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Laura sells encyclopaedias, and her monthly income has two

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Laura sells encyclopaedias, and her monthly income has two [#permalink] New post 14 Feb 2007, 15:42
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Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.

OPEN DISCUSSION OF THIS QUESTION IS HERE: laura-sells-encyclopaedias-and-her-monthly-income-has-two-64646.html
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Re: DS: Laura's Income [#permalink] New post 15 Feb 2007, 08:40
kevincan wrote:
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.



E?

lets say she sells K sets

now her income = 1000 + (k-n) *C

A) states that 3C = 600 => c = 200 ... nothing about the income

B) 1000 + (10-n) *C >4000

=> if n=0 C>300 ????

even if you cobine both n = - 5

I think there is some thing missing in the question? or the Ans is E
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Re: DS: Laura's Income [#permalink] New post 16 Feb 2007, 21:24
kevincan wrote:
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.



Stat1:
Laura could have sold n+1 or n+2 or n+3 sets so in each case we would get a different value for the variable amount C and we dont have any information to find how much she earned

Stat2:
(10-n)*C > 3000

we dont know C and we dont know n


so I'm going with E
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 [#permalink] New post 19 Feb 2007, 05:48
Focus on what (1) and (2) tell us about C
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 [#permalink] New post 19 Feb 2007, 09:08
Definetley E. we have 3 unknowns and only 1 equation and 1 inequality
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 [#permalink] New post 20 Feb 2007, 07:23
from (1)

3C = 600 => c = 200

insufficient



from (2)

1000 + (10 -n)C > 4000

insufficient


taking (1) and (2)together

n comes negative..

E
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 [#permalink] New post 21 Feb 2007, 02:34
(1) tells us that there are three possible values for C
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 [#permalink] New post 21 Feb 2007, 08:41
the passage says that there is a variable component of $C for each set of encyclopaedias so there is no way of finding the variable component.

E
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 [#permalink] New post 21 Feb 2007, 13:15
adub35 wrote:
the passage says that there is a variable component of $C for each set of encyclopaedias so there is no way of finding the variable component.

E


If Laura sold only 1 more than n ency. C=600
If she sold two more than n, C=...
If she sold 3 or more than n, C=...
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 [#permalink] New post 21 Feb 2007, 23:25
kevincan wrote:
If Laura sold only 1 more than n ency. C=600
If she sold two more than n, C=...
If she sold 3 or more than n, C=...


Could you explain C=600 in case of n=1?
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 [#permalink] New post 22 Feb 2007, 20:34
kevincan wrote:
Focus on what (1) and (2) tell us about C

still dont get it ...
ok given the 3 possible values of C we can rule out n+2, n+3, n+4 and so on since stat2 (10-n)*C > 3000 works only for n+1
so C = 600
which means n < 5
so how do we get the amt she earns in march from this ?
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 [#permalink] New post 24 Feb 2007, 10:18
The answer must be E. Mostly because we need more concrete data...2. states that she makes more than 4000 dollars if she sells ten books. THis means she could make 400 000 dollars on 10 books since it is more than 4000.

1. insuff not enough information
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 [#permalink] New post 25 Feb 2007, 11:24
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.


Ans:
Let's assume Laura sold n copies and commision earned will be $(n-a)c, where 'n' is target number of copies, 'a' is number of actual copies sold and 'c' is commision/encyclopaedia

in case 'a' copies sold, total earnings would be 1000+ (n-a)c.

1)Laura sold 3 copies less than target so, (n-a-3)C=600, alone is not enough to solve the equation.


2) we can not formulate any equation from ths option.

Therefore the answer is 'E'
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 [#permalink] New post 25 Feb 2007, 13:08
kevincan wrote:
Read (1) carefully


Does this PS have any relation to the real GMAT question?
Where did u find it?
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Re: DS: Laura's Income [#permalink] New post 25 Feb 2007, 14:15
kevincan wrote:
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.


Income in March = 1000+C(x-n) if x>n, but Income =1000 if x is less than or equal to n
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 [#permalink] New post 26 Feb 2007, 16:41
I strongly feel answer is E.
Or the question posted in incomplete or missing some key words.
(Kevincan, u sure u typed it right?)

Also does anyone know correct explanation if answer is really C.
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Re: DS: Laura's Income [#permalink] New post 26 Feb 2007, 23:17
kevincan wrote:
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.


If Laura sold x sets, there are four cases:
x-n is not postive, in which case her earnings for March were 1000
x-n=1, in which case her earnings for March were 1000+C
x-n=2, in which case her earnings for March were 1000+2C
x-n is at least 3, in which case her earnings for March were 1000+(x-n)C
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Re: DS: Laura's Income [#permalink] New post 27 Feb 2007, 03:56
kevincan wrote:
kevincan wrote:
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.


If Laura sold x sets, there are four cases:
x-n is not postive, in which case her earnings for March were 1000
x-n=1, in which case her earnings for March were 1000+C
x-n=2, in which case her earnings for March were 1000+2C
x-n is at least 3, in which case her earnings for March were 1000+(x-n)C


Man this is def a Q60 question :wink:

I understand that from 1 you have that C can be: 600,300 or 200, right?
But alone insuff as is 2 alone.

Taking both together: From 2, putting n=1, we find out that for C=200 or C=300, there is no way to be over $4000 her earnings (maximung of 1000 + 9*200 or 1000 + 9*300).

Ok we discovered that C can only be $600, but here is where I get confused. I really do not know how to find the n number so that we could fix the value of her earnings...
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 [#permalink] New post 27 Feb 2007, 13:51
Kevincan

I just dont get how there are 4 scenarios in case 1)

If you have $10 and I have 2 fewer than you , then do I have $8?

If answer is YES, then why is that different from case 1 where its given that 3 fewer sets were sold and 600 less dollars were received...thus indicating C=200

I am not sure if you can go down few notches and explain...but I am really not getting this one.

Also,
How does your analysis still helps find out actual income.
What $ value did you calaculate as final income for March and how did you do that?
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Re: DS: Laura's Income [#permalink] New post 28 Feb 2007, 00:11
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kevincan wrote:
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000, and a variable component of $C for each set of encyclopaedias that she sells in that month over a sales target of n sets, where n>0. How much did she earn in March?

(1) If Laura had sold three fewer sets in March, her income for that month would have been $600 lower than it was.
(2) If Laura had sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000.


If Laura sold x sets, there are four cases:
x-n is not postive, in which case her earnings for March were 1000
x-n=1, in which case her earnings for March were 1000+C
x-n=2, in which case her earnings for March were 1000+2C
x-n is at least 3, in which case her earnings for March were 1000+(x-n)C[

(1) If Laura had sold three fewer sets, she would have earned either C, 2C or 3C less, so C must be either 200, 300 or 600. For example, if she sold one more than n sets, her comission was C. If she had sold three fewer sets than this, she would have lost this commission.

NOT SUFF


(2) 1000+(10-n)C>4000
C>3000(10-n)

As n>0, C>300

NOT SUFF

Together C=600 and x-n=1, so her income can be calculated


SUFF
Re: DS: Laura's Income   [#permalink] 28 Feb 2007, 00:11
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