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Let Z be a function satisfying Z(ab) = Z(a)/bfor all positive real num

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Joined: 02 Sep 2009
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Let Z be a function satisfying Z(ab) = Z(a)/bfor all positive real num  [#permalink]

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13 Feb 2020, 06:59
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Difficulty:

35% (medium)

Question Stats:

65% (01:45) correct 35% (02:31) wrong based on 46 sessions

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Let Z be a function satisfying $$Z(ab) = \frac{Z(a)}{b}$$ for all positive real numbers a and b. if Z(500) = 3, then what is the value of Z(600) ?

A. 1
B. 2
C. 5/2
D. 3
E. 18/5

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Joined: 25 Jul 2018
Posts: 712
Let Z be a function satisfying Z(ab) = Z(a)/bfor all positive real num  [#permalink]

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13 Feb 2020, 07:53
1
$$Z(ab)=\frac{Z(a)}{b}$$ (a,b --positive real numbers)

--> $$Z(5*100)= \frac{Z(100)}{5}=3$$ --> $$Z(100)= 15$$

$$Z(6*100)= \frac{Z(100)}{6}= \frac{15}{6} = \frac{5}{2}$$

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Joined: 22 Jan 2019
Posts: 8
Let Z be a function satisfying Z(ab) = Z(a)/bfor all positive real num  [#permalink]

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13 Feb 2020, 16:58
Z ( 50*10)=

Z50/10=3

Z=3/5

So, Z(60*10)=

Z60/10=> 3/5*60/10= 18/5

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Re: Let Z be a function satisfying Z(ab) = Z(a)/bfor all positive real num  [#permalink]

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21 Feb 2020, 08:06
1
Bunuel wrote:
Let Z be a function satisfying $$Z(ab) = \frac{Z(a)}{b}$$ for all positive real numbers a and b. if Z(500) = 3, then what is the value of Z(600) ?

A. 1
B. 2
C. 5/2
D. 3
E. 18/5

Are You Up For the Challenge: 700 Level Questions

Solution:

Since Z(500) = Z(a)/b = 3 for some positive real numbers a and b, we see that ab = 500 and Z(a) = 3b. However, since b = 500/a, we have:

Z(a) = 3 * 500/a

Z(a) = 1500/a

Therefore, Z(600) = 1500/600 = 15/6 = 5/2.

Alternate Solution:

Notice that 600 = 500 * (6/5). Thus, if we take a = 500 and b = 6/5, we obtain:

Z(600) = Z(500 * 6/5) = Z(500)/(6/5) = 3/(6/5) = (3*5)/6 = 5/2.

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Re: Let Z be a function satisfying Z(ab) = Z(a)/bfor all positive real num   [#permalink] 21 Feb 2020, 08:06