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

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112<br />

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

Practice Test One Answers and Explanations<br />

(A) step 1 : 6.127<br />

step 3: 7<br />

step 2: 700<br />

step 4: 712<br />

This does not match with what we found in 4F 15, so it does<br />

not work.<br />

(B) step 1 : 6.127<br />

step 4: 18.127<br />

step 3: 19<br />

step 2: 1900<br />

This does not match with what we found in 4F 1 5, so it does<br />

not work.<br />

(C) step 1 : 6. 127<br />

step 2: 612.7<br />

step 3: 625<br />

step 4: 624.7<br />

This seems to work. Notice that since we're adding an<br />

integer, it doesn't matter whe<strong>the</strong>r we add first or round first.<br />

Choice (D) doesn't work since (A) and (B) don't work.<br />

Choice (E) is not true, since (C) works.<br />

18. c<br />

Difficulty: High<br />

Strategic Advice: In a substitution problem like this, look <strong>for</strong><br />

<strong>the</strong> common term that will help you express one equation<br />

in terms of ano<strong>the</strong>r. Substitute carefully!<br />

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

Start with <strong>the</strong> smallest equation. Express it in terms of c.<br />

b=-c+4<br />

-b + 4 =C<br />

Substitute this into <strong>the</strong> o<strong>the</strong>r equation:<br />

a= 3(2c2 + 3c + 4)<br />

a= 3[2(-b + 4)2 + 3(-b + 4) + 4]<br />

a= 3[2(b2 - 8b + 16) + -3b + 12 + 4]<br />

a= 3[2b2 - 16b + 32 + -3b + 12 + 4]<br />

a= 3[2b2 - 19b + 48]<br />

a= 6b2 - 57b + 144<br />

19. c<br />

Difficulty: High<br />

Strategic Advice: Visualizing a three-dimensional diagram<br />

can be difficult. Redraw this diagram if necessary. Be sure to<br />

label what you know as you work through <strong>the</strong> problem.<br />

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

3<br />

We're essentially trying to work out two Pythagorean<br />

<strong>the</strong>orem problems, once <strong>for</strong> <strong>the</strong> XF base of <strong>the</strong> cube, and<br />

once again <strong>for</strong> BF. Looking at <strong>the</strong> base of <strong>the</strong> cube, since<br />

XY and YF are both 3, <strong>the</strong> base <strong>for</strong>ms a 45-45-90 right<br />

triangle with sides in <strong>the</strong> ratio of x:x:xv2. There<strong>for</strong>e, <strong>the</strong><br />

hypotenuse is 3Yl.<br />

Now let's find BF. This diagonal is a hypotenuse of a<br />

right triangle with legs of 3 and 3Yl, so let's apply <strong>the</strong><br />

Pythagorean <strong>the</strong>orem:<br />

32 + (3\/2)2 = (BF)2<br />

9 + (9)(2) = (BF)2<br />

9 + 18 = (BF)2<br />

27 = (BF)2<br />

BF = V27 = v'9 \/3 = 3\/3<br />

20. c<br />

Difficulty: High<br />

Strategic Advice: When in<strong>for</strong>mation is given in <strong>the</strong> question<br />

stem, add it to <strong>the</strong> figure and work from <strong>the</strong>re.

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