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by Divya Ann Chacko » Wed Jun 11, 2014 5:41 pm
Hai,

Can anyone explain me through the techniques applied in Gmat prep now counting topics?
Here is the question:

Right triangle RST can be constructed in the xy-plane such that RS is perpendicular to the y-axis and the right angle is at R. The x and y-coordinates of R, S, and T are to be nonzero integers that satisfy the inequalities −3 ≤ x ≤ 4 and −7 ≤ y ≤ 3. Given these restrictions, how many different triangles can be constructed?
A. 3780
B. 4200
C. 4900
D. 6160
E. 7744

Answer is A.

Thanks in advance
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by [email protected] » Wed Jun 11, 2014 6:09 pm
Hi Divya Ann Chacko,

This question is about following graphing rules and keeping track of individual possibilities. Here's how I'd solve it:

We're told a number of facts:
1) The X and Y co-ordinates must be NON-ZERO INTEGERS.
2) RS is perpendicular to the Y-axis; this means that points R and S will have the SAME Y-coordinate.
3) Angle R is a right angle; this means that points R and T will have the SAME X-coordinate.
4) The X co-ordinates are in the following ranges: -3 ≤ x ≤ 4 and -7 ≤ y ≤ 3

Let's start with point R. It can have any of 7 different X-coordinates (-3,-2,-1, 1, 2, 3, 4) and and of the 10 different Y-coordinates (-7,-6,-5,-4,-3,-2,-1,1,2,3).

That is 7x10 different possibilities = 70 different spots for point R

Once we FIX R in one spot, S MUST have the SAME Y-coordinate, BUT a DIFFERENT X-coordinate. We already used one of the X-coordinates when we placed point R, which means....

There are 7-1 = 6 possible spots for point S

With R still in the same spot, we know that point T must have the SAME X-coordinate, but a DIFFERENT Y-coordinate. We already used one of the Y-coordinates when we place point R, which means....

There are 10-1 = 9 possible spots for Point T

We multiply the options: (70)(6)(9) = 3780 different triangles.

Final Answer: A

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by GMATGuruNY » Wed Jun 11, 2014 6:29 pm
I posted a solution to a very similar problem here:

https://www.beatthegmat.com/how-many-tri ... 13159.html
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by Brent@GMATPrepNow » Wed Jun 11, 2014 7:21 pm
Here's another similar question to practice with: https://www.beatthegmat.com/how-many-tri ... 28974.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Divya Ann Chacko » Thu Jun 12, 2014 12:09 pm
Thank you Guruny for providing a similar example. I understood the explanation given.
Thank you.

Thanks
Divya
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by Divya Ann Chacko » Thu Jun 12, 2014 12:27 pm
[email protected] wrote:Hi Divya Ann Chacko,

This question is about following graphing rules and keeping track of individual possibilities. Here's how I'd solve it:

We're told a number of facts:
1) The X and Y co-ordinates must be NON-ZERO INTEGERS.
2) RS is perpendicular to the Y-axis; this means that points R and S will have the SAME Y-coordinate.
3) Angle R is a right angle; this means that points R and T will have the SAME X-coordinate.
4) The X co-ordinates are in the following ranges: -3 ≤ x ≤ 4 and -7 ≤ y ≤ 3

Let's start with point R. It can have any of 7 different X-coordinates (-3,-2,-1, 1, 2, 3, 4) and and of the 10 different Y-coordinates (-7,-6,-5,-4,-3,-2,-1,1,2,3).

That is 7x10 different possibilities = 70 different spots for point R

Once we FIX R in one spot, S MUST have the SAME Y-coordinate, BUT a DIFFERENT X-coordinate. We already used one of the X-coordinates when we placed point R, which means....

There are 7-1 = 6 possible spots for point S

With R still in the same spot, we know that point T must have the SAME X-coordinate, but a DIFFERENT Y-coordinate. We already used one of the Y-coordinates when we place point R, which means....

There are 10-1 = 9 possible spots for Point T

We multiply the options: (70)(6)(9) = 3780 different triangles.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich
Thank you rich for the wonderful explanation
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by Divya Ann Chacko » Thu Jun 12, 2014 12:38 pm
Brent@GMATPrepNow wrote:Here's another similar question to practice with: https://www.beatthegmat.com/how-many-tri ... 28974.html

Cheers,
Brent

Thank you Brent for your systematic approach to the questions. The example link helped me a lot.
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