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

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10Bl<br />

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

Practice Test Twelve Answers and Explanations<br />

13. 48<br />

Difficulty: Medium<br />

Strategic Advice: Making a table is always an excellent way<br />

to approach rate questions.<br />

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

Let d be <strong>the</strong> distance to/from <strong>the</strong> friend's house and solve<br />

<strong>for</strong> <strong>the</strong> time using t = 1-.<br />

r<br />

Rate<br />

To friend's 40<br />

From friend's 60<br />

Overall<br />

Time<br />

d<br />

40<br />

d<br />

60<br />

d d<br />

40 + 60<br />

Distance<br />

Now solve <strong>for</strong> d using <strong>the</strong> fact that <strong>the</strong> overall time was l<br />

hour:<br />

d d<br />

40 + 60 = l<br />

3d 2d<br />

120 + 120 = l<br />

5d<br />

120 =1<br />

d<br />

d<br />

2d<br />

d = 1 ( 1 ;o) = 24 (This is <strong>the</strong> distance each way.)<br />

We're interested in <strong>the</strong> total distance of 2(24) = 48.<br />

14. 192, 480, or 960<br />

Difficulty: Medium<br />

Strategic Advice: The best way to do this problem is to use<br />

trial and error. Just choose three consecutive even integers<br />

and see if <strong>the</strong>ir product is a 3-digit number. Start with 2, 4,<br />

and 6 and go from <strong>the</strong>re.<br />

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

2 x 4 x 6=48<br />

That doesn't work, so now try 4, 6, and 8:<br />

4 x 6 x 8 = 192<br />

That combination works, so grid in 192. O<strong>the</strong>r possible<br />

combinations are 6 x 8 x 10 = 480 and 8 x 10 x 12 = 960.<br />

15. 75<br />

Difficulty: High<br />

Strategic Advice: Since you're starting off with a group of<br />

10 numbers and <strong>the</strong>n removing 2 of <strong>the</strong>m, <strong>the</strong> highest<br />

and <strong>the</strong> lowest, express <strong>the</strong> sum of <strong>the</strong> 10 numbers as <strong>the</strong><br />

sum of <strong>the</strong> highest number, <strong>the</strong> lowest number, and <strong>the</strong><br />

o<strong>the</strong>r 8 numbers. The sum of <strong>the</strong> 10 numbers is equal to<br />

<strong>the</strong> highest number + <strong>the</strong> lowest number + <strong>the</strong> sum of <strong>the</strong><br />

o<strong>the</strong>r 8 numbers.<br />

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

If you call <strong>the</strong> highest number h, <strong>the</strong> lowest number I, and<br />

<strong>the</strong> sum of <strong>the</strong> o<strong>the</strong>r 8 numbers 5, you can express <strong>the</strong><br />

sum of <strong>the</strong> l 0 numbers as h +I+ S. Then <strong>the</strong> average of<br />

<strong>the</strong> 10 numbers is h + 1 6+ 5 , which you're given is 87.<br />

The average of <strong>the</strong> 8 numbers is just , and you're<br />

given that that's 90. You want to know <strong>the</strong> average of <strong>the</strong><br />

2 grades that were removed, or h ; 1. If = 90, <strong>the</strong>n<br />

S = 90 x 8 = 720. And if h + 16 + 5 = 87, <strong>the</strong>n<br />

h + I + S = 87 x 10 = 870. But you know that S = 720,<br />

so h + I + 720 = 870, and h + I= 150. Then <strong>the</strong><br />

average of h and I is 1 0 , or 75.<br />

1<br />

16. o t· So pick a number <strong>for</strong> b that is<br />

between % and 1 , such as i· If b = i· <strong>the</strong>n 2a + i = 1,<br />

2a = i· and a = l' which is greater than 0 and <strong>the</strong>re<strong>for</strong>e<br />

meets <strong>the</strong> only o<strong>the</strong>r requirement you're given <strong>for</strong> a.

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