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Laura sells encyclopaedias, and her monthly income has two [#permalink]
14 Feb 2007, 15:42

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (01:21) correct
0% (00:00) wrong based on 11 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

Re: DS: Laura's Income [#permalink]
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

Re: DS: Laura's Income [#permalink]
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

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 ?

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.

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.

Re: DS: Laura's Income [#permalink]
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

Re: DS: Laura's Income [#permalink]
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

Re: DS: Laura's Income [#permalink]
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

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...

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?

Re: DS: Laura's Income [#permalink]
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

gmatclubot

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