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

I solved it twice and am getting $1320 as the ans...

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

[spoiler] OA: B[/spoiler]

Tough question man. Was taking too long to solve it, found the answer on https://www.beatthegmat.com/percentages-t88524.html

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

Rahul,

Dude, please check the OA of your question again. It had me confused. If possible, could you cite the source of this question ?

Any other solution

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]

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

Another approach is to guess and check.

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.