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A saleswoman s monthly income has two components, a fixed

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New post 09 Jul 2006, 03:25
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A saleswoman’s monthly income has two components, a fixed component of $1000 per month, and a variable component, which is $C for each set of encyclopaedias that she sells in that month over a sales target of n sets. How much did she earn in March?

(1) If she had sold three fewer sets, her March income would have been $600 lower.
(2) If she had sold 8 sets of encyclopaedias, her income in March would have been over $4000
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New post 09 Jul 2006, 14:57
Any takers? :)
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New post 09 Jul 2006, 16:06
kevincan wrote:
A saleswoman’s monthly income has two components, a fixed component of $1000 per month, and a variable component, which is $C for each set of encyclopaedias that she sells in that month over a sales target of n sets. How much did she earn in March?

(1) If she had sold three fewer sets, her March income would have been $600 lower.
(2) If she had sold 8 sets of encyclopaedias, her income in March would have been over $4000


ummm, not sure but i'll take shot at it.
either E or C, i'll pick E.
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New post 09 Jul 2006, 17:03
I would have picked (E) too...

I know for sure I can eliminate (A), (B), and (D). Without comlpetely solving it, I am leaning toward (E)
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New post 09 Jul 2006, 17:10
kevincan wrote:
A saleswoman’s monthly income has two components, a fixed component of $1000 per month, and a variable component, which is $C for each set of encyclopaedias that she sells in that month over a sales target of n sets. How much did she earn in March?

(1) If she had sold three fewer sets, her March income would have been $600 lower.
(2) If she had sold 8 sets of encyclopaedias, her income in March would have been over $4000


(C)
I. implies, either C = 600, or 2C = 600, or 3C = 600
II. C(8-n) > 4000 --insufficient

Assuming, n to be a non-zero no. we get n = 1, C = 600, => sufficient
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New post 09 Jul 2006, 18:53
Answer: E

Given Salary S = 1000 + C(n-t)

where t = target,
n = # of sets sold

S1: S' = S - 600

or 1000 + C(n-t) - 600 = 1000 + C(n-3-t)

Can compute C. Still don't know n, t

Not sufficient.

S2: If sold 8 sets, income over 4000
C(8-t) > 3000 Still don't know t or C. Not sufficient.

S1 & S2:

From S1 : can compute C
From S2 : Can compute t.

But we still don't know how many sets were *actually sold*. S2 just says "if 8 were sold"...
Insufficient.

Answer: E
  [#permalink] 09 Jul 2006, 18:53
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