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Joe and the lamp post

Expert replies
by Brent@GMATPrepNow » Sun Dec 28, 2008 11:20 am
When Joe stands 12 meters away from a lamp post, his shadow is 3 meters long. If the lamp post is 9 meters high, what is Joe’s height in meters?
A) 13/8
B) 5/3
C) 9/5
D) 16/9
E) 9/4
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Source: — Problem Solving |

by cramya » Sun Dec 28, 2008 11:22 am
12 / 3 = 9/x

x=9/4 = 2.25

Choose E)
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by vivek.kapoor83 » Sun Dec 28, 2008 11:26 am
same as cramya...it is E.
Cramya...pls reply my PM..I m waiting.

9/12=x/3
x = 9/4
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by parallel_chase » Sun Dec 28, 2008 11:29 am
I totally agree with cramya's solution as posted above. Just want to elucidate the fact:

after drawing the figure, we get two similar triangles, with the same base and having 1 angle of 90 degrees each, one with base 12 and another with base 3.
The triangle with base 12 has a height 9.

Therefore,

12/9 = 3/x

x = 9/4.

Hence E.
No rest for the Wicked....
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by Brent@GMATPrepNow » Sun Dec 28, 2008 11:40 am
The answer isn't E, but you are correct about the similar triangles.
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by parallel_chase » Sun Dec 28, 2008 11:47 am
Hey Brent is it 9/5?

I hope this is correct.
No rest for the Wicked....
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by vivek.kapoor83 » Sun Dec 28, 2008 11:49 am
is the ans C...i solved it by trignometry.
after drawing fig and solving by Tangent x = p/b =9/15
and then using similar trinangle concept and using the vaule of tan x
i got it to C...is it the rite choice.9/5
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by Brent@GMATPrepNow » Sun Dec 28, 2008 11:57 am
Nice work.

There are two similar triangles:

Triangle 1: Height = 9 (lamp post) and base = 15 (12 meters PLUS 3 meter shadow)

Triangle 2: Height = x (Joe) and base = 3 (shadow).

We get 9/15 = x/3 (or 9/x = 15/3 . . . etc)

Solve for x to get 9/5 (C)
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by vivek.kapoor83 » Sun Dec 28, 2008 11:59 am
was trigonometry d rite approach here..coz we use to solve it in school classes..similar ques i rem..
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by mental » Sun Dec 28, 2008 10:50 pm
tried to give the solution

Its same as Brent's.
Attachments
Sol shadow.jpg
Similar triangles
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