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12.Practice.Tests.for.the.SAT_2015-2016_1128p

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

Practice Test Seven Answers and Explanations 661<br />

I<br />

3. A<br />

Difficulty: Low<br />

Strategic Advice: Try to reduce two variables to one<br />

variable.<br />

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

This is a problem in which you are working with a system of<br />

equations. Subtract <strong>the</strong> second equation from <strong>the</strong> first:<br />

2x+ 3y = 1 1<br />

2x -y=7<br />

4y =4<br />

y =l<br />

Plug this value <strong>for</strong> y back into <strong>the</strong> first equation:<br />

2x+3(1)=11<br />

4. D<br />

2x =8<br />

X=4<br />

Difficulty: Medium<br />

Strategic Advice: This is a problem using common percent<br />

and fractional equivalents.<br />

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

Find -} of 60 and <strong>the</strong>n figure what percent of 50 that<br />

number is. First, -} x 60 = 20. To find out what percent of<br />

50 is 20, first find what fraction of 50 is 20: 20 is ;g or i<br />

of 50. To convert i to a percent, multiply i by 100% : ix<br />

100% = 40%.<br />

5. A<br />

Difficulty: Medium<br />

Strategic Advice: Solve this linear equation by letting r =<br />

<strong>the</strong> cost of <strong>the</strong> radio in dollars.<br />

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

A television set costs $25 less than twice <strong>the</strong> cost of <strong>the</strong><br />

radio, or 2r - 25. The two items total to $200, so r + 2r -<br />

25 = 200, 3r = 225, r = 75. So a radio costs $75 and a<br />

television set $125, making <strong>the</strong> cost of <strong>the</strong> television $125<br />

- $75 = $50 more than a radio.<br />

6. B<br />

Difficulty: Medium<br />

Strategic Advice: The key to this problem is drawing a<br />

diagram.<br />

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

30°<br />

20<br />

60°<br />

Since <strong>the</strong> top angle of <strong>the</strong> right triangle is 30 degrees, this is<br />

a 30-60-90 right triangle. You're asked to find <strong>the</strong> distance<br />

from <strong>the</strong> bottom of <strong>the</strong> ladder to <strong>the</strong> base of <strong>the</strong> building,<br />

which is <strong>the</strong> leg opposite <strong>the</strong> 30-degree angle and <strong>the</strong>re<strong>for</strong>e<br />

half of <strong>the</strong> hypotenuse, or f of 20, which is 1 0.<br />

7. D<br />

Difficulty: Medium<br />

Strategic Advice: Draw a diagram of <strong>the</strong> circle and <strong>the</strong> line<br />

segment to solve this problem.<br />

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

Notice that drawing in two radii gives you two right<br />

triangles, each with hypotenuse 2 and a leg of 1. Using <strong>the</strong><br />

Pythagorean <strong>the</strong>orem, you can find that <strong>the</strong> length of <strong>the</strong><br />

o<strong>the</strong>r leg is v3. Since each of <strong>the</strong>se legs makes up half <strong>the</strong><br />

line segment, <strong>the</strong> length of <strong>the</strong> line segment is 2\/3.

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