New SAT Math Workbook Episode 2 part 8 pps

B Substitute the variable expression provided in each answer choice, in turn, for x in the function, and you’ll find that only choice B provides an expression that transforms the functio

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Additional Geometry Topics, Data Analysis, and Probability 305

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8 (B) Substitute the variable expression

provided in each answer choice, in turn, for x in

the function, and you’ll find that only choice

(B) provides an expression that transforms the

function into one whose graph matches the one

in the figure: f(x + 1) = x + 1 To confirm that

the line in the figure is in fact the graph of y = x

+ 1, substitute the two (x,y) pairs plotted along

the line for x and y in the equation The

equation holds for both pairs: (3) = (2) + 1; (–2)

= (–3) + 1

9 The correct answer is 25 The total number of

bowlers in the league is 240, which is the total

of the six numbers in the frequency column

The number of bowlers whose averages fall

within the interval 161–200 is 60 (37 + 23)

These 60 bowlers account for 60

240, or 25%, of the league’s bowlers

10 (C) In the scatter plot, B and D are further to

the left and further up than all of the other three

points (A, C, and E), which means that City B

and City D receive less rainfall but higher

temperatures than any of the other three cities

Statement (C) provides an accurate general

statement, based on this information

11 (D) Of six marbles altogether, two are blue

Hence, the chances of drawing a blue marble

are 2 in 6, or 1 in 3, which can be expressed as

the fraction 1

3

12 The correct answer is 3/16 Given that the ratio

of the large circle’s area to the small circle’s

area is 2:1, the small circle must comprise 50%

of the total area of the large circle The shaded

areas comprise 38 the area of the small circle,

and so the probability of randomly selecting a

point in one of these three regions is

3

8

1

2

3

16

× = .

Exercise 1

1 (B) Since the figure shows a 45°-45°-90°

triangle in which the length of one leg is known, you can easily apply either the sine or cosine function to determine the length of the hypotenuse Applying the function sin45° =

2

2 , set the value of this function equal to

x

7

opposite hypotenuse







, then solve for x:

2

2 =x ; 2x= 2 ;x=

7 2

2

2 (C) Since the figure shows a 30°-60°-90°

triangle, you can easily apply either the sine or the cosine function to determine the length of the hypotenuse Applying the function cos60° =

1

2, set the value of this function equal to

x

10

adjacent hypotenuse







1

2 = x

10; 2x

= 10 ; x = 5.

3 (B) The question describes the following 30°-60°-90° triangle:

Since the length of one leg is known, you can easily apply the tangent function to determine

the length of the other leg (x) Applying the

function tan30° = 3

3 , set the value of this function equal to 6x oppositeadjacent





, then solve for

x: 3

3 =x ; x= ;x=

6 3 6 3 2 3

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Chapter 16

306

4 (D) The area of a triangle = 1

2 × base × height Since the figure shows a 30°-60°-90°

triangle with base 3, you can easily apply the

tangent function to determine the height (the

vertical leg) Applying the function tan60° =

3, let x equal the triangle’s height, set the

value of this function equal to x3 oppositeadjacent





,

then solve for x: 3

1 =x ;x=

3 3 3 Now you can determine the triangle’s area:

1

2 3 3

9

2 3

9 3 2

× × 3 = , or

5 (A) The two tracks form the legs of a right

triangle, the hypotenuse of which is the shortest

distance between the trains Since the trains

traveled at the same speed, the triangle’s two

legs are congruent (equal in length), giving us a

1:1: 2 with angles 45°, 45°, and 90° , as the

next figure shows:

To answer the question, you can solve for the

length of either leg (x) by applying either the

sine or cosine function Applying the function

cos45° = 22, set the value of this function

equal to 2

x

adjacent hypotenuse







2

2 = x ; 2x= 2 ;x= 2

asks for an approximate distance in miles.

Using 1.4 as the approximate value of 2: x ≈

(35)(1.4) = 49

Exercise 2

1 The correct answer is 1 AB is tangent to PO; therefore, AB⊥PO Since the pentagon is

regular (all sides are congruent), P bisects AB Given that the perimeter of the pentagon is 10,

the length of each side is 2, and hence AP = 1.

2 (D) Since AC is tangent to the circle, m∠OBC

= 90° ∠BCO is supplementary to the 140° angle

shown; thus, m∠BCO = 40° and, accordingly,

m ∠BOE = 50° Since ∠BOE and ∠DOE are supplementary, m∠DOE = 130° (This angle measure defines the measure of minor arc DE.)

