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v_schame
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Hi,
I'm having a hard time with this question. I'm not sure if my approach was correct.
Question
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) Comapny R's revenue for September of next year is $30,000 greater than its revenue target for June of next year.
My Approach
Let a = the revenue target for January
x = the amount that the revenue target is greater than the month before (as stated in question)
Revenue targets per month
J: a
F: a+x
M: a+2x
A: a+3x
M: a+4x
J: a+5x
J: a+6x
A: a+7x
S: a+8x
O: a+9x
N: a+10x
D: a+11x
S1) December: a+11x = 310,000 -> 2 variables and 1 equations, cannot solve, S1 is insufficient
S2) September = 30,000 + June -> a+8x = 30,000+a+5x -> x=10,000
Replace x value in March -> Revenue target=a+4(10,000) -> 2 variables and 1 equations, cannot solve, S2 is insufficient
S1+S2) Replace the value of X in S1 December equation to find value of a. With values of a and x known, they can be replace in the March equation to find its revenue target.
Answer: C
Thanks!
Vanessa
I'm having a hard time with this question. I'm not sure if my approach was correct.
Question
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) Comapny R's revenue for September of next year is $30,000 greater than its revenue target for June of next year.
My Approach
Let a = the revenue target for January
x = the amount that the revenue target is greater than the month before (as stated in question)
Revenue targets per month
J: a
F: a+x
M: a+2x
A: a+3x
M: a+4x
J: a+5x
J: a+6x
A: a+7x
S: a+8x
O: a+9x
N: a+10x
D: a+11x
S1) December: a+11x = 310,000 -> 2 variables and 1 equations, cannot solve, S1 is insufficient
S2) September = 30,000 + June -> a+8x = 30,000+a+5x -> x=10,000
Replace x value in March -> Revenue target=a+4(10,000) -> 2 variables and 1 equations, cannot solve, S2 is insufficient
S1+S2) Replace the value of X in S1 December equation to find value of a. With values of a and x known, they can be replace in the March equation to find its revenue target.
Answer: C
Thanks!
Vanessa













