BTGmoderatorLU wrote:Source: GMAT Club Tests

If the vertices of a triangle have coordinates (x, 1), (5, 1), and (5, y) where x < 5 and y > 1, what is the area of the triangle?

(1) x = y.
(2) The angle at the vertex (x, 1) is equal to the angle at the vertex (5, y).

The OA is C.

On a coordinate plane, put the point (5, 1). Since we cannot fix the other two points (x, 1) and (5, y), we can draw the straight line passing through the vertex (5, 1).

Somewhere on the line parallel to X-axis the vertex (x, 1) would lie and somewhere on the line parallel to Y-axis the vertex (5, y) would lie. This implies that it is a right-angled triangle.

Thus, the area of the triangle = 1/2*(5 - x)*(y - 1)

Let's take each statement one by one.

(1) x = y.

Can't get the value of 1/2*(5 - x)*(y - 1). Insufficient.

(2) The angle at the vertex (x, 1) is equal to the angle at the vertex (5, y).

=> This implies that it is an isosceles right-angled triangle.

=> (5 - x) = (y - 1)

Can't get the value of 1/2*(5 - x)*(y - 1). Insufficient.

(1) and (2) together

From x = y and (5 - x) = (y - 1), we get that x = y = 2

Thus, the area of the triangle = 1/2*(5 - x)*(y - 1) = 1/2*(5 - 2)*(2 - 1) = 3 unit

Sufficient.

The correct answer: C

Hope this helps!

-Jay
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BTGmoderatorLU wrote:Source: Official Guide

If the set S consists of five consecutive positive integers, what is the sum of these five integers?

(1) The integer 11 is in S, but 10 is not in S.
(2) The sum of the even integers in S is 26

The OA is D.

This seems to be a different question than the one Jay has quoted. Did you accidentally change it when editing?

BTGmoderatorLU wrote: ↑

Mon Aug 06, 2018 3:13 pm

Source: Official Guide

If the set S consists of five consecutive positive integers, what is the sum of these five integers?

(1) The integer 11 is in S, but 10 is not in S.
(2) The sum of the even integers in S is 26

The OA is D.

Solution:
Question Stem Analysis:


We need to determine the sum of the 5 consecutive integers in set S.

Statement One Alone:

Since 11 is in S and 10 is not, 11 must be the smallest integer in S. Therefore, the sum of the 5 integers in S is 11 + 12 + 13 + 14 + 15 = 65. Statement one alone is sufficient.

Statement Two Alone:

Since there are 5 consecutive integers, either 2 of them are even (if the smallest integer is odd) or 3 of them are even (if the smallest integer is even). If 2 of them are even, we can let the smallest even integer be x, so the larger one is x + 2. We can create the equation:

x + x + 2 = 26

2x = 24

x = 12

This means the integers are 11, 12, 13, 14, and 15, and their sum is 65. If 3 of them are even, we can let the smallest even integer be y, so the larger two are y + 2 and y + 4, respectively. We can create the equation:

y + y + 2 + y + 4 = 26

3y = 20

y = 20/3

Since 20/3 is not an integer (let alone an even integer), it means there can’t be 3 even integers in S. So the only possible sum of the integers in S is 65. Statement two alone is sufficient.

Answer: D