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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 Nine Answers and Explanations<br />

833<br />

I<br />

5. c<br />

Difficulty: Medium<br />

Strategic Advice: This question involves solving an algebraic<br />

equation and working with decimals, so if you had trouble<br />

with it, you might need to review one or both of those<br />

topics. All you have to do to solve this problem is work with<br />

<strong>the</strong> equation step-by-step until you get x alone on one side:<br />

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

0.4x + 2 = 5.6 Subtract 2 from both sides.<br />

6. B<br />

0.4x = 3.6 Multiply both sides by 10 to get rid of<br />

<strong>the</strong> decimals.<br />

4x =36<br />

X=9<br />

Difficulty: Medium<br />

Divide by 4.<br />

And you've found x, (C).<br />

Strategic Advice: If a rectangle has a perimeter of 12, <strong>the</strong>n<br />

2(W + L) = 12, where Wis <strong>the</strong> width of <strong>the</strong> rectangle and L<br />

is its length.<br />

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

If 2(W + L) = 12, <strong>the</strong>n W + L = 6. If <strong>the</strong> width is 2 less than<br />

<strong>the</strong> length, <strong>the</strong>n W = L - 2. You can plug L - 2 <strong>for</strong> W into<br />

<strong>the</strong> equation W + L = 6, so W + L = 6 becomes (L - 2) +<br />

L = 6 and so 2L - 2 = 6, 2L = 8, and L = 4. If <strong>the</strong> length is<br />

4, <strong>the</strong>n <strong>the</strong> width, which is 2 less, must be 2. The area of a<br />

rectangle with length 4 and width 2 is 4 x 2 = 8, (B).<br />

7. c<br />

Difficulty: Medium<br />

Strategic Advice: The easiest way to do this one is to just<br />

draw lines from each point and <strong>the</strong>n add up <strong>the</strong> number of<br />

line segments drawn.<br />

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

From point A, draw one line to each of points 8, C, 0, and<br />

E. From point 8, you already have a line to point A, so just<br />

draw a line to each of points C, 0, and E. From point C, you<br />

have already drawn lines to A and 8, so draw in a line to<br />

point 0 and one to point E. Finally, draw a line from point<br />

0 to point E. (You've drawn a star inside a pentagon-very<br />

artistic!) Can you see any point that is unconnected to any<br />

o<strong>the</strong>r point? No, so just add up <strong>the</strong> number of lines you've<br />

already drawn-<strong>the</strong>re are 4 + 3 + 2 + <strong>for</strong> 10 of <strong>the</strong>m, (C).<br />

8. B<br />

Difficulty: High<br />

Strategic Advice: Whenever you have a percent word<br />

problem that doesn't give you a definite amount and asks<br />

you a question like What fraction of <strong>the</strong> total . .. ?, you<br />

should pick a number <strong>for</strong> <strong>the</strong> total.<br />

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

Since you're dealing with percents here and will be<br />

converting <strong>the</strong> percents to fractions, a good number <strong>for</strong><br />

<strong>the</strong> total is 100. So, say that 100 people were polled:<br />

80 percent of <strong>the</strong> 1 00 people were registered voters, so 80<br />

people were registered voters; 75 percent of <strong>the</strong> registered<br />

voters voted in <strong>the</strong> last election, so 750/o x 80, or 60 people<br />

voted in <strong>the</strong> last election. If 60 of <strong>the</strong> 80 registered voters<br />

actually voted in <strong>the</strong> last election, <strong>the</strong>n 80 - 60 = 20 of <strong>the</strong><br />

registered voters didn't vote in <strong>the</strong> last election. The fraction<br />

of <strong>the</strong> people surveyed who were registered but didn't vote<br />

. 20 1<br />

IS ( )<br />

100' or s. , B.<br />

9. E<br />

Difficulty: High<br />

Strategic Advice: Mixture problems are very tricky. The<br />

important thing to look <strong>for</strong> in a mixture problem is this:<br />

Which quantities stay <strong>the</strong> same, and which quantities<br />

change? Here, <strong>the</strong> alcohol is evaporating but <strong>the</strong> iodine is<br />

not. There<strong>for</strong>e, <strong>the</strong> quantity of iodine will be unchanged.<br />

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

So, start with 4 ounces of iodine and 1 6 ounces of alcohol<br />

and end with 4 ounces of iodine and an unknown quantity<br />

of alcohol, which you can call x ounces. That means that<br />

in <strong>the</strong> end, <strong>the</strong> whole solution has a total of 4 + x ounces,<br />

since <strong>the</strong> solution is made up of only iodine and alcohol.<br />

The final quantities of iodine and total solution are in <strong>the</strong><br />

ratio of 2 to 3. That means that <strong>the</strong> ratio of 4 to 4 + x is<br />

equal to <strong>the</strong> ratio of 2 to 3, or 4<br />

! x<br />

= 1-. which is just an<br />

algebraic equation that can be easily solved:<br />

4 2<br />

4+x - 3<br />

Cross-multiply.<br />

2( 4 + x) = 4 x 3 Multiply out <strong>the</strong> left side.<br />

8+2x= 12 Subtract 8.<br />

2x =4 Divide both sides by 2.<br />

x=2 That's <strong>the</strong> amount of alcohol left.

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