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1074<br />

1 12 Practice <strong>Tests</strong> <strong>for</strong> <strong>the</strong> <strong>SAT</strong><br />

Practice Test Twelve Answers and Explanations<br />

15. D<br />

Difficulty: Medium<br />

Strategic Advice: First find how many graduates went on<br />

to a four-year college in 1985. Looking at <strong>the</strong> second graph,<br />

you can see that 45 percent of <strong>the</strong> 700 graduates went to<br />

a four-year college; that is<br />

1 0 x 700 = 315 graduates.<br />

In 1975, 380/o of <strong>the</strong> 600 graduates went to a four-year<br />

college; that is,<br />

1 3 0 8 0 x 600 = 228 graduates.<br />

Getting to <strong>the</strong> Answer: The difference is 3 15 - 228 = 87<br />

graduates.<br />

16. c<br />

Difficulty: High<br />

Strategic Advice: Don't let problems like this slow you<br />

down. If you can't figure it out at first glance, mark it to come<br />

back to if you have time at <strong>the</strong> end.<br />

Getting to <strong>the</strong> Answer:<br />

Since we are adding <strong>the</strong> integers toge<strong>the</strong>r, <strong>the</strong> negative<br />

integers will negate <strong>the</strong> value of <strong>the</strong> positive integers up<br />

through 15. So we start with 16 to find numbers that will<br />

total 51. We only need to add 16 + 17 + 18 to get 51, so<br />

our answer is 18.<br />

17. D<br />

Difficulty: High<br />

Strategic Advice: If <strong>the</strong> remainder is 2, what number does x<br />

always divide into?<br />

Getting to <strong>the</strong> Answer:<br />

Because 20 divided by x has a remainder of 2, x divides by<br />

(or is a factor of) 20 - 2 = 18.<br />

The factors of 18 are 1, 2, 3, 6, 9, and 18.<br />

But if you divide 20 by 1 or 2, <strong>the</strong>re's no remainder. Cross<br />

<strong>the</strong>se off <strong>the</strong> list, leaving 3, 6, 9, and 18, so <strong>the</strong> answer is<br />

(D).<br />

18. A<br />

Difficulty: High<br />

Strategic Advice: Add <strong>the</strong> in<strong>for</strong>mation in <strong>the</strong> question stem<br />

to <strong>the</strong> diagram. What can you infer from <strong>the</strong> area of ADGK ?<br />

What does it mean that <strong>the</strong> triangles are isosceles?<br />

Getting to <strong>the</strong> Answer:<br />

Since <strong>the</strong> area of square ADGK is 9, each side of <strong>the</strong> square<br />

is 3 units long. Since AB = BC = CD, AB = BC = CO = 1 .<br />

Each of <strong>the</strong> triangles is a right isosceles triangle, so <strong>the</strong> ratio<br />

between <strong>the</strong> sides of each triangle is x:x:x\1'2. Add this<br />

in<strong>for</strong>mation to <strong>the</strong> diagram:<br />

A<br />

x<br />

_ 1_<br />

V2<br />

D<br />

c V2 E<br />

_ 1_<br />

\/2 1 L \/2 J<br />

w 1<br />

K<br />

_1 _<br />

V2<br />

F<br />

V2<br />

H<br />

1<br />

V2<br />

y<br />

1<br />

V2<br />

_1 _<br />

V2<br />

As you can see, <strong>the</strong> total area of <strong>the</strong> figure is <strong>the</strong> area of square<br />

ADGK plus <strong>the</strong> area of <strong>the</strong> four triangles with legs of ·<br />

Each of <strong>the</strong>se triangles has an area of -H )( ) = t• so<br />

<strong>the</strong> total area of <strong>the</strong> figure is 9 + 4( t) = 10.<br />

19. c<br />

Difficulty: High<br />

Strategic Advice: Don't let tough algebra problems<br />

intimidate you. Often, getting <strong>the</strong> problem set up is <strong>the</strong><br />

most difficult part. Then solving <strong>the</strong> problem goes quickly.<br />

Getting to <strong>the</strong> Answer:<br />

You know _x_ = 2x2 - 2 and _x_ = x2, so set <strong>the</strong>m equal to<br />

each o<strong>the</strong>r.<br />

20. D<br />

2x2 - 2 =x2<br />

2x2 =x2 + 2<br />

x2 =2<br />

Difficulty: High<br />

X =±\1'2<br />

Strategic Advice: A great beginning to a question like this<br />

one is to map out a path that will get you to <strong>the</strong> answer. By<br />

comparing <strong>the</strong> area of <strong>the</strong> entire square to <strong>the</strong> areas of <strong>the</strong><br />

circles inside, you can find <strong>the</strong> requested probability.<br />

z<br />

G

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