) 61%
3. C) 124%
4. D) 156%

18

The “renewability index” is defined as the fraction of total U.S. energy consumption that comes from non-nuclear renewable energy. What was the percent increase in the renewability index from 1970 to 2010?

1. A) 17%
2. B) 37%
3. C) 47%
4. D) 99%

19

For how many of the years shown above did fossil fuels account for less than 90% of the annual U.S. energy consumption?

1. A) One
2. B) Two
3. C) Three
4. D) Four

20

Between 1950 and 2010, the average annual rate of increase in the consumption of non-nuclear renewable energy was closest to

1. A) 0.085 QBTU/yr
2. B) 0.128 QBTU/yr
3. C) 1.70 QBTU/yr
4. D) 2.27 QBTU/yr

21

Between 1970 and 2010, the annual consumption of fossil fuels in the U.S. increased nearly linearly. If this linear trend were to continue, which of the following is closest to the level of U.S. fossil fuel consumption we would expect for 2035 (in quadrillion BTUs)?

1. A) 90
2. B) 91
3. C) 92
4. D) 93

1 . A According to the table, the population of culture A increases by 200 bacteria every minute, indicating a linear relationship with a slope of 200. (Remember that the slope of a function is equivalent to its unit rate of change .) Choice (A) is the only option that indicates a line with slope 200.

2 . B The table indicates that culture B is not increasing linearly, since the population difference from minute to minute is not constant, but increasing. This rules out choices (A) and (D). By substituting t = 0, t = 1 and t = 2, we can see that only the function in (B) gives the correct populations. Notice that the base of the exponential, 1.2, indicates that the population grows by 20% each minute.

3 . C At the 2 minute mark, the populations are 920 and 720, respectively, so culture A has a population that is (920 − 720)/720 × 100% = 27.8% greater.

4 . B If we use the population equation (see question 1), we can solve for t . Plugging in 2,000 for P gives us 2,000 = 200t + 520, which gives a solution of t = 7.4 minutes. Since 0.4 minutes equals 0.4 × 60 = 24 seconds, the time elapsed is 7 minutes 24 seconds.

5 . C In the first 3 minutes, culture B grew from 500 to 864 bacteria, which is an increase of (864 − 500)/500 × 100% = 72.8%.

6 . B The total revenue for each ticket type equals the price per ticket times the number of tickets sold. Therefore, the price for each senior ticket is \$96 ÷ 16 = \$6.

7 . D One adult ticket costs \$630 ÷ 84 = \$7.50, and one student ticket costs \$495 ÷ 110 = \$4.50, so each adult ticket costs \$3 more.

8 . C The average price per ticket equals the total revenue for all tickets divided by the number of tickets: (\$630 + \$200 + \$96 + \$495)/250 = \$5.684.

9 . A The median price of 250 tickets is the average of the prices of the 125th and 126th tickets, if the price for each ticket is listed in increasing order. The ticket prices, in increasing order, are \$4.50 for students (110 tickets), \$5.00 for children (40 tickets), \$6.00 for seniors (16 tickets), and \$7.50 for adults (84 tickets). With this ordering, the 125th and 126th price are both \$5.00.

10 . B The earliest the first Tuesday could be is the 1st, so the earliest the third Tuesday could be is the 15th.

11 . C The latest the first Tuesday could be is the 7th, so the latest the third Tuesday could be is the 21st.

12 . C In 1970, nuclear energy consumption was 0.24, and in 1990 it was 6.10. This represents an increase of (6.10 − 0.24)/0.24 × 100% = 2,442%.

13 . D In 2010, non nuclear renewables accounted for 8.09/97.63 × 100% = 8.3% of consumption, which would correspond to a 0.083 × 360° = 29.88° central angle.

14 . B In 1970, the total “non-greenhouse” energy was 0.24 + 4.07 = 4.31. Therefore the percent that was nuclear is 0.24/4.31 × 100% = 5.6%.

15 . D In 2010, this percent was 8.43/(8.43 + 8.09) × 100% = 51%.

16 . C The total non nuclear renewable energy consumption for the four years is 2.98 + 4.07 + 6.04 + 8.09 = 21.18, and the total energy consumption is 34.61 + 67.84 + 84.47 + 97.63 = 284.55. Therefore the percent is 21.18/284.55 × 100% = 7.4%.

17 . D In 1950, fossil fuel consumption was 31.63, and in 2010 it was 81.11. This is an increase of (81.11 − 31.63)/31.63 × 100% = 156%.

18 . B In 1970, the renewability index was 4.07/67.84 = 0.060, and in 2010 it was 8.09/97.63 = .082. This is a percent increase of (0.082 − 0.060)/0.060 × 100% = 37%.

