sample DAT test ordered from the ADA website

[dhk5]

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Hi, so I just took this sample DAT test that I bought from ADA and the QR (which is usually my best section) was difficult, mostly because the question topics were not covered by topscore, Kaplan, Barrons or anything. I would really appreciate if someone could help me figure out these problems:

4. the horizontal incident ray 1 strikes the circular mirror, x^2+y^2 = 25, resulting in the reflected ray R. What is the y coordinate of the intersection point of the ray R and the live x=12? you may need to use the identity tan20 = 2tan0/1(1-tan^20). (i substituted 0 for theta, and i uploaded the picture below)
http://i50.tinypic.com/205v9ed.jpg

5. An elliptical mirror has foci at (+1,0), a major axis length of 20^ (1/2), and a minor axis of length 4 units. A beam of light originating from the focus at (-1,0) is reflected of the elliptical mirror once before arriving at the other focus. What is the total lenght of path followed by the beam of light? (no picture)

8. Determine the set of points whose distances from (-2,2) and (3,-3) are in the ratio of 2:3.

40. Two sides of an isosceles triangle have length 10 with a third side shorter than these two. If the area is 48, what is the longest possible third side? (I thought this was obviously 10, since the third side is shorter)?

ANSWER:
4. 30 and 3/7
5. square root of 20
8. circle with center (-6,6) and radius 6(2)^(1/2)
40. 16

 

[dhk5]

please?

 

[erk005]

40. Sorry about this enormous picture.
note: you must divide this triangle in half. You could think of side x as HALF the bottom line of this triangle. Side y (shown here as "m") goes from the center of the bottom line to the tip of the two sides that are each a length of 10.

(side x)^2 * (side y)^2 = 10^2 (pythagorean theorem)
(16/2)^2 * (side y)^2 = 100
8^2 * (side y)^2 = 100
64 * (side y)^2 = 100
(side y)^2 = 100-64
(side y)^2 = 36
(side y) = 6

(side x)*(side y) = 48
6*8 = 48
This equation finds the area of two halves of the triangle, so it is the whole area.

 

[UBDent19]

I agree the question is faulty. None of the answers are right because the third side has to be shorter than 10. Even if it were 10, the triangle would be equilateral, not isosceles

That sample test has so many typos it's ridiculous.

 

[dat2010]

Please if you are done with your sample dat quizzes and are willing and ready to part with it. I would like to ask for them.
Thanks

 

[basskleff]

Hi, so I just took this sample DAT test that I bought from ADA and the QR (which is usually my best section) was difficult, mostly because the question topics were not covered by topscore, Kaplan, Barrons or anything. I would really appreciate if someone could help me figure out these problems:

4. the horizontal incident ray 1 strikes the circular mirror, x^2+y^2 = 25, resulting in the reflected ray R. What is the y coordinate of the intersection point of the ray R and the live x=12? you may need to use the identity tan20 = 2tan0/1(1-tan^20). (i substituted 0 for theta, and i uploaded the picture below)
http://i50.tinypic.com/205v9ed.jpg

5. An elliptical mirror has foci at (+1,0), a major axis length of 20^ (1/2), and a minor axis of length 4 units. A beam of light originating from the focus at (-1,0) is reflected of the elliptical mirror once before arriving at the other focus. What is the total lenght of path followed by the beam of light? (no picture)

8. Determine the set of points whose distances from (-2,2) and (3,-3) are in the ratio of 2:3.

40. Two sides of an isosceles triangle have length 10 with a third side shorter than these two. If the area is 48, what is the longest possible third side? (I thought this was obviously 10, since the third side is shorter)?

ANSWER:
4. 30 and 3/7
5. square root of 20
8. circle with center (-6,6) and radius 6(2)^(1/2)
40. 16

Hi, so I just took this sample DAT test that I bought from ADA and the QR (which is usually my best section) was difficult, mostly because the question topics were not covered by topscore, Kaplan, Barrons or anything. I would really appreciate if someone could help me figure out these problems:

4. the horizontal incident ray 1 strikes the circular mirror, x^2+y^2 = 25, resulting in the reflected ray R. What is the y coordinate of the intersection point of the ray R and the live x=12? you may need to use the identity tan20 = 2tan0/1(1-tan^20). (i substituted 0 for theta, and i uploaded the picture below)
http://i50.tinypic.com/205v9ed.jpg

5. An elliptical mirror has foci at (+1,0), a major axis length of 20^ (1/2), and a minor axis of length 4 units. A beam of light originating from the focus at (-1,0) is reflected of the elliptical mirror once before arriving at the other focus. What is the total lenght of path followed by the beam of light? (no picture)

8. Determine the set of points whose distances from (-2,2) and (3,-3) are in the ratio of 2:3.

40. Two sides of an isosceles triangle have length 10 with a third side shorter than these two. If the area is 48, what is the longest possible third side? (I thought this was obviously 10, since the third side is shorter)?

ANSWER:
4. 30 and 3/7
5. square root of 20
8. circle with center (-6,6) and radius 6(2)^(1/2)
40. 16

Wow, I know this is a few years old, but I searched the same set of questions as I have the same 2009 sample DAT from the ADA.
Question # 4 is a nasty, tricky little time-sucker! What a pita!!
I did figure this out, and hopefully this will help others googling this problem for their DAT studies.

This is somewhat a hybrid physics geometric optics and trig problem. But it's much more biased towards the later.
You have to tune out and forget all the mirror equation crap and most of concomittant physics theory , equations, etc. Like I said, it is mostly a trig problem in cheap Physics clothing.
What you need to know is that the angle of incidence is equal to the angle of reflection with a convex mirror (or any mirror for that matter).

I just did a nice writeup and will try to upload the image.