3 (B) Since AB is tangent to circle O at C, you can draw a radius of length 6 from O to C ,

forming two congruent 45°-45°-90° triangles (∆ACO and ∆BCO), each with sides in the ratio 1:1: 2 Thus, OA = OB = 6 2, and AC = CB

= 6 The perimeter of ∆ABO = 12 + 6 2 +

6 2 = 12 + 12 2

4 (D) To find the circle’s area, you must first

find its radius Draw a radius from O to any of

the three points of tangency, and then construct

a right triangle—for example, ∆ABC in the

following figure:

Since m∠BAC = 60°, m∠OAD = 30°, and

∆AOD is a 1: 3:2 triangle Given that the perimeter of ∆ABC is 18, AD = 3 Letting

x = OD: 3 3

3

x= ; x= ;x= , or The circle’s radius = 3 Hence, its area =

π 3 3 π 2

( ) =

5 (A) Since AC is tangent to the circle,

AC⊥BC Accordingly, ∆ABC is a right

triangle, and m∠B = 50° Similarly, AB⊥DO,

∆DBO is a right triangle, and m∠DOB = 40°.

∠DOC (the angle in question) is supplementary

to ∠DOB Thus, m∠DOB = 140° (x = 140.)

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Exercise 3

1 (C) For all values of x, y = –92 Thus, the

equation describes a horizontal line with

y-intercept –9

2 The slope of a horizontal line is

0 (zero)

2 (A) Apply the formula for determining a

line’s slope (m):

x x

= −

− = − −− − = − = −

2 1

6 4

3 1

10 4

5 2 ( )

3 (B) In the general equation y = mx + b, the

slope (m) is given as 3 To determine b,

substitute –3 for x and 3 for y, then solve for b:

3 = 3(–3) + b; 12 = b.

4 (D) Line P slopes upward from left to right at

an angle less than 45° Thus, the line’s slope (m

in the equation y = mx + b) is a positive fraction

less than 1 Also, line P crosses the y-axis at a

positive y-value (above the x-axis) Thus, the

line’s y-intercept (b in the equation y = mx + b)

is positive Only choice (D) provides an

equation that meets both conditions

5 (E) First, find the midpoint of the line

segment, which is where it intersects its

perpendicular bisector The midpoint’s

x-coordinate is 4 3

2

1 2

− = , and its y-coordinate is

− + = 2 5

2

3

2 Next, determine the slope of the

line segment: 5 2

3 4

7

7 1

− −

− − =− = −

( )

Since the slope of the line segment is –1, the slope of its

perpendicular bisector is 1 Plug (x,y) pair

( , ) 1 3

2 and slope (m) 1 into the standard form of

the equation for a line (y = mx + b), then solve

for b (the y-intercept):

3

2 1

1

2

1

= 

 +

=

( ) b

b

You now know the equation of the line:

y = x + 1.

Exercise 4

1 The correct answer is 4 For every value of x, f(x) is the corresponding y-value By visual inspection, you can see that the maximum

y-value is 4 and that the graph attains this y-value twice, at (–8,4) and (4,4) Similarly, the

minimum value of y is –4 and the graph attains

this value twice, at (–4,–4) and (8,–4); in both

instances, the absolute value of y is 4 Thus, the absolute value of y is at its maximum at four different x-values.

2 (E) The figure shows the graph of y = 2 For any real number x, f(x) = 2 Thus, regardless of what number is added to or subtracted from x,

the result is still a number whose function is 2

(y = 2).

3 (D) To determine the features of the transformed line, substitute x− 2

2 for x in the

function:

f x( )− 2 =( )x− − = − − = −x x

2

The correct figure should show the graph of the

equation y = x – 4 Choice (D) shows the graph

of a line with slope 1 and y-intercept –4, which

matches the features of this equation No other answer choice provides a graph with both these features

4 (D) Substitute (x+ 1) for x in the function:

f(x + 1) = [(x + 1) – 1]2 + 1 = x2 + 1

In the xy-plane, the equation of f(x + 1) is y = x2

+ 1 To find the y-intercept of this equation’s graph, let x = 0, then solve for y:

y = (0)2 + 1 = 1

5 (A) The graph of x = –(y2) is a parabola opening to the left with vertex at the origin

(0,0) The function f(y) = –(y2 + 1) is equivalent

to f(y) = –y2 – 1, the graph of which is the graph

of x = –(y2), except translated one unit to the

left, as the figure shows [Since (–y)2 = y2 for

any real number y, substituting –y for y in the function f(y) = –(y2 + 1) does not transform the

function in any way.]

