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In your explanation , If X is your total set sold in March....then you have a bad assumtion that max sets sold is 3 more than 'n' (which is threshold for comission of C per additional set)
I am not sure what correct answer is.
But I feel your approach is incorrect.
I feel it will be great if you can complete your answer
as in give us value of N and C and X and Final answer which is income in March.
Perhaps you are frustrated trying to explain everything....but I do believe there is something missing in your explanation....
In your explanation , If X is your total set sold in March....then you have a bad assumtion that max sets sold is 3 more than 'n' (which is threshold for comission of C per additional set)
I am not sure what correct answer is. But I feel your approach is incorrect.
I feel it will be great if you can complete your answer as in give us value of N and C and X and Final answer which is income in March.
Perhaps you are frustrated trying to explain everything....but I do believe there is something missing in your explanation....
I still believe answer is E. :
If we combine (1) and (2), We don't know the value of n, but we do know that x-n=1, C=600 and thus her income for March was 1600
Re: DS: Laura's Income [#permalink]
28 Feb 2007, 15:02
kwam wrote:
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...
You were really close. If C=600, we know that the number of emcycl. she sold was 1 more than n
Re: DS: Laura's Income [#permalink]
05 Aug 2008, 21:26
2
This post received KUDOS
very good question.... i made it a point not look at solution at second page of thsi thread and tried solving ... after 5 minutes here is what i got
statement 1 : difference is 600 when x is between 1 and n (included) sale = 1000 when x > n sale = 1000 + (x-n)C
x --- sale n-2---1000 n-1---1000 n ---- 1000 n+1 --- 1000 + C ---- differenec from last 3 = C = 600 n+2 --- 1000 + 2C --- difference from last 3 = 2C = 600. C = 300 n+3 --- 1000 + 3C ---- difference from last 3 = 3C = 600. C = 200 n+4 --- 1000 + 4C ---- difference from last 3 = 3C = 600. C = 200
statement 1 is telling us that C could be 600, 300 or 200 ... not suff
statement : if x = 10, total sale > 4000 variable sale > 3000 (10-n) * C > 3000
Not suff
Combine both
if C = 600 (10-n) * 600 > 3000 10-n > 5 n < 5
if C = 300 (10-n) * 300 > 3000 10-n > 10 n < 0 ... not possible similarly C = 200 is also not possible
possible values of n = 1,2,3,4.... and C = 600, total number of books sold in march n+1 = 2,3,4,5
here it looks like that we cant answer the question and E is the right answer but wait a minute... we dont need to find exact n... we only have to find out her total sale in March.. that will be 1000 + C = 1600 Suff
Re: DS: Laura's Income [#permalink]
06 Aug 2008, 22:02
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.
good one:
point to be noted: the marginal revenue loss of $600 is for how many copies of encyclopedias? 1 or 2 or 3?
remember for :
x - n = 1 x - n = 2 x - n = 3
where x = no of copies of encyclopedias sold _________________
Re: DS: Laura's Income [#permalink]
07 Aug 2008, 17:09
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.
lets say laura sold x sets in march and n is target C is cost per x-n sales hence (1) 1000+(x-n)*C=1000+(x-3-n)*C+600 INSUFFI since x ,n cannot be evaluated (2)this says 1000+(10-n)*C >4000 INSUFFI since no clue about what value above 4000
I wish I'd seen this thread sooner- if you designed this question, nicely done! (+1) _________________
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