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If x is the decimal 8.1d5, with d as an unknown digit, and x

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iikarthik
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If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   Updated on: 19 Oct 2014, 07:17
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If x is the decimal 8.1d5, with d as an unknown digit, and x rounded to the nearest tenth is equal to 8.1, which digits could not be the value of d?

Solution:
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In order for x to be 8.1 when rounded to the nearest tenth, the right-digit-neighbor, d, must be less than
5. Therefore, d cannot be 5, 6, 7, 8 or 9.


But i feel that d Cannot be 4 as well,bcos 8.145,when rounded off will become 8.15 and then 8.2

Pls correct if my understanding is incorrect

Originally posted by iikarthik on 20 Apr 2013, 22:59.
Last edited by Bunuel on 19 Oct 2014, 07:17, edited 2 times in total.
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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   20 Apr 2013, 23:14
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iikarthik wrote:
Hi,

This is a question from MGMAT Study Guides.

If x is the decimal 8.1d5, with d as an unknown digit, and x rounded to the nearest
tenth is equal to 8.1, which digits could not be the value of d?

Solution:
In order for x to be 8.1 when rounded to the nearest tenth, the right-digit-neighbor, d, must be less than
5. Therefore, d cannot be 5, 6, 7, 8 or 9.

But i feel that d Cannot be 4 as well,bcos 8.145,when rounded off will become 8.15 and then 8.2

Pls correct if my understanding is incorrect


When rounding, you only look at the closest digit to the cutoff point, not any of the ones past it.

Rounding .145 to .2 because you first round the 4 to a 5 would be the same as rounding the number 145 up to 200. 145 is closer to 100 than to 200.. For another example, if we had .4999999 we would never round that number up to 1, because it's closer to 0 than to 1... and that is why we only look at the closest digit.
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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   21 Apr 2013, 04:14
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iikarthik wrote:
Hi,

This is a question from MGMAT Study Guides.

If x is the decimal 8.1d5, with d as an unknown digit, and x rounded to the nearest tenth is equal to 8.1, which digits could not be the value of d?

Solution:
In order for x to be 8.1 when rounded to the nearest tenth, the right-digit-neighbor, d, must be less than
5. Therefore, d cannot be 5, 6, 7, 8 or 9.

But i feel that d Cannot be 4 as well,bcos 8.145,when rounded off will become 8.15 and then 8.2

Pls correct if my understanding is incorrect


Rounding rules

Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, ROUND UP the last digit that you keep. If the first dropped digit is 4 or smaller, ROUND DOWN (keep the same) the last digit that you keep.

Example:
5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5.
5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5.
5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:
8.1d5 rounded to the nearest tenth is equal to 8.1 --> the dropped digit is d, thus we are only concerned with it: d cannot be 5, 6, 7, 8, or 9.

Hope it helps.
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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   21 Apr 2013, 02:32
Hi iikarthik,

before posting any question, please check our rules for posting: you have to provide 5 possible answers.

The solution to your question is (of course) correct

\(8,1d5\) rounded becomes \(8,1\): so d can be range from 0 to 4.

X is rounded to the nearest tenth, this means that all the possible values of d (0,1,...,9) will be rounded to the "closest" tenth.

\(8,145\), in particular is rounded to \(8.1\). Distance to \(8.1\) is \(0.045\), distance to \(8.2\) is \(0.055\). Will be rounded to \(8.1\)

Hope this clarifies, let me know
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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   19 Oct 2014, 05:38
If x is the decimal 8.1d5, with d as an unknown digit, and x rounded to the
nearest tenth is equal to 8.1, which digits could not be the value of d?

In my view answer should be 4,5,6,7,8,9 but the original answer is 5,6,7,8,9. Am i interpreting the question wrong ?
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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   19 Oct 2014, 07:18
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akshaychandna wrote:
If x is the decimal 8.1d5, with d as an unknown digit, and x rounded to the
nearest tenth is equal to 8.1, which digits could not be the value of d?

In my view answer should be 4,5,6,7,8,9 but the original answer is 5,6,7,8,9. Am i interpreting the question wrong ?


Merging topics. Please refer to the discussion above.
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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink] New post   11 Apr 2018, 19:53
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Hi All,

If we're rounding the number 8.1d5 to the nearest TENTH and ending up with 8.1, then here's what we know about the digit d:

It COULD be 0, 1, 2, 3, or 4 because these numbers would round DOWN to the nearest tenth, giving us 8.1

It CANNOT be 5, 6, 7, 8, 9 because these numbers would round UP to the nearest tenth, giving us 8.2, which is NOT a match for what we're told.

In this question, it's worth noting that since we're rounding to the nearest TENTH, the "5" in the thousanths place has NO EFFECT on the rounding.

The question asks for which digits COULD NOT be the value of d, so the answer is 5, 6, 7, 8, 9

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Re: If x is the decimal 8.1d5, with d as an unknown digit, and x [#permalink]
11 Apr 2018, 19:53
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