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A firm's annual revenue grows twice as fast as its costs. In
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27 Mar 2013, 08:15
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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 found this problem quite difficult to be solved under 2 mins. Will be really keen to know if there are any shortcuts to handle such Questions. Would you rate this problem a 700 + ??
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Re: A firm's annual revenue grows twice as fast as its costs. In
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16 Aug 2013, 19:23
1. Let the revenues and costs in 2007 be r1 and c1 respectively. We have r1c1=1000 2. Let revenues in 2008 be r2 and c2 respectively. We have r2c2=0 3. Let revenues and costs in 2009 be r3 and c3 respectively. we have r3=1.44r1. 4. Since the growth rate of sales is constant, growth rate per year is 20%. growth rate of costs is half of that which is 10% 5. (2) can be written as 1.2 r11.1 c1=0 6. Solving for (1) and (5), we have c1=12000 andr1=11000 7. r3=1.44*11000= 15840 and c3=1.21*12000=14520 8. Therefore profit in 2009=1584014520=1320 The answer should be $1320
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Re: A firm's annual revenue grows twice as fast as its costs
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27 Mar 2013, 12:48
mridulparashar1 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 First solve for how much revenue is increasing each year: R 2008 = R 2007*(1+x) R 2009 = R 2007*(1+x)^2=1.44*R 2007(1+x)^2 = 1.44 1+x=1.2 x=0.2 aka revenue increases 20% each year and cost increases 10% annually Next solve for R 2007 and C 2007: R 2007 = C 2007  1000 1.2*R 2007  1.1*C 2007 = 0 1.2*[C 2007  1000]  1.1*C 2007 = 0 0.1*C 2007 = 1,200 C 2007 = 12,000 R 2007 = 11,000 Finally find 2009 profits: Profit 2009 = 1.44*11,000  1.21*12,000 Profit 2009 = 15,840  14,520 Profit 2009 = 1320 Answer: C



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Re: A firm's annual revenue grows twice as fast as its costs. In
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16 Aug 2013, 17:36
2013gmat 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
can anyone explain above question ? thanks By strict definitions, revenue can not be negative. But in the corrupt world of finance, revenue is a number between positive infinity and negative infinity depending on how the upper management instruct the finance people how to lie on the spreadsheets in their submissions to the Securities and exchange commission.
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Re: A firm's annual revenue grows twice as fast as its costs. In
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17 Aug 2013, 02:13
SravnaTestPrep wrote: 1. Let the revenues and costs in 2007 be r1 and c1 respectively. We have r1c1=1000 2. Let revenues in 2008 be r2 and c2 respectively. We have r2c2=0 3. Let revenues and costs in 2009 be r3 and c3 respectively. we have r3=1.44r1. 4. Since the growth rate of sales is constant, growth rate per year is 20%. growth rate of costs is half of that which is 10% 5. (2) can be written as 1.2 r11.1 c1=0 6. Solving for (1) and (5), we have c1=12000 andr1=11000 7. r3=1.44*11000= 15840 and c3=1.21*12000=14520 8. Therefore profit in 2009=1584014520=1320
The answer should be $1320 How did you deduce that "growth rate per year is 20%."????
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Re: A firm's annual revenue grows twice as fast as its costs. In
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17 Aug 2013, 03:32
jaituteja wrote: SravnaTestPrep wrote: 1. Let the revenues and costs in 2007 be r1 and c1 respectively. We have r1c1=1000 2. Let revenues in 2008 be r2 and c2 respectively. We have r2c2=0 3. Let revenues and costs in 2009 be r3 and c3 respectively. we have r3=1.44r1. 4. Since the growth rate of sales is constant, growth rate per year is 20%. growth rate of costs is half of that which is 10% 5. (2) can be written as 1.2 r11.1 c1=0 6. Solving for (1) and (5), we have c1=12000 andr1=11000 7. r3=1.44*11000= 15840 and c3=1.21*12000=14520 8. Therefore profit in 2009=1584014520=1320
The answer should be $1320 How did you deduce that "growth rate per year is 20%."???? The rate can be calculated from the formula (1+r/100) ^2 = 1.44 , which will give r=20
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A firm's annual revenue grows twice as fast as its costs.
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13 Feb 2014, 09:06
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 Also whats the level of this question ?



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Re: A firm's annual revenue grows twice as fast as its costs.
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13 Feb 2014, 10:26
akankshasoneja 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 Also whats the level of this question ? Merging similar topics. Please refer to the solutions above. As for the difficulty level, I'd say it's ~650700.
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Re: A firm's annual revenue grows twice as fast as its costs. In
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16 Apr 2014, 08:14
I am lost. Can't solve it.



