Two rays P1 and P2 originate from point O, such that there are a point

Question

Two rays P1 and P2 originate from point O, such that there are a points on one-ray and b points on the other, excluding O. How many triangles can be formed with these points as vertices if point O is not a vertex of any such triangle?

A. ½ (ab) (a + b – 1)
B. ½ ab
C. ½ (ab) (a + b)
D. ½ (ab) (a + b – 2)
E. ½ (ab) (a + b – 3)

A. ½ (ab) (a + b – 1)
B. ½ ab
C. ½ (ab) (a + b)
D. ½ (ab) (a + b – 2)
E. ½ (ab) (a + b – 3)

shridhar786 · Answer

suppose P1 contains a points and P2 contains b points
to form triangle we need three points
1st case 2 points from P1 and 1 point from P2 
 ac2*bc1
2nd case 2 points from P2 and 1 point from P1
 bc2*ac1

adding both
 ac2*bc1 + bc2*ac1
  \frac{a!}{2!(a-2)!} *b +  \frac{b!}{2!(b-2)!} *a
  \frac{(a-1)a*b}{2}  +  \frac{b(b-1)}{2} *a
 \frac{{a^2}b-ab+{b^2}a-ab}{2} 
 \frac{{a^2}b+{b^2}a-2ab}{2} 
 \frac{ab(a+b-2)}{2} 
D is the answer

firas92 · Answer

I think the quickest way would be by assigning values for a & b

Let a=2, b=3

A total of 9 triangles are possible in this case - with the base at P1, 3 triangles are possible and with the base at P2, 6 triangles are possible

Go back to the answer choices and plug in a=2, b=3 to see which option yields 9

Answer is  (D)

Hit Kudos if this helped!

CrackverbalGMAT · Answer

The simplest approach is the best approach to solve this question. If we take a=2 and b=2, we can practically count the number of triangles we can form. Whichever option corresponds to this number, when we plug in the values of a and b, should be the answer.

The diagram, corresponding to the question should look like this:

Attachment:

13th June 2019 - Reply 2.JPG [ 13.96 KiB | Viewed 2401 times ]

Let us name the two points on P1 as 1 and 2; let the points on P2 be 3 and 4. Then, the possible triangles are:

1-3-4, 2-3-4, 1-2-3 and 1-2-4.

So, we have a total of 4 triangles.

When we plug in the values of a and b in option A, we get 6; so option A is not the right answer. When we carry on in a similar fashion, we can eliminate options B, C and E. Option D is the right answer.

Of course, this question can also be solved using the method of combinations, as highlighted by the other experts on this thread. For those of you who are not able to think of that method, taking values is more viable, as we have shown.

Hope this helps!

saukrit · Answer

For every point amongst the  a  points on p1, we need to choose 2 from b points
Triangles with one pointfrom P1 and and 2 pts from P2 = a* (bC2) = a*b*(b-1)/2

Similarly for the other side(P2), No of tiangles= b*(aC2)=b*a*(a-1)/2

Total points = ab(a-1)/2 + ab(b-1)/2 = ab(a-1+b-1)= ab/2(a+b-2)

The answer is D

PSKhore · Answer

Take the easiest numbers and plugin them in the options
Do not take 1 & 2 points because they'll fit in in 2-3 options.
Take P1=1 P2=3 points so that it'll be easier to calculate the triangles as well as easier to check the options
With these values, we get 3 triangles (you can draw and verify)

Option D satisfies our values.

Two rays P1 and P2 originate from point O, such that there are a point

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GMAT Problem Solving (PS) Questions

Two rays P1 and P2 originate from point O, such that there are a point

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

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