SAT Mc Graw Hill 2011

CHAPTER 16 / PRACTICE TEST 3 755

Section 2

1.A (x+4)+ 7 = 14
Subtract 7: x+ 4 = 7
Subtract 4: x= 3
(Chapter 8, Lesson 1: Solving Equations)

2.B Write out a mathematical equation for how
you would actually find the cost for the month: $.95
×31. Answer choice B, $1.00 ×30, is closest to that
amount.
(Chapter 7, Lesson 1: Numbers and Operations)

3.D A linear angle measures 180°. Write an
equation:
w+x+ 50 = 180 °
Subtract 50°: w+x= 130 °
(Chapter 10, Lesson 1: Lines and Angles)

4.D g(x)= 3 x+ 4

Substitute 5 for x: g(5)=3(5)+ 4

Simplify: g(5)= 15 + 4 = 19

(Chapter 11, Lesson 2: Functions)

5.D The difference between xandyis (x−y).

The sum of xandyis (x+y).

The product of those two is equal to 18:

(x−y)(x+y)= 18

FOIL: x^2 −xy+xy−y^2 = 18

Combine like terms: x^2 −y^2 = 18

(Chapter 8, Lesson 5: Factoring)

6.E

Add 7:
Divide by 3:
Square both sides: x= 81
(Chapter 8, Lesson 4: Working with Roots)

7.C Letb=cost of chocolate bar and g=cost of gum.

b+g=$1.75

Chocolate bar is $.25 more: b=$.25+g

Substitute for b: $.25 +g+g= $1.75

Combine like terms: $.25 + 2 g= $1.75

Subtract $.25: 2 g= $1.50

Divide by 2: g= $.75

(Chapter 8, Lesson 1: Solving Equations)

8.C First find 40% of 80: .40 ×80 = 32

Now find what percent of 96 is 32.

Translate:

Multiply by 100: 96 x=3,200

Divide by 96: x= 331 ⁄ 3

(Chapter 7, Lesson 5: Percents)

x

100

×=96 32

x= 9

327 x=

3720 x−=

9.A Iflm=21 and both landmare integers, then

mmust be either 1, 3, 7, or 21. If mn=39, however,

thenmmust also be a factor of 39, so it must be 3.

Therefore,l=21/3=7 and n=39/3=13, so n>l>m.

(Chapter 8, Lesson 6: Inequalities, Absolute Values,

and Plugging In)

10.D There’s no need to do a lot of calculation here.

Look for the two adjacent bars with the greatest pos-

itive difference between them. Since 1999 shows the

least profits of all the years on the graph and 2000

shows the greatest profits of any year on the graph,

1999–2000 must have the greatest change in profit.

(Chapter 11, Lesson 5: Data Analysis)

11.B A Venn diagram can help you with this prob-

lem: Imagine that the

4 students who play

two sports play soccer

and tennis. (It doesn’t

matter which specific

pair of sports they

play.) This means that

12 − 4 =8 students play

just soccer, 7 − 4 =3 students

play just tennis, and

9 students play just lacrosse.

This shows that there is a total

of 9 + 8 + 4 + 3 =24 students.

(Chapter 9, Lesson 5: Counting Problems)

12.B To solve this problem,

you need to find the distance

between the center of the

circle (14, 14) and the point

on the circle (2, 9). To do this,

you can use the distance

formula.

You can also draw a right triangle

connecting the two points.

It gives you a triangle with

one leg of 5 and one leg of 12.

Set up the Pythagorean

theorem and solve for r.

52 + 122 =r^2

Simplify: 25 + 144 =r^2

Combine like terms: 169 =r^2

Take square root: 13 =r

The diameter is twice the radius =2(r)=2(13)=26.

(Chapter 10, Lesson 3: The Pythagorean Theorem)

(14,14)

(2,9)

r 5

12

(14,14)

(2,9)

r 5

12

(^843)

9

0

0

0

lacrosse

soccer tennis