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DS: Laura's Income

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DS: Laura's Income

by kevincanspain » Thu Mar 22, 2007 6:46 am
Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000 per month, and a variable component, which is $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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Re: DS: Laura's Income

by gabriel » Thu Mar 22, 2007 12:24 pm
kevincanspain wrote:Laura sells encyclopaedias, and her monthly income has two components, a fixed component of $1000 per month, and a variable component, which is $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.
hmm.... i am not sure but i think the answer shuld be E...

from the q stem we get that lauras income is 1000 + C*x.... where x is the no. of encyclopedias she sold sbove the target of n

from the first statement we get ... 1000+C*x- ( 1000 + C*(x-3))=600... that wuld giv the value of C as 200 $... since we dont know the value of x.. so insufficient

from the second statement.... 1000 + C*x>4000... C*x>3000... but since we dont know the value of either C or x... therefore insufficient...

combining both we get x > 15...so insufficient... so E

... but i think i am missing something...
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by f2001290 » Sun May 13, 2007 6:05 am
I will go with A.

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by Nisha1218 » Sun May 13, 2007 7:10 am
I'm going with E because each statement doesn't give us "n" and if we were to combine the statements, then we still can't solve it because we don't have 2 distinct equations with one unknown.
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Re: DS: Laura's Income

by gviren » Mon May 21, 2007 5:15 pm
C cannot be $200 as in that case, statement (2) will not hold true

I think the answer is C.......

Taking 1 , we know that if Laura sells 3 less then the profits are $600 less.
So if all the 3 books are above the sales target of n sets, then c= $200
if 2 books are above the sales target of n sets, then c= $400
if only one is above the sales target of n sets, then c= $600

Now taking 2, if sold 10 sets of encyclopaedias in March, her income for that month would have been over $4000. To support this statement, c can only be $600 as in case of $200 and $400, income for March would be $3000 ($1000 + $200*10) and $4000 ($1000 + $400*10) (assuming that sales target of n sets = 0)



So that means Laura sold only one book in the month of March and her income was $1600

Please let me know if my understanding is wrong
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by Cybermusings » Tue May 22, 2007 12:10 am
Statement I: If Laura sold x-3 sets in March her income would be y – 600 (where x is sets sold by Laura and y is the income earned by Laura). We also know that y is made of two components (1000 + variable component). And we can also make out that x>n (else in both cases i.e. when she sells “x-3” sets and “x” sets Laura’s income would be the same i.e. 1000). However, it is insufficient to determine Laura’s income
Statement II: We can deduce from this statement that n<10, and 10C + 1000>4000 or 10C > 3000 or C > 300. However, we do not know for sure how much C is. Hence insufficient
Statement I and II:
From the two statements we know the following
C>300
n<10
and that after selling x-3 copies her income would be y-100
I think I would go with E
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