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Can someone explain me what is the meaning of terminating decimal.
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trafficspinners
Can someone explain me what is the meaning of terminating decimal.

A decimal number that has digits that do not go on forever.

Examples:

0.25 (it has two decimal digits)
0.123456789 (it has nine decimal digits)

In contrast a Recurring Decimal has digits that go on forever

Example of a Recurring Decimal: 1/3 = 0.333... (the 3 repeats forever)
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Rock750
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer.

(2) 28x is an integer.

I have a bit of difficulty in understanding the intended meaning of "is it possible" part of the question.

The answer can be yes and ofcourse no, but just that there is a possibility that the answer could be yes confuses me a bit. Had the question been framed like this " is x a terminating decimal?", then it would have been clearer. The use of the term "possible" makes it just a bit ambiguous.

Put up for guidance please.
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Rock750
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer.

(2) 28x is an integer.

I have a bit of difficulty in understanding the intended meaning of "is it possible" part of the question.

The answer can be yes and ofcourse no, but just that there is a possibility that the answer could be yes confuses me a bit. Had the question been framed like this " is x a terminating decimal?", then it would have been clearer. The use of the term "possible" makes it just a bit ambiguous.

Put up for guidance please.

The question basically asks: if x is written as a decimal will it be a terminating decimal?

Hope it's clear.
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since the question asks "is it possible", wouldn't the answer be D since .5 is a terminating decimal and 24*.5=12, and 28*.5=24?
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thebloke
since the question asks "is it possible", wouldn't the answer be D since .5 is a terminating decimal and 24*.5=12, and 28*.5=24?

You misinterpret the question. The question asks: if x is written as a decimal will it be a terminating decimal? Thus the correct answer is C, not D.
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we know that x is a proper positive fraction. we need to check whether x has powers of 5 or 2 in the denominator or not.

1. 24(x)=INT ---> \(x=Int/24\) if our integer is 3 then x is can be written as a terminating decimal otherwise x will be a non-terminating decimal

2. 28(x)=INT ----> same story here if our int is 7 then x can be written as a terminating decimal, otherwise x will be a non-terminating decimal

1+2 \(Int/3(2^3)=Int/7(2^2)\) -----> 7(4)Int=8(3)Int the expression has to be equal on both sides thus on the right hand side we need a 7 and on the right hand side we need a 3 and a two. We now know that our integer a terminating decimal because we can get rid of both 7 and 3 in the denominator.

C.

Hope it helps.
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Bunuel
I am sorry
I can not get the combining statements.
We do not need to prove that n is a multiple of 7 ?
another question regarding the denominator, is it enough to be 2s or 5s to terminate X as decimal or 2s * 5s is a must.
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hatemnag
Bunuel
I am sorry
I can not get the combining statements.
We do not need to prove that n is a multiple of 7 ?
another question regarding the denominator, is it enough to be 2s or 5s to terminate X as decimal or 2s * 5s is a must.

From (1) we have that \(x=\frac{m}{24}=\frac{m}{2^3*3}\). If m is a multiple of 3, then 3 in the denominator will be reduced and x will be a terminating decimal.

Similarly, from (2) we have that \(x=\frac{n}{28}=\frac{n}{2^2*7}\). If n is a multiple of 7, then 7 in the denominator will be reduced and x will be a terminating decimal.

The answer to your other question is yes, if a fraction has only 2's or 5's in the denominator it'll terminate.

Check Terminating and Recurring Decimals Problems in our Special Questions Directory.

Hope it helps.
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Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> \(24x=m\), where m an integer --> \(x=\frac{m}{24}=\frac{m}{2^3*3}\), If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> \(28x=n\), where n an integer --> \(x=\frac{n}{28}=\frac{n}{2^2*7}\), If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

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 \(2*5^2\). 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 the 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.

