SAT Mc Graw Hill 2011

756 MCGRAW-HILL’S SAT

13.B The population doubles every 18 months.
Start with January of 2000 and start doubling.

January 2000 12,000
18 months later: July 2001 24,000
18 months later: January 2003 48,000
18 months later: July 2004 96,000
(Chapter 9, Lesson 3: Numerical Reasoning Problems)

14.C Use the Fundamental Counting Principle from
Chapter 9, Lesson 5. To arrange these students,
five choices must be made. First select the students for
each end. Since one of the five (the tallest) cannot go on
either end, you have four students to choose from for
one end, and then, once that choice has been made,
three students to choose from for the other end:

4 3
**
Now fill the remaining spots. There are three students
left to choose from for the second spot:

4 3 3
**
Then, once that selection has been made, there are
two for the next spot, then one for the remaining spot:

4 3 2 1 3

To find the total number of possible arrangements,
simply multiply: 4 × 3 × 2 × 1 × 3 =72.
(Chapter 9, Lesson 5: Counting Problems)

15.B From the diagram, we know that a+b+c=
180, and we know that b=c+3.

If you want bto be as large as possible, then you need
to make the sum of aand cas small as possible. The
smallest integer value of apossible is 91. So let’s say
that a=91.
Substitute 91 for a: 91 +b+c= 180
Substitute c+3 for b: 91 +c+ 3 +c= 180
Combine like terms: 94 + 2 c= 180
Subtract 94: 2 c= 86
Divide by 2: c= 43
So 43 is the largest possible value of c; this means that
43 + 3 =46 is the largest possible value of b.
(Chapter 10, Lesson 2: Triangles)

16.B Begin by finding the area
of the big equilateral triangle.
An equilateral triangle with
sides of length 4 has a height of
, because the height divides the
triangle into two 30°-60°-90°triangles.

23

Area =^1 ⁄ 2 (base)(height) =^1 ⁄ 2 (4)( ) =

The big triangle is divided into four equal parts, three

of which are shaded, so the shaded area is^3 ⁄ 4 of the

total area.

Shaded area =^3 ⁄ 4 (4 ) =

(Chapter 10, Lesson 5: Areas and Perimeters)

17.D Just look at the graph and draw a line at y=1.

The y-values of the graph are at or above that line

from x= −4 to x= −2 and from x=2 to x=4.

(Chapter 11, Lesson 2: Functions)

18.C This table shows all of the 5 ×5 = 25 possible

values of ab:

Of those, only the seven shaded values are greater

than 20 and less than 50, so the probability is 7/25.

(Chapter 9, Lesson 6: Probability Problems)

19.E (wa)(w^5 ) = w^15

Simplify: w5 + a= w^15

Equate the exponents: 5 + a= 15

Subtract 5: a= 10

(w^4 )b= w^12

Simplify: w^4 b= w^12

Equate the exponents: 4 b= 12

Divide by 4: b= 3

So a+ b= 10 + 3 = 13.

(Chapter 8, Lesson 3: Working with Exponentials)

20.D The graph of y= f(x– 2) is the graph of y= f(x)

shifted to the right two units without changing its

shape. Therefore, the “peak” at point (6, 4) should shift

to (8, 4).

(Chapter 11, Lesson 3: Transformations)

3 33

23 43

23

4

2

60 °

30 ° 30 °

60 °

2

6

4

2

-2

-2 2 4 6 x

y = 1

y

-4

x 1 3 5 7 9

2

2

6

10

14

18

4

4

12

20

28

36

6

6

18

30

42

54

8

8

24

40

56

72

10

10

30

50

70

90