Question: There are m points on straight line AB & n points on the line AC none of them being the point A and A, B, and C not being on one line. Triangles are formed with these points as vertices, when

(i) A is excluded
(ii) A is included.

The ratio of number of triangles in the two cases is:

(A) \(\frac{m + n - 2}{m + n}\)
(B) \(\frac{m + n - 2}{m + n - 1}\)
(C) \(\frac{m + n - 2}{m + n + 2}\)
(D) \(\frac{n(n - 1)}{(m + 1) (n + 1)}\)
(E) \(\frac{n(n - 1)}{ (m + 1)(n - 1)}\)

Correct Answer: A
Solution and Explanation:
Approach Solution 1:

The problem statement states that:

Given:

• There are m points on the straight line AB
• There are n points on the line AC
• none of them being the point A
• A, B, and C not being on one line.
• Triangles are formed with these points as vertices, when A is excluded and A is included

Find out:

• The ratio of the number of triangles in the two cases.

As per the conditions of the question, we need to derive the two cases to find the ratio of the number of triangles.

Case 1: when A is excluded
We need to select 2 points from 1 line and 1 from the other.

Hence, the total number of ways
= mC2 * nC1 + nC2 * mC1
= m(m − 1)/2 ∗ n + n(n − 1)/2 ∗ m
= mn/2 ∗ (m − 1 + n − 1)
= mn/2 ∗(m + n − 2)

Case 2: when A is included
In addition to the triangles we can form in the above case, we can create a triangle that must include point A.
Therefore, the total number of triangles that can be created, when A must be included
= mC1 * nC1 = mn

Hence, the total number of triangles possible in this case
= mn/2 ∗ (m − 1 + n − 1) + mn
= mn/2 ∗ (m − 1 + n − 1 + 2)
= mn/2 ∗ (m + n)

Therefore, the ratio of the number of triangles in the two cases is:

= \(\frac{mn/2 ∗(m + n − 2)}{mn/2 ∗ (m + n)}\)

= \(\frac{m + n - 2}{m + n}\)

Approach Solution 2:

The problem statement informs that:

Given:

• There are m points on the straight line AB
• There are n points on the line AC
• none of them being the point A
• A, B, and C not being on one line.
• Triangles are formed with these points as vertices, when A is excluded and A is included

Find out:

• The ratio of the number of triangles in the two cases.

The problem can be solved with a quicker and easier approach.

As per the conditions of the questions:
When A is not included, the number of triangles formed= mC2 * nC1 + mC1 * nC2
When A is included, the number of triangles formed = mC2 * nC1 + mC1 * nC2 + mC1 * nC1

Ratio = \(\frac{mC2 * nC1 + mC1 * nC2}{mC2 * nC1 + mC1 * nC2 + mC1 * nC1}\)

= \(\frac{mn/2 (m−1+n−1)}{mn/2 (m−1+n−1+2)}\)

= \(\frac{m+n−2}{m+n}\)

Hence, the ratio of the number of triangles = \(\frac{m+n−2}{m+n}\)

Approach Solution 3:

The problem statement implies that:

Given:

• There are m points on the straight line AB
• There are n points on the line AC
• none of them being the point A
• A, B, and C not being on one line.
• Triangles are formed with these points as vertices, when A is excluded and A is included

Find out:

• The ratio of the number of triangles in the two cases.

1st case: Triangles without vertex A
It is required to select 2 vertices from line AB and one from line AC.
Or two lines AC and one vertice from line AB.
vertices
= mC2 × nC1 + nC2 × mC1
= m(m−1)/2 × n + n(n−1)/2 × m
= mn/2 [m−1+n−1]
= mn/2 (m+n−2)

2nd case: when A is included
(a) we have already 1 vertex (A)
Therefore, we must choose 1 vertex from line AB
and 1 vertex from line AC.
= 1× mC1 × nC1
= mn

(b) If point A is not selected, then the selection is the same as 1st case.
Therefore, the total no. of ways would be = mn(m + n – 2)/2
Hence, the total possibility = mn + mn(m + n – 2)/2
= mn [1 + (m + n – 2)/2]
= {mn(m + n)}/2

Therefore, the ratio of the number of triangles = \(\frac{mn(m+n−2)/2}{mn(m + n)/2} = \frac{m + n - 2}{m + n}\)

“There are m points on straight line AB & n points on the line AC”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This question has been taken from the book “GMAT For Dummies 2021”. To solve the GMAT Problem Solving questions, the candidates must hold a basic concept of arithmetic, geometry and algebra. The candidates can analyse GMAT Quant practice papers to get acquainted with varieties of questions in order to score better in the exam.

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