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Re: If x,y are numbers >0 and 2^x*5^y=400 [#permalink]
04 May 2013, 17:32

1

This post received KUDOS

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 _________________

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

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

IMO Answer: D _________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

Re: If x,y are numbers >0 and 2^x*5^y=400 [#permalink]
21 Jul 2015, 07:47

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