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29 May 2008, 13:54
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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.
(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.
13 Feb 2010, 08:24
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kevincan
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.
(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.
27 Feb 2013, 20:17
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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.
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
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!
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!
