  # PSAT Math Help

<p>Please explain these questions from the Form S, 2007 PSAT, if you can (answers are bolded):</p>

<p>MULTIPLE CHOICE:</p>

<p>(16) x^2 + kx +4= 4 = (x + c)^2</p>

<p>In the equation above, k and c are positive constants. If the equation is true for all values of x, what is the value of k?</p>

<p>(A) 2
<a href="B">b</a> 4<a href="C">/b</a> 8
(D) 16
(E) 32</p>

<p>(27) The temperature at 6:00 A.M. on a certain day was 11 degrees C. It increased at a constant rate until 3:00 P.M., and then it decreased at the same constant rate until 9:00 P.M. If the temperature at 9:00 P.M. was 17 degrees C, what was the temperature at 2:00 P.M. that day?</p>

<p>(A) 23 degrees
(B) 24 degrees
<a href="C">b</a> 27 degrees<a href="D">/b</a> 30 degrees</p>

<p>FREE RESPONSE:</p>

<p>(34) If x and y are positive odd integers and the square root of x + the square root of y = 10, what is one possible value of x + y?</p>

<p>I said 52, because the square root of 36 (6) + the square root of 16 (4) = 10, and 36 +16 =52. But answers are listed as 50, 58, or 82. Explain?</p>

<p>(35) Jane made a 7-foot by 5-foot rectangular patchwork quilt consisting of red, blue, yellow, and green squares that each measure 1 foot by 1 foot on the finished quilt. She used twice as many green squares as red and 1 more blue square than yellow. If Jane used 9 blue squares, how many green squares were used?</p>

<p>I said 6; correct answer is 12. Explain?</p>

<p>(37) If a /=/ 0, a/x = b/y, and a/2b = c (3 [times] x/y), what is the value of c?</p>

<p>c = 1/6 (?)</p>

<p>(38) What is the length, in inches, of an edge of a cube that has the same volume as a rectangular solid with length 3 inches, width 3/4 inches, and height 3/2 inches?</p>

<p>3/2 or 1.5</p>

<p>Thanks in advance for the help :)</p>

<ol>
<li>You need to first find the volume of the rectangular solid: </li>
</ol>

<p>V=lwh=3<em>3/4</em>3/2=27/8</p>

<p>Then, you know that the volume of a cube is s<em>s</em>s (s=length of one side).</p>

<p>So, s<em>s</em>s=s^3=27/8</p>

<p>Take the cube root of 27/8 to get s.</p>

<p>Cube root of 27/8= (Cube root of 27)/(Cube root of 8)= 3/2=1.5</p>

<ol>
<li>Your answer is incorrect because the values you chose for x and y (36 and 16) are even, when the problem states that x and y are positive odd integers.</li>
</ol>

<p>The answer can be 50 because sqrt(25)+sqrt(25)=5+5=10. In this case, x=y=25.</p>

<p>I'm not sure how you can obtain 58 or 82, but I'm sure it's possible.</p>

<p>Thanks :) You explained that very well, it makes sense to me now.</p>

<p>I'm not sure how I missed that in 34. =/ It's stupid mistakes like that that keep me from achieving in the Math section (although I have gone 62-->65 in two weeks, so I'm improving). Grrr. 80 W twice now, 74-77 CR, but math is holding me back...</p>

<p>Also, for 34 - </p>

<p>There are 35 squares total, and the colors of the squares are explained by 2G = R and B = Y + 1. If B = 9 and Y = 8, that's 17 total squares, which leaves 18 to fill green and red. Therefore, 18 = 3G, and G = 6. Yet CB tells me that the answer is 12, not 6. What am I doing wrong?</p>

<p>You have the first part right, so we'll start there:</p>

<p>There are 18 remaining squares left to be filled by green and red. You also know that there are twice as many green squares as red. So, split 18 into three equal parts: 6, 6, 6. Two of those parts have to be green and the remaining part has to be red. So, green squares=6+6-12.</p>

<p>I think your problem is that you're trying to "solve for G" when in reality, you need to divide the remaining squares evenly based on the information given in the question.</p>

<p>Oh! I get it. For some reason, I was forgetting that there were twice as many green squares as red, plus what you mentioned. Thanks again for explaining. :)</p>

<p>Is 16 typed correctly?</p>

<ol>
<li>The temperature increased for 9 hour-long intervals and then decreased for 6, meaning it increased for 3 hours more than it decreased.<br></li>
</ol>

<p>The temperature at 9:00 p.m. was 6 degrees warmer than at 9:00 a.m., so the change in temperature must have been 2 degrees/hour.</p>

<p>Finally, at 2:00 p.m., the temperature had increased for 8 hour-long intervals, meaning it was 16 degrees warmer than at 9:00 a.m. 11 + 16 = 27.</p>

