SAT Means, Mediums, Modes

 Arithmetic Means, Weighted Averages, Medians, and Modes

In this section, you will learn how to:

• Calculate arithmetic means

• Find the value of an unknown score when given all the other scores and an average

• Calculate medians

• Calculate modes

There are three types of averages: arithmetic means, medians, and modes.  When you find your semester grade by adding all of your scores together and dividing by the number of scores, you are finding an arithmetic mean. 

Suppose that you earned the following grades during a semester:  86, 98, 92, 79, and 97.  To find your final grade, you add all the scores together and divide by 5, which is the number of scores. 

Suppose, however, you want to earn at least a 90% for the semester and your first four test grades are 91, 87, 88, and 94.  What must you score on the fifth test to earn an average of 90% for the entire semester?

Most SAT problems require knowing how to solve the second type of problem.   

Example 1: Charlie’s girlfriend, Rhonda, a bright girl who spent too much time practicing her cheerleading, scored 48, 62, 59 and 74 on each of 4 algebra tests.  What grade must she make on her fifth and last test in order to pass algebra with a D –?  Assume that 60 is a D –.  

This is an average problem.  The formula for computing an arithmetic mean is: 

Average =  

In this problem we do not know the sum of the scores directly, since we do not know the score on the fifth test.  However, if we rearrange the equation above, we find that

Average   Number of scores = Sum of scores. 

When working with averages, if you are asked to find an unknown score, always, always find the sum of scores first by multiplying the average by the number of scores. 

In this case, the desired average is 60. The number of scores is 5, including the test Rhonda hasn’t taken yet.  The sum of scores = 60  5 = 300.  That means that all the scores for the test she has already taken plus the score of the test that she hasn’t taken yet must add to 300. 

48 + 62 + 59 + 74 + x = 300. 

243 + x = 300.

x = 57

The SAT likes to test your ability to work with variables instead of numbers. 

Example 2:  The average of four numbers, a, b, c, and d is 12.  The average of c and d is 6. What is the average of a and b? 

First, find the sum of scores. 

  

 Find the sum of scores by multiplying 12   4 and get 48. 

 Next find the sum of scores for c and d. 

  .  Therefore, c + d = 12.  

Therefore, a + b = 48 – 12 = 36.  

Find the average of a and b:  

Another useful thing to know about averages is that if the scores being averaged are different, the average must be somewhere between the highest and lowest scores.  Returning to Rhonda’s 4 scores, since the lowest score is 48 and the highest score is 74, her average for the 4 tests must be somewhere between 48 and 74.  If you were given a multiple choice question and asked to find a reasonable average score, you would not chose a number like 40 or 76 because these numbers are not between the highest and lowest scores. Similarly, 48 would not be a reasonable average because the other, higher scores, would boost the average above 48. 

Weighted Averages:

If you have ever had a teacher count tests more heavily than quizzes and homework in determining your final grade, you have an intuitive idea of what a weighted average is.  Some scores are counted more—or weighted more—than other scores.  Weighted averages are typically used when averaging test scores or finding the average rate on a trip in which people travel at different speeds for different lengths of time. 

Example 3:  In an SAT review class, ten students took a practice SAT math test.  Their scores are summarized in the table below. Find the average score. 

Test Score Number of Students Earning That Score

480 1

520 4

540 1

590 2

610 1

650 1

The wrong way to find the score is to add 480 + 520 + 540 + 590 + 610 + 650 and divide by 10.  This is wrong because while only 1 student scored 480, 4 students scored 520. The 520 must be weighted more because more students got this score. 

The correct way to solve this problem is to multiply each score by the number of times someone got this score and then divide by 10, the total number of scores there are. 

480(1) + 520( 4) + 540(1) + 590(2) + 610(1) + 650(1).  

 = 554

Example 4:  Charlie drives to Sea World at 90 mph for 2 hours.  Then Rhonda drives at 60 miles per hour for 3 hours.  What is the average rate for the entire trip? 

The wrong way to solve this problem is to add 90 and 60 and divide by 2.  This is wrong because the rates must be weighted by the amount of time spent driving at each rate. 

Average speed =  =

 

Distance Charlie drives = 90(2) = 180

Distance Rhonda drives = 60(3) = 180

Total Distance driven = 360. 

