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# If x,y are numbers >0 and 2^x*5^y=400

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If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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04 May 2013, 12:42
3
9
00:00

Difficulty:

85% (hard)

Question Stats:

44% (01:35) correct 56% (01:30) wrong based on 257 sessions

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If $$x,y$$ are numbers $$>0$$ and $$2^x*5^y=400$$, what could be the value of $$x$$?

I)2
II)3
III)4

A)Only I
B)Only II
C)I and II
D)Only III
E)I, II and III

Hi guys, this is my first PS question hope you like it, any feedback is appreciated.

(BTW: it's not as easy as it seems, it mays tricks you...)

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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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04 May 2013, 23:15
4
3
Good job hfbamafan!

Official Explanation

If x,y are numbers >0 and $$2^x*5^y=400$$, what could be the value of x?

I)2
II)3
III)4

Read the question carefully! I said numbers not integers

$$2^x*5^y=2^4*5^2$$

III is true for sure. But pay attention that y (because could be a non integer) makes all the possible answers correct:

I)$$2^2*5^y=2^4*5^2$$, $$5^y=2^2*5^2$$, $$5^y=100$$, so y is a number $$2<y<3$$
II)$$2^3*5^y=2^4*5^2$$, $$5^y=2^1*5^2$$, $$5^y=50$$ , $$2<y<3$$

The correct answer is E) I, II and III
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##### General Discussion
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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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04 May 2013, 17:32
1
Zarrolou wrote:
If $$x,y$$ are numbers $$>0$$ and $$2^x*5^y=400$$, what could be the value of $$x$$?

I)2
II)3
III)4

A)Only I
B)Only II
C)I and II
D)Only III
E)I, II and III

Hi guys, this is my first PS question hope you like it, any feedback is appreciated.

I'll post the OA and the OE after some discussion. Kudos to the first ones who got it right!
(BTW: it's not as easy as it seems, it mays tricks you...)

I am getting E

You can make it into \sqrt{2^x} * \sqrt{5^y} =20

given that all numbers are >0

Therefor x=2 x^2*5^1=20 2 works

and when x=4 gives you 16*25=400

if x=3 it comes out to 2\sqrt{2}*5\sqrt{2}=400
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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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04 May 2013, 20:42
Zarrolou wrote:
If $$x,y$$ are numbers $$>0$$ and $$2^x*5^y=400$$, what could be the value of $$x$$?

I)2
II)3
III)4

A)Only I
B)Only II
C)I and II
D)Only III
E)I, II and III

Hi guys, this is my first PS question hope you like it, any feedback is appreciated.

I'll post the OA and the OE after some discussion. Kudos to the first ones who got it right!
(BTW: it's not as easy as it seems, it mays tricks you...)

First we represent 400 in its prime factored form.
$$400 = 4 * 100 = 2^2 * 2^2 * 5^2 = 2^4 * 5^2$$
Hence $$x = 4$$ is the only possible value.
Correct option is D
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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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04 May 2013, 21:48
Zarrolou wrote:
If $$x,y$$ are numbers $$>0$$ and $$2^x*5^y=400$$, what could be the value of $$x$$?

I)2
II)3
III)4

A)Only I
B)Only II
C)I and II
D)Only III
E)I, II and III

$$2^x*5^y=400$$
Substituting x=2,3,4 in the given equation
I) 2 --> $$2^2*5^y=400$$; $$5^y=100$$ --> Not possible
II) 3 --> $$2^3*5^y=400$$; $$5^y=50$$ --> Not possible
III) 4 --> $$2^4*5^y=400$$; $$5^y=25$$ , so y =2--> Correct

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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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21 Jul 2015, 07:57
$$2^x * 5^y = 400$$

(i) if $$x = 2$$, then $$5^y = \frac{400}{2^2} = 100$$. So, $$5^y = 100$$ => $$2 < y < 3$$.

(ii) if $$x = 3$$, then $$5^y = \frac{400}{2^3} = 50$$. So, $$5^y = 50$$ => $$2 < y < 3$$.

(iii) if $$x = 4$$, then $$5^y = \frac{400}{2^4} = 25$$, hence $$y = 2$$.

