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
3 Questions!
This topic has expert replies
- AIM TO CRACK GMAT
- Senior | Next Rank: 100 Posts
- Posts: 80
- Joined: Sat Sep 15, 2012 1:07 am
- Thanked: 1 times
- Followed by:1 members
- sahilchaudhary
- Master | Next Rank: 500 Posts
- Posts: 153
- Joined: Mon Apr 11, 2011 7:13 am
- Location: India
- Thanked: 22 times
- Followed by:7 members
- GMAT Score:540
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!
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!
Sahil Chaudhary
If you find this post helpful, please take a moment to click on the "Thank" icon.
https://www.sahilchaudhary007.blocked
If you find this post helpful, please take a moment to click on the "Thank" icon.
https://www.sahilchaudhary007.blocked
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
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
Cheers,
Brent
- AIM TO CRACK GMAT
- Senior | Next Rank: 100 Posts
- Posts: 80
- Joined: Sat Sep 15, 2012 1:07 am
- Thanked: 1 times
- Followed by:1 members
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 advancesahilchaudhary 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.
- sahilchaudhary
- Master | Next Rank: 500 Posts
- Posts: 153
- Joined: Mon Apr 11, 2011 7:13 am
- Location: India
- Thanked: 22 times
- Followed by:7 members
- GMAT Score:540
When you increase x to 800%, it becomes 8xAIM TO CRACK GMAT wrote: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 advancesahilchaudhary 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.
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.
Sahil Chaudhary
If you find this post helpful, please take a moment to click on the "Thank" icon.
https://www.sahilchaudhary007.blocked
If you find this post helpful, please take a moment to click on the "Thank" icon.
https://www.sahilchaudhary007.blocked
- sahilchaudhary
- Master | Next Rank: 500 Posts
- Posts: 153
- Joined: Mon Apr 11, 2011 7:13 am
- Location: India
- Thanked: 22 times
- Followed by:7 members
- GMAT Score:540
Right @Brent.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
Sahil Chaudhary
If you find this post helpful, please take a moment to click on the "Thank" icon.
https://www.sahilchaudhary007.blocked
If you find this post helpful, please take a moment to click on the "Thank" icon.
https://www.sahilchaudhary007.blocked
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Let the original area = π.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%
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Let the tank = 6 gallons.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
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Let C = the cost price.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
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3