 

Medians and Modes

Another kind of average is the median:  It is the middle score in a group of scores when the scores are arranged from smallest to largest (or largest to smallest). If we have an odd number of scores, we arrange the scores from smallest to largest and take the middle score.  If we have an even number of scores, we arrange the scores from smallest to largest and take the average of the two middle scores. 

Yet a third kind of average is the mode.  The mode is the most common score.  

For example, if a teacher gave a ten-point quiz and the scores were 2, 5, 3, 4, 7, 9, and 5, to find the median, we would arrange the scores in order from smallest to largest and get 

2, 3, 4, 5, 5, 7, 9.  Because we have an odd number of scores, we use the middle number. The middle score is 5 because half of the scores—2, 3, and 4 are below the median of 5 while the other scores—5, 7, and 9 are above the median.  The mode is also 5 because it is the score that occurs most frequently. 

Example 5: Given the scores 2, 1, 5, 9, 13, 21, 4, 1, 18, 1, 11, 1

a. Find the median.

Arrange the scores from smallest to largest: 1, 1, 1, 1, 2, 4, 5, 9, 11, 13, 18, 21

Because there is an even number of scores, we take the average of the 2 middle scores. The two middle scores are 4 and 5.  Their average is 4.5.  Therefore, the median score is 4.5. 

b. Find the mode.

The number occurring most frequently is 1. Therefore, 1 is the mode. 

Example 6:  The expressions 2x –1, x + 3, and 3x – 2 represent positive integers written in order from smallest to largest. If x is a positive integer greater than 1 and x + 3 is the median, find a possible value of x. 

A. 1

B. 2

C. 3

D. 4

E. 5

You can do one of two things: you can look at the multiple choice answers and plug each one into the algebraic expressions until you get answers that fit the conditions of the problem. 

You can also play with inequalities. 

Compare the inequalities two at a time.  If we assume that the three expressions are written in order from smallest to largest, we have

2x – 1 < x + 3

Solving for x, we get x < 4.  

Comparing the other two inequalities, we get x + 3 < 3x – 2.  Solving for x, we get

  – 2x < – 5 or x > .  

The only integer between   and 4 is 3. 

Example 7:  There are 5 consecutive integers, the median of which is 30. What is the largest of the integers? 

_   _  30 _ _ 

If there are five consecutive integers, two of them are larger than the median and two of them are smaller.  The largest integer will be larger than the median by 2. The answer is 32. 

 

Try the following problems.

1. Refer to Example 1 in this section.  If 70% is a C in this class, what must Rhonda score on the fifth test to get a C?

2. If the test has only 100 points, can Rhonda hope for a B?

3. Rhonda’s twin sister, Nadine, is a math whiz.  Her scores on the 4 tests are 93, 98, 89, and 96.  If 90% if an A, what is the lowest score Nadine can make on the fifth test and keep her A average? 

4. The average of w, x, y, and z is 36.  The average of w and x is 12. What is the average of y and z? 

5. The average of e, f, g, and h is 100.  The average of g and h is 25.  What is the average of e and f? 

6. If Rhonda’s lowest score on a French exam was 78 and her highest score was 94, what is a reasonable estimate of her average? 

A. 78

B. 94

C. 82

D. 76

E. 98

7. Charlie’s lowest score on a math test was 77 and his highest score was 91.  What is a reasonable estimate for his average? 

A. 77

B. 92

C. 91

D. 72

E. 88

8. Josh plays hockey on a travel team.  He plays a series of 10 games.  In 3 of the games, he scores no goals.  In 4 games, he scores 3 goals, and in 3 of the games, he scores 5 goals.  What is the average number of goals he scores per game? 

9. Grandpa used to build race cars and once drove in a NASCAR race.  He loves to speed.  Grandma has never built race cars, never driven in a NASCAR race, and hates to speed.  Grandpa drives for 2 hours at 92 miles per hour.  After 2 hours of listening to grandma say “Lowell, slow down!” Grandpa gets mad and tells her to drive.  She drives at 60 mph for 4 hours.  What is the average speed for the entire trip? 