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Chapter 16

308

Exercise 5

1 (D) Division D’s income accounted for 30%

of $1,560,000, or $468,000 Income from

Division C was 20% of $1,560,000, or

$312,000 To answer the question, subtract:

$468,000 – $312,000 = $156,000

2 (A) Visual inspection reveals that the

aggregate amount awarded in 1995 exceeded

that of any of the other 3 years shown During

that year, minority awards totaled

approximately $730,000 and non-minority

awards totaled approximately $600,000 The

difference between the two amounts is

$130,000

3 (E) The two greatest two-month percent

increases for City X were from 1/1 to 3/1 and

from 5/1 to 7/1 Although the temperature

increased by a greater amount during the latter

period, the percent increase was greater from 1/

1 to 3/1:

January–March: from 30 degrees to 50 degrees,

a 66% increase

May–July: from 60 degrees to 90 degrees, a

50% increase

During the period from 1/1 to 3/1, the highest

daily temperature for City Y shown on the

chart is appoximately 66 degrees

4 (A) To answer the question, you can add

together the “rise” (vertical distance) and the

“run” (horizontal distance) from point O to

each of the five lettered points (A–E) The

shortest combined length represents the

fastest combined (total) race time Or, you

can draw a line segment from point O to each

of the five points—the shortest segment

indicating the fastest combined time As you

can see, OA is the shortest segment, showing

that cyclist A finished the two races in the

fastest combined time

5 (C) You can approximate the (race 1):(race 2) time ratio for the ten cyclists as a group by

drawing a ray extending from point O through

the “middle” of the cluster of points—as nearly

as possible Each of the five answer choices suggests a distinct slope for the ray Choice (C) suggests a ray with slope 1 (a 45° angle), which does in fact appear to extend through the middle of the points:

(Although six points are located above the ray, while only four are located below the ray, the ones below the ray, as a group, are further from the ray; so the overall distribution of values is fairly balanced above versus below the ray.) Any ray with a significantly flatter slope (answer choice A or B) or steeper slope (answer choice D or E) would not extend through the “middle” of the ten points and therefore would not indicate an accurate average (race 1):(race 2) ratio

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Exercise 6

1 (D) There are two ways among five possible

occurrences that a cherry candy will be

selected Thus, the probability of selecting a

cherry candy is 2

5

2 (B) In each set are three distinct member

pairs Thus the probability of selecting any pair

is one in three, or 13 Accordingly, the

probability of selecting fruit and salad from the

appetizer menu along with squash and peas

from the vegetable menu is 1

3

1 3

1 9

× =

3 (E) You must first calculate the chances of

picking the same student twice, by multiplying

together the two individual probabilities for the

student: 1

7

1 7

1 49

× = The probability of picking

the same student twice, added to the probability

of not picking the same student twice, equals 1

So to answer the question, subtract 1

49 from 1

4 (C) Let x = the number of quarters in the

bank (this is the numerator of the probability

formula’s fraction), and let x + 72 = the total

number of coins (the fraction’s denominator)

Solve for x:

1

72 4

72 3

24

=

+

+ =

=

x

5 The correct answer is 1/6 Given that the ratio

of the large circle’s area to the small circle’s

area is 3:1, the area of the “ring” must be twice

that of the small circle Hence the probability of

randomly selecting a point in the outer ring is

2

3 The shaded area accounts for 14 of the ring,

and so the probability of selecting a point in the

shaded area is 2

3

1 4

2 12

1 6

× = =

Retest

1 (E) Since the figure shows a 45°-45°-90°

triangle in which the length of the hypotenuse

is known, you can easily apply either the sine

or cosine function to determine the length of either leg Applying the function cos45° = 2

2

x

adjacent hypotenuse







2

2 = x2 ; 2x=( )( )2 2 ; 2x= ;x=

2 (C) The two flight paths form the legs of a right triangle, the hypotenuse of which is the shortest distance between the trains (40 miles)

As the next figure shows, a 120° turn to either the left or right allows for two scenarios (point

T is the terminal):

As the figures show, the two flight paths, along with a line segment connecting the two planes, form a 30°-60°-90° triangle with sides in the ratio 1: 3:2 To answer the question, solve for

the length of the longer leg (x), which is

opposite the 60° angle One way to solve for x

is by applying either the sine or cosine function Applying the function sin60° = 3

x

40

opposite hypotenuse







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Chapter 16

310

3

2 = x ; 2x= 3 ;x= 3

asks for an approximate distance in miles.

