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Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3
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21 Sep 2017, 20:35
goalMBA1990 wrote: Bunuel wrote: goalMBA1990 wrote: Thank you very much for clarification. So answer we would be B for above question. Correct? Sent from my XT1663 using GMAT Club Forum mobile appIn the below question: Is P/Q a terminating decimal? (1) P is a factor of 100 (2) Q is a factor of 100 Does the solution mention that p can be 1/6 for the first statement or for the second? If for (2), then yes p could be 1/6 and in this case (2) is not sufficient. When combined, since p is a factor of 100, then it must be an integer, thus p/q will be terminating decimal and the answer would be C. Whether it is mentioned or not, how any fraction become factor of any integer? How can we assume fraction for P or Q? Sent from my XT1663 using GMAT Club Forum mobile appFor (1) since p is a factor of 100, then it must be an integer. But q can be any real number: fraction, integer, irrational, ... For (2) since q is a factor of 100, then it must be an integer. But p can be any real number: fraction, integer, irrational, ...
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Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3
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23 Oct 2017, 23:19
Bunuel wrote: enigma123 wrote: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d
Any idea what is the concept behind this question to get a answer B? 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 nonnegative 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 2s and/or 5s 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. BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d? (1) a > c. Not sufficient. (2) b > d. Sufficient. Answer: B. Hope it helps.  as you said that both the numerator and denominator should be reduced to the lower term ,,, but as in the question, it was clearly seen that 2^a and 2^c is present in both numerator and denominator so the term is not reduced to its lowest term... now from statement 1, we can clearly state that 2 will not present in the denominator and we have to have only 2 and 5 in the denominator to say that the term would be terminating. Hence statement 1 looks sufficient to me could you please help me explain this ?



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Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3
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24 Oct 2017, 00:06
priyanknema wrote: Bunuel wrote: enigma123 wrote: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d
Any idea what is the concept behind this question to get a answer B? 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 nonnegative 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 2s and/or 5s 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. BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d? (1) a > c. Not sufficient. (2) b > d. Sufficient. Answer: B. Hope it helps.  as you said that both the numerator and denominator should be reduced to the lower term ,,, but as in the question, it was clearly seen that 2^a and 2^c is present in both numerator and denominator so the term is not reduced to its lowest term... now from statement 1, we can clearly state that 2 will not present in the denominator and we have to have only 2 and 5 in the denominator to say that the term would be terminating. Hence statement 1 looks sufficient to me could you please help me explain this ? If the denominator of a reduced fraction has only 2's OR only 5's OR only 2's and 5's the fraction will be a terminating decimal.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3
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15 Apr 2018, 21:50
Bunuel wrote: enigma123 wrote: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d
Any idea what is the concept behind this question to get a answer B? 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 nonnegative 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 2s and/or 5s 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. BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d? (1) a > c. Not sufficient. (2) b > d. Sufficient. Answer: B. Hope it helps. Hi Bunuel, what if c and e are negative? Since the questions says they are integers so they could be negative as well.



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Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3
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15 Apr 2018, 22:46
vedanshimurarka wrote: Bunuel wrote: enigma123 wrote: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal? (1) a > c (2) b > d
Any idea what is the concept behind this question to get a answer B? 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 nonnegative 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 2s and/or 5s 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. BACK TO THE ORIGINAL QUESTION: If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?Question: is \(\frac{2^a*3^b}{2^c*3^d*5^e}\) a terminating decimal? The question basically asks whether we cans reduce 3^d in the denominator so to have only powers of 2 and 5 left, which can be rephrased is b (the power of 3 in the nominator) greater than or equal to d (the power of 3 in the denominator): is b>=d? (1) a > c. Not sufficient. (2) b > d. Sufficient. Answer: B. Hope it helps. Hi Bunuel, what if c and e are negative? Since the questions says they are integers so they could be negative as well. Hello Good point I think. But then again if the powers can be negative integers, we should only be concerned with 'b' and 'd' here (since a, c, e are powers of 2's/5's and whether their powers are positive or negative doesn't matter in this question as with any integer powers of 2's/5's it will still remain a terminating decimal. So our concern is with powers of 3 here, because after cancelling all terms, after reducing this fraction to its lowest form, if there still remains a '3' or a higher power of '3' in the denominator then it will turn out to be a nonterminating decimal, but if '3' is completely removed from the denominator then it will be a terminating decimal). So now even if b and d are say negative integers (or one of them is negative and other is nonnegative): still second statement tells us that b > d. Since b > d, in this case we will have 3^(bd) in the numerator. No matter what signs b/d have, if b > d, then bd will be > 0 and thus we will have 3^(positive integer) in the numerator. This means 3 will be completely eliminated from the denominator, and thus this will be a terminating decimal indeed. Hence B answer. (You can try taking negative values of b/d but where b > d, you will find that there will always be a terminating decimal as 3 will be removed from the denominator)




Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3 &nbs
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