10. Rhonda’s math teacher gives a series of 5 tests.  On two of the tests, Rhonda scores a 78.  On two of the tests, she scores an 87.  On one test, she scores a 97.  What is Rhonda’s test average? 

11. What is the median of the following numbers: 2, 5, 3, 7, 17, 4, and 2. 

12. In the problem above, what is the mode? 

13. Ten houses sell in a small town for the following prices:

Selling Price Number of Houses Sold at that Price

30,000 2

35, 000 3

120,000 2

200,000 1

250,000 2

 

a. What is the median price of the houses? 

b. What is the mode? 

c. What is the arithmetic mean of the prices? 

d. If you are trying to convince the federal government that your community is impoverished and needs federal funds, what statistic should you use—the mean, the median, or the mode? 

e. If you are trying to convince investors that your community is prosperous and a good investment, should you use the mean, the median, or the mode? 

14. If  , x + 1, and 3x – 5 are integers written in order from smallest to largest and x + 1 is the median, what is a possible value of x. 

A. 3

B. 4

C. 7

D. 8

E. 9

15. The median of 7 consecutive integers is 9.  What is the smallest of the integers? 

Answers:

1. 107 (Maybe the test has extra-credit questions.)

2. No. 

3. 74

Nadine’s scores of 93, 98, 89, and 96 add to 376.  To get at least a 90, all of her scores must add to 450.  The difference between 450 and 376 is 74. 

4. 60

The sum of w, x, y, and z is 36(4) = 144.  The sum of w and x is 24. Therefore, the sum of y and z is 144 – 24 = 120.  Because there are two terms, divide 120 by 2 to get an average of 60. 

5. 175

The sum of e, f, g, and h is 100(4) = 400.  The sum of g and h is 50.  The sum of e and f is therefore 350 and the average is 350 divided by 2, which is 175. 

6. C

7. E

8. 2.7 goals per game.   

9. .  The weighted average is  

10. 85.4.  Take 2(78) + 2(87) + 97. This sums to 427. Divide this sum by 5 and get 85.4

11. The median is 4. 

12. The mode is 2. 

13. a.  Median: $77, 500

      b. Mode:  $35,000

      c. Arithmetic mean = $110,500

      d. Mode: it is the smallest value and makes the town look the poorest. 

      e. Arithmetic mean: it is the largest value and makes the town look prosperous. 

14.      B.  A possible value for x is 4. 

2x  – 5 <  x + 1 < 3x – 5

Compare two inequalities at one time:  2x – 5 < x + 1

This tells us that x < 6.  

Now compare x + 1 < 3x – 5.  

Rearranging the terms of the inequality, we get: 

– 2x < – 6 or x > 3.

There are only two integers that fit these requirements: 4 and 5.  However, 5 is not one of the multiple choice answers while 4 is. 

15.       6 is the smallest integer.  Look at the diagram below. 

_   _   _  9   _   _    _ 

The smallest integer will be 3 less than 9.  That will be 6. 

 

 

Independent Probability

In this section, you will learn how to

• Calculate the probability of a single event occurring

• Calculate the probability of several events occurring

Suppose you toss a die.  One of 6 things can happen: you can get a 1, 2, 3, 4, 5, or 6.  Everything that could happen is called the “total possible outcomes.”  Now, suppose that you want to roll an even number.  There are 3 ways that this can occur: by rolling a 2, 4, or 6. Because rolling an even number is what you want to happen, let’s call this result the “desired outcome.”  We can also call the desired outcome A and call the probability of obtaining the desired outcome P(A), which is read “P of A” or “the probability of A.”

The probability of rolling the die and obtaining an even number is 

P(A) =     

            =  =  

To use another example, if you toss a coin, there are two possible outcomes:  head or tails.  There is only one way you can get a head.  The probability of getting heads, H, is 

P(H) =  =  .  

For similar reasons, the probability of getting tails, T, is 

P (T ) =  .

Tossing a coin and rolling a die are examples of independent probabilities. When probabilities are independent, the outcome of one event does not affect the outcome of other events. The coin does not know and is not influenced by what happened on previous tosses.  If you toss a coin 9 times and it comes up heads all 9 times, the probability of getting a head on the 10th toss is still  

To find the probability of two independent events occurring, find the probability of each event and multiply the two probabilities together. 

