DS - GMATPrep 12 questions attached in doc

Question

PFA doc with 12 question, please explain your answers, two questions are on PS.

I am not able to find how to paste individual question sceen shot in post, so have to attach all questions in single doc. incase anyone knows how to paste single screen shot in post.

thanks

coelholds · Answer

manojgmat, You can take each question, and save in a jpg file in the paint program and post in the forum.

Attachment:

circle.JPG

From the equation \(h^2 = a^2 + b^2\) we have
\(radius^2 = -\sqrt{3}^2 + 1^2 = \sqrt{4}\)
\(radius = 2\).

Again, using the same equation you have a triangle formed with the two radius and the distance between the two points P and Q. The distance is the hypotenuse.

Thus, you can avoid some calculation.
Once \(s\) is the x-coord. from the point Q and we have the distance from the x-coord. of the point P to point Q, and the value of x-coord of point P you just need to calculate the difference.

Mikko · Answer

Hi Coelholds, 
How can we know if PQ parallels with x-axis? U have this only when Oy is the bisector of angle POQ

aagar2003 · Answer

I have a different answer and hopefully the correct one. I disagree with Mikko's answer

Drop Perpendiculars from P and Q on x-axis as P' and Q'.

Attachment:

Untitled.jpg [ 3.11 KiB | Viewed 2479 times ]

Let the angle POP' = alpha = OQQ'
We know using pythagoras theorem as Mikko did in earlier post, radius = \sqrt{3+1} = 2

In triangle, OQQ', we need to calculate OQ'= s.
Since OQ is obtained already from above as radius,
we can just use
Sin(alpha) = OQ'/OQ = s/2

Also, from triangle, OPP' we have Sin(alpha) = PP'/OP = 1/2

Thus, s=1

Explanation looks more lengthy but it seems pretty straight forward unless I am missing something.
Please note I have not assumed PQ to be paralled to X-axis, as this is not given and I do not even require this... hehehe

sfeiner · Answer

I have been trying to figure this question out for soooo long. I always aproached it from the unit circle perspective but I have no idea. Is there a simpler way to solve without multiple calculations?

deepakraam · Answer

For Q no.5,my workout

1. symbol can be + or * .Nt suff
2 .symbol can be / or *. Nt suff

Combining both we get the symbol as *.So I wud go with option C

deepakraam · Answer

For q no.4

Statement 1:
==========
1.Insuff. Z = 1 & m=4 satisfies.Also m =0 & z = -1/3 also satisfies.

Statement 2;
==========
1. Insuff z = 0 m = -1 ; m =3;z=1 satisfies.

combining both the statements we can ans the q.

i wud go with option C

deepakraam · Answer

For q no.10

Statement 1 is insuff. Since S = 3 and r =2 can also satisfy the condition.

Statement 2 is suff bcos only when r =-s the condition will get satisfied.So when r = -s then origin will be in between r & s.

I wud go with option 2 for the above question

deepakraam · Answer

For question no.6

Statement 1: insuff as it doesn't say particular value for n

statement 2 : suff . 7/16 - 3/8 = n which gives unique value for n.so suff

I wud go with option 2

sunny4frenz · Answer

Solution to problem 1 ....

Slope of line OP = (1-0) / (-(sqrt3)-0) = 1/ -sqrt3
Now slope of line OQ = sqrt3 (both line are perpendicular)
So equation of line OQ is y = sqrt3 x .....

but as we now both pts are on circle so .... sqrt(s^2 + t^2) = sqrt( 3 + 1)
s^2 + t^2 = 4
Now from eq of line s^2 + (sqrt 3s)^2 = 4 ===> 4 s^2= 4 => s= 1,-1 but s is in 1st quadrant so s = 1...

my solution doesn't look beautifully written becoz .. I don't know how to put exact symbols for sqrt ..
Please if you like the solution ...don't forget to give me kudos ... :-)

sunny4frenz · Answer

Hi Guys .... I created links most of the problems we are discussing with my version of solution ... I hope it will helpful for everyone .... Here are links 1. semi-circle-problem-84123.html#p630426 Ans = 1 2. gcd-of-m-and-n-84124.html Ans = c 3. does-line-contain-point-84125.html Ans = C 4. x-plus-z-gt-than-zero-problem-84126.html Ans = C 5. delta-operation-84127.html Ans = A 6 No links :-(