Questions testing this concept:
https://gmatclub.com/forum/does-the-deci ... 89566.html
https://gmatclub.com/forum/any-decimal-t ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-an ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a- ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the- ... 88937.html

Hope it helps.


hi man

since 0<x<1, say x is a proper fraction ....

taking 2 statements together,

x cannot be 1/3, as 3 can divide 24, but cannot divide 28...
in the same line of reasoning, x cannot be 1/7, as 7 can divide 28, but cannot divide 24...
So, for both of the two statements to hold true, x cannot be 3 and/or 7, thus we are left with only 2s in the denominator. Sufficient ....

please say to me whether the reasoning is okay.... :(

thanks in advance, man ...
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gmatcracker2017
Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> \(24x=m\), where m an integer --> \(x=\frac{m}{24}=\frac{m}{2^3*3}\), If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> \(28x=n\), where n an integer --> \(x=\frac{n}{28}=\frac{n}{2^2*7}\), If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

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 \(2*5^2\). 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 the 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.

Questions testing this concept:
https://gmatclub.com/forum/does-the-deci ... 89566.html
https://gmatclub.com/forum/any-decimal-t ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-an ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a- ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the- ... 88937.html

Hope it helps.


hi man

since 0<x<1, say x is a proper fraction ....

taking 2 statements together,

x cannot be 1/3, as 3 can divide 24, but cannot divide 28...
in the same line of reasoning, x cannot be 1/7, as 7 can divide 28, but cannot divide 24...
So, for both of the two statements to hold true, x cannot be 3 and/or 7, thus we are left with only 2s in the denominator. Sufficient ....

please say to me whether the reasoning is okay.... :(

thanks in advance, man ...

Yes, x, when reduced to its simplest form must have 2 or 2^2 in the denominator.
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OFFICIAL EXPLANATION
All terminating decimals share one property: the denominator of the decimal (written as a fraction) must contain only 2’s and 5’s as prime factors. For example, 0.25 written as a fraction is 1/4. The denominator contains only 2’s.

(1): INSUFFICIENT. For 24x to be an integer, the prime factors of 24 must completely cancel out the factors in the denominator of x. 24 = 23 × 3. There is no way to know which prime factors the denominator of x actually contains. If it contains a 3, the decimal will not terminate. If the denominator only contains 2’s, then it will terminate.

For example, if x is 1/2, 24x = 12, and the decimal terminates (0.5). If, however, x = 1/3, 24x = 8, but the decimal does not terminate (0.3333….). The statement is insufficient.

(2): INSUFFICIENT. For 28x to be an integer, the prime factors of 28 must completely cancel out the factors in the denominator of x. 28 = 22 × 7. There is no way to know which prime factors the denominator of x actually contains. If it contains a 7, the decimal will not terminate. If the denominator only contains 2’s, then it will terminate.

For example, if x is 1/2, 28x = 14, and the decimal terminates (0.5). If, however, x = 1/7, 28x = 4, but the decimal does not terminate (0.14285….). The statement is insufficient.

(1) & (2): SUFFICIENT. For both 24x to be an integer and 28x to be an integer, the denominator of x can only contain prime factors that both 24 and 28 share. The only prime factors that 24 and 28 share are 2 and 2(22). Therefore, the denominator of x must only contain 2’s, and the decimal must terminate.

The correct answer is C.
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Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> \(24x=m\), where m an integer --> \(x=\frac{m}{24}=\frac{m}{2^3*3}\), If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> \(28x=n\), where n an integer --> \(x=\frac{n}{28}=\frac{n}{2^2*7}\), If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

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 \(2*5^2\). 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 the 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.

Questions testing this concept:
https://gmatclub.com/forum/does-the-dec ... 89566.html
https://gmatclub.com/forum/any-decimal- ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-a ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the ... 88937.html

Hope it helps.

Hi, I have understood how you have explained the answer. But I solved the question in a different way. Please help me understand if my logic is right.

i. 24x = Int so x = 1/3 or x = 1/4. So NS

ii. 28x = Int so x = 1/4 or x = 1/7 NS

To combine, I did not list out the values of x. Instead, I did 28x - 24x = Int. So 4x = Int
x= 1/2 or x=1/4 in any case x will be non-terminating. Because there is no presence of 3 or 7 in the denominator.

