In a recent street fair students were challenged...

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In a recent street fair students were challenged to hit one of shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular region is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 2/3

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

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by EconomistGMATTutor » Sun Nov 19, 2017 10:07 am
Hello LUANDATO.

Here is the key. Let's suppose that the length of the side of the large triangle is 3, so each side of the smalls triangles is equal to 1.

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By the other hand, the unshaded region is a regular hexagon and it can be splited in 6 equilateral triangles (side equal 1).

So, the large triangle has been splited in 9 triangles, all with the same dimensions.

It implies that there are 9 posibilities and only 3 are favoable.

In conlcusion, the probability of hitting a shaded triangle is 3/9=1/3.

The correct answer is C .

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Regards.
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BTGmoderatorLU wrote:
Sat Nov 18, 2017 2:38 pm
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In a recent street fair students were challenged to hit one of shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular region is an equilateral triangle whose side is a third of the length of the large triangle board. If the ping pong ball hits the large triangular region, what is the probability of hitting a shaded triangle?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 2/3

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
We see that each of the smaller shaded equilateral triangles has the same area. Furthermore, the unshaded region is a regular hexagon that can divide into 6 equilateral triangles, each equalling the area of a shaded triangle. Thus there are 3 + 6 = 9 equilateral triangles of the same area in the largest triangle, and the probability of hitting a shaded triangle is therefore 3/9 = 1/3.

Alternate Solution:

Let’s assume that each side of the large triangle is 6 units. The area of the large triangle is thus (1/2)(6)(6√3) = 18√3.

A side of any of the shaded triangles is 2. The area of one shaded triangle is (1/2)(2)(2√3) = 2√3. There are 3 shaded triangles, so their total area is 6√3.

The probability of hitting any shaded triangle is the total area of the shaded triangles divided by the total area of the entire large triangle: 6√3 / 18√3 = 1/3.

Answer: C

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