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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]

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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.
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New post 29 May 2008, 20:10
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It feels like a total trap, especially with that inequality. I want to say E soooo bad...but with the inequality it's gotta be C somehow.

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New post 29 May 2008, 20:32
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.


These conditions don't fit with each other

The best case scenario for Laura is n = 1, her base minimum sale. So according to the statement 2, she made $1000 base salary and over $3000 for the 9 additional books that she sold. That would make her average commission on each additional book at more than $334/book.

Statement 1 (and 2 together), it says that the average commission for sold books #8, 9, 10 is only $200/book.

So it appears that (up to a certain number of books) the more books laura sells, the lesser her commission rate per book. Her commission rate depends on the exact number of books she sells, and that rate is not given.

I vote for E

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Re: Inequalities [#permalink]

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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.


This is very good question, +1. And I think answer is not E, it's C.

Laura's income \(I=1000+c(s-n)\), where \(s\) is number of sets she sold and \(n\) is target number (\(s-n\leq{0}\) --> \(I=1000\)).

(1) Three cases:
\(s-n=1\) --> \(c=600\) (surplus of 600$ was generated by 1 set);
\(s-n=2\) --> \(c=300\) (surplus of 600$ was generated by 2 set);
\(s-n\geq3\) --> \(c=200\) (surplus of 600$ was generated by 3 set).

If \(s-n\) equals to 1, 2, or 3, then income for March will be 1600$, BUT if \(s-n>3\), then income will be more then 1600. Or another way: if we knew that c=300 or 600, then we couldd definitely say that \(I=1600\) BUT if \(c=200\), then \(I=1600\) (for \(s-n=3\)) or more (for \(s-n>3\)). Not sufficient.

(2) \(1000+c(10-n)>4000\) --> \(c>\frac{3000}{10-n}\) --> \(0<n<10\) --> as the lowest value of \(n=1\), then \(c>333.(3)\). Not sufficient.

(1)+(2) as from (2) \(c>333.(3)\), then from (1) \(c=600\) --> \(I=1600\). Sufficient.

Answer: C.
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Re: Inequalities [#permalink]

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New post 18 Feb 2010, 03:14
Great trap....Nice one.....

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Re: Inequalities [#permalink]

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New post 18 Feb 2010, 03:47
I'm with Bunuel on this one, I did something similar.

BTW Bunuel if you don't mind sharing, what did you get on the GMAT (or specifically on the quant section)?

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Re: Inequalities [#permalink]

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Bunuel 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.


This is very good question, +1. And I think answer is not E, it's C.

Laura's income \(I=1000+c(s-n)\), where \(s\) is number of sets she sold and \(n\) is target number (\(s-n\leq{0}\) --> \(I=1000\)).

(1) Three cases:
\(s-n=1\) --> \(c=600\) (surplus of 600$ was generated by 1 set);
\(s-n=2\) --> \(c=300\) (surplus of 600$ was generated by 2 set);
\(s-n\geq3\) --> \(c=200\) (surplus of 600$ was generated by 3 set).


Bunuel.... not too clear so as to why you consider 3 cases here. In the S1 one (highlighted part in red) it clearly states that the Laura did sell 3 fewer sets. So if Laura previous sold S sets... as per statement one, it should be S-3 only.... why do you consider.... 3 or less than 3 sets! The statement doesn't say this at all! Please comment.
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Re: Inequalities [#permalink]

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jeeteshsingh wrote:
Bunuel 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.


This is very good question, +1. And I think answer is not E, it's C.

Laura's income \(I=1000+c(s-n)\), where \(s\) is number of sets she sold and \(n\) is target number (\(s-n\leq{0}\) --> \(I=1000\)).

(1) Three cases:
\(s-n=1\) --> \(c=600\) (surplus of 600$ was generated by 1 set);
\(s-n=2\) --> \(c=300\) (surplus of 600$ was generated by 2 set);
\(s-n\geq3\) --> \(c=200\) (surplus of 600$ was generated by 3 set).


Bunuel.... not too clear so as to why you consider 3 cases here. In the S1 one (highlighted part in red) it clearly states that the Laura did sell 3 fewer sets. So if Laura previous sold S sets... as per statement one, it should be S-3 only.... why do you consider.... 3 or less than 3 sets! The statement doesn't say this at all! Please comment.


"If Laura had sold three fewer sets in March..." --> Laura sold \(s\) and had \(income=I\), but if she had sold \(s-3\) then her income would have been \(I-600\).

Now:
If \(s\) is 1 more than \(n\) (or as I wrote \(s-n=1\)), then it would mean that 600$ was generated by only 1 set;
If \(s\) is 2 more than \(n\) (or as I wrote \(s-n=2\)), then it would mean that 600$ was generated by 2 sets;
If \(s\) is more than 3 more than \(n\) (or as I wrote \(s-n\geq3\)), then it would mean that 600$ was generated by all 3 sets;

In first two cases income for March will be 1600$, BUT for third case income can be 1600$ or more. So this statement is not sufficient.

Hope it's clear.
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Re: Inequalities [#permalink]

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New post 12 Dec 2010, 05:49
i completely agree with Bunuel's explanation.

I actually had this question wrong, and would appreciate some comments on my approach:

Laura's income was: I=1000+c(s-n)

If she had sold 3 fewer set, that is s'=s-3, she would have earned 600$ less, that is I'=I-600.

