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30 Sep 2010, 04:28
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C
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37%
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Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?
(1) k = 3
(2) j is an odd multiple of 3.
(1) k = 3
(2) j is an odd multiple of 3.
30 Sep 2010, 04:37
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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.)
Questions testing this concept:
700-question-94641.html?hilit=terminating%20decimal
is-r-s2-is-a-terminating-decimal-91360.html?hilit=terminating%20decimal
pl-explain-89566.html?hilit=terminating%20decimal
which-of-the-following-fractions-88937.html?hilit=terminating%20decimal
BACK TO THE ORIGINAL QUESTION:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?
(1) \(k = 3\) --> now, if \(j=3p\) (j is a multiple of 3) then \(\frac{j}{k}=\frac{3p}{3}=p=integer=terminating \ decimal\) but if \(j\) is not a multiple of 3 then reduced fraction \(\frac{j}{k}=\frac{j}{3}\) won't be a terminating decimal, as denominator has primes other than 2 and/or 5. Not sufficient.
(2) \(j\) is an odd multiple of 3 --> \(j=3(2k+1)\), clearly insufficient as no info about the denominator \(k\).
(1)+(2) \(\frac{j}{k}=\frac{3(2k+1)}{3}=2k+1=integer=terminating \ decimal\). Sufficient.
Answer: C.
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.)
Questions testing this concept:
700-question-94641.html?hilit=terminating%20decimal
is-r-s2-is-a-terminating-decimal-91360.html?hilit=terminating%20decimal
pl-explain-89566.html?hilit=terminating%20decimal
which-of-the-following-fractions-88937.html?hilit=terminating%20decimal
BACK TO THE ORIGINAL QUESTION:
Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?
(1) \(k = 3\) --> now, if \(j=3p\) (j is a multiple of 3) then \(\frac{j}{k}=\frac{3p}{3}=p=integer=terminating \ decimal\) but if \(j\) is not a multiple of 3 then reduced fraction \(\frac{j}{k}=\frac{j}{3}\) won't be a terminating decimal, as denominator has primes other than 2 and/or 5. Not sufficient.
(2) \(j\) is an odd multiple of 3 --> \(j=3(2k+1)\), clearly insufficient as no info about the denominator \(k\).
(1)+(2) \(\frac{j}{k}=\frac{3(2k+1)}{3}=2k+1=integer=terminating \ decimal\). Sufficient.
Answer: C.
General Discussion
07 Oct 2010, 01:19
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But it says the ratio j/k is expressed as a decimal, so how come j=3p ?
I thought answer is A since J has to be a non-multiple of 3 since if it is a multiple of 3, j/k cannot be expressed as a decimal. COrrect me if i am wrong!
I thought answer is A since J has to be a non-multiple of 3 since if it is a multiple of 3, j/k cannot be expressed as a decimal. COrrect me if i am wrong!
