SAT Mc Graw Hill 2011

760 MCGRAW-HILL’S SAT

Section 5

1.B There are 180°on the side of a line.
2 x+ 3 x= 180 °
Combine like terms: 5 x= 180 °
Divide by 5: x= 36 °
Multiply by 2: 2 x= 72 °
(Chapter 10, Lesson 1: Lines and Angles)

2.B The equation states that some number, when
squared, equals 36. That number can be either 6 or –6.
Taking the square root of both sides of the equation
gives:
x− 4 =±6
Add 4: x=10 or − 2
Therefore, the answer is (B) −2.
(Chapter 8, Lesson 1: Solving Equations)

3.C There are 180°in a triangle. Set up equations
for the two triangles in the figure.
a+b+ 52 = 180
Subtract 52: a+b= 128
c+d+ 52 = 180
Subtract 52: c+d= 128
a+b+c+d=
Substitute: 128 + 128 = 256
(Chapter 10, Lesson 2: Triangles)

4.B f(x) =x^2 − 4
Set f(x) equal to 32: x^2 − 4 = 32
Add 4: x^2 = 36
Take positive square root: x= 6
(Chapter 11, Lesson 2: Functions)

5.C The ratio of the nuts is a part-to-part-to-part-
to-part ratio. Adding these numbers gives the total
number of parts: 2 + 4 + 5 + 7 =18. Since four of these
parts are almonds, the fraction of the mixture that is
almonds is 4/18, or 2/9.
(Chapter 7, Lesson 4: Ratios and Proportions)

6.D If 20 students scored an average of 75 points,
then the sum of their scores is 20 × 75 =1,500 total points.

If 12 of those students scored an average of 83 points,
then the sum of their scores is 12 × 83 =996 points.

Therefore, the remaining 8 students scored 1,500 −
996 =504 points altogether, so their average score is
504 ÷ 8 =63 points.

(Chapter 9, Lesson 2: Mean/Median/Mode Problems)

7.A The sides of square EFGHall have length

. A diagonal of this square can be found with the

Pythagorean theorem: (8 )^2 +(8 )^2 =EG

— 2

.

Simplify: 128 + 128 =EG

— 2

256 =EG

— 2

Take square root: 16 =EG

—

2 2

82

(Or, more simply, you can remember that the length of

the diagonal of a 45°-45°-90°triangle is the length of

the side times. So the diagonal is .)

By the same reasoning, since the sides of square

ABCDall have length 14 : AD

—

= 14 ×=28.

Notice that AC

—

= AE

—

+ EG

—

+ CG

—

; therefore,

28 =AE

—

+ 16 +CG

—

, so AE

—

+CG

—

=12. By the same rea-

soning, BF

—

+DH

—

=12, so AE

—

+BF

—

+CG

—

+DH

—

=24.

(Chapter 10, Lesson 3: The Pythagorean Theorem)

8.D Although you were probably taught to add the

“rightmost” digits first, here the “leftmost” digits pro-

vide more information about the number, so it’s best

to start there.

RS

+SR

TR 4

The largest possible 3-digit number that can be

formed by adding two 2-digit numbers is 99 + 99 =

198. Therefore, Tmust be 1.
     RS
     +SR
     1 R 4
     Therefore, there must be a “carry” of 1 from the ad-
     dition of R+Sin the 10s column. Looking at the units
     column tells us that S+Ryields a units digit of 4, so
     S+R=14. The addition in the 10s column tells us that
     R+S+ 1 =R+10. (The “+10” is needed for the carry
     into the 100s column.)
     R+S+ 1 =R+ 10
     Substitute R+S=14: 14 + 1 =R+ 10
     Subtract 10: 5 =R
     So 2R+T=2(5) + 1 =11.
     (Chapter 9, Lesson 3: Numerical Reasoning Problems)


199. 1 n=

If it helps, you can think of this as f(n) =.

Find the value of (f(4))^2

Plug in 4 for n: f(4) =

Plug in 1 for f(4): ((f(4))^2 =(1)^2 = 1

(Chapter 9, Lesson 1: New Symbol or Term Problems)

4

16

16

16

1

2

==

n^2

16

n^2

16

(^222)

2 82 2 16×=

14

DC

AB

EF

H G

2

66

6 6

8

8

8

8

14 2

14 2 82 82 14 2

82

82