Example 1:     What is the probability of tossing a coin and getting tails while also rolling a die and getting a 5?  

The probability of tossing a coin and getting tails is  

The probability of rolling a die and getting a 5 is 

 

The probability of the two events occurring together is  

The sum of probabilities that completely describe all possibilities in a given situation is 1. 

Tossing a coin gives only 2 possible outcomes: heads or tails.  The probability of getting a head is  and the probability of getting a tail is also  .  

P(H) + P(T) =   

Now consider the probabilities involved in tossing a die: 

Probability of rolling a 1 = 

Probability of rolling a 2 =  

Probability of rolling a 3 =  

Probability of rolling a 4 =  

Probability of rolling a 5 =  

Probability of rolling a 6 =  

 If you add all the probabilities of all the possible outcomes—rolling a 1, 2, 3, 4, 5, or 6—the probabilities add to 1. 

You can use this knowledge to find the probability of an event if you know all the probabilities of all the other outcomes. 

Example 2:     Suppose you have a jar filled with an unknown number of marbles. The marbles can be either red, blue, or green.  Half of the marbles are blue while one-fourth of the marbles are red.  What is the probability of drawing a green marble?             

We can find the probability of drawing a green marble even if we didn’t know how many marbles there were. 

We do know that the probability of drawing a blue marble is  while the probability of drawing a red marble is   What is the probability of drawing a green marble? 

P(drawing a blue marble) + P(drawing a red marble) + P(drawing a green marble) = 1

P(drawing a blue marble) =  

P (drawing a blue marble) =  

P(drawing a green marble) is unknown, so far. Call it x. Therefore, we have

 

The probability of drawing a green marble is  

Example 3:     The spinner is shown below.  What is the probability of getting prime 

                         numbers on two successive spins? 

 

Prime numbers are 2, 3, 5, and 7.  There are 4 ways to get a prime number and 8 possible outcomes.  The probability of getting a prime number on the first spin is  .  The probability of getting a prime number on the second spin is also  The probability of getting a  prime numbers on two successive spins is  .

 

Try the following problems.

1.     Charlie is doing card tricks with a standard deck of cards.  What is the probability that he will draw the ace of clubs? 

2.     What is the probability that Charlie will draw a diamond from a 52-card deck? 

3.     A couple plans to have 3 children.  Assume that the probability of having a boy   is equal to the probability of having a girl and that each probability =   What is the probability of the couple having a girl followed by two boys? 

4.     What is the probability of rolling an even number on a die followed by drawing the jack of hearts? 

5.     A jar contains 3 kinds of candy: miniature bags of chocolate kisses, peppermints, and caramels.  The odds getting a chocolate kiss are   The odds of getting a peppermint are   What are the odds of getting a caramel? 

6.     A jar contains red marbles, blue marbles, and orange marbles.  If the odds of     getting a red marble are   and the odds of getting a blue marble is   what are the odds of drawing an orange marble? 

7.   On the spinner shown below, what is the probability of getting 2 numbers on 2 spins whose product is less than 8?  Assume that we can use the same number twice.  For example, the pointer can land on 6 on both spins and the product would be 36.  Also assume that the order of numbers the pointer lands on matters.  For example, getting a 1 followed by a 2 is different from getting a 2 followed by a 1. 

 

 

Answers

1. Probability of drawing the ace of clubs = 

     

2. Probability of drawing a diamond = 

   

3.   

Probability of the first child being a girl =  

Probability of second child being a boy =  

Probability of the third child being a boy is  

Multiply the probabilities of the 3 independent events together and get:  

4. Probability of rolling an even number on a die =  .

Probability of drawing the jack of hearts =  

Multiplying the two independent probabilities together to get the probability of both events occurring gives us  

5.   .  Remember, you can subtract the two known probabilities from 1 to find the unknown probability. 

 

6.    Again, you can subtract the known probabilities from 1 and get 

 

7.  

