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# M01 27

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GMAT Tutor
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M01 27 [#permalink]

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21 Aug 2008, 09:29
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This question, and its explanation, do not make sense.

A shopper came to a store to purchase 4 items. What were the prices of the four items?

1) The total cost was $5.11 2) The product of the prices was$5.11

The explanation says:

This problem has 4 variables and two equations. It is impossible to solve it.

This is not a two-equation, four-variable situation. If we call the prices a, b, c and d, we know:

a+b+c+d = 5.11
abcd = 5.11

But we also know:

* a, b, c and d are all positive;
* 100a, 100b, 100c and 100d are all integers- they are prices in a shop, after all. You can't see a price of an item in a shop that is equal to $$\sqrt{2}$$, $$\Pi$$ or $$3.44417$$.

If we let A, B, C and D be the prices in cents, so that A = 100a, B = 100b, C = 100c and D = 100d, we have:

A+B+C+D = 511
(A/100)(B/100)(C/100)(D/100) = 511/100 or ABCD = 511*100^3 = (2^3)(5^3)(7)(73)

Thus, A, B, C and D need to be positive integer factors of (2^3)(5^3)(7)(73), their product needs to be (2^3)(5^3)(7)(73), and they need to add to 511. These are extreme restrictions- this is not a 2-equation, 4-variable problem. In fact, there are no possible values for A, B, C and D, if my calculations are correct.
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Re: M01 27 [#permalink]

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21 Aug 2008, 12:34
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So you're saying E is what you choose for your answer?

IanStewart wrote:
This question, and its explanation, do not make sense.

A shopper came to a store to purchase 4 items. What were the prices of the four items?

1) The total cost was $5.11 2) The product of the prices was$5.11

The explanation says:

This problem has 4 variables and two equations. It is impossible to solve it.

This is not a two-equation, four-variable situation. If we call the prices a, b, c and d, we know:

a+b+c+d = 5.11
abcd = 5.11

But we also know:

* a, b, c and d are all positive;
* 100a, 100b, 100c and 100d are all integers- they are prices in a shop, after all. You can't see a price of an item in a shop that is equal to $$\sqrt{2}$$, $$\Pi$$ or $$3.44417$$.

If we let A, B, C and D be the prices in cents, so that A = 100a, B = 100b, C = 100c and D = 100d, we have:

A+B+C+D = 511
(A/100)(B/100)(C/100)(D/100) = 511/100 or ABCD = 511*100^3 = (2^3)(5^3)(7)(73)

Thus, A, B, C and D need to be positive integer factors of (2^3)(5^3)(7)(73), their product needs to be (2^3)(5^3)(7)(73), and they need to add to 511. These are extreme restrictions- this is not a 2-equation, 4-variable problem. In fact, there are no possible values for A, B, C and D, if my calculations are correct.

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings GMAT Tutor Joined: 24 Jun 2008 Posts: 1183 Followers: 421 Kudos [?]: 1507 [0], given: 4 Re: M01 27 [#permalink] ### Show Tags 21 Aug 2008, 13:42 jallenmorris wrote: So you're saying E is what you choose for your answer? Logically, there is no good answer: C is 'correct', because the situation is impossible, and therefore we know with both statements that the prices are impossible to find. That is, using both statements, we know the answer is 'no solution'. But E is also 'correct', because even with both statements, we can't find the prices. The statements in a DS question should never be contradictory. Still, more importantly, the explanation provided is a gross oversimplification of the problem, which is what I was pointing out. _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com SVP Joined: 30 Apr 2008 Posts: 1887 Location: Oklahoma City Schools: Hard Knocks Followers: 40 Kudos [?]: 570 [0], given: 32 Re: M01 27 [#permalink] ### Show Tags 21 Aug 2008, 13:53 I agree with your answer of E. I think it would be C if the question were a yes/no answer where we have to find out if the answer is always Yes, or always no, but when the question asks for values and we still have multiple possible values...that is E and not C. Besides, I've never heard of a "no solution" on the real GMAT. IanStewart wrote: jallenmorris wrote: So you're saying E is what you choose for your answer? Logically, there is no good answer: C is 'correct', because the situation is impossible, and therefore we know with both statements that the prices are impossible to find. That is, using both statements, we know the answer is 'no solution'. But E is also 'correct', because even with both statements, we can't find the prices. The statements in a DS question should never be contradictory. Still, more importantly, the explanation provided is a gross oversimplification of the problem, which is what I was pointing out. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: M01 27 [#permalink]

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21 Aug 2008, 14:31
jallenmorris wrote:
I agree with your answer of E. I think it would be C if the question were a yes/no answer where we have to find out if the answer is always Yes, or always no, but when the question asks for values and we still have multiple possible values...that is E and not C. Besides, I've never heard of a "no solution" on the real GMAT.

