In a recent street fair students were challenged to hit one

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

In a recent street fair students were challenged to hit one

by BTGmoderatorLU » Wed Nov 07, 2018 2:32 pm

Timer

00:00

Your Answer

Global Stats

Source: Economist GMAT

In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions 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

Quote

Source: Beat The GMAT — Problem Solving | Wed Nov 07, 2018 2:32 pm

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pm; Thanked: 59 times; Followed by:33 members

by fskilnik@GMATH » Wed Nov 07, 2018 4:11 pm

BTGmoderatorLU wrote:Source: Economist GMAT

In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions 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 in a random point, what is the probability of hitting a shaded triangle?

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

\[? = P\left( {{\text{hit}}\,\,{\text{shaded}}\,\,{\text{region}}} \right)\]
\[\frac{{{S_{{\text{each}}\,\Delta {\text{shaded}}}}}}{{{S_{\Delta {\text{large}}}}}} = {\left( {\frac{1}{3}} \right)^2} = \frac{1}{9}\,\,\,\,\,\,\,\left[ {\,{\text{each}}\,\,\Delta {\text{shaded}}\,\,{\text{is}}\,\,{\text{similar}}\,\,{\text{to}}\,\,{\text{the}}\,\,\Delta {\text{large}}\,} \right]\]
\[? = 3 \cdot \frac{1}{9} = \frac{1}{3}\,\,\,\,\,\,\left( {{\text{geometric}}\,\,{\text{probability}}} \right)\]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

Quote

GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8083; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Nov 09, 2018 12:19 pm

BTGmoderatorLU wrote:Source: Economist GMAT

In a recent street fair students were challenged to hit one of the shaded triangular regions on the large equilateral triangular board below with a ping pong ball. Each of the triangular regions 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

We see that each of the smaller shaded equilateral triangle has the same area. Furthermore, the unshaded region is a regular hexagon that can divide into 6 equilateral triangles each equalling to the area of a shaded triangle. Thus there are 3 + 6 = 9 equilateral triangles of the same area and the probability hitting a shaded triangle is 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)(6âˆš3) = 18âˆš3.

A side of any of the shaded triangles is 2. The area of one shaded triangle is (1/2)(2)(2âˆš3) = 2âˆš3. There are 3 shaded triangles, so their total area is 6âˆš3.

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: 6âˆš3 / 18âˆš3 = 1/3.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]