All the possible results are 

1, 1 1,2 1,3 1,4 1,5 1,6 1,7 1,8

2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8

3,1 3,2 3,3 3,4 3,5 3,6 3,7 3,8

4,1 4,2 4,3 4,4 4,5 4,6 4,7 4,8

5,1 5,2 5,3 5,4 5,5 5,6 5,7 5,8

6,1 6,2 6,3 6,4 6,5 6,6 6,7 6,8

7,1 7,2 7,3 7,4 7,5 7,6 7,7 7,8

8,1 8,2 8,3 8,4 8,5 8,6 8,7 8,8

There are 64 possible outcomes.  Of these 64 outcomes, 16 of them have products less than 8. The ratio is  

 

Dependent Probability

• In this section, you will learn how to compute the probabilities of dependent events. 

Dependent probability differs from independent probability in this way: in dependent probability problems, the outcome of a prior event affects the outcome of another event. 

Example 1:  Charlie is doing card tricks.  First, he draws an ace of clubs from a 52-card deck.  If  he doesn’t replace the ace of clubs, what is the probability of getting another club when he draws from the deck the second time? 

Probability of drawing an ace of clubs =   

Probability of getting a second club: this is the tricky part.  If he doesn’t replace the ace of clubs, he has only 12 clubs left in the deck instead of 13.  Also, he has only 51 cards of any kind left in the deck instead of 52.  His probability of drawing a second club is therefore  .

The probability of drawing an ace of clubs followed by another club  = probability of drawing the ace club the first time   the probability of drawing a club the second time.  

Multiplying the two probabilities together, we get  

Generally, you will use rules for dependent probability when something is being removed and NOT replaced. 

Example 2:  A jar contains slips of paper numbered 1 – 10.  What is the probability of drawing two even numbers in a row, assuming the slips of paper are not replaced after being drawn? 

Assume that you draw an even number on the first trial. 

Probability of getting an even number on the first draw:   .

Probability of getting an even number on the second draw:  Assume that you have drawn an even number on the first draw and not replaced it.  You will therefore have four even numbers left to draw from a total of 9 numbers.  The odds of getting an even number the second time are    

The odds of drawing two even numbers are  .

 

Try the following problems.

1.   Nadine is working as a summer camp counselor and has to divide children into teams.   To do this fairly, she goes to a child and draws a number from a box.  If the number is  even, the child is on one team.  If the number is odd, the child is on the other team.  There are 16 numbers in the box.  What are the odds that the first two numbers she draws will both be even numbers? 

2.  Children are allowed to select bags of candy from a jar.  The jar contains 5 bags of chocolate, 3 bags of peppermint, 7 bags of candy corn, and 4 bags of caramels.  What are the odds of getting 2 bags of peppermint in a row?  Hint:  How many bags of candy are in the jar? 

3.   A jar contains 7 dimes, 3 pennies, 4 quarters, and 2 half-dollars.  What are the odds of selecting 1 quarter and then 1 half-dollar? 

4.  Jessica selects cards one at a time from a deck.  What are the odds of her selecting 3 clubs in a row? 

5.  What are the odds of selecting a club followed by a diamond from a deck of cards?  Assume there is no replacement. 

 

Answers

1.   

2.   

3.    

4.     

5.      

 

Geometric Probability

In this section, you will learn how to

• compute probabilities using geometric figures

For example, you will learn to compute the probability of hitting a bull’s eye on a target. 

Geometric probability problems often involve tossing a ball, a die, or a dart and hitting a certain area on the target. Obviously, the probability of, say,  a certain area hitting a target depends on the size of the area you want to hit.  If the area is larger, there is a greater probability of the object landing within the area. If the area is smaller, there is a lesser probability of the object landing within the area.  

Remember that probability is defined as P = 

 

For geometric probabilities, the outcome is usually the area of a geometric figure. 

Therefore, for geometric probability, the probability, 

P =  

Example 1: A circle of radius 1 is inside a square with a side length of 3.  What is the probability of hitting an area inside the circle? 

 

In this case, the number of outcomes is an area.  Since the desired outcome is hitting an area inside the circle, the numerator is the area of the circle, which is A =  .

The denominator is the total number of outcomes.  This is the area of the square, which is 9.

The probability of hitting the circle is therefore .

Example 2: A circle with radius r is inscribed in a square. What is the probability of a dart landing inside the area that lies inside the square but outside the circle? Hint: What is the side length of the square?

 

The side length of the square is twice the radius. Hence the side length is 2r. 

