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# If a, b, c, d and e are integers and p=2^a3^b and q=2^c3

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Joined: 02 Sep 2009
Posts: 52254
Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3  [#permalink]

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21 Sep 2017, 19:35
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 app

In 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 app

For (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  [#permalink]

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23 Oct 2017, 22:19
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 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 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.

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

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Joined: 02 Sep 2009
Posts: 52254
Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3  [#permalink]

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23 Oct 2017, 23:06
1
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 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 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.

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

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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Re: If a, b, c, d and e are integers and p=2^a3^b and q=2^c3  [#permalink]

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15 Apr 2018, 20:50
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 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 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.

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  [#permalink]

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15 Apr 2018, 21:46
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 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 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.

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 non-terminating 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 non-negative):- still second statement tells us that b > d. Since b > d, in this case we will have 3^(b-d) in the numerator. No matter what signs b/d have, if b > d, then b-d 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 [#permalink] 15 Apr 2018, 21:46

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# If a, b, c, d and e are integers and p=2^a3^b and q=2^c3

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