All three cases are possible. Hence Ans (E).
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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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26 Feb 2016, 18:32
i understand that y can be a non-integer
but who can give an exact answer when 2^2 or 2^3 * 5^y, where y is positive, is equal to 400?
please, 1 single example, how 5^y = 100 or 50.
what power y should be? what fraction? with exact steps..
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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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26 Feb 2016, 18:52
mvictor wrote:
i understand that y can be a non-integer
but who can give an exact answer when 2^2 or 2^3 * 5^y, where y is positive, is equal to 400?
please, 1 single example, how 5^y = 100 or 50.
what power y should be? what fraction? with exact steps..

A suggestion: when you ask a question, ask in such a way that shows that you want to learn something. Be humble about it.

The question statement should have told you that the answer must be all of them as you can definitely calculate some sort of a value (using logs, not required for GMAT!). You are not supposed to get the exact value. All you need is to make sure that you get a real value, something that you will surely get for the question asked. As for the range, you can clearly see that 5^2<100<5^3, thus 2<y<3

As for the academic part of this discussion, I will use logs to get the value:

5^y=100 ---> take logs to the base 10 on both the sides:

y log 5 = 2 log 10 , log 10 =1 and log 5 = log 10 - log 2 = 1-0.301 = 0.699.

Thus y = 2.86245 , giving you $$5^y \approx 100$$

Again, this question is not GMAT like but is a good practice to understand how far you can go in 'solving' a particular question.

Hope this helps.
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Re: If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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20 Sep 2017, 14:52
1
Zarrolou wrote:
If $$x,y$$ are numbers $$>0$$ and $$2^x*5^y=400$$, what could be the value of $$x$$?

I)2
II)3
III)4

A)Only I
B)Only II
C)I and II
D)Only III
E)I, II and III

Let’s analyze each Roman numeral:

I. 2

If x = 2, then:

4 x 5^y = 400

5^y = 100

Since 5^2 = 25 and 5^3 = 125, y must be some number between 2 and 3, which is a number greater than 0. So x could be 2.

II. 3

If x = 3, then:

8 x 5^y = 400

5^y = 50

Since 5^2 = 25 and 5^3 = 125, y must be some number between 2 and 3, which is a number greater than 0. So x could be 3.

III. 4

16 x 5^y = 400

5^y = 25

Since 5^2 = 25, y = 2. So x could be 4.

(Note: When analyzing Roman numerals I and II, do not think that y needs to be an integer and reject I and II. The problem stem does not say that y has to be an integer; it just says that y has to be a positive number.)

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If x,y are numbers >0 and 2^x*5^y=400  [#permalink]

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23 Sep 2017, 14:42
Engr2012 wrote:
As for the academic part of this discussion, I will use logs to get the value:

5^y=100 ---> take logs to the base 10 on both the sides:

y log 5 = 2 log 10 , log 10 =1 and log 5 = log 10 - log 2 = 1-0.301 = 0.699.

Thus y = 2.86245 , giving you $$5^y \approx 100$$

Again, this question is not GMAT like but is a good practice to understand how far you can go in 'solving' a particular question.

Question is definitely out of scope for GMAT and actually it's D not E, I solved it for E as well,as looking at 95% hard it was obvious that the solution can't be so simple
And dude is right even for academic purposes you will never ever get pure "100" out of exponential function of 5
So E is by default wrong answer per se, try and play here https://www.desmos.com/calculator/auubsajefh
log and exp functions are tricky, because they are injective non-surjective functions, there are always some values in the range which you can't obtain by using a certain domain, in our case 5

And finally, your example of using log even is not as much precise as it can be because you used common log:
5^2.86245 gives you 100.17(669256356374644322057572814) ~0.17

better to use log base 5 to get rid of log on the left side and work with the right side only
5^y=100
logbase5 of (5^y) = logbase5 of 100
y= logbase5 of 100 (you can play with the power but no real sense)
y = 2.86135
5^2.86135 = 99.99(9498476779805789638948585901) ~0.01

That kind of questions should not be posted in Gmat forum, wrong from math standpoint or at least the question should be mentioning approximation
If x,y are numbers >0 and 2^x*5^y=400 &nbs [#permalink] 23 Sep 2017, 14:42
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