19 . B In 1990, the percent of consumption from fossil fuels was 72.33/84.47 × 100% = 85.6%, and in 2010 it was 81.11/97.62 × 100% = 83.1%.

20 . A The annual rate of increase is the total increase divided by the time span in years. The total increase is 8.09 − 2.98 = 5.11. Over a 60-year span, this gives a rate of 5.11/60 = 0.085.

21 . C In the 40 year span from 1970 to 2010, fossil fuel consumption increased at a rate of (81.11 − 63.52)/40 = 0.44 QBTU/Yr. In 25 more years at this rate, the consumption should be 81.11 + 25(0.44) = 92.11 QBTU.

Skill 4: Working with Graphs of Data

Lesson 13: Working with scatterplots

Scatterplots are graphs of ordered pairs that represent data points. They are very useful for showing relationships between variables that do not vary in a highly predictable way.

Particulate matter is a class of air pollutants. The scatterplot above shows 40 readings for particulate matter concentration, in micrograms per cubic meter, for a particular metropolitan area over 9 years. Based on the line of best fit shown, which of the following is closest to the average yearly decrease in particulate matter concentration?

1. A) 0.32 micrograms/m3per year
2. B) 0.64 micrograms/m3per year
3. C) 3.2 micrograms/m3per year
4. D) 6.4 micrograms/m3per year

A line of best fit is simply a line that “hugs” the data in a scatterplot better than any other line. In a statistics course, you will learn how to calculate a line of best fit precisely (or your calculator can do it for you), but the SAT will only expect you to use lines of best fit to make inferences about general trends in data or to find “expected values.”

(Medium ) In this problem, the line of best fit shows a general downward trend in the data, even though the data are fairly widely scattered. To find the “average yearly decrease” in particulate matter concentration, we simply have to find the slope of this line. To get the most accurate estimate for slope, we should take points on the line that are far apart. The leftmost endpoint seems to be at (2003, 12), and the rightmost endpoint seems to be at (2013, 9). This gives us a slope of (9 − 12)/(2013 − 2003) = −3/10 = −0.3 micrograms/m3 per year, so the correct answer is (A).

According to the line of best fit to the data above, which of the following is closest to the percent decrease in average particulate matter concentration from 2007 to 2012?

1. A) 9%
2. B) 18%
3. C) 36%
4. D) 60%

(Medium-hard ) This question is similar to the previous one, but notice the two important differences: first, it is asking us to compare two specific years, and it is asking us to calculate the percent decrease rather than the rate of decrease , so we will need the percent change formula from Lesson 8. The line of best fit gives a value of about 11 in 2007 and about 9 in 2012. Therefore the percent change is (9 − 11)/11 × 100% = −18%, and the correct answer is (B).

If the Environmental Protection Agency”s air quality standard is 12 micrograms of particulate matter per cubic meter, as shown with the dotted line, what percent of these data fall above this standard?

1. A) 11%
2. B) 14.5%
3. C) 27.5%
4. D) 35%

(Easy ) This question is simply asking for a part-to-whole ratio expressed as a percentage. There are 40 total data points (as the problem states), and 14 of them lie above the line: 14/40 × 100% = 35%, so the correct answer is (D).

Lesson 14: Linear and non linear relationships

Sometimes the SAT Math test will ask you to draw inferences from graphs that are non-linear. Even if a graph is not linear, you should still be able to draw inferences based on the points on that graph. For instance, you should be able to find the average rate of change between two points by finding the slope of the line segment connecting them, or compare the rate of growth of one curve to that of another curve.

The graph above shows the number of cells in two separate bacterial cultures as a function of time. How much time elapsed between the point when the two cultures had equal population and the point when the population of culture A was twice that of culture B?

1. A) 2 minutes
2. B) 3 minutes
3. C) 4 minutes
4. D) 5 minutes

(Medium ) The time when the two populations are equal is easy to find: it”s where the two curves intersect, at the 4-minute mark. The four answer choices tell us where to look next on the graph. Since choice (A) is 2 minutes, we should find the populations at the 4 + 2 = 6-minute mark. At 6 minutes, culture B has a population of 700 and culture A has a population of 1,400, which of course is twice as great, and therefore the answer is (A).

How much longer did it take culture B to double its original population than it took culture A?

1. A) 2 minutes
2. B) 3 minutes
3. C) 4 minutes
4. D) 5 minutes

(Medium ) Culture A has a starting population of 200 and doubles its population to 400 at the 3-minute mark. Culture B has a starting population of 400 and doubles its population to 800 at the 8-minute mark. Therefore, culture B took 8 − 3 = 5 more minutes to double its population, and the correct answer is (D).

Lesson 15: Drawing inferences from graphs

If culture B were to continue its linear growth, how many more minutes (beyond the 10 minutes shown in the graph) would be required for culture B to reach 1,500 cells?

