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# DS

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Director
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05 Jul 2007, 20:46
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Any help why C is wrong..
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Senior Manager
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05 Jul 2007, 21:43
E. e,g,w,t- first letters of expenses.

Let's say T-total home expenses
P- what they paid so far.

Then T=e+g+w+t
P=1/2*e+1/2*g+1/4*w+3/4*t

P/T doesn't give any #. Because variables will not totally cancel off.
GMAT Club Legend
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06 Jul 2007, 00:53
St1:
Insufficient. We don't know what other expenses the Dufaas have.

St2:
Insufficient. We don't know how much of each item they have paid off.

Using St1 and St2:
Expenses = e+g+w+t

Have paid off:
p = e/2+g/2+w/4+3t/4
= 2(e+g)+w+3t/4

We still cant' work out what percentge is paid off.

Ans E
Director
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06 Jul 2007, 15:29
I think it is C.

We know we have 4 bills part of the annual expenses. Paying a bill in full represents a 1. So paying 100% should equal 4.

I have paid the following of bill A,B,C,D

A = 1/2
B = 1/2
C = 1/4
D = 3/4

That totals 8/4.

(8/4) / 4 = 2/4 = .5 * 100 = 50% of the annual bills paid.

Thoughts?
Manager
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06 Jul 2007, 21:11
E

You need to know the relationship between the 4 expenses. To help visualize this, think of two scenarios:

1. the electricity bill is huge and is most of the annual expenses

2. the electricity bill is only 1 dollar

The two scenarios render two very different answers.
Director
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06 Jul 2007, 21:33
Amit05 wrote:
Any help why C is wrong..

it cannot be C cuz you can not add up them. you can if only allexpeses are equal but we do not know.

should be E.
VP
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06 Jul 2007, 23:39
jimmyjamesdonkey wrote:
I think it is C.

We know we have 4 bills part of the annual expenses. Paying a bill in full represents a 1. So paying 100% should equal 4.

I have paid the following of bill A,B,C,D

A = 1/2
B = 1/2
C = 1/4
D = 3/4

That totals 8/4.

(8/4) / 4 = 2/4 = .5 * 100 = 50% of the annual bills paid.

Thoughts?

This Is very clever problem.

The problem is testing your understanding of averages. As long as you choose the same numbers as plug-ins for C,D the outcome will be 50%, since 1/4 and 3/4 are completing. (i.e. C = 100, D = 100)

Only when you start pluging in diffrent numbers for C,D the outcome will be different then 50%. (i.e. C = 150, D = 100)

The writer of this problem assumes that a test taker with a good grip on averages will choose (E) either by useing algebra or logic, but a weaker test taker will plug in numbers (and in a hurry) and will choose the same numbers for C,D and eventually choose (C).

Director
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07 Jul 2007, 05:28
So basically, we cannot add these and because we do not know what the total bill amount is, correct? If we know the total bill was $100 then this could be solved with C? VP Joined: 08 Jun 2005 Posts: 1146 Followers: 7 Kudos [?]: 187 [0], given: 0 [#permalink] ### Show Tags 07 Jul 2007, 06:19 jimmyjamesdonkey wrote: So basically, we cannot add these and because we do not know what the total bill amount is, correct? If we know the total bill was$100 then this could be solved with C?

Knowing the total amount won't help. You need to know the distributions between the different bills - see attachment:

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Bills.JPG [ 15.69 KiB | Viewed 593 times ]

CIO
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07 Jul 2007, 09:26
jimmyjamesdonkey wrote:
So basically, we cannot add these and because we do not know what the total bill amount is, correct? If we know the total bill was \$100 then this could be solved with C?

Agree here with the others. Put simply, this is a weighted average question but you don't know the weights. Essentially, you need to know the ratio of all the bills, and you could do it. But without that, you just can't.
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