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J = x rounded to the nearest TENTH, and K = x

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J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 31 May 2018, 06:40
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J = x rounded to the nearest TENTH, and K = x rounded to the nearest THOUSANDTH. If 0 < x < 1, what is the greatest possible value of J – K?

A) 0.005
B) 0.0055
C) 0.05
D) 0.0505
E) 0.0555

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Re: J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 25 Jun 2018, 15:57
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GMATPrepNow wrote:
J = x rounded to the nearest TENTH, and K = x rounded to the nearest THOUSANDTH. If 0 < x < 1, what is the greatest possible value of J – K?

A) 0.005
B) 0.0055
C) 0.05
D) 0.0505
E) 0.0555


Since J = x rounded to the nearest tenth, J = 0.a (where a is a digit 0-9).
Since K = x rounded to the nearest thousandth, K = 0.bcd (where b, c and d are any digits 0-9).
The value of K implies that J-K can have at most three digits to the right of the decimal.
Eliminate B, D and E.

To maximize J-K, we want x to round UP to the nearest tenth (so that J is as big as possible) but round DOWN to the nearest thousandth (so that K is as small as possible).
One option:
x = 0.0501. with the result that J = 0.1 and K = 0.050.
In this case, J-K = 0.1 - 0.05 = 0.05.
Since option C can be the value of J-K, eliminate A, which is smaller.


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J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 01 Jun 2018, 07:32
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Hi chandra004

The difference between an integer(x) rounded to its nearest tenth and thousandth must be maximum.
Also, the integer(x) needs to be in between 0 & 1. For this to be true, the tenth must be the maximum
possible (which is 0.9**) & thousandth must be the minimum possible(which is be 0.**0)

Therefore, the integer x has to be 0.850. The reason the hundredth' digit is 5 as we need the maximum difference.

Hope this helps you!
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Re: J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 31 May 2018, 06:51
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GMATPrepNow wrote:
J = x rounded to the nearest TENTH, and K = x rounded to the nearest THOUSANDTH. If 0 < x < 1, what is the greatest possible value of J – K?

A) 0.005
B) 0.0055
C) 0.05
D) 0.0505
E) 0.0555

*kudos for all correct solutions


Since 0 < x < 1, x = 0.8502(Since we need the greatest possible value of J - K)

When x is rounded to the nearest tenth, we will get 0.9(which is the value of J)
When x is rounded to the nearest thousandth, we will get 0.850(which is the value of K)

Therefore, the greatest possible difference between J and K is 0.900 - 0.850 = 0.05(Option C)
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Re: J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 01 Jun 2018, 05:57
pushpitkc wrote:
GMATPrepNow wrote:
J = x rounded to the nearest TENTH, and K = x rounded to the nearest THOUSANDTH. If 0 < x < 1, what is the greatest possible value of J – K?

A) 0.005
B) 0.0055
C) 0.05
D) 0.0505
E) 0.0555

*kudos for all correct solutions


Since 0 < x < 1, x = 0.8502(Since we need the greatest possible value of J - K)

When x is rounded to the nearest tenth, we will get 0.9(which is the value of J)
When x is rounded to the nearest thousandth, we will get 0.850(which is the value of K)

Therefore, the greatest possible difference between J and K is 0.900 - 0.850 = 0.05(Option C)


Hello pushpitkc
I didn't understand your solution. How did you choose x as 0.8502. Please elaborate.
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Re: J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 25 Jun 2018, 03:45
Hi
I agree that J should be 0.9 but I'm still not clear about K... Why Can't K be 0.001 if K is rounded up to the nearest thousandth? 0.850 seems rounded up to the nearest hundredth to me since it has zero in the thousandth place. Could you please help me out here. I'm confused.

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Re: J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 25 Jun 2018, 15:43
i didn't understand the solution, please explain more. thank you
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Re: J = x rounded to the nearest TENTH, and K = x  [#permalink]

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New post 04 Aug 2019, 00:58
laknaouiwarda1 wrote:
i didn't understand the solution, please explain more. thank you



The greatest possible value of J-K is when J = x is considered to be the highest possible number between 0 and 1 i.e. 0.9 (rounded to the nearest 10th) and when K = x is considered to be the lowest possible number between 0 and 1 i.e. 0.001 (rounded to the nearest 1000th). This gives as J-K = 0.899. This tells us two things:

1. There can be a maximum of three digits to the right of the decimal.
2. Max possible value of the difference is 0.899. So the real difference in the options can be any number less or equal to 0.899


Let's eliminate the options now. B, D, and E have more than 3 digits to the right of the decimal so eliminate these options. Compare and select the highest possible value between option A and C to get the right answer basis what is asked. Hence, Option C is the right answer.


Hope this resolves your doubt.
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Re: J = x rounded to the nearest TENTH, and K = x   [#permalink] 04 Aug 2019, 00:58
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