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x=1+0.01*d, d is a positive integer and d<10 what is the

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VP
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x=1+0.01*d, d is a positive integer and d<10 what is the [#permalink] New post 24 Sep 2007, 15:20
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D
E

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x=1+0.01*d, d is a positive integer and d<10
what is the value of d?

1) 2<=d<=4
2) The thousandth digit of 10(x^2)= the hundredth digit of x^2
VP
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Re: DS [#permalink] New post 24 Sep 2007, 17:30
x=1+0.01*d, d is a positive integer and d<10
what is the value of d?

1) 2<=d<=4
2) The thousandth digit of 10(x^2)= the hundredth digit of x^2[/quote]

1. doesnt give deff. answer, so insuff. alon
2. not enough info again, d can be max. 9 then can 10(x^2) be 1000 or a greater number? the staement says 10(x^2) has a thousands digit.

i would go with E.
or something is missing
or my reasoning is wroing
VP
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 [#permalink] New post 25 Sep 2007, 13:27
OA=B

x=1+0.01*d, d is a positive integer and d<10
what is the value of d?

1) 2<=d<=4
2) The thousandth digit of 10(x^2)= the hundredth digit of x^2

We know from the question that d can be 1,2,3,...,9
For (1) This is obviously INSUFFICIENT since d can be 2, 3, 4
For (2) we know that
If d=1,2,3,4,...,9, then x equals 1.01, 1.02, 1.03, 1.04...,1.09
Then we know that x^2 equals 1.0201, 1.0404, 1.0609, 1.0816, etc...
Therefore, 10*(x^2) will be 10.201, 10.404, 10.609, 10.816, etc...

Now, we can create a table:
(d, thousandth digit of 10*(x^2), hundredth digit of x^2)
(1, 1, 2)
(2, 4, 4)
(3, 9, 6)
(4, 6, 8)
etc...
If you look at the pattern, the question is asking when is the digit unit of d^2 ever equals to 2d knowing that 1<=d<=9. This is only true when d=2. Therefore, B is sufficient.

Cheers.
VP
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Posts: 1105
Location: London
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Re: DS [#permalink] New post 25 Sep 2007, 14:50
bkk145 wrote:
x=1+0.01*d, d is a positive integer and d<10
what is the value of d?

1) 2<=d<=4
2) The thousandth digit of 10(x^2)= the hundredth digit of x^2


Oooops, i messed thousands with thousandth

thanks Bkk
Re: DS   [#permalink] 25 Sep 2007, 14:50
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x=1+0.01*d, d is a positive integer and d<10 what is the

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