So C
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Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> \(24x=m\), where m an integer --> \(x=\frac{m}{24}=\frac{m}{2^3*3}\), If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> \(28x=n\), where n an integer --> \(x=\frac{n}{28}=\frac{n}{2^2*7}\), If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

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 \(2*5^2\). 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 the 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.

Questions testing this concept:
https://gmatclub.com/forum/does-the-dec ... 89566.html
https://gmatclub.com/forum/any-decimal- ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-a ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the ... 88937.html

Hope it helps.

Hi, I have understood how you have explained the answer. But I solved the question in a different way. Please help me understand if my logic is right.

i. 24x = Int so x = 1/3 or x = 1/4. So NS

ii. 28x = Int so x = 1/4 or x = 1/7 NS

To combine, I did not list out the values of x. Instead, I did 28x - 24x = Int. So 4x = Int
x= 1/2 or x=1/4 in any case x will be non-terminating. Because there is no presence of 3 or 7 in the denominator.

So C

Yes, you are correct. For equation (1) though, x also can be 1/2, 1/6, 1/8, and 1/12. For equation (2), x can also be 1/2, 1/8, and 1/14. Additionally, for (1)+ (2), we obtain 4x = integer, which implies x = integer/4, which means that x will terminate because if the denominator of a fraction contains only 2's and/or 5's, then the fraction will be a terminating decimal.
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Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.


Hi Bunuel , sorry but how does \(\frac{m}{n}=\frac{2*3}{7}\) prove that m is a multiple of 3? shouldnt it be M is a multiple of 6? i say this because if m is just a multiple of 3, then if we cross multiply, it will be 7(3)/6 = 3.5 which is not an integer.
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Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.


Hi Bunuel , sorry but how does \(\frac{m}{n}=\frac{2*3}{7}\) prove that m is a multiple of 3? shouldnt it be M is a multiple of 6? i say this because if m is just a multiple of 3, then if we cross multiply, it will be 7(3)/6 = 3.5 which is not an integer.

Yes, \(\frac{m}{n}=\frac{2*3}{7}\) implies m is a multiple of 6. However, if it's a multiple of 6, then it must also be a multiple of 3. As we got in (1) if m is a multiple of 3, then x will be a terminating decimal because 3 in the denominator of \(x=\frac{m}{2^3*3}\) will get reduced.

Hope it's clear.
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Hi Bunuel, could you please explain how is this happening, I couldn't understand the below point, how are we concluding m to be a factor of 6, and n to be a factor of 7

m/n = 6/7, Since m and n are integers, m will be a multiple of 6 (and thereby of 3 too) and n will be a multiple of 7

Bunuel
If 0 < x < 1, is it possible to write x as a terminating decimal?

(1) 24x is an integer --> \(24x=m\), where m an integer --> \(x=\frac{m}{24}=\frac{m}{2^3*3}\), If m is a multiple of 3, then the answer is YES, else the answer is NO. Not sufficient.

(2) 28x is an integer --> \(28x=n\), where n an integer --> \(x=\frac{n}{28}=\frac{n}{2^2*7}\), If n is a multiple of 7, then the answer is YES, else the answer is NO. Not sufficient.

(1)+(2) \(x=\frac{m}{2^3*3}=\frac{n}{2^2*7}\) --> \(\frac{m}{n}=\frac{2*3}{7}\) --> m IS a multiple of 3 (as well as n IS multiple of 7). Sufficient.

Answer: C.

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 \(2*5^2\). 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 the 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.

Questions testing this concept:
https://gmatclub.com/forum/does-the-dec ... 89566.html
https://gmatclub.com/forum/any-decimal- ... 01964.html
https://gmatclub.com/forum/if-a-b-c-d-a ... 25789.html
https://gmatclub.com/forum/700-question-94641.html
https://gmatclub.com/forum/is-r-s2-is-a ... 91360.html
https://gmatclub.com/forum/pl-explain-89566.html
https://gmatclub.com/forum/which-of-the ... 88937.html

Hope it helps.
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