Area of square =  

Area of circle =  

Area of the region lying inside the square but outside the circle =  .

Probability of hitting the region lying inside the square but outside the circle =    

 

Try the following problems.

1. A blindfolded man is tossing darts randomly at a target.  The target consists of concentric circles.  The innermost circle has a radius of 1. The next circle has a radius of 2.  The outer circle has a radius of 3. All of his tosses hit the target.  What is the probability that he will get a dart in the shaded area?

 

2. A blindfolded man tosses a die on the area pictured below. The inner square has a side length of 1.  The outer square has a side length of 4.   What is the probability that the die will land in the shaded area?

 

3.   What is the probability of a randomly tossed ball landing in the shaded area if one assumes that all balls will land on the figure?  The rectangle measures 6 ft by 8 ft while the circle has a radius of 2 feet. 

 

4. If darts are thrown randomly at the target below, what is the probability of a dart hitting the shaded area?  Assume that all darts hit the target. Assume that the circles have radii of 1, 2, and 3.

 

5.  A circle with a radius of 5b is inscribed in a square.  Find the probability of a dart hitting the region that lies inside the square but outside the circle. 

  

Answers

1.  The entire circle has an area of  while the shaded area has an area of  

The probability of hitting the shaded region is  

2.  The area of the entire square is 16.  The area of the unshaded square is 1.  The area of the shaded region is the difference between the two, which is 15.  The probability of hitting the shaded region is  .

3.  The area of the entire rectangle is including the circle is 6   8 = 48. 

The area of the circle is .  The probability of the hitting the shaded region is

   

4.  To find the shaded area, find the area of the second circle, with a radius of 2. To get the area of the shaded region, subtract the area of the innermost circle. The area of the second circle is  The area of the shaded region is therefore   The area of the entire target is the area of    the largest circle, which is   The probability of hitting the shaded area is   

5. 

a. Area of square =  

b. Area of circle =  

c. Area of region inside the square but outside the circle =  

d. Probability of hitting region inside the square but outside the circle =  

 

Proportions

In this section, you will learn how to:

• Find the missing term in a proportional equation

• Set up and solve a proportional equation

• Set up and solve proportional equations using variables instead of numbers

Proportions are two fractions set equal to each other.  

Example 1: If 3 packages of chewing gum cost 75 cents, how much do 7 packages of chewing gum cost? 

To solve this problem using proportions, we need to find two fractions.  Consider the expression “three packages of chewing gum cost 75 cents.”  We could write this as 

 .  The second expression “how much do 7 packages of chewing gum cost?” could be written as  

The equation becomes  = .

It is crucial that the ratio of units be the same in both fractions. In both parts of the equation, the units are   .  We could have had a proportion in which both units were expressed as   We could never have an equation in which one fraction has units of   while the other fraction had a ratio of  .  Cross multiply. The equation becomes 3x = 75(7);  3x = 525 cents.  Dividing both sides by three gives 175 cents or $1.75

The following problem is similar:

Example 2: If 3 cans of apple sauce sell for $1.25, how many cans of applesauce can you buy for $20? 

 

Again, notice the consistency of units in both fractions. 

Cross multiply and solve. 

20(3) = 1.25x

60 = 1.25x

 

x = 48. 

 

Many SAT tests will have at least one problem asking you to solve a proportion problem using variables instead of numbers. 

Example 3: If Charlotta bought 5 CDs for c cents, how much did she pay for 11 CDs?

 

Solve for x by cross-multiplying. 

5x = 11c. 

x =  

This problem can be made more difficult by having a unit conversion.  Imagine the problem above is changed slightly as follows:

Example 4: If Charlotta bought 5 CDs for c cents, how much did she pay in dollars for 11 CDs? 

The easiest way to solve this problem is to solve for x in terms of cents and then convert this figure to dollars by dividing the answer by 100. 

 

5x = 11c

x =  c

Dividing by 100 is the same as multiplying by  , which gives us  c

 

Try the following problems.

1. If it takes 3 ounces of lavender oil mixed with 8 ounces of sesame oil to make a skin cream, how much lavender oil must be used if the manufacturer starts with 56 ounces of sesame oil? 