We do not have multiple possible values here- there is no possible solution, if you use both statements. And yes, as you say, that never happens on the real GMAT (in a real GMAT DS question that asks for a value, there is always at least one solution when you combine both statements), for precisely the reason I pointed out above- it leaves both C and E as logically 'correct' answers. This is one of the problems with the question that I am pointing out.
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Re: M01 27 [#permalink]

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22 Aug 2008, 02:19
Thanks guys, +2 for you both. If you have any suggestions on how to improve the question we encourage you to share your ideas. Any help will be greatly appreciated.

Thanks again.
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Re: M01 27 [#permalink]

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03 Nov 2008, 06:00
I solved it in a different way:

1. In Suff

2. product of abcd in cents is 511
511 = 7*73

So, the prices for the 4 items can only be 7, 73, 1,1
The order doesn't matter, because the question asks only for the prices all together.

I know the answer is E
Can someone explain where I might be going wrong
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Re: M01 27 [#permalink]

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09 Jan 2009, 01:27
511, 1, 1, 1 cannot be an answer? So 73, 7, 1, 1 is not the definite answer.
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Re: M01 27 [#permalink]

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14 Jan 2009, 00:54
LiveStronger wrote:
I solved it in a different way:

1. In Suff

2. product of abcd in cents is 511
511 = 7*73

So, the prices for the 4 items can only be 7, 73, 1,1
The order doesn't matter, because the question asks only for the prices all together.

I know the answer is E
Can someone explain where I might be going wrong

I think the answer is 'E', because it's not possible to determine prices for goods with given information. Just keep it simple!
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Re: M01 27 [#permalink]

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30 Jan 2010, 12:21
IanStewart wrote:
This question, and its explanation, do not make sense.

A shopper came to a store to purchase 4 items. What were the prices of the four items?

1) The total cost was $5.11 2) The product of the prices was$5.11

The explanation says:

This problem has 4 variables and two equations. It is impossible to solve it.

This is not a two-equation, four-variable situation. If we call the prices a, b, c and d, we know:

a+b+c+d = 5.11
abcd = 5.11

But we also know:

* a, b, c and d are all positive;
* 100a, 100b, 100c and 100d are all integers- they are prices in a shop, after all. You can't see a price of an item in a shop that is equal to $$\sqrt{2}$$, $$\Pi$$ or $$3.44417$$.

If we let A, B, C and D be the prices in cents, so that A = 100a, B = 100b, C = 100c and D = 100d, we have:

A+B+C+D = 511
(A/100)(B/100)(C/100)(D/100) = 511/100 or ABCD = 511*100^3 = (2^3)(5^3)(7)(73)

Thus, A, B, C and D need to be positive integer factors of (2^3)(5^3)(7)(73), their product needs to be (2^3)(5^3)(7)(73), and they need to add to 511. These are extreme restrictions- this is not a 2-equation, 4-variable problem. In fact, there are no possible values for A, B, C and D, if my calculations are correct.

HAs the forum been updated? as i have totally different question for M01 number 27

Max's current monthly salary is $2,000, and will be raised by 25% next month. If he decides to save exactly 25% of his salary each month, how many months will it take for Max to save$5,500?

* 4
* 5
* 6
* 7
* 9
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Re: M01 27 [#permalink]

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01 Feb 2010, 03:36
Yes, this question was replaced. I will update the master thread with all M01 discussions.
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Re: M01 27 [#permalink]

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14 Jan 2012, 21:54
Right Questions for M01 Q 27 is this :

Max's current monthly salary is $2,000, and will be raised by 25% next month. If he decides to save exactly 25% of his salary each month, how many months will it take for Max to save$5,500?

 4
 5
 6
 7
 9
Re: M01 27   [#permalink] 14 Jan 2012, 21:54
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# M01 27

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