1. A) 9 minutes
2. B) 12 minutes
3. C) 13.5 minutes
4. D) 15 minutes

(Medium ) By taking any two points on the line for culture B, for instance (0,400) and (4,600), we can calculate the slope of the line, which equals the unit rate of growth : (600 − 400)/(4 − 0) = 50 cells per minute. Since culture B contains 900 cells at the 10 minute mark, it would take 600/50 = 12 more minutes to reach 1,500 cells, and the correct answer is (B).

A particular experiment requires that both culture A and culture B contain between 400 and 800 cells each. The time period in which the cell population for both cultures is within this range is called the “experimental window.” Which of the following is closest to the “experimental window” for the two cultures shown in the graph above? (1 minute = 60 seconds)

1. A) 95 seconds
2. B) 120 seconds
3. C) 165 seconds
4. D) 240 seconds

(Medium-hard ) The “experimental window” is the period when both populations are between 400 and 800. The population of culture A hits 400 at the 3-minute mark and exceeds 800 cells after roughly the 4.6-minute mark. The population of culture B starts off (at the 0-minute mark) at 400 cells, and exceeds 800 cells after the 8-minute mark. The overlapping period is between 3 minutes and 4.6 minutes, for a period of roughly 1.6 minutes or 1.6 × 60 = 96 seconds; therefore, the correct answer is (A).

Lesson 16: Working with pie graphs

We”ve all seen pie graphs. They are convenient ways of representing part-to-part and part-to-whole relationships. On the SAT Math test, you may be asked to analyze the features of pie graphs in some detail, or to discuss the features of a hypothetical pie graph.

When analyzing pie graphs, remember this helpful formula:

In the diagram above, which of the following is closest to the measure of the central angle of the sector representing total minority ownership of U.S. broadcast television media in 2007?

1. A) 3°
2. B) 8°
3. C) 12°
4. D) 15°

(Easy ) According to the graphic, the total minority ownership of television media in 2007 was 0.6% + 1.3% + 0.9% + 0.4% = 3.2%. Therefore the central of the sector representing this portion measures 0.032 × 360° = 11.52°, so the correct answer is (C).

Maria is constructing a pie graph to represent the expenses for her project, consisting of three expense categories: marketing, design, and development. She knows that the marketing expenses are \$12,000 and the design expenses are \$30,000, but the development expense could range anywhere from \$30,000 to \$48,000. Based on this information, which of the following could be the measure of the central angle of the sector representing marketing expenses?

1. A) 36°
2. B) 54°
3. C) 62°
4. D) 70°

(Medium-hard ) The measure of the central angle of the sector depends on the part-to-whole ratio, so we need to calculate the maximum and minimum part-to-whole ratio for the marketing expenses, the marketing expenses are fixed at \$12,000, but the total expenses could range from \$12,000 + \$30,000 + \$30,000 = \$72,000 to \$12,000 + \$30,000 + \$48,000 = \$90,000. This means that the part-to-whole ratio for marketing could range from 12,000/90,000 = 0.133 to 12,000/72,000 = 0.167. Therefore the central angle for the marketing sector can measure anywhere from 0.133 × 360° = 48° to 0.167 × 360° = 60°. The only choice in this range is (B) 54°.

Exercise Set 5 (Calculator)

Questions 1–4 refer to the following information

1

The scatterplot above shows the annual revenue for all of the individual retail stores operated by a clothing company for each year from 2004 through 2012. Based on the line of best fit to the data shown, which of the following is closest to the percent increase in revenue per store from 2005 to 2011?

1. A) 50%
2. B) 100%
3. C) 120%
4. D) 300%

2

In 2006, the total combined revenue for all stores was closest to

1. A) \$350,000
2. B) \$480,000
3. C) \$700,000
4. D) \$950,000

3

From 2009 to 2010, the total combined revenue for all stores increased by approximately

1. A) \$50,000
2. B) \$200,000
3. C) \$400,000
4. D) \$600,000

4

Between 2006 and 2012, what was the percent increase in the total number of retail stores for this company?

1. A) 45%
2. B) 50%
3. C) 100%
4. D) 200%

Questions 5–9 refer to the following information

5

The chart above shows the allocation of \$3.5 trillion in U.S. federal expenses for 2010. What were the total 2010 expenditures on Defense?

1. A) \$700 billion
2. B) \$70 billion
3. C) \$7 billion
4. D) \$700 million

6

What is the measure of the central angle for the sector representing Medicare expenses?

1. A) 13.0°
2. B) 45.5°
3. C) 46.8°
4. D) 48.2°

7

If Interest on National Debt expenses were to decrease by \$20 billion from their 2010 levels, this would represent a percent decrease of approximately

1. A) 6%
2. B) 10%
3. C) 12%
4. D) 15%

8

How much more did the United States spend in 2010 on Interest on National Debt than on Education?

