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by kaf » Tue Apr 14, 2009 1:50 pm
For each month of next year, Company R's monthly revenue target is x dollars greater than its monthly revenue target for the preceding month. What is Company R's revenue target for March of next year?

1 Company R's revenue target for December of next year is 310,000

2 Company R's revenue target for september of next year is 30,000 greater than its revenue target for June of next year.

OA C
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by Sher1 » Tue Apr 14, 2009 2:10 pm
C because


St 1

1 Company R's revenue target for December of next year is 310,000

Don't know anything about x

2 Company R's revenue target for september of next year is 30,000 greater than its revenue target for June of next year.

Don't know the original number . x is 30k

Combined

December is 310,000 and x is 30,000 so march (3 months from december) is 310,000+ 3*30000 = 400,000
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by gmat740 » Tue Apr 14, 2009 7:51 pm
I. March + 8x = Dec

March + 8x = 310k

Insuff

(II) Sept-June = 30k
but Sept - June = 3x

so 3x = 30k
x =10

But we have to find for March

Which can be calculated with (I)

So Both (I) and (II) are required

Hope this helps

Karan
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by cubicle_bound_misfit » Wed Apr 15, 2009 6:22 am
Why we can not assume that say as next year started for january so beginiing was 'J' $ and after that a series with CD X. Anything wrong with that approach? stil getting answer C as we have to solve for two varibales J and x
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