A firm's annual revenue grows twice as fast as its costs. In 2007 it operated at a $1000 loss, it broke even in 2008, and in 2009 its revenues were 44% higher than in 2007. If the firm's revenues and costs grew at a constant rate over this period, what was its profit in 2009?
A $700
B $1000
C $1300
D $1600
E $2000
[spoiler] OA: B[/spoiler]
Firm's revenues
This topic has expert replies
- rijul007
- Legendary Member
- Posts: 588
- Joined: Sun Oct 16, 2011 9:42 am
- Location: New Delhi, India
- Thanked: 130 times
- Followed by:9 members
- GMAT Score:720
I solved it twice and am getting $1320 as the ans...A firm's annual revenue grows twice as fast as its costs. In 2007 it operated at a $1000 loss, it broke even in 2008, and in 2009 its revenues were 44% higher than in 2007. If the firm's revenues and costs grew at a constant rate over this period, what was its profit in 2009?
A $700
B $1000
C $1300
D $1600
E $2000
-
- Senior | Next Rank: 100 Posts
- Posts: 90
- Joined: Thu Nov 05, 2009 9:14 am
- Thanked: 5 times
- Followed by:3 members
Tough question man. Was taking too long to solve it, found the answer on https://www.beatthegmat.com/percentages-t88524.htmlrahulvsd wrote:A firm's annual revenue grows twice as fast as its costs. In 2007 it operated at a $1000 loss, it broke even in 2008, and in 2009 its revenues were 44% higher than in 2007. If the firm's revenues and costs grew at a constant rate over this period, what was its profit in 2009?
A $700
B $1000
C $1300
D $1600
E $2000
[spoiler] OA: B[/spoiler]
First is to figure out that if the revenue increases 44% over two years that the yearly increas is 20% (20% of 120% is 144%) Which means that costs increase by 10% each year. we have the profit (R - C) for each of the first two years. I found it helpful to create a chart
R7 1.2R7 1.44R7
C7 1.1C7 1.21C7
-1000 0 ???
Now it is just about doing the algebra:
you know that 1.2R7 = 1.1C7 and that R7 - C7 = -1000
You can rewrite R7 = C7 - 1000 which means that 1.2(C7 - 1000) = 1.1C7 which simplifies to
1.2C7 - 1200 = 1.1C7
.1C7 = 1200
C7 = 12000
which means R7 = 11000.
Now you have to put that into the final column:
1.44(11000) - 1.21 (12000) = ?
15840 - 14520 = 1320
so the closest answer is 1300
-
- Senior | Next Rank: 100 Posts
- Posts: 90
- Joined: Thu Nov 05, 2009 9:14 am
- Thanked: 5 times
- Followed by:3 members
Rahul,
Dude, please check the OA of your question again. It had me confused. If possible, could you cite the source of this question ?
Dude, please check the OA of your question again. It had me confused. If possible, could you cite the source of this question ?
- Jim@StratusPrep
- MBA Admissions Consultant
- Posts: 2279
- Joined: Fri Nov 11, 2011 7:51 am
- Location: New York
- Thanked: 660 times
- Followed by:266 members
- GMAT Score:770
Let's set some variables first
R = Revenue
C = Costs
x =1 + (constant growth rate of revenue (thus .5x = growth of costs))
2007
C - R = 1000
2008
(1 + .5x)C - (1 + x) R = 0
2009
Rx^2 = 1.44R
the R's cancel leaving: x ^2 = 1.44 or x = 1.2
Plugging this into the 2008 equation you get:
(1 +.5(.2))C - 1.2R = 0
1.1C = 1.2R
C = (12/11)R
Using this you can go back to the first equation:
(12/11)R - R = 1000
(1/11)R = 1000
R = 11000
Revenue in 2009 is thus:
11,000 * 1.44 = 15,840
Costs in 2009 = 2008 cost * 1.1 and 2008 costs = 2007 Rev * 1.2
2007 Rev * 1.2 = 11000 * 1.2 = 13,200
13,200 * 1.1 = 14,520
The answer is then 15,840 - 14,520, which is 1,320 so the answer is C.
Hope that helps...[/u]
R = Revenue
C = Costs
x =1 + (constant growth rate of revenue (thus .5x = growth of costs))
2007
C - R = 1000
2008
(1 + .5x)C - (1 + x) R = 0
2009
Rx^2 = 1.44R
the R's cancel leaving: x ^2 = 1.44 or x = 1.2
Plugging this into the 2008 equation you get:
(1 +.5(.2))C - 1.2R = 0
1.1C = 1.2R
C = (12/11)R
Using this you can go back to the first equation:
(12/11)R - R = 1000
(1/11)R = 1000
R = 11000
Revenue in 2009 is thus:
11,000 * 1.44 = 15,840
Costs in 2009 = 2008 cost * 1.1 and 2008 costs = 2007 Rev * 1.2
2007 Rev * 1.2 = 11000 * 1.2 = 13,200
13,200 * 1.1 = 14,520
The answer is then 15,840 - 14,520, which is 1,320 so the answer is C.
Hope that helps...[/u]
GMAT Answers provides a world class adaptive learning platform.
-- Push button course navigation to simplify planning
-- Daily assignments to fit your exam timeline
-- Organized review that is tailored based on your abiility
-- 1,000s of unique GMAT questions
-- 100s of handwritten 'digital flip books' for OG questions
-- 100% Free Trial and less than $20 per month after.
-- Free GMAT Quantitative Review
-- Push button course navigation to simplify planning
-- Daily assignments to fit your exam timeline
-- Organized review that is tailored based on your abiility
-- 1,000s of unique GMAT questions
-- 100s of handwritten 'digital flip books' for OG questions
-- 100% Free Trial and less than $20 per month after.
-- Free GMAT Quantitative Review
- 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
Another approach is to guess and check.rahulvsd wrote:A firm's annual revenue grows twice as fast as its costs. In 2007 it operated at a $1000 loss, it broke even in 2008, and in 2009 its revenues were 44% higher than in 2007. If the firm's revenues and costs grew at a constant rate over this period, what was its profit in 2009?
A $700
B $1000
C $1300
D $1600
E $2000
Revenues:
An increase of 44% over 2 years implies an increase of 20% each year.
To illustrate:
100 + .2(100) = 120.
120 + .2(120) = 144.
Percent increase = (144-100)/100 = 44%.
Costs:
Since the revenues grow twice as fast, the costs increase 10% each year.
Given that the loss in 2007 is $1000, the revenues in 2007 are almost certainly a multiple of 1,000.
Let's start with a nice, round number.
Case 1: Revenues in 1007 = 10,000.
Since the loss = 1000, the costs = 11,000.
2008:
Revenues = 10,000 + .2(10,000) = 12,000.
Costs = 11,000 + .1(11,000) = 12,200.
In order to break even, the revenues needs to increase just a bit.
Case 2: Revenues in 2007 = 11,000.
Since the loss = 1000, the costs = 12,000.
2008:
Revenues = 11,000 + .2(11,000) = 13,200.
Costs = 12,000 + .1(12,000) = 13,200. Success!
2009:
Revenues = 13,200 + .2(13,200) = 15,840.
Costs = 13,200 + .1(13,200) = 14,520.
Profit = 15,840 - 14,520 = 1320.
The closest 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