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3 Questions!

by AIM TO CRACK GMAT » Sun Oct 20, 2013 10:49 am
1)The area of a circle is increased by 800%. By what percent has the diameter of the circle increased?
(A) 100%
(B) 200%
(C) 300%
(D) 600%

B

2) Pipe A can fill a tank in 3 hours. Pipe B can empty the same tank in 2 hours. If the tank is initially two thirds full then in how many hours will the tank be completely empty in if both the pipes are turned on simultaneously?

(A)2 hours (B) 3 hours (C) 4 hours (D)5 hours

C


3)The profit earned when an article is sold for $ 600 is double the loss incurred when the article is sold for $150. At what price should the article be sold if a profit of 20% is to be earned?
(A) $150 (B) $300 (C) $360 (D) $400

C
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by sahilchaudhary » Sun Oct 20, 2013 11:22 am
1. Let initial diameter = d1, final diameter = d2
Initial area = ((pi)d1^2)/4
Final area = ((pi)d2^2)/4
Final area = initial area + (800/100 * initial area) = 9*initial area = 9*((pi)d1^2)/4
Equate the above 2 final areas.
((pi)d2^2)/4 = 9*((pi)d1^2)/4
d2 = 3*d1
% increase = ((d2 - d1)*100)/d1 = ((3*d1 - d1)*100)/d1 = 200%

So, B.

2. General Formula for completely filling 2 tanks simultaneously = A*B/(A+B). If A fills and B drains, take A +ve and B -ve.
If the answer is +ve, then it will fill.
If the answer is -ve, then it will empty.

Applying this formula here
Time taken to fill/empty the full tank = 3*(-2)/(3+(-2)) = -6 hours
So, it implies, it will empty in 6 hrs.

1 full tank can be emptied in 6 hrs.
2/3 tank can be emptied in 4 hrs.

So, C.

3. Let C.P. = x
Profit, P = 600 - x
Loss, L = x - 150

Given, P = 2*L
600 - x = 2 * (x - 150)
x = 300

Now, to get 20% profit
Actual Profit = (20/100) * 300 = 60
Therefore, S.P. = 300 + 60 = 360

So, C.

Hope that helps!
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by Brent@GMATPrepNow » Sun Oct 20, 2013 11:24 am
In the future, please post only one question per thread. Otherwise things can become pretty complicated when there are discussions on multiple questions.

Cheers,
Brent
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by AIM TO CRACK GMAT » Sun Oct 20, 2013 11:28 am
sahilchaudhary wrote:1. Let initial diameter = d1, final diameter = d2
Initial area = ((pi)d1^2)/4
Final area = ((pi)d2^2)/4
Final area = initial area + (800/100 * initial area) = 9*initial area = 9*((pi)d1^2)/4
Equate the above 2 final areas.
((pi)d2^2)/4 = 9*((pi)d1^2)/4
d2 = 3*d1
% increase = ((d2 - d1)*100)/d1 = ((3*d1 - d1)*100)/d1 = 200%

So, B.
Thanks a ton!!! i really needed these explanations! Ill b glad if u cud xplain me the 1st one in a simpler way! plz..Thanks in advance
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by sahilchaudhary » Sun Oct 20, 2013 11:48 am
AIM TO CRACK GMAT wrote:
sahilchaudhary wrote:1. Let initial diameter = d1, final diameter = d2
Initial area = ((pi)d1^2)/4
Final area = ((pi)d2^2)/4
Final area = initial area + (800/100 * initial area) = 9*initial area = 9*((pi)d1^2)/4
Equate the above 2 final areas.
((pi)d2^2)/4 = 9*((pi)d1^2)/4
d2 = 3*d1
% increase = ((d2 - d1)*100)/d1 = ((3*d1 - d1)*100)/d1 = 200%

So, B.
Thanks a ton!!! i really needed these explanations! Ill b glad if u cud xplain me the 1st one in a simpler way! plz..Thanks in advance
When you increase x to 800%, it becomes 8x
When you increase x by 800%, it becomes x + 8x = 9x

I think this is the easiest way I could explain this.
Please ask a specific question and I would be more than happy to help you.
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by sahilchaudhary » Sun Oct 20, 2013 11:52 am
Brent@GMATPrepNow wrote:In the future, please post only one question per thread. Otherwise things can become pretty complicated when there are discussions on multiple questions.

Cheers,
Brent
Right @Brent.
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by GMATGuruNY » Sun Oct 20, 2013 2:57 pm
AIM TO CRACK GMAT wrote:1)The area of a circle is increased by 800%. By what percent has the diameter of the circle increased?
(A) 100%
(B) 200%
(C) 300%
(D) 600%
Let the original area = π.
Here:
πr² = π
r² = 1
r = 1
d = 2.

Increased by 800%, the new area = π + (800/100)π = 9π.
Here:
πr² = 9π
r² = 9
r = 3
d = 6.

Percent increase in the diameter = difference/original * 100 = (6-2)/2 * 100 = 200%.

The correct answer is B.
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by GMATGuruNY » Sun Oct 20, 2013 3:06 pm
AIM TO CRACK GMAT wrote:Pipe A can fill a tank in 3 hours. Pipe B can empty the same tank in 2 hours. If the tank is initially two thirds full then in how many hours will the tank be completely empty in if both the pipes are turned on simultaneously?

(A)2 hours (B) 3 hours (C) 4 hours (D)5 hours
Let the tank = 6 gallons.

Since A can fill the tank in 3 hours, A's rate = w/t = 6/3 = 2 gallons per hour.
Since B can empty the tank in 2 hours, B's rate = w/t = 6/2 = 3 gallons per hour.
Since A pumps IN 2 gallons each hour, while B pumps OUT 3 gallons each hour, the NET LOSS each hour = 1 gallon.

When the tank is 2/3 full, the volume = (2/3) * 6 = 4 gallons.
Time for the tank to empty = (volume in the tank)/(net loss rate) = 4/1 = 4 hours.

The correct answer is C.
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by GMATGuruNY » Sun Oct 20, 2013 3:57 pm
AIM TO CRACK GMAT wrote:The profit earned when an article is sold for $ 600 is double the loss incurred when the article is sold for $150. At what price should the article be sold if a profit of 20% is to be earned?

(A) $150 (B) $300 (C) $360 (D) $400
Let C = the cost price.

The profit earned when an article is sold for $ 600 is double the loss incurred when the article is sold for $150.
Represented on a number line:
150<----loss of x---->C<----profit of 2x---->600
The number line indicates that the distance between 150 and 600 is equal to x+2x.
Thus:
600-150 = x+2x
450 = 3x
x = 150.

Implication:
C = 150 + x = 150 + 150 = 300.
Selling price required to yield a 20% profit = 300 + (20/100)(300) = 360.

The correct answer is C.
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