1. A) \$2 billion
2. B) \$7.0 billion
3. C) \$20 billion
4. D) \$70 billion

9

If 50% of the budget for Federal Pensions were to be reallocated as Social Security expenses, the size of the Social Security budget would increase by what percent?

1. A) 1.75%
2. B) 8.75%
3. C) 17.75%
4. D) 21.75%

Questions 10–17 refer to the following information

Source: Rick Nevin, Lead Poisoning and the Bell Curve , 2012

10

According to the graph above, in 1970 the number of violent crimes per 100,000 capita in the United States was closest to

1. A) 25
2. B) 375
3. C) 700
4. D) 750

11

In 1970 the average preschool blood lead level, in mcg/dL, was closest to

1. A) 10
2. B) 12
3. C) 23
4. D) 25

12

The percent decline in violent crime from 1993 to 2013 is closest to

1. A) 11%
2. B) 35%
3. C) 47%
4. D) 88%

13

From 1970 to 1990, the average annual rate of decline in preschool blood lead levels, in mcg/dL per year, was approximately

1. A) 1
2. B) 5
3. C) 15
4. D) 17

14

Which of the following 10-year spans saw the greatest percent increase in preschool blood lead levels?

1. A) 1945–1955
2. B) 1955–1965
3. C) 1965–1975
4. D) 1975–1985

15

Which of the following five-year spans saw the greatest percent increase in violent crime?

1. A) 1963–1968
2. B) 1968–1973
3. C) 1973–1978
4. D) 1978–1983

16

Approximately how many years did it take for average preschool blood lead levels to return to their 1950 levels?

1. A) 25
2. B) 30
3. C) 35
4. D) 40

17

For approximately how many years between 1963 and 2013 was the violent crime rate in the United States greater than 375 crimes per 100,000 capita?

1. A) 25
2. B) 30
3. C) 37
4. D) 42

1 . C In 2005, the revenue per store, according to the line of best fit, was about \$300,000, and in 2012 it was about \$650,000, so the percent change is (650,000 − 300,000)/300,000 × 100% = 116.67%, which is closest to (C) 120%.

2 . D In 2006, the data points show that there were 3 stores, with revenue of roughly \$200,000, \$330,000, and \$420,000, for a total of \$950,000.

3 . D In 2009, the combined revenue for the three stores was approximately \$420,000 + \$450,000 + \$550,000 = \$1,420,000. In 2010, the combined revenue for four stores was approximately \$220,000 + \$520,000 + \$600,000 + \$675,000 = \$2,015,000, for an increase of about \$595,000.

4 . C In 2006 there were 3 stores and in 2012 there were 6 stores, which is an increase of (6 − 3)/3 × 100% = 100%.

5 . A The chart shows that 20% of the expense budget went to defense, which equals 0.2 × \$3,500,000,000,000 = \$700 billion.

6 . C Medicare accounts for 13% of expenses, so the sector angle is 0.13 × 360° = 46.8°.

7 . B The Interest on National Debt in 2010 was 0.057 × \$3.5 trillion = \$199.5 billion, so a decrease of \$20 billion would be 20/199.5 × 100% = 10%.

8 . D The difference between Interest on National Debt and Education is 5.7% − 3.7% = 2%, and 0.02 × \$3.5 trillion = \$70 billion.

9 . B The Social Security budget in 2010 was 0.20 × \$3.5 trillion = \$700 billion. 50% of the Federal Pensions budget is 0.5 × 0.035 × \$3.5 billion = \$61.25 billion. This would be an increase of 61.25/700 × 100% = 8.75%.

10 . B The vertical axis label on the left shows that the violent crime trend is indicated by the solid curve and the bottom time series (1963–2013). For this curve, 1970 is slightly to the left of the vertical line at 1973, which shows values clearly between 300 and 450.

11 . C The vertical axis label on the left shows that the preschool blood lead trend is indicated by the dashed curve and the top time series (1940–1990).

12 . C In 1993, the violent crime rate was 750, and in 2013 it was about 400. The percent decrease is therefore (400 − 750)/750 × 100% = 46.7%

13 . A In 1970, the blood lead levels were about 23 and in 1990, they were about 3. The rate of decline is therefore (23 − 3)/(1990 − 1970) = 1 mcg/dL per year.

14 . A From 1945–1955 preschool blood lead levels increased from about 5 to about 17, a percent increase of (17 − 5)/5 × 100% = 240%.

15 . A The question asks for the greatest percent increase, not the greatest net increase in violent crime. Notice that the net increase from 1963–1968 (from roughly 150 to 250) seems to be slightly less than net increase from 1968–1973 (from roughly 250 to 375), the percent increase from 1963–1968 (+67%) is clearly greater than that from 1968–1973 (+50%).