16.03.2017 Views

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

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

12 Practice <strong>Tests</strong> <strong>for</strong> t h e <strong>SAT</strong><br />

Practice Test Seven Answers and Explanations I<br />

647<br />

17. A<br />

Difficulty: Hig h<br />

Strategic Advice: T h e interior ang l es of any triang l e sum to<br />

1 80°, so a + b + c = 1 80.<br />

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

T h at gives you d 3 ifferent equations containing t h e 3<br />

unknowns a, b, an d c-you can so l ve t h ese <strong>for</strong> b. One way<br />

to d o t h is is to combine t h e equations: a + b = 80 an d<br />

b + C= 140.<br />

T h en:<br />

a + b + b + c = 80 + 140<br />

a+ 2b + c=220<br />

Now subtract t h e t h ir d equation:<br />

a+ 2b + c -(a + b + c) = 220 - 180<br />

a+ 2b + c - a -b - c = 220 - 180<br />

a - a + 2b - b + c - c = 40<br />

b=40<br />

18. c<br />

Difficulty: Hig h<br />

Strategic Advice: If you know t h e s l ope of RS, you can fin<br />

t h ey-coor d inate of d<br />

R using t h e s l ope <strong>for</strong>mu l a. T h s l ope of<br />

a l ine is t h e c h ange in y-coor d inates d ivi d e d by t h e c h ange<br />

in x-coor d inates.<br />

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

Since t h e l ine a l so passes t h roug h t h e origin, (0, 0), t h e<br />

s l ope of RS = t h e s l ope of SO = 1..<br />

6 - 0 = .§. =<br />

9 3·<br />

P l ug t h is into your s l ope-intercept equation to so l ve <strong>for</strong> y<br />

w h en x = -6, given t h ey-intercept= 0.<br />

y= 1.. (-6) + 0<br />

3<br />

y=-4<br />

9-0<br />

19. D<br />

Difficulty: Hig h<br />

Strategic Advice: Since t h e ra d io station can be receive d in<br />

a ll d irections, its region of reception is a circ l e wit h t h e ra d io<br />

station at t h e center.<br />

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

Be<strong>for</strong>e t h e c h ange, t h e signa l cou l d be receive d <strong>for</strong> 60<br />

mi es, so t h e area of reception was a circ l e wit ra d ius 60<br />

mi l es. Now it can be receive d <strong>for</strong> 40 mi l es furt h er, or <strong>for</strong> a<br />

tota l of 60 + 40, or 100 mi l es. T h e area is now a circ l e wit<br />

ra d ius h<br />

l 00 mi l es:<br />

New region of reception<br />

Ra d io station<br />

O l d region of reception<br />

T e increase is just t h e d ifference in t h ese areas; t h at is, t h e<br />

s h a d e d region on t h e above d iagram.<br />

Increase New area - O l d area<br />

7t(l 00) 2 - 7t(60) 2<br />

l 0,0007t - 3,6007t<br />

= 6,4007t<br />

T e va l ue of 7t is a bit more t h an 3, s o 6,4007t is a bit more<br />

t h an 3 x 6,400, or just over 1 9,200. T h e on l y c h oice c l ose<br />

to t h is is (0), 20,000.<br />

20. A<br />

Difficulty: Hig h<br />

Strategic Advice: T h e tota l number of wor d s in t h e paper is<br />

e number of wor d s per page times t h e number of pages,<br />

t h at is, wp .<br />

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

If Bi ll types atx wor d s per minute, h e can type wp wor d s in<br />

wp -+-x minutes, or an average rate of .!'Y.e. minutes. Now to x

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!