Search found 87 matches


Hey Realanoop (hmmmm....i wonder wat your name could be....its so difficult to guess)

which center were you in?

In was in Mumbai, on 13th itself.

Got a 740 (49, 43)......

by rajataga

Tue Jan 13, 2009 7:08 am
Forum: I just Beat The GMAT!
Topic: 760 - Q50 V44
Replies: 4
Views: 2457

I agree with the OA..... MUST BE true is used here to denote that the option selected must be true for all possible values of x that can be derived from the given equation. Hence, if we choose A, there are values of x, such that -1 < x < 0, that are not satisfied by option A. Hence, option A would n...

by rajataga

Sun Jan 11, 2009 1:23 pm
Forum: Problem Solving
Topic: Xs good one.
Replies: 16
Views: 3901

Came across it on 4gmat.com....

Actually very true, the question is kinda worded wrong if u think about the first part...

by rajataga

Sun Jan 11, 2009 10:47 am
Forum: Problem Solving
Topic: Probability of once in four
Replies: 6
Views: 4321

i guess i know the answer mathematically....

each of these events are exclusive, and the outcome of one doesn't affect the outcome of another.....However, i am not yet convinced about this....

by rajataga

Sun Jan 11, 2009 8:35 am
Forum: Problem Solving
Topic: Probability of once in four
Replies: 6
Views: 4321

Also, would you consider y=0 as one of the possibilities???

Even though it falls in the given range, y=0 would essentially mean that that edge lies on the X axis itself, and hence not parallel to the X-axis as given in the question.

by rajataga

Sun Jan 11, 2009 8:26 am
Forum: Problem Solving
Topic: No. of rectangles
Replies: 6
Views: 1672

Thanks Stuart, you are correct.

However, what is the reason that in this scenario, you cannot think of them as 4 exclusive events, and say that if he takes 4 shots at the target, he will hit it atleast once?

by rajataga

Sun Jan 11, 2009 8:21 am
Forum: Problem Solving
Topic: Probability of once in four
Replies: 6
Views: 4321

Probability of once in four

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

A. 1

B. 255/256

C. 175/256

D. 1/4

E. 1/2

by rajataga

Sun Jan 11, 2009 8:07 am
Forum: Problem Solving
Topic: Probability of once in four
Replies: 6
Views: 4321

Now that i write this, i realize why this difference is occurring. each of the points is being taken as unique. Hence, points ABCD could be interchanged by points DABC, and that is being counted as a separate rectangle. However, shouldn't this hold true? would rectangle ABCD be the same as DABC, bec...

by rajataga

Sun Jan 11, 2009 7:27 am
Forum: Problem Solving
Topic: No. of rectangles
Replies: 6
Views: 1672

5c2 * 5c2 Just a minor correction - 4c2 * 5c2 which is 60. I got this. But when i did it in a slightly different way, i got a different answer. Can you just point out the mistake. Total no. of points in this area = 20 Point A can now take up any of these points. Now, point B, assuming AB is parallel...

by rajataga

Sun Jan 11, 2009 7:20 am
Forum: Problem Solving
Topic: No. of rectangles
Replies: 6
Views: 1672

No. of rectangles

The coordinates (x,y) of each corners of rectangle ABCD are such that x and are integers and satisfy the equations 1 < x < 6 and -3 < y < -3. The edges of the rectangle are parallel to the X and Y axes. How many distinct rectangles could be formed which would satisfy these requirements?

by rajataga

Sun Jan 11, 2009 6:05 am
Forum: Problem Solving
Topic: No. of rectangles
Replies: 6
Views: 1672

ch017, wat you were talking about is called 'bisection' that is dividing the line into half.

Intersection is just a common point/s between 2 figures.

by rajataga

Sat Jan 10, 2009 5:55 am
Forum: Data Sufficiency
Topic: line intersection??
Replies: 11
Views: 4495

Thanks.

I had thought about the same point, but then went ahead with it, thinking that there wasnt a time factor involved.

Thanks again.

by rajataga

Fri Jan 09, 2009 7:53 am
Forum: Problem Solving
Topic: Average
Replies: 7
Views: 6778

Even wrtau followed the same method, and has the same question.

Has PMed to Ian to check it.

by rajataga

Fri Jan 09, 2009 5:46 am
Forum: Problem Solving
Topic: Average
Replies: 7
Views: 6778

I applied weighted averages,

[25(10) + 40(50)]/(10+50) = 37.5

what am i doing wrong?

by rajataga

Fri Jan 09, 2009 3:07 am
Forum: Problem Solving
Topic: Average
Replies: 7
Views: 6778

Just one minor correction,

it is....

11a1 + 47

by rajataga

Thu Jan 08, 2009 9:42 am
Forum: Problem Solving
Topic: LOST
Replies: 2
Views: 1469