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 =
- 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) - 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