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by arfabe16 » Tue Oct 05, 2010 7:16 pm
Here are two GMAT math problems that I don't know how to solve. I would greatly appreciate your help in figuring out the process

1. A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?
a) Ï€r^2
b) Ï€r^2 + 10
c) Ï€r^2 + (1/4)(Ï€^2)(r^2)
d) Ï€r^2 + (40 - 2Ï€r)^2
e) Ï€r^2 + (10 - .5Ï€r)^2

2. If 2^x - 2^(x-2) = 3(2^13), what is the value of x?
a) 9
b) 11
c) 13
d) 15
e) 17

THANK YOU!

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by arfabe16 » Tue Oct 05, 2010 7:31 pm
Never mind about the first one. I just figured it out. Still trying to figure out a quick way to solve the 2nd one though. Thanks!

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by deepakb » Tue Oct 05, 2010 7:45 pm
If 2^x - 2^(x-2) = 3(2^13), what is the value of x?

LHS = 2^x - 2^(x-2) = 2^x - 2^x *2^-2 = 2^x - 2^x /2^2 = 2^x - 2^x /2^2
= 2^x - 2^x /4 = (4*2^x - 2^x)/4

2^x(4-1) = 4*3(2^13)
2^x(3) = 4*3(2^13)
2^x = 4(2^13)
2^x = 2^15
x = 15

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by saurabhmahajan » Wed Oct 06, 2010 5:28 am
whats the OA for 1st question ?
Thanks and regards,
Saurabh Mahajan

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by GMATGuruNY » Wed Oct 06, 2010 6:17 am
arfabe16 wrote:Here are two GMAT math problems that I don't know how to solve. I would greatly appreciate your help in figuring out the process

1. A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in terms of r?
a) Ï€r^2
b) Ï€r^2 + 10
c) Ï€r^2 + (1/4)(Ï€^2)(r^2)
d) Ï€r^2 + (40 - 2Ï€r)^2
e) Ï€r^2 + (10 - .5Ï€r)^2
We can plug in values and ballpark.

Let r = 2, let pi = 3.
Area of circle = 4pi = 12
Circumference = 4pi = 12
Remaining wire = 40-12 = 28. This is the perimeter of the square, so s = 28/4 = 7.
Area of square = 7^2 = 49.
Combined areas = 12+49 = 61. This is our target answer.

Ï€r^2 + (10 - .5Ï€r)^2 = 3*(2^2) + (10 - .5*3*2)^2 = 12+49 = 61.
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by GMATGuruNY » Wed Oct 06, 2010 6:26 am
arfabe16 wrote:Here are two GMAT math problems that I don't know how to solve. I would greatly appreciate your help in figuring out the process

2. If 2^x - 2^(x-2) = 3(2^13), what is the value of x?
a) 9
b) 11
c) 13
d) 15
e) 17
We can plug in the answer choices for x.

2^13 - 2^(13-2) = 3(2^13)
2^13 - 2^11 = 3(2^13)
2^11(2^2 - 1) = 3(2^13)
2^11(3) = 3(2^13)

Plugging in x=13 made the exponent on the left 2^11.
To match 2^13 on the right side of the equation, the exponent needs to be increased by 2.
Thus, x = 13+2 = 15.

Mitch Hunt
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by Yanat » Wed Oct 06, 2010 6:42 am
For Question 1, it is E

and for For Question 2 it is D, i.e. x = 15

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