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# Challenge 1 - Q27

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Senior Manager
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Challenge 1 - Q27 [#permalink]  09 Jun 2006, 03:18
I do not agree to the OA, so please explain your answers.
Thanks.
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VP
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[#permalink]  09 Jun 2006, 03:34
I would go with E. Given that 511 is a prime number, there are no other factors, which means there is no way to split $5.11 to satisfy 1 & 2. Director Joined: 16 Aug 2005 Posts: 947 Location: France Followers: 1 Kudos [?]: 10 [0], given: 0 [#permalink] 09 Jun 2006, 03:42 E... But partly because I cant find a way to figure out how to split$5.11

What was your solution deowl? (dont post OA yet, lets see what others say)
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SVP
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[#permalink]  09 Jun 2006, 03:46
Lets say we are dealing with two items.
I think we all agree that using only the product or the sum we cannot find the individual prices.
But we can do that (in case of 2 items ) when we know both.......

Now with 4 items.... I think it can be done....... but the equation would be complex. Since I am not sure..... I will go with E.
VP
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[#permalink]  09 Jun 2006, 04:22
I would also have picked E.
Senior Manager
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[#permalink]  09 Jun 2006, 04:36
From the given we can make two equations:

a+b+c+d = 511
a*b*c*d=511

511 is not a prime number. It is a multiple of 73 and 7.

Since it is imposible to pay by less then a penny , all of them should be
integers.

So if a and b are 73 and 7 , c*d should be equal to 1.

so we get two equations:

c*d=1
c+d=431

Two equations with two variables could be solved. So it should be C.

However the OA is E.

Any ideas?
VP
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[#permalink]  09 Jun 2006, 04:48
You are right... 511 is NOT a prime.

Too early in the morning... no coffee!

deowl wrote:
From the given we can make two equations:

a+b+c+d = 511
a*b*c*d=511

511 is not a prime number. It is a multiple of 73 and 7.

Since it is imposible to pay by less then a penny , all of them should be
integers.

So if a and b are 73 and 7 , c*d should be equal to 1.

so we get two equations:

c*d=1
c+d=431

Two equations with two variables could be solved. So it should be C.

However the OA is E.

Any ideas?
Director
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[#permalink]  09 Jun 2006, 05:06
deowl wrote:

c*d=1
c+d=431

Did you try solving for c and d to check if you get values that are in cents and not in fractions?
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Senior Manager
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[#permalink]  09 Jun 2006, 05:56
gmatmba wrote:
deowl wrote:

c*d=1
c+d=431

Did you try solving for c and d to check if you get values that are in cents and not in fractions?

Obviously they can't be whole numbers. The only positive integers
that satisfy the first equations are 1-s. But that clearly contradicts the second one. However it doesn't mean that there is not enough info to get an answer as E states. The answer is that there is no answer. So it has to be C.
Director
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[#permalink]  09 Jun 2006, 06:10
deowl wrote:
gmatmba wrote:
deowl wrote:

c*d=1
c+d=431

Did you try solving for c and d to check if you get values that are in cents and not in fractions?

Obviously they can't be whole numbers. The only positive integers
that satisfy the first equations are 1-s. But that clearly contradicts the second one. However it doesn't mean that there is not enough info to get an answer as E states. The answer is that there is no answer. So it has to be C.

I would assume because you cannot have an answer, that's why choice E is there. C means when you can show valid answers from the information provided. Otherwise, for all questions, we can assume that answer is we cannot get a valid answer from the info provided and hence C. There is always an answer available, the question here is whether we have enough info to deduce them.
_________________

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Senior Manager
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[#permalink]  09 Jun 2006, 06:19
gmatmba wrote:
deowl wrote:
gmatmba wrote:
deowl wrote:

c*d=1
c+d=431

Did you try solving for c and d to check if you get values that are in cents and not in fractions?

Obviously they can't be whole numbers. The only positive integers
that satisfy the first equations are 1-s. But that clearly contradicts the second one. However it doesn't mean that there is not enough info to get an answer as E states. The answer is that there is no answer. So it has to be C.

I would assume because you cannot have an answer, that's why choice E is there. C means when you can show valid answers from the information provided. Otherwise, for all questions, we can assume that answer is we cannot get a valid answer from the info provided and hence C. There is always an answer available, the question here is whether we have enough info to deduce them.

Nope, this is not correct. The DS question doesn't ask you what is the answer to the given question. The only thing the it asks is whether there is enough info to get ANY definite answer. The "no solution" answer is a definite answer. So this particular question provides just enough info to conclude that it doesn't have a solution so no additional information can change a thing. So that clearly cannot be E. That what I think.
Director
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[#permalink]  09 Jun 2006, 06:24
deowl wrote:
gmatmba wrote:
deowl wrote:
gmatmba wrote:
deowl wrote:

c*d=1
c+d=431

Did you try solving for c and d to check if you get values that are in cents and not in fractions?

Obviously they can't be whole numbers. The only positive integers
that satisfy the first equations are 1-s. But that clearly contradicts the second one. However it doesn't mean that there is not enough info to get an answer as E states. The answer is that there is no answer. So it has to be C.

I would assume because you cannot have an answer, that's why choice E is there. C means when you can show valid answers from the information provided. Otherwise, for all questions, we can assume that answer is we cannot get a valid answer from the info provided and hence C. There is always an answer available, the question here is whether we have enough info to deduce them.

Nope, this is not correct. The DS question doesn't ask you what is the answer to the given question. The only thing the it asks is whether there is enough info to get ANY definite answer. The "no solution" answer is a definite answer. So this particular question provides just enough info to conclude that it doesn't have a solution so no additional information can change a thing. So that clearly cannot be E. That what I think.

Yes you do have a valid point. I guess in this case Gmat (or gmatclub) is thinking that since we cannot identify 4 prices in cents or dollars, answer is E. This is probably not an original gmat question as Gmat tends to stay away from controversies.
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Senior Manager
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[#permalink]  09 Jun 2006, 06:28
gmatmba wrote:
Yes you do have a valid point. I guess in this case Gmat (or gmatclub) is thinking that since we identify 4 prices in cents or dollars, answer is E. This is probably not an original gmat question as Gmat tends to stay away from controversies.

I'm sure it isn't. Just wanted to make sure I don't miss anything here.
Anyway thank you all for participation
VP
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[#permalink]  09 Jun 2006, 09:15
from i, we can get 511 cents in many ways. so insuff.
from ii, the four prices could be, (511, 1, 1, and 1) or 73, 7, 1, and 1.

togather, its not possible. if two statements lead to different results, the question is poorly structured . IMO, OA doesnot make any sense.

good news : we donot have to deal with such a disputablpblems in real test.
SVP
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[#permalink]  10 Jun 2006, 17:38
Actually, I would say the answer is B, since we can determine that the price for the four items are 1,1,7 and 73 cents, respectively. However, this is a bad questions since (1) and (2) are not consistent. In real GMAT a rule of thumb is if there are solutions for (1) and (2), the solutions should be the same for both (although you are not required to actually provide the solutions). For example, it is not likely that you would get x=1 for (1) and x=3 for (2) and get D as an answer.
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Director
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[#permalink]  10 Jun 2006, 18:30
deowl,

how did you figure out the 511 = 73 * 7? Is there a quick way?

I am just thinking in the gmat.. with 2 mins/problem.. i would have assumed it was prime!
[#permalink] 10 Jun 2006, 18:30
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# Challenge 1 - Q27

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