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2 GMATPrep questions [#permalink] New post 21 Jul 2007, 07:39
I need some help. The first problem I just do not know who to solve. The second question I assumed the answer, but I'd like to understand the reasoning behind it. If I can get some explanations, that will be very helpful. Thanks in advance!

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Re: 2 GMATPrep questions [#permalink] New post 21 Jul 2007, 17:14
Question 1: D
Question 2: E (x=100, y=-1)
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 [#permalink] New post 22 Jul 2007, 05:27
Please, can someone provide explanations? Thanks!
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 [#permalink] New post 23 Jul 2007, 11:23
Bump...can anyone help?
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 [#permalink] New post 23 Jul 2007, 11:53
Where are the questions?
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 [#permalink] New post 23 Jul 2007, 11:58
I attached them in the first post. I believe you need to be registered to see them. Do you need me to post them here?
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 [#permalink] New post 23 Jul 2007, 12:03
Now I see them.

Is the hundredths digit of the decimal d greater than 5?

(1) The tenths digit of 10d is 7.
(2) The thousandths digit of d/10 is 7.

So from 1, we see that the hundredths digit is 7, ie d=0.07 since (1) means that (0.07)10=0.7. So from this, we know that the hundredths digit is greater than 5. sufficient.

From 2, same thing basically. We know that the hundredths digit is 7 because (0.07)/10=0.007 which would make the thousandths digit of d/10 equal to 7.

So the answer is D. Does this make sense?
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 [#permalink] New post 23 Jul 2007, 12:04
Question 1:
Is the hundredths digit of the decimal d greater than 5?

[1] The tenths digit of 10d is 7.
[2] The thousandths digit of (d/10) is 7.


Question 2:
If x and y are integers and x>0, is y>0?

[1] 7x-2y>0
[2] -y<x
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 [#permalink] New post 23 Jul 2007, 12:06
THANK YOU! Yes, that makes sense. Any ideas on question 2?
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 [#permalink] New post 23 Jul 2007, 12:08
If x and y are integers and x>0, is y>0?

(1) 7x-2y>0
(2) -y<x>2y, which means x>(2/7)y. So y is not necessarily greater than 0. Take the case of x=5 and y=-1. So (1) is insufficient.

From (2), we see that x+y>0. Again, y is not necessarily greater than 0. Let x=2 and y=-1.

Together, we know that x>(2/7)y and x>-y. Not sufficient. Let x=10 and y=-1. Then the conditions hold. But they also hold for x=10 and y=1. So y could be neg or pos.

So E.
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 [#permalink] New post 23 Jul 2007, 12:11
Ah, the formatting is all messed up on that last explanation.

After the (2) it is supposed to say just -y<x.

Then it's supposed to say, from 1, 7x>2y, which means x>(2/7)y.... etc.

It won't let me edit for some reason!
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 [#permalink] New post 23 Jul 2007, 12:15
I've had the same problem with the editing, so don't worry about it.

Thank you for your explanations, you've been very helpful.
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 [#permalink] New post 23 Jul 2007, 12:52
briks123 wrote:
Now I see them.

Is the hundredths digit of the decimal d greater than 5?

(1) The tenths digit of 10d is 7.
(2) The thousandths digit of d/10 is 7.

So from 1, we see that the hundredths digit is 7, ie d=0.07 since (1) means that (0.07)10=0.7. So from this, we know that the hundredths digit is greater than 5. sufficient.

From 2, same thing basically. We know that the hundredths digit is 7 because (0.07)/10=0.007 which would make the thousandths digit of d/10 equal to 7.

So the answer is D. Does this make sense?


I'm having some trouble following this. Isn't it asking for the hundredths and we were able to see the tenths and thousandths? Or are we saying its a repeating decimal?
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 [#permalink] New post 23 Jul 2007, 13:05
If I understand correctly, what we need to calculate is for what value of d, when multiplied by 10, do we get a number with a tenth digit of 0.7. Here is the equation:
10d=0.7 --> to find d, simply divide 0.7 by 10 and we get 0.07. Therefore, the hundredths digit is greater than 5.

Statement 2 requires the same process. Does this help?
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 [#permalink] New post 23 Jul 2007, 13:50
Bluebird wrote:
If I understand correctly, what we need to calculate is for what value of d, when multiplied by 10, do we get a number with a tenth digit of 0.7. Here is the equation:
10d=0.7 --> to find d, simply divide 0.7 by 10 and we get 0.07. Therefore, the hundredths digit is greater than 5.

Statement 2 requires the same process. Does this help?


got it. Thanks for the clarification for the math impaired :)
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 [#permalink] New post 23 Jul 2007, 18:29
St1:
10*d = 7

If d = 0.07, then d*10 = 0.7 -> tenth digit of 10d is 7, 100th digit of d is 7. Sufficient.

St2:
If d = 0.07, then d/10 = 0.007 -> 1000th digit of d/10 is 7, 100th digit of d is 7
Sufficient

Ans D
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 [#permalink] New post 23 Jul 2007, 18:34
St1:
7x-2y > 0

Insufficient. x could be 2 and y could be 1 and 7x-2y > 0. It colud also be x = 2, y = -2, then 7x-2y > 0.

St2:
0 < x+y
Since x is positive, this inequality only tells us y is either positive, or a negative that can be cancelled out by a bigger x.
E.g If x = 1000, y = -1, then x+y > 0. If x = 3, y = 3, x+y > 0. Insufficient.

Using both st1 and st2, still insufficient.

y could be positive or negative.
  [#permalink] 23 Jul 2007, 18:34
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