Using 1.7 as the approximate value of 3: x ≈

(20)(1.7) = 34

3 (D) The entire area between the two circles is

the area of the larger minus the area of the

smaller Letting that area equal A:

r

= ( ) −

=

π

2 4 3

2 2

2 2 2

Drawing a line segment from C to O forms two

right triangles, each with hypotenuse 2r Since

OC = r, by the Pythagorean Theorem, the ratios

among the triangle’s sides are 1: 3:2, with

corresponding angle ratios 90°:60°:30° ∠A and

∠B each = 30° Accordingly, interior ∠AOB

measures 120°, or one third the degree measure

of the circle Hence, the area of the shaded

region is two thirds of area A and must equal

2πr2

4 (E) The line shows a negative y-intercept (the

point where the line crosses the vertical axis)

and a negative slope less than –1 (that is,

slightly more horizontal than a 45° angle) In

equation (E), −2

3 is the slope and –3 is the

y-intercept Thus, equation (E) matches the graph

of the line

5 (B) Points (5,–2) and (–3,3) are two points on

line b The slope of b is the change in the

y-coordinates divided by the corresponding

change in the x-coordinate:

m b= − −

− − =− −

3 2

3 5

5 8

( )

, or

6 (E) Put the equation given in the question

into the form y = mx + b:

2 3

x y

− =

= −

The line’s slope (m) is 2 Accordingly, the

slope of a line perpendicular to this line is –1

2

Given a y-intercept of 3, the equation of the

perpendicular line is y = –1

2x + 3 Reworking

this equation to match the form of the answer

choices yields 2y + x = 6.

7 (D) The figure shows the graph of y = 2x,

whose slope (2) is twice the negative reciprocal

of − 1

2, which is the slope of the graph of f(x) =

− 1

2x You obtain this slope by substituting –4x for x in the function: f(–4x) = − 1

2(–4x) = 2x.

8 (E) Substitute the variable expression given

in each answer choice, in turn, for x in the function f(x) = –2x2 + 2 Substituting − x

2

(given in choice E) for x yields the equation

y= −x2 +

4 2:

x

−

( )= −( ) ( )− + = −( )

 + =

2

8 2

2 ,, or −x2 +

4 2

The graph of y= −x2

4 is a downward opening parabola with vertex at the origin (0,0) The figure shows the graph of that equation, except translated 2 units up To confirm that (E) is the

correct choice, substitute the (x,y) pairs (–4,–2) and (4,–2), which are shown in the graph, for x and y in the equation y= −x2+

4 2, and you’ll find that the equation holds for both value pairs

9 The correct answer is 6 By multiplying the number of chickens by the number of eggs they lay per week, then adding together the

products, you can find the number of eggs laid

by chickens laying 9 or fewer eggs per week: (2)(9) + (4)(8) + (5)(7) + (3)(6) + (2)(5) + (0)(4) + (2)(3) = 119 eggs

To find the number of chickens that laid 10 eggs during the week, subtract 119 from 179 (the total number of eggs): 179 – 119 = 60 Then divide 60 by 10 to get 6 chickens

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10 (E) For each year, visually compare the

difference in height between Country X’s white

bar and Country Y’s dark bar (For each year,

the left-hand bars represent data for Country X,

while the right-hand bars represent data for

Country Y.) A quick inspection reveals that

only for the year 1990 is Country Y’s dark bar

approximately twice the height of Country X’s

white bar Although you don’t need to

determine dollar amounts, during 1990,

Country Y’s imports totaled about $55 million,

while Country X’s exports totaled about $28

million

11 (D) Regardless of the number of marbles in

the bag, the red : blue : green marble ratio is

4:2:1 As you can see, blue marbles account for

2

7 of the total number of marbles Thus, the

probability of picking a blue marble is 2

7

12 (A) The probability that the left-hand die will

NOT show a solid face is 3 in 6, or 12 The

probability that the right-hand die will NOT

show a solid face is 2 in 6, or 1

3 To calculate the combined probability of these two

independent events occurring, multiply:

1

2

1

3

1

6

× =

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17

Practice Tests

PRACTICE TEST A

Answer Sheet

Directions: For each question, darken the oval that corresponds to your answer choice Mark only one

oval for each question If you change your mind, erase your answer completely.

Section 1

Section 2

Note: Only the answers entered on the grid are scored Handwritten answers at the top of the column are not scored.

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