Putting this words in a formula, we have:

I'=1000+c(s-3-n)=1000+c(s-n)-600

From here you can easily obtain that c=200, so I dont know neither n nor s, so I dont know I, so therefore A is out.

B) same as Bunuel: c>3000/10-n Not sufficient

(A)+(B) Since c=200, I can substitute above and I obtain that n<-5, which makes no sense and I therefore chose E.

Any thoughts on my approach?

Thanks in advance!
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Re: Inequalities [#permalink]

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New post 15 Dec 2010, 19:37
I agree that the answer should be C. Very tricky question as I wanted to pick E at first. Same train of thought as Bunuel.

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Re: Inequalities [#permalink]

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New post 16 Dec 2010, 06:13
I got the same answer C. Tricky Q.
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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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After I spent 45 min at midnight - I got the answer and agree with Bunuel that the correct ans is C .

Let me simplify the trap . The trap is we don’t know the threshold -n . Hence selling 3 fewer sets does not readily translate in distribution of 600 into 3 sets .

Say if the threshold is 10 and the no. of sale were 11 , the extra variable contribution is actually coming from 1 set . So even if we sell 3 fewer sets the less income of 600 is actually because of one set . On the other hand if the threshold is 10 and the initial sale were 12 - selling 3 fewer sets ->corresponding reduction in variable part will be because of 2 sets .
Hope this will clarify the future reader!

Thanks,
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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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New post 09 Jan 2012, 11:47
Wow! a trap question.

1. 1000 + (t-n-3))c = 1000 + (t-n)c - 600, gives us c. However, we don't even know if the 3 insufficient is below n. insuff.
2. 1000 + (10-n)c >= 4000, clearly insuff since we don't know n and c

The S2 ineq makes things complicated to provide a clear soution. Guess is E.
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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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New post 09 Jan 2012, 12:03
Definitely C; nice explanation by Bunuel.

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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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New post 24 Feb 2013, 23:26
Hi all,

I would be grateful if you could pls tell answer my query. I totally understand Buneul's approach but i wanted to ask, why is it done that ways?as in why do we have to take three cases? why cant we do it like the question says her income would be equal to 1000+c(s-n)

as per statement 1:
assuming the no of sets she sold in march as M the equation of net income turns out to be as
--> 1000 + c(M-n) - [1000 + c(M-3-n)]=600
=>3c=600
=>c=200

which doesnt seem sufficient to estimate her sales in March Hence insufficient.

As per statement 2:
1000 + c(10-n)>4000

again insufficient

Using both statemnt 1 &2 we get value of n as < -5, which is invalid as per given conditions. Hence answer is E

kindly tell me where i m going wrong in this approach!

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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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Hi Swarman,

There are two things wrong with this method.

Firstly, when you combine both the statements together, you get the value of n t be less than -5. However, we know that n denotes the number of set of books and thus, cannot be negative. So, the answer does not stand. However, what this negative value does tell you is that the value of c is not 200.

Secondly, while evaluating the first statement you got c=200 because you considered all three sets of encyclopedias to be above the sales target of n sets. However, there are three possibilities.

1. All three of them are above the sales target in which case 3c=600 implying c=200
2. Two of them are above the sales target in which case 2c=600 implying c=300
3. Only one of them is above the sales target in which case c=600

Since, one is already proved to be wrong, the answer has to be one out of 2 and 3. Substituting c=300 in the statement 2 gives us a value of n=0. Hence, not possible.
The answer is the c=600 and hence we need to use this value to work further with this problem.

Hope this helped! Let me know in case of any further doubts/concerns.

swarman wrote:
Hi all,

I would be grateful if you could pls tell answer my query. I totally understand Buneul's approach but i wanted to ask, why is it done that ways?as in why do we have to take three cases? why cant we do it like the question says her income would be equal to 1000+c(s-n)

as per statement 1:
assuming the no of sets she sold in march as M the equation of net income turns out to be as
--> 1000 + c(M-n) - [1000 + c(M-3-n)]=600
=>3c=600
=>c=200

which doesnt seem sufficient to estimate her sales in March Hence insufficient.

As per statement 2:
1000 + c(10-n)>4000

again insufficient

Using both statemnt 1 &2 we get value of n as < -5, which is invalid as per given conditions. Hence answer is E

kindly tell me where i m going wrong in this approach!

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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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New post 28 Feb 2013, 10:05
oh god YES!! you are absolutely right!!
perfect.. thank u sooo much..

God Bless :)

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Re: Laura sells encyclopaedias, and her monthly income has two [#permalink]

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In part 1, we know that 1 extra book is equal to either $200, $300, or $600 because she is either 1 book, 2 books, or 3 books over the threshold.

In part 2, we know that her total revenue from selling 10 encyclopedias is GREATER THAN 4000. A fixed part of her revenue is $1000, so we subtract that to get the variable part:

Her Variable revenue for 10 encyclopedias is GREATER THAN 3000.

Therefore, her variable revenue for 1 encyclopedia is GREATER THAN 300.

Because from part 1 we know that the variable revenue from each book could be one of only three options, and because only one of those options is GREATER THAN $300 per book, we know that she gets paid a base of $1000 and then $600 per book for each book over the threshold.

However, here's the trick... We know that if she had sold THREE fewer in march, then she would have made $600 less.. which means that she sold only ONE encyclopedia over the threshold. That means she only gets paid for ONE encyclopedia, plus her base pay. Even though we don't know what the threshold is, we know that $1000 + $600 = $1600, her salary in March.

C does it.

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