SAT math+writing questions!

Test Preparation SAT Preparation
1

<p>This book only has the answers, no explanations, so can someone explain them!?</p>

<p>MATH
Multiple Choice:
12. An object started at 0 on a number line. It moved 1 unit to the left and then 3 units to the right, and this patter of movement continued until the object stopped at 30. How many times did the object move 1 unit to the left?
(D) 25</p>

<ol>
<li>The five digits 1,2,3,4, and 5 are used to form five-digit numbers in which no digit is repeated. How many such five-digit numbers greater than 40,000 are possible?
(B) 45</li>
</ol>

<p>Student Produced Response:
35. Gift certificates were sold by an ice-cream parlor in the month of July. Each gift certificate was worth either $2, $3, or $5. Twice as many $2 gift certificates were sold as $3 gift certificates, and twice as many $3 gift certificates were sold as $5 gift certificates. The total value of all the gift certificates sold was $57. How many $3 gift certificates were sold in July?</p>

<p>WRITING
28. (ALTHOUGH) the 500-mile covers the (HARSHEST) part of the Alaskan tundra, (SCORES OF) dogsled teams (COMPLETE) it successfully each winter since 1925.
Answer: (COMPLETE)</p>

<ol>
<li><p>In the eighteenth century a (SIMPLE) method of musical notation (HELPED TO MAKE) (MUSIC MORE POPULAR) than in (ANY CENTURY)
Answer: (ANY CENTURY) </p></li>
<li><p>(HAVING HEARD) both local candidates give speeches, Time (WAS CONVINCED) that neither of them (WERE LIKELY) (TO BE ELECTED) in the primary.
Answer: (WERE LIKELY)</p></li>
</ol>

<p>Thanks!</p>

2

<p>Wait, mistake! On number 12 on math, the answer is (B) 15, not (D) 25</p>

3

<p>ahh another mistake, On number 16 on math, the answer should be 48, not 45!</p>

4

<p>12) If I were pckeller, I would say this is a displacement problem :p. In physics, displacement is simply the distance from the starting point. Distance, on the other hand, is the total distance traveled. But that’s besides the point. </p>

<p>Just know that the displacement of the point in this problem is 2. Each time the point does its pattern of movement, the point is displaced 2 to the right. The point is moving 3 units. But it’s net movement is 2 to the right. </p>

<p>So 30/2 = 15.</p>

<p>16) Your looking for digits greater than 40,000. That excludes the digits 1-3 from the first spot. You can only pick 4 or 5 for the first spot. </p>

<p>2 * 4 * 3 * 2 * 1 = 45</p>

<p>You may be tempted to say FIVEEE!!!, or 5! (5 factorial). But don’t fall for this. Your looking for digits greater than 40,000.</p>

<p>35) </p>

<p>4x * 2 + 2x * 3 + 5x = 57 </p>

<p>Six $3 dollar certificates were sold.</p>

<p>28) Use the present perfect tense to indicate that the action began in the past and is still going on in the present, to quote Gruber’s Complete SAT Guide. </p>

<p>In this case, it would be “have completed.” </p>

<p>32) Comparison error - the sentence is comparing music’s popularity with “any century.” </p>

<p>33) “Neither” is singular. Thus, “were” is incorrect; switch it with the word “was.” </p>

<p>But in a “neither … nor …” construction, the verb agrees with the closest subject. Example:</p>

<p>“Neither Kelly nor Lucy finished her homework.” </p>

<p>“Neither Bob nor the plumbers knew what was wrong.”</p>

5

<p>^Correction:</p>

<p>“Neither Bob nor the plumbers knew what they were doing.”</p>

6

<p>I’m not convinced of IceCube’s explanation for the reason “any century” is incorrect in problem 35. I do however agree that it’s incorrect.</p>

<p>The sentence makes sense if “any century” is replaced with “any other century”. It’s idiomatic … in effect you don’t want to include the 18th century in the comparison, and if you write “any century” that’s what you’re doing.</p>

<p>The comparison itself seems implicit to me, and the word “in” suggests that the corrected sentence would specify a time period.</p>

7

<p>fogcity: You are correct, #35 is wrong, and so is my explanation. </p>

<p>It should indeed read “any other century” for the reason you outlined.</p>

8

<p>thanks so much, all of you guys!!</p>

<p>I still don’t understand the math one about the ice cream coupons lol</p>

9

<p>bump abcdef</p>

10

<p>The problem is:</p>

<ol>
<li>Gift certificates were sold by an ice-cream parlor in the month of July. Each gift certificate was worth either $2, $3, or $5. Twice as many $2 gift certificates were sold as $3 gift certificates, and twice as many $3 gift certificates were sold as $5 gift certificates. The total value of all the gift certificates sold was $57. How many $3 gift certificates were sold in July?</li>
</ol>

<p>Proceed slowly and methodically. What makes the problem complicated are the multiple relations. Pick for the unknown the number of $3 certificates sold in July. Call that number “x”. We’ll solve for it.</p>

<p>The first statement: twice as many $2 gift certificates were sold as $3 gift certificates. So the number of $2 gift certificates sold in July is 2x.</p>

<p>And the second statement: twice as many $3 gift certificates were sold as $5 gift certificates. So the number of $5 gift certificates sold in July is x/2.</p>

<p>And the third statement: the total value of all the gift certificates sold was $57. The value of the $3 certificates sold is price<em>of</em>certificates times number sold or ($3)<em>(x) which is 3x. The value of the $2 certificates sold is ($2)</em>(2x) which is 4x, and the $5 certificates is ($5)*(x/2) which is 2.5x. Add the three values together since the problem asks for “all”. That equals $57. So: 3x+4x+2.5x=57. Add and get 9.5x=57. Solve for x to get x=57/9.5=6.</p>