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GMAT Prep - triangle and aveage problem

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GMAT Prep - triangle and aveage problem

by lorivogue » Sat Mar 14, 2009 9:38 am
Some assistance on two GMATPrep problems would be appreciated!

1) Two triangles are shown: Not drawn to scale. The triangle on the left is smaller, called s. The triangle on the right is larger, called S. The angles are, x, y, and z for both.

Question: If the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S=

A) Sq.rt 2/2 *s

B) Sq.rt 3/2 * s

C) Sq.rt 2

D) Sq.rt 3

E) 2s



2) A certain car averages 25 miles per gallon of gas in the city and 40 miles per gallon of gas on highway. According to these rates, which of the following is the closest to the number of miles per gallon that the car averages when it is driver 10 miles in the city and 50 miles on the highway.

A)28
B)30
C)33
D)36
E)38
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by DanaJ » Sat Mar 14, 2009 10:07 am
Q1. Remember that Area of triangle = (base) * (height)/2.
Now, for the smaller triangle, you get that Area = s*h/2
For the larger triangle, you get Area = S*H/2
Since the area of the small triangle is half the area of the big one, you get that S*H/2 = s*h, or that S*H = 2s*h, or that S = 2sh/H

Now, you need to use the fact that the two triangles are similar. This happens because all the angles are equal. In similar triangles, proportions are kept between corresponding lengths, whether it's lengths of sides or lengths of heights, so you can safely conclude that S/s = H/h.
This means that S = (H/h)*s. Now, equate this with what we already have to get that:
sH/h = 2sh/H
H/h = 2h/H
This last one is equivalent to H^2 = 2h^2, making H = h*sqrt(2).
Now, what you should do is replace H in any of the two equations regarding S in order to simplify h: S = 2sh/H = 2sh/[h*sqrt(2)] - simplify h to get:

S = 2s/[sqrt(2)] - and since 2 = sqrt(2)*sqrt(2), you get the final answer:
S = s*sqrt(2).


Q2. Miles per gallon in the city = 25, meaning you get 1 mile in the city by using up 1/25 gallons.

Miles per gallon on the highway = 40, meaning you get 1 mile on the highway by using 1/40 gallons.

Now, drive 10 miles in the city and 50 miles on the highway and use:
10*1/25 + 50*1/40 = 33/20 gallons.
Your total distance was = 10 miles + 50 miles = 60 miles and you used 33/20 gallons of gas, which gives you an average of 60/(33/20) = 36.3 miles per gallon.
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by lorivogue » Sat Mar 14, 2009 10:42 am
Thank you! Very clear!
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by franciskyle » Sat Mar 14, 2009 6:34 pm
I arrived at a different answer for the triangle question. I looked at it the following way... is there a mistake in my logic?

eqn 1: S = (B*H)/2

and

eqn 2: 2s = (B*H)/2

Therefore

Rearrange eqn 2: 2s = (B*H)/2 --> B*H = 4s

Plug eqn 2 into eqn 1

S = (4s)/2

Therefore

S = 2s
k. Francis
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by DanaJ » Sat Mar 14, 2009 11:11 pm
franciskyle: The problem is that S and s are both basis of the triangles respectively, they are not surfaces or areas.
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