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Tough DS word prob

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Tough DS word prob

by satishchandra » Sat Dec 03, 2011 1:12 am
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.

Please explain in detail
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by sam2304 » Sat Dec 03, 2011 2:01 am
1000 + (x-n)C. X being the total sets sold and n being the sales target.

1. 1000 + (x-n)c - (1000 + (x-n-3)c) = 600
1000 + (x-n)c - 1000 - (x-n-3)c = 600
xc - nc - xc + nc + 3c = 600
c = 200 INSUFF as no info on x and n.

2.1000 + (10 - n)C > 4000. INSUFF as we cannot find total amount earned in march.

Combining both (10 - n)200 > 3000. We cannot find a specific value for n using this eqn though n < 10. So INSUFF

IMO E.
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by GmatMathPro » Sat Dec 03, 2011 9:20 pm
satishchandra 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.

Please explain in detail
1. This tells us that she exceeded her sales goal of n encyclopedias for March, but we don't know by how much. If she exceeded it by one, then C=600, if she exceeded it by 2, then C=300, and if she exceeded it by 3 or more, then C=200. So, besides not fixing the value of C, we also still have no idea how many encylopedias she sold. INSUFFICIENT.

2. This is purely hypothetical and tells us nothing about what actually happened in March. INSUFFICIENT.

Statements 1&2: This allows us to rule out c=300 and c=200 from statement 1, because the most she could make selling 10 encyclopedias in that case would be if n=1 and c=300, in which case she would have gotten 1000+ 9(300)=3700, which is not more than 4000. Thus, c=600. However, based on the analysis in statement 1, c is only 600 if she sold exactly one more set of encyclopedias than n. Thus, she made 1000+600=1600 dollars in March.

Ans: C
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by satishchandra » Sun Dec 04, 2011 6:35 am
If she exceeded it by one, then C would become 600 and n= 2;

However, (2) does not satisfy with the same.
1000+8(600) = 5800 which is not equal to 4000


Am i missing something here?
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by GmatMathPro » Sun Dec 04, 2011 6:58 am
satishchandra wrote:If she exceeded it by one, then C would become 600 and n= 2;

However, (2) does not satisfy with the same.
1000+8(600) = 5800 which is not equal to 4000


Am i missing something here?
1. We can't deduce that n=2, only that she exceeded it by exactly one.

2. Statement 2 doesn't say it's supposed to be EQUAL to 4000, it says it's supposed to be MORE THAN 4000.
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by satishchandra » Sun Dec 04, 2011 7:15 am
GmatMathPro wrote: 2. Statement 2 doesn't say it's supposed to be EQUAL to 4000, it says it's supposed to be MORE THAN 4000.
Got it. :)
GmatMathPro wrote: 1. We can't deduce that n=2, only that she exceeded it by exactly one.
Why can't we deduce that n=2?
If c=600, can you explain how 'n' takes other numbers, with (1) and (2) being satified
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by GmatMathPro » Sun Dec 04, 2011 7:43 am
If n=3, then if she sells 10, she exceeds n by 7, in which case her income is 1000+7(600)=5200

If n=4, then if she sells 10, she exceeds n by 6, in which case her income is 1000+6(600)=4600
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