<ol>
<li><p>The explanation of 50 above is correct. 82 = 1^2 + 9^2. 58 = 3^2 + 7^2.</p></li>
<li><p>You seem to have gotten on your own to the conclusion that red + green = 18. There are twice as many green as red, and the question asked you for the number of green. There are 12 green and 6 red.</p></li>
<li><p>If a/x = b/y, then ay = xb. Divide both sides of that equation by (yb) and get a/b = x/y.
Now, a/(2b) = (1/2)(a/b) = (1/2)(x/y).
So (1/2)(x/y) = c * (3x/y), which can be rewritten as (c<em>3)(x/y).
If (1/2)(x/y) = (c</em>3)(x/y), then (1/2) = c*3. Divide both sides by 3, and 1/6 = c.</p></li>
</ol>

<p>Heck, I've spent so long typing that by now, maybe they've ALL been explained!</p>

<p>For # 35: "There are 35 squares total, and the colors of the squares are explained by 2G = R and B = Y + 1."</p>

<p>Not 2G = R, but rather G = 2R. The number of greens is twice the number of reds; there are more green than red.</p>

<p>Thanks to much for all of the help :) Math is my weakness, and these explanations are helping like you wouldn't believe.</p>

<p>
[quote]
Is 16 typed correctly?

[/quote]
</p>

<p>I double-checked; it is typed correctly.</p>

<p>
[quote]
Not 2G = R, but rather G = 2R. The number of greens is twice the number of reds; there are more green than red.

[/quote]
</p>

<p>This is another one of those "how could I miss that?" moments. :P</p>

<p>Are there any suggestions on working through this section more efficiently? Time is a huge issue for me; I'm always rushing to finish, and that's with leaving 1 or more blank.</p>

<p>For 16, I think you meant: x^2 + kx +4 = (x + c)^2 ... Use FOIL for the right hand side. If the equation must work for <em>all</em> values of x, then the coefficients in front of the corresponding terms on each side must be the same. Hope this helps!</p>

<p>So could I literally plug in any value for x in order to solve the equation? Ie, 2^2 + 4(2) = (2 + c)^2...?</p>

<p>Yes, you can plug in numbers to solve equations.</p>

<p>You could, I suppose, but assuming fignewton is right about the actual equation, I wouldn't.</p>

<p>I'd expand the right side, yielding:</p>

<p>x^2 + kx + 4 = x^2 + 2cx + c^2.</p>

<p>Now, the coefficients of the first-power terms must be equal, and the constants must be equal.</p>

<p>4 = c^2 means c = 2 or -2, but the problem tells you c>0, so c = 2.</p>

<p>If kx = 2cx, then k = 2c; since c = 2, k = 2(2) = 4.</p>

<p>So I get that from x^2 + kx + 4 = x^2 + 2cx + c^2, you could then subtract x^2, and get to kx + 4 = x^2 + c^2, but how do you know that 4=c^2? (Sorry; I've forgotten a lot from Algebra I and Algebra II is next year.)</p>

<p>I think I'm forgetting something about constants, actually. If a constant is true for every value, then wouldn't x be a constant is well?</p>

<p>Miss, when they say that c and k are constants, they mean that they're fixed numbers, like 7 or -12.8 or pi, that don't change value. They represented these numbers with letters because the point of the problem is to hide from you exactly WHICH fixed numbers they're talking about. x in this problem is a variable and not a constant because the question calls for you to find c and k so that the equation will remain true no matter what real number you substitute for x.</p>

<p>This is a little bit like, let's say, the slope-intercept form of the equation of a line: y = mx +b. They never give you that many letters in a single exercise. The equations you deal with actually look like y = 2x + 1, or y = (-4/3)x - 2. And in each equation the "m" spot and the "b" spot are filled by numbers that don't change. The point of the exercise is to find (and then graph) all the (x,y) pairs that make the equation y = 2x + 1, or y = (-4/3)x - 2 true.</p>

<p>Now, let's go back to x^2 +kx + 4 = x^2 + 2cx + c^2.</p>

<p>First, subtract x^2 from both sides, leaving kx + 4 = 2cx + c^2.</p>

<p>Now, if these two expressions are to be equal for all values of x, then the x terms will have to be equivalent, meaning the coefficients of the x's are equal, and the constants must be equal.</p>

<p>That's how I jumped to 4 = c^2, and then c = 2. Then, having found c, I reasoned that k = 2c, because kx must equal 2cx for all values of x.</p>

<p>Clear as mud, right?</p>

<p>I FINALLY understand it! Thanks for that very detailed explanation. :)</p>

<p>You're welcome. I'm an algebra teacher. This is what I do!</p>

<p>Haha cool! It's SO nice of you to hang around these boards. Algebra can be confusing.</p>

<p>Believe me, Miss, it's not just public service. I have a daughter going into 12th grade.</p>