Combinations: Please help?

Question

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

Rich.C@EMPOWERgmat.com · Answer

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

Contact Rich at [email protected]

GMATGuruNY · Answer

I posted a solution to a very similar problem here: https://www.beatthegmat.com/how-many-tri ... 13159.html

Brent@GMATPrepNow · Answer

Here's another similar question to practice with: https://www.beatthegmat.com/how-many-tri ... 28974.html Cheers, Brent

Divya Ann Chacko · Answer

Thank you Guruny for providing a similar example. I understood the explanation given. Thank you. Thanks Divya

Divya Ann Chacko · Answer

Rich.C@EMPOWERgmat.com 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

Divya Ann Chacko · Answer

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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