2. If three gallons of gas cost $5, how much do 12 gallons of gas cost? 

3. If mailing 5 letters to Senegal costs $4.35, how much does it cost to mail 8 letters to Senegal? 

4. If dinner for 4 costs $72, how much does dinner for 10 people cost? 

5. Jeff bought mice to feed his pet snake.  If 3 mice cost c cents, how much did 7 mice cost? 

6. If 8 ipods cost c cents, how much do 12 ipods cost in dollars? 

7. If Charlotta bought 2 tubes of lipstick for d dollars, how much did 5 tubes cost in cents? 

8. If it costs d dollars to buy 1 tube of lipstick, how many dollars does it cost to buy t tubes of lipstick? 

9.   If it takes h hours to complete p pages of homework, how many minutes does it take to complete m pages of homework? 

 

Answers

1.  21 ounces of lavender oil.   .  Cross-multiplying gives 3(56) = 8x; 168 = 8x; x = 21. 

2.  $20.    .   Cross multiplying gives 3x = 60; x = 20.

3. x = $6.96;   .  Cross-multiplying gives 5x = 34.80; x = 6.96

4. $180.   

5.  .  The proportion is set up this way:   ; 3x = 7c. x =  .

6.  .    .  8x = 12c. x =  . This answer is in cents but we need it in dollars. To convert it to dollars we divide by 100, which is the same as multiplying by    Thus,  

7. 250d cents.    To convert to cents, multiply by 100 and get  

8. It costs dt dollars to buy d tubes of lipstick.   . Cross-multiply to get dt = 1x.  Solve for x and get x = dt.  Notice that we don’t have to convert between cents and dollars for this problem because both the given information and the required answer are in dollars. 

9.  It will take   .  Cross multiplying, we get hm = xp.  Solving for x, we get  .  However, this answer is in hours.  We need the answer in minutes.  To convert to minutes, multiply by 60 and get  

 

Similar triangles

If you are working the chapters in order, you have just finished the chapter on proportions. You can cement that knowledge by solving similar triangle problems, which also require the use of proportions. Furthermore, similar triangles play an important role in differential calculus, so if you are taking calculus or planning to take it in the future, this section should help you. 

In this chapter, you will learn to: 

• Recognize similar triangles

• Use proportions to solve for unknown side lengths

• Solve word problems using similar triangles

Congruent triangles have the same size and shape.  Similar triangles, on the other hand, are the same shape but have different sizes. If two triangles are similar, all three pairs of corresponding angles are congruent but the side lengths are different.  In the figure below,  and the two angles are called corresponding angles.  Similarly,   and those two angles are said to be corresponding angles.

How do you know if two triangles are similar?

1. Two pairs of corresponding angles are congruent.  If two pairs of angles are congruent, the third pair of corresponding angles will necessarily be congruent and the triangles will have the same shape.  The two triangles below are similar.  All three pairs of corresponding angles are congruent. 

 

 

Note, we are not told directly that the third pair of angles = 40° but we may infer it because the angles of a triangle must add to 180°.  If the first two angle measures, 80° and 60°, add to 140°, the third pair of corresponding angles must necessarily equal 40°. 

2.  The ratios of corresponding sides of the triangles are equal to the same number.  

 

 

For all pairs of corresponding sides, the ratio is 3; hence, the triangles are similar.  Of course, for different pairs of triangles, the ratios could be something other than 3.  It could be 2 or  or any positive real number. 

Some drawings are used to represent similar triangles. If CD // AB, the triangles in the figure are similar because  by alternate interior angles.  For the same reason,  .  Because two pairs of corresponding angles are congruent, the triangles are similar. Note that  are vertical angles and therefore congruent.

 

In the diagram above, .  (Note: in this context, the symbol   indicates similarity.  

Sometimes, similar triangles can be drawn inside of each other. 

 

If ED // AB, the  because   and by corresponding angles. 

A common SAT question gives students a pair of similar triangles and asks them to find the length of an unknown side by setting up a proportion and solving it. 

Example 1:  Given the following pair of similar triangles, what is the value of x? 

 

To solve for x, set up a proportion. 

 .

Cross multiply:

3(4) = 2x

12 = 2x

6 = x. 

