Company R's monthly revenue....

[rumni]

For each month of next year, Company R's monthly revenue target is x dollars greater than its monthly target for the preceding month. What is Company R's revenue target for March of next year?

1 Co R''s revenue target for December of nexxt year is $310,000.

(2) Co R's revenue target for September of next year is $30,000 greater than its revenue target for June of next year.

OA D

[JonathanSchneider]

Can you please indicate a particular question/difficulty you had with this problem?

[rumni]

I don't know how to solve it.

[RonPurewal]

you've got the wrong oa. go back and check your source; the answer to this problem, unless you've transcribed the problem incorrectly, should be (c).

statement (1):
you've got the target for december. therefore, if you had any information about the monthly progression of the target, you'd be able to count backwards, month by month, to arrive at the figure for march.
unfortunately, you have no such information. therefore, we have no way of knowing definite revenue-target figures for any month except december.
insufficient.

statement (2):
this information is good enough for us to know everything about the monthly progression of the revenue target. specifically, we know that it grows by $30,000 over three months (from june to september); therefore, the revenue target grows by $10,000 per month.
unfortunately, that's all we know; this isn't enough information to arrive at any definite figure, since we have no actual numbers from which to subtract, or to which to add, this $10,000 per month.
insufficient.

together:
the december revenue target is $310,000, and the target grows by exactly $10,000 per month.
therefore, we can count back by $10,000 per month until we get the figure for march. (note: no need to actually do this)
sufficient.

ans = (c)

[hema.laxmanan]

Hi, I got a different solution for this question than the one that GMATPrep states. Can you let me know where i am wrong?

I imagined myself to be at December of this year and my revenue for November was $1000. So, For Jan of next year, the problem statement is 1000 + $x.For Feb, its $1000 + 2($x) thus For June its $1000 + 6($x) , Sept = $1000 + 9x and Dec = $1000 + 12x

So, now with Option1, Dec which is $1000+12x = $31000 when simplified its x = 2500, then for March its $1000 + 3x which is $7600.

Option 2: Sept target = $30000 + June's target
1000 + 9x = 30000 + 1000 + 6x => 3x=30000 x = 10000 thus March can be arrived at.

I am arriving at Option D for the solution, but I want to know where I went wrong here.

[jnelson0612]

Hi, I got a different solution for this question than the one that GMATPrep states. Can you let me know where i am wrong?

I imagined myself to be at December of this year and my revenue for November was $1000. So, For Jan of next year, the problem statement is 1000 + $x.For Feb, its $1000 + 2($x) thus For June its $1000 + 6($x) , Sept = $1000 + 9x and Dec = $1000 + 12x

So, now with Option1, Dec which is $1000+12x = $31000 when simplified its x = 2500, then for March its $1000 + 3x which is $7600.

Option 2: Sept target = $30000 + June's target
1000 + 9x = 30000 + 1000 + 6x => 3x=30000 x = 10000 thus March can be arrived at.

I am arriving at Option D for the solution, but I want to know where I went wrong here.

Hi there, nice try! You went wrong when you assumed the $1000 number for the November revenue. We just don't know the Revenue starting in January. All we know is that:
January=Revenue (we'll call it R)
February=Revenue + x (amount to gain every month)
March=Revenue + 2x
April= Revenue + 3x
all the way to
December=Revenue + 11x

To get the March figure, we need both the base Revenue (R) and the increase each month (x).

Statement 1 gives us R+11x=$310,000. I can't use this to solve for R or x; I have two variables and one equation. Insufficient.

Statement 2 tell us that Company R's revenue target for September of next year is $30,000 greater than its revenue target for June of next year. Insufficient.

Okay! June Revenue target=R+5x
September Revenue target=R+8x
There is a difference of 3x between the two, equal to $30,000. Thus, I learn that x=$10,000. But I still don't know R!

Together, remember that from statement 1 we learn that R+11x=$310,000. We now know that x=$10,000 and can solve for R. Sufficient!