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A terminating decimal is defined as a decimal that has a

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shekar123
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A terminating decimal is defined as a decimal that has a [#permalink] New post   22 May 2010, 14:35
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A terminating decimal is defined as a decimal that has a finite number of nonzero digits. Examples of terminating decimals are 0.24, 52, and 6.0314. x and y are positive integers. If x/y is expressed as a decimal, is it a terminating decimal?

(1) 40 < x < 45

(2) y = 8
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Re: 700+ question [#permalink] New post   22 May 2010, 16:46
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Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

In original question statement (2) says that denominator equals to 2^3=8, hence x/8 will be terminating decimal no matter what the value of x is.
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   24 Apr 2012, 04:20
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Found this helpful explanation on Mgmat site. :good
Author- Emily Sledge

This rule took a while for me to internalize. It's tough to picture a decimal terminating when the denominator is so huge, such as DWG's example of 43/256. I found it helped me to think about the basic patterns:

1/2^1 = 0.5
1/2^2 = 0.25
1/2^3 = 0.125
1/2^4 = 0.0625
1/2^5 = 0.03125
1/2^6 = 0.015625
1/2^7 = 0.0078125

1/5^1 = 0.2
1/5^2 = 0.04
1/5^3 = 0.008
1/5^4 = 0.0016
1/5^5 = 0.00032
1/5^6 = 0.000064
1/5^7 = 0.0000128

Every one of these terminates, and the pattern indicates that would continue to be true for higher powers. The number of decimal places increases along with the powers of 2 or 5, but the number of decimal places will always be finite.

In contrast, any factors other than 2 or 5 in the denominator can quickly be shown to be non-terminating, even for the most basic case (exponent of 1). Higher powers would be even messier:
1/3 = 0.33333(3 repeating)
1/6 = 0.16666(6 repeating)
1/7 = 0.142857(142857 repeating)
1/9 = 0.11111(1 repeating)
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Re: 700+ question [#permalink] New post   22 May 2010, 14:54
(1) Definitely isn't enough, because you can tweak the denominator any way you like to arrive at fraction that simplifies down to say, 2/3 (0.66666...)
(2) Is the same thing.

At first glance this question seems really too easy. You need to know what's going on with both x and y in order be able to answer it.

C) jumps out immediately. It's obvious that if you know even the ranges of both x and y you could test each combo out and eventually arrive at an answer so clearly taken together they are sufficient.

It's just a matter of seeing if A, B or D is feasible. I don't think they are for the reasons I gave above. If I only know either the nominator or denominator, I can find a match somewhere down the line that allows to make it so the simplified result ends up to be 2/3 (recurring decimal) or 1/1 (obviously a terminating one).

So: C
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Re: 700+ question [#permalink] New post   22 May 2010, 15:08
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dalmba wrote:
(1) Definitely isn't enough, because you can tweak the denominator any way you like to arrive at fraction that simplifies down to say, 2/3 (0.66666...)
(2) Is the same thing.

At first glance this question seems really too easy. You need to know what's going on with both x and y in order be able to answer it.

C) jumps out immediately. It's obvious that if you know even the ranges of both x and y you could test each combo out and eventually arrive at an answer so clearly taken together they are sufficient.

It's just a matter of seeing if A, B or D is feasible. I don't think they are for the reasons I gave above. If I only know either the nominator or denominator, I can find a match somewhere down the line that allows to make it so the simplified result ends up to be 2/3 (recurring decimal) or 1/1 (obviously a terminating one).

So: C



OA is B

Statement (2): Any number divided by 8 results in a terminating decimal. This is because when a number is divided by 2, the only possible remainders are or 1, 2, 3, 4, 5, 6, and 7 (actually 1/8, 2/8, etc.). These remainders are expressed as .125, .25, .375, .5, .625, .75, and .875, respectively. Therefore x/y is a terminating decimal; SUFFICIENT.

I did not understand this...

could u please help me
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Re: 700+ question [#permalink] New post   22 May 2010, 15:33
Haha, this is why I'm bad at DS.

You're supposed to intuitively know that anything ever divided by 8 will result in a terminal decimal.

I messed up in assuming I had control over the denominator (when I said (2) is the same logic as why (1) doesn't work), when it clearly said it was 8 and nothing else.

It's funny, the last DS question I answered here I got right but made a big deal about feeling reluctant to toss out B. I was right to be feeling that way, but just not for the correct question. *sigh* :roll:
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Re: 700+ question [#permalink] New post   22 May 2010, 15:50
shekar123 wrote:
A terminating decimal is defined as a decimal that has a finite number of nonzero digits. Examples of terminating decimals are 0.24, 52, and 6.0314. x and y are positive integers. If x/y is expressed as a decimal, is it a terminating decimal?