Example 2:  Solve for x.  Find the lengths of AE and EC. 

 

Set up a proportion and solve. 

 

4x = 3(x + 2)

4x = 3x + 6

x = 6

AE = x + 2 = 6 + 2 = 8.

EC = x = 6. 

Example 3:  

 

With these kinds of similar triangles, it is important to understand exactly what the corresponding sides are.  If it is assumed that AB // DC, the pairs of corresponding angles are   Also,   and  ADC are similar.  Thus, EB and EC are corresponding sides as are EA and ED.  In general, corresponding sides will be parts of the same line segment. 

Set up the proportion.  Note how the proportion is set up.  The side of the triangle on the left is in the numerator while the corresponding side of the triangle on the right is in the denominator. 

 

4(2) = 3x

8 = 3x

 

Example 4: A variation on similar triangle problems is shown in the figure below. 

 

In this problem, the entire side CB has a length of 10.  We don’t know the lengths of either CF or FB.  We only know that they add to 10.  In this case, we must use a slightly different technique.  

We must find the entire length of AC, which is 8. 

 

Notice that we added the two pieces of the left side and put that sum, 8, in the denominator.  We now have a proportion that is  

Cross multiply:5(10) = 8x.  

50 = 8x

 

6.25 = x. 

 

Similar triangles are often used to solve problems involving shadows, blueprints, maps, and scale models. 

Example 5: If a man who is 5’8” tall casts a 7-foot shadow, how long is the shadow cast by a 6-foot tall man? 

While it is not essential to draw similar triangles to solve this problem, doing so will help you see what is going on. 

 

Why is a shadow problem solved using similar triangles?  Because two pairs of corresponding angles are congruent.  The two right angles formed by each person and the ground are congruent.  The angles formed by the person and the imaginary line connecting each person’s head with the end of the shadow are also congruent.  Because triangles with two pairs of corresponding, congruent angles are similar triangles, we can use methods involving similar triangles. 

Notice that we have different units in this problem.  Some of the measurements are given in feet while one measurement is given in a measure of feet and inches.  Proportions must have the same units.  Therefore, convert everything to inches. 

5 ft 8 inches = 68 inches. To convert 5 feet to inches, multiply by 12, the number   of inches in a foot. The product is 60 inches. Add the 8 inches for a sum of 68. 

6 feet = 72 inches 

 

68x = 72(72)

68x = 5184

x = 76.24 in. which equals approximately 6 feet and 4.24 inches. 

 

Try the following problems.  

1.  In the figure below, ED // AB. Solve for x. 

 

Note: Figure not drawn to scale. 

2. Using the figure from Problem 1 above, find the length of CD. 

3. Again using the figure from Problem 1, find the length of BC.

4. If a 12-foot flagpole casts a 15-foot tall shadow, how long is the shadow cast by a 16-foot flag pole? 

5. Given that AB // CD, Find the length of x in the figure below. 

 

 

6. Find the length of x.  Note that CB is equal to 12. 

 

7. Charlie stands 6’4” tall.  At a certain time of day, he casts an 8-foot shadow.  If his sister is 5’8” tall, how long is her shadow?   

8.  Given the similar triangles below, find the length of x. 

  

Answers

1. x = 10

Note that the fact that ED and AB are parallel tells us that the two triangles are similar because the corresponding angles will be similar. 

 

3x = 2x + 10. 

 

2. 15

x + 5 = 10 + 5 = 15

3. 25

The sum of the two sides is 10 + 15 = 25. 

4. 20 feet 

 

5. 2.4 units 1

 

 

 

6.  

To solve this equation, find the value of the left side of the larger triangle by adding the pieces 4 and 5 to get 9.  Set up a proportion using the values 9 and 12 in the denominators because these are the lengths of the entire sides. 

   

7. The shadow is 7 feet, 2 inches long. 

Convert the feet and inches to inches only to keep the units the same. A height of 6’4” is 76 inches while a height of 5’8” is 68 inches.  Similarly, 8 feet is equal to 96 inches. Setting up a proportion, we get  .  Cross-multiplying gives 76x = 68(96);

 76x = 6528; rounding to the nearest inch, x = 86 inches.  The shadow is 7 feet, 2 inches long. 

8. x = 1.5