Exponents

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Exponents

by avenus » Sun May 31, 2009 10:51 am
If x, y, and z are integers and (2^x)(5^y)z = 0.00064, what is the value of xy?
(1) z = 20
(2) x = –1
Source: — Data Sufficiency |

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Re: Exponents

by iamcste » Sun May 31, 2009 12:34 pm
avenus wrote:If x, y, and z are integers and (2^x)(5^y)z = 0.00064, what is the value of xy?
(1) z = 20
(2) x = –1
IMO A

(2^x)(5^y)z = 64/(10^5)

(2^x)(5^y)z=(2^6)/[(2^5)*(5^5)]..splitting 10^5 into 2^5*5^5

(2^x)(5^y)z=2/(5^5)=(2)*(5^-5)


1. Z=(2^2)*5

substitute z in above eqtn

(2^x)(5^y)=[(2)*(5^-5)]/[ (2^2)*5]=(2^-1)*(5^-6)

2^1 in Numberator and 2^2 in denominator hence effectively 2^(1-2)=2^-1
similarly 5^-5 in numr and 5^1 in denominator hence 5^(-1-1)=5^-6

x=-1, y=-6 calculate xy sufficient

2. we know x=-1

(2^x)(5^y)z==(2)*(5^-5)


z=(2^2)*[5^(-5+y)]

for z to be an integer, y can be any integer >=5

hence xy could change

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by aj5105 » Mon Jun 01, 2009 5:25 am
Statement (1)

2^x * 5^y * Z = 2/3125

z = 20

2^x * 5^y = 2/62500

We can find out values of x and y. Hence xy.

Sufficient


Statement (2)

2^x * 5^y * Z = 2/3125

x = -1

5^y * Z = 1/3125

y value can vary.

Insufficient

(A)


P.S : Any suggestions on the method I used? welcome!

OA, please

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by avenus » Tue Jun 02, 2009 12:57 am
OA is A