A couple of Math Questions

ACESAT_1 · March 18, 2011, 5:50am

[Image</a> - ■■■■■■■ - Free Image Hosting, Photo Sharing & Video Hosting](<a href=“http://■■■■■■■.com/view.php?pic=2djck0j&s=7]Image”>http://■■■■■■■.com/view.php?pic=2djck0j&s=7)

The Answer is C

[Image</a> - ■■■■■■■ - Free Image Hosting, Photo Sharing & Video Hosting](<a href=“http://■■■■■■■.com/view.php?pic=vbzma&s=7]Image”>http://■■■■■■■.com/view.php?pic=vbzma&s=7)

The Answer is E

[Image</a> - ■■■■■■■ - Free Image Hosting, Photo Sharing & Video Hosting](<a href=“http://■■■■■■■.com/view.php?pic=jkc96f&s=7]Image”>http://■■■■■■■.com/view.php?pic=jkc96f&s=7)

The Answer is C

[Image</a> - ■■■■■■■ - Free Image Hosting, Photo Sharing & Video Hosting](<a href=“http://■■■■■■■.com/view.php?pic=2ivzj9z&s=7]Image”>http://■■■■■■■.com/view.php?pic=2ivzj9z&s=7)

The Answer is A

Can anyone explain these?

Mirage2000 · March 18, 2011, 7:32am

I assume that “Princeton Review 11 practice tests” has its own explanations.

DrSteve · March 18, 2011, 7:42am

For the first one, let’s plug in a value for x, say x=-2 (using choice B as a guide). Then we get the following for each answer choice:

(A) x=-2, y=-2, xy=4
(B) x=-2, y=anything (let’s let y=-5), xy=10
(C) x=-2, y=4, xy=-8
(D) x=-2, y=2, xy=-4
(E) x=-2, y=-6, xy=12

So we can eliminate choices (A), (B) and (E). Now let’s pick another value. A positive number will not help, so let’s choose a negative fraction, say x=-1/2=-.5:

(C) x=-.5, y=1, xy=-.5
(D) x=-.5, y=-1, xy=.5

So we can now eliminate choice (D) and the answer is choice (C).

Notes: (1) I know you have a copy of my book. Notice that used strategy 4 of picking a number, and then I used (8) on page 51 to decide how to pick my next numbers when the first one didn’t eliminate all choices.

(2) As always, picking numbers only eliminates answer choices.

Method (2):

We start with choice (C) and multiply x and y to get

Since x^2 is always nonnegative, -2x^2 is always nonpositive. Thus the answer is choice (C).

DrSteve · March 18, 2011, 7:52am

For the second, let’s list all the factors of 30:

1, 2, 3, 5, 6, 10, 15, 30.

These are the possible lengths of a side of x (since the side is an integer)

We square these numbers to get the possible areas:

1, 4, 9, 25, 36, 100, 225, 900

So the answer is either 9 or 25. We can eliminate 25 because if the side of a square is 5, then the other side of each rectangle is 6. So there is only 1 unit left for 2 pieces of the rectangle.

Thus the answer is 9, choice (E).

Note: My explanation here is brief because I have to get going to work now. If you need further clarification on this one let me know and I’ll answer a bit later. I’ll also answer the others when I have a few minutes.

ACESAT_1 · March 18, 2011, 8:11am

Dr.Steve:

Thank you very much =D

O and i just figured the last 1

DrSteve · March 18, 2011, 10:41am

OK great. So I guess you still need the third one explained. This has become a pretty standard “very difficult” type of SAT question. By “very difficult” I mean that it usually appears as one of the last 2 questions on the SAT.

In this type of question you want to locate some “key points.” These points are generally places where 2 or more things intersect. In this picture, the key points are L and N.

L is the y-intercept of the graph. If we set x=0 in the given equation we get y=4. So L is the point (0,4).

N is an x-intercept of the graph. Now, note that it DOES NOT say “figure not drawn to scale.” So we can assume it is. Which means N is the point (2,0). (N is half the distance from the origin as L is). You can check this easily by substituting 2 in for x in the equation.

Now it is easy to see that the triangle has base 2 and height 4, so that A=1/2*bh=4, choice (C).

A couple of Math Questions

CONNECT WITH US