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The hundredths digit

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The hundredths digit

by sanju09 » Mon Feb 02, 2009 5:51 am
If 2 < n < 3, is the hundredths digit of the decimal fraction part of n equal to 9?

(1) n + 0.001 > 3

(2) n + 0.0001 < 3

IMO A
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by DanaJ » Mon Feb 02, 2009 8:21 am
By using 1 you get that 2.999 < n < 3. This is of course possible only when the hundredths digit of n is 9 and the thousandths digit is somewhere between 0 and 9.
So 1 alone is sufficient.
Now let's look at 2: it basically tells us that n< 2.9999 < 3, which is no useful "news" since n could be either 2.987 or 2.345 or any other such number between 2 and 3.

Answer A.
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by naaga » Mon Feb 02, 2009 9:48 am
miss Danaj , can you plz give clear explanation, I didnt understand your logic
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Poon Poon

by nattatob » Mon Feb 02, 2009 11:38 am
She means that === you have to solve this question by use inequality equation.

Only thing you have to dealing with this prob is moving 0.001 to another side.

Get ?
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by peddisetty » Mon Feb 02, 2009 5:08 pm
How can you get 2.999+0.001 > 3 ? how is this possible? Please explain.

Thanks,
Raj Peddisetty
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by DanaJ » Mon Feb 02, 2009 11:35 pm
You have stmt 1:
n + 0.001 > 3. Subtract 0.001 from each side and you get that n > 2.999.
Now, you are initially told that 2 < n < 3. By using this and the inequality above, you get that 2.999 < n < 3.
Then you get stmt 2:
n + 0.0001 < 3. Again, subtract 0.0001 from each side and you get that n < 2.9999. But since we know that 2 < n < 3, this only shortens the range for n from 2 - 3 to 2 - 2.9999, which doesn't help much.
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