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Re: A firm's annual revenue grows twice as fast as its costs. In
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17 Apr 2014, 22:26
tough one for sure. 3rd year revenue going up by 44% = 1st year revenue * (12^2/10^2) so 2nd year revenue is 1st year revenue * (12/10) > 1st year revenue + 1st year revenue (2/10) so growth rate of revenue is 20% and costs is 10%
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Re: A firm's annual revenue grows twice as fast as its costs. In
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17 Sep 2014, 14:29
SravnaTestPrep wrote: 1. Let the revenues and costs in 2007 be r1 and c1 respectively. We have r1c1=1000 2. Let revenues in 2008 be r2 and c2 respectively. We have r2c2=0 3. Let revenues and costs in 2009 be r3 and c3 respectively. we have r3=1.44r1. 4. Since the growth rate of sales is constant, growth rate per year is 20%. growth rate of costs is half of that which is 10% 5. (2) can be written as 1.2 r11.1 c1=0 6. Solving for (1) and (5), we have c1=12000 andr1=11000 7. r3=1.44*11000= 15840 and c3=1.21*12000=14520 8. Therefore profit in 2009=1584014520=1320
The answer should be $1320 Something is messing up my calculations. I substituted the amount by which C and R go up with X. This meant that at 2009 we needed to find out : R*x^4  C*X^2. Since X^4 = 1.44 > X^2 = 1.2 And that's why I multiplied C by 1.2 and not 1.21. Why is this wrong?



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Re: A firm's annual revenue grows twice as fast as its costs. In
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17 Sep 2014, 22:15
ronr34 wrote: SravnaTestPrep wrote: 1. Let the revenues and costs in 2007 be r1 and c1 respectively. We have r1c1=1000 2. Let revenues in 2008 be r2 and c2 respectively. We have r2c2=0 3. Let revenues and costs in 2009 be r3 and c3 respectively. we have r3=1.44r1. 4. Since the growth rate of sales is constant, growth rate per year is 20%. growth rate of costs is half of that which is 10% 5. (2) can be written as 1.2 r11.1 c1=0 6. Solving for (1) and (5), we have c1=12000 andr1=11000 7. r3=1.44*11000= 15840 and c3=1.21*12000=14520 8. Therefore profit in 2009=1584014520=1320
The answer should be $1320 Something is messing up my calculations. I substituted the amount by which C and R go up with X. This meant that at 2009 we needed to find out : R*x^4  C*X^2. Since X^4 = 1.44 > X^2 = 1.2 And that's why I multiplied C by 1.2 and not 1.21. Why is this wrong? C and R do not go up by the same rate X. You are given that revenue grows twice as fast as cost. So if rate of revenue growth is X, rate of cost growth is X/2.
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Re: A firm's annual revenue grows twice as fast as its costs. In
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17 Sep 2014, 22:24
WoundedTiger 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
I found this problem quite difficult to be solved under 2 mins. Will be really keen to know if there are any shortcuts to handle such Questions. Would you rate this problem a 700 + ?? Revenues increased by 44% in 2 years. Rate of revenue increase per year is the same. So \((1 + x)^2 = 1.44\) \(x = 20%\) So revenue increases by 20% per year and since cost increases at half the rate, cost increases by 10% every year. Say 2007 revenue is R and 2007 cost is C. We know C = R + 1000 ....(I) In 2008, they break even. So 1.1*C = 1.2R Substituting from (I), 1.1*(R + 1000) = 1.2R R = 11,000 C = 12,000 In 2009, Profit = (1.2)^2*R  (1.1)^2C = 1.44*11,000  1.21*12000 = 1320
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A firm's annual revenue grows twice as fast as its costs. In
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11 Jul 2016, 10:30
WoundedTiger 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
I found this problem quite difficult to be solved under 2 mins. Will be really keen to know if there are any shortcuts to handle such Questions. Would you rate this problem a 700 + ?? First of all..+1 for the question. I got the right answer..but the only thing I felt eerie about this problem is that it has a lot of calculations to be done under the given timing constraints..Experts..please suggest something about the calculations this has. And what level would you classify this problem as..
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A firm's annual revenue grows twice as fast as its costs. In
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24 Jul 2016, 20:41
Profit=RevenueCost 2007: 1000=R1C1 2008: 0=(1+2X%)R1(1+X%)C1 2009: P3=(1+2X%)^2*R1(1+X%)^2*C1 we've already known that R3=1.44R1, so (1+2X%)^2=1.44, X=10 we put X=10 back to the equation of 2008: 0=1.2R11.1C1, and we have 1.2R1=1.1C1, which we could combine with the equation of 2007. Then we have C1=12000, R111000 Finally, take C1 and R1 into the equation of 2009 and we got P3=1584014520=1320 Thanks,
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