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# The value of a machine decreases by 'r' percent every year at the end

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Intern
Joined: 13 Oct 2019
Posts: 21
Location: India
GPA: 4
The value of a machine decreases by 'r' percent every year at the end  [#permalink]

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Updated on: 18 Jan 2020, 22:39
1
00:00

Difficulty:

55% (hard)

Question Stats:

64% (02:38) correct 36% (02:01) wrong based on 25 sessions

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The value of a machine decreases by 'r' percent every year at the end of each year where 'r' remains the same each year. If the value of the machine on 1 January, 2000 was 'x' dollars and the value of the machine on 1 January, 2002, was 'y' dollars, what was the value of the machine on 1 January, 2003, in terms of 'x' and 'y'?

A. $$y+ (y-x)/2$$
B. $$y+ (y-x)y/2x$$
C. $$y\sqrt{(y/x)}$$
D. $$x^2/(2y)$$
E. $$y\sqrt{(x/y)}$$

Originally posted by ann1111 on 18 Jan 2020, 05:29.
Last edited by ann1111 on 18 Jan 2020, 22:39, edited 1 time in total.
Intern
Joined: 11 Aug 2019
Posts: 4
Re: The value of a machine decreases by 'r' percent every year at the end  [#permalink]

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18 Jan 2020, 11:03
1
2000 x
2001 x(1-r/100)
2002 y=x(1-r/100)2
2003 y(1-r/100)=x(1-r/100)3

By the equation from 2002,
(1-r/100)2 = y/x
(1-r/100) = √(y/x)
Therefore in 2003, the value will be
y*√(y/x)
OA says y/√(y/x).
I am not sure where i went wrong

Posted from my mobile device
Intern
Joined: 13 Oct 2019
Posts: 21
Location: India
GPA: 4
The value of a machine decreases by 'r' percent every year at the end  [#permalink]

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19 Jan 2020, 01:22
1
Pramod1246 wrote:
2000 x
2001 x(1-r/100)
2002 y=x(1-r/100)2
2003 y(1-r/100)=x(1-r/100)3

By the equation from 2002,
(1-r/100)2 = y/x
(1-r/100) = √(y/x)
Therefore in 2003, the value will be
y*√(y/x)
OA says y/√(y/x).
I am not sure where i went wrong

Posted from my mobile device

Your solution is right, there was a typo error.
Thanks for letting me know.
The value of a machine decreases by 'r' percent every year at the end   [#permalink] 19 Jan 2020, 01:22
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