(1) 40 < x < 45

(2) y = 8


B.

any positive integer number divided by 8 gives terminate decimal equal to 5.
1) is very alluring, cause we read 1) and then 2), having in mind "I dont know x, so I cant find out what is the decimal point, thus I need to know the range of numbers for X"
Thus c is wrong.
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Re: 700+ question [#permalink] New post   22 May 2010, 17:05
Bunuel wrote:
For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

That is sweet. A nifty little tool that's not too horrible to remember.

Also, in order to preserve some sort of semblance of competency in this realm I will quite proudly point out that there was a typo in your post here:

Bunuel wrote:
as \(250\) (denominator) equals to \(2*5^2\).

As it should read - "\(250\) (denominator) equals to \(2*5^3\)" :wink:
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Re: 600 + question [#permalink] New post   22 May 2010, 21:58
shekar123 wrote:
A terminating decimal is defined as a decimal that has a finite number of nonzero digits. Examples of terminating decimals are 0.24, 52, and 6.0314. x and y are positive integers. If x/y is expressed as a decimal, is it a terminating decimal?

(1) 40 < x < 45

(2) y = 8



1) we don't know anything about y, so Insufficient

2) when you divide anything by 8 the answer will be either an integer, a fraction of a multiple of 0.125. For example 201/8 = 25.125 and 203/8 = 25.375

So in any case, this will always lead to a terminating decimal. Sufficient

My Answer: B
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   27 Apr 2016, 14:43
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In statement 1 x can be 41,42,43 or 44. insufficient as we don't have value of y

In statement 2, clearly y is given as 8 which is the denominator of the fraction x/y. we know that any number which has denominator as 8 will be a terminating decimal

so correct answer - B
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   21 Dec 2020, 10:36
Hi Bunuel

My question is can we consider an integer as a terminating decimal?

Ex: for statement 2,

if y = 8 and x is a multiple of 8 such as 8 , 16, 24 etc., we get x/y as an integer rather than a decimal.

So, if an integer cannot be considered a terminating decimal, then C would be the correct answer as option C excludes this possibility.

Can you please clarify?
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   20 Nov 2022, 12:32
Bunuel wrote:
Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

In original question statement (2) says that denominator equals to 2^3=8, hence x/8 will be terminating decimal no matter what the value of x is.



Do we consider an integer a terminating decimal too???
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   20 Nov 2022, 12:35
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Stanindaw wrote:
Bunuel wrote:
Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

In original question statement (2) says that denominator equals to 2^3=8, hence x/8 will be terminating decimal no matter what the value of x is.



Do we consider an integer a terminating decimal too???


Yes because every integer has finite number of nonzero digits.
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   20 Nov 2022, 18:11
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shekar123 wrote:
A terminating decimal is defined as a decimal that has a finite number of nonzero digits. Examples of terminating decimals are 0.24, 52, and 6.0314. x and y are positive integers. If x/y is expressed as a decimal, is it a terminating decimal?

(1) 40 < x < 45

(2) y = 8


Question Stem Analysis:

We need to determine whether x/y is a terminating decimal, where x and y are positive integers. Recall that the decimal equivalent of a fraction is a terminating decimal only if the denominator of the most reduced form of the fraction contains only prime factors of 2 and 5.

Statement One Alone:

\(\Rightarrow\) 40 < x < 45

Without any information about y, we cannot determine whether x/y terminates. If x = 41 and y = 3, then x/y does not terminate. On the other hand, if x = 41 and y = 10, then x/y = 4.1 is a terminating decimal. Since there are more than one possible answers, statement one alone is not sufficient.

Eliminate answer choices A and D.

Statement Two Alone:

\(\Rightarrow\) y = 8

If y = 8, then regardless of the value of x, the denominator of the most reduced form of x/y = x/8 cannot contain any prime factors other than 2. This means that x/y is a terminating decimal. Statement two alone is sufficient.

Answer: B
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Re: A terminating decimal is defined as a decimal that has a [#permalink] New post   21 Nov 2022, 00:33
Bunuel wrote:
Stanindaw wrote:
Bunuel wrote:
Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

In original question statement (2) says that denominator equals to 2^3=8, hence x/8 will be terminating decimal no matter what the value of x is.



Do we consider an integer a terminating decimal too???


Yes because every integer has finite number of nonzero digits.



Thanks for such a quick reply!

Posted from my mobile device
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