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If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 58% (01:12) correct 42% (01:10) wrong based on 102 sessions

### HideShow timer Statistics If $$k=\frac{x}{5^az^by^c}$$, and k is expressed as a decimal, will that decimal terminate?

(1) y = 3
(2) z = 2

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If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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Bunuel wrote:
If $$k=\frac{x}{5^az^by^c}$$, and k is expressed as a decimal, will that decimal terminate?

(1) y = 3
(2) z = 2

I will go with C since...any number if it has dr in the form 2^a*5^b results in terminating decimal.
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Math Expert V
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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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2
Balajikarthick1990 wrote:
Bunuel wrote:
If $$k=\frac{x}{5^az^by^c}$$, and k is expressed as a decimal, will that decimal terminate?

(1) y = 3
(2) z = 2

I will go with C since...any number if it has dr in the form 2^a*5^b results in terminating decimal.BUT THERE IS A 3 IN DR

Always be careful, when we do not know anything about the variables..
combined...
we do not know anything about x, a,b, or c.. if any of them is 0 or fraction or -ive..

Insuff
E
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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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[quote="Bunuel"]If $$k=\frac{x}{5^az^by^c}$$, and k is expressed as a decimal, will that decimal terminate?

(1) y = 3
(2) z = 2

I think the answer should be A because if any fraction having 3 in its denominator. That should be non terminating.
I can be wrong also but i will go with A.
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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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Completely agree with chetan2u

The answer has to be (E) , you simply can not solve this problem without knowing the other variables...

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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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[quote="Abhishek009"]Completely agree with chetan2u

[b][color=#ed1c24]The answer has to be (E) , you simply can not solve this problem without knowing the other variables...
[/colorHimanshu

but it is given in the question that k is expressed as a decimal
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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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Himanshu9818 wrote:

but it is given in the question that k is expressed as a decimal

Have a look at this discussion here about terminating decimals :

http://gmatclub.com/blog/2012/04/gmat-m ... -decimals/

Quote:
if any fraction having 3 in its denominator.

You have assumed that since the denominator has 3 , it will be a terminating decimal, thats true but not always..

Consider it this way -

12/3 =4 { Terminating whole number }

1/3 = 0.3333 { Non Terminating decimal }

Here in this case we need to have the value of denominator as well as the numerator....

Say x = 9 , y = 3 {as given in option 1} and c = 2

The we can have a terminating decimal...

Hope it helps !!
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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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Himanshu9818 wrote:
Bunuel wrote:
If $$k=\frac{x}{5^az^by^c}$$, and k is expressed as a decimal, will that decimal terminate?

(1) y = 3
(2) z = 2

I think the answer should be A because if any fraction having 3 in its denominator. That should be non terminating.
I can be wrong also but i will go with A.

Yes if a 3 is in denomnator, answer should be a non terminating decimal, BUT the fraction has to be in simplest form..
examples
$$k=\frac{x}{5^az^by^c}$$..
a) x= 9, c=2..
$$k=\frac{9}{5^az^b3^2}$$ = $$k=\frac{1}{5^az^b}$$.. so 3 is cancelled out. we do ot know what z is. so CAN'T say..
b) x=12, c=2
$$k=\frac{12}{5^az^b3^2}$$ = $$k=\frac{4}{5^az^b*3}$$.. so 3 isin denominator. again MAY be, But if z is 1/3 and b=1...NO
you can actually make so many combinations of these variables..
Neither it is given that these are Integers or Fraction ; -ive or +ive
Clearly Insuff
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Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t  [#permalink]

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_________________ Re: If k=x/5^az^by^c, and k is expressed as a decimal, will that decimal t   [#permalink] 23 Aug 2018, 11:58
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