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positive integer

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positive integer [#permalink] New post 21 Jul 2009, 11:36
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If x and y are positive integers, and x<y, which of the following could be the value of (x/x+y)*10+(y/x+y)*20?
(a) 10
(b) 14
(c) 16
(d) 21
(e) 30
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Re: positive integer [#permalink] New post 21 Jul 2009, 12:48
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(x/x+y)*10+(y/x+y)*20 = K

solving above we get 10(x+2y)/(x+y)

we need to find K

from the answer choices we see that K is a positive integer

This can happen only if x+y which is the denominator is a factor of 10

Factors of 10 are 1,2,5

We cannot get 1 and 2 from x+y as x and y are +ve integers and x<y

so lets take 5 where x=2 and y = 3

we substitute these values we get

10(x+2y)/(x+y) = 10(2+6)/2+3 = 2(8) = 16 so answer is C

The other possible value of x and y is x=1 and y =4, we see that we wont get any of the given answers, this is not needed as we got the solution already but just gave as an extra step.

Please confirm the answer
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Re: positive integer [#permalink] New post 21 Jul 2009, 14:06
Good explanation gmanjesh, I did the same method.
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Re: positive integer [#permalink] New post 21 Jul 2009, 19:55
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gmanjesh wrote:
(x/x+y)*10+(y/x+y)*20 = K

solving above we get 10(x+2y)/(x+y)


Can you please tell us how did you solve the equation.
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Re: positive integer [#permalink] New post 21 Jul 2009, 20:08
(x/x+y)*10+(y/x+y)*20 = K

take x+y as the common denomintor

10x+20y/(x+y)

take common factor 10 out

10(x+2y)/(x+y)

hope that helps
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Re: positive integer [#permalink] New post 21 Jul 2009, 22:37
Duh! I read * as ^. :shock: I should call it a day. Thanks gmanjesh for your patient elaboration.
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Re: positive integer [#permalink] New post 22 Jul 2009, 01:18
OA given is A i.e. 10.. i am not sure how..
Also, factors of 10 are 1, 2, 5, and 10.
By the method explained above in a post answer comes to be 16 but OA is 10..
please lemme know, is there any other method to solve this question..
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Re: positive integer [#permalink] New post 22 Jul 2009, 05:13
sorry I missed 10, but yeah still the answer would be 16 which matches one of the answers.

I guess OA was wrong or a typo.
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Re: positive integer [#permalink] New post 22 Jul 2009, 08:15
well if you consider the final equation

10(x+2y)/(x+y)
or 10((X+y)+Y/X+Y)
= 10(1+(Y/X+Y))
i.e. 10+(10Y/X+Y)

In all probability it should be more than 10.
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Re: positive integer [#permalink] New post 22 Jul 2009, 09:50
i simplified to 10(y+2)/x+y= ?

then plugged in the first answer choice

10(y+2)/x+y=10 y's cancel and x=2, so y's can be any value. I picked A as the answer.
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Re: positive integer [#permalink] New post 22 Jul 2009, 15:43
Can you pls provide the steps
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Re: positive integer [#permalink] New post 23 Jul 2009, 04:58
bigtreezl wrote:
i simplified to 10(y+2)/x+y= ?

then plugged in the first answer choice

10(y+2)/x+y=10 y's cancel and x=2, so y's can be any value. I picked A as the answer.

please elaborate some more on ur solution..
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Re: positive integer [#permalink] New post 25 Jul 2009, 11:49
guys, i confirmed this question with the experts and the answer to this question is definitely C.. oa is incorrect..
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Re: positive integer [#permalink] New post 29 Jul 2009, 21:36
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Another way of solving this problem

let's say c = [10x/(x+y)] +[20y/(x+y)] = (10x+20y)/(x+y) = 10 + {10y/(x+y)} = 10 + [10/((x/y)+1)]
so ans should be > 10
now, x<y
hence, x/Y <1
hence, (x/Y) + 1 <2
so, 10/ [(x/y)+ 1] >5
so, c> 15
now, since both x and y are positive and (1+(x/y)) > 1
therefore, 10 /(1+(x/y) must be <10

so, c < 10 + 10
c<20
so, 15<c<20
and only 16 satisfy this condition. hence ans is C.
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Re: positive integer [#permalink] New post 30 Jul 2009, 12:57
@sudiptak
(10x+20y)/(x+y)
= 10 + {10y/(x+y)}

How did you get this (10 + {10y/(x+y)})? Please explain

regards,
hhk
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Re: positive integer [#permalink] New post 31 Jul 2009, 08:33
C. 16

As follows:

(x/x+y)*10+(y/x+y)*20

=10*(x/(y+x))+10*(y/(y+x))+10*(y/(y+x))

=10*(x/(y+x)+y/(y+x)+y/(y+x))

=10*((x+y+y)/(x+y))

=10+10*(y/(y+x))

only possibility is 16.
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Re: positive integer [#permalink] New post 31 Jul 2009, 16:06
10+10*(y/(y+x))
From qn,
x/y<1


A.Not possible
B.
Substitute 10(y/y+x) =4
X/y>1 Hence, it is false.
C.x/y=2/3 Hence, it is correct
D& E. As per the qn, x and y are positive integers.

regards,
hhk
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Re: positive integer [#permalink] New post 31 Jul 2009, 17:02
nice work sudiptak.
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Re: positive integer [#permalink] New post 01 Aug 2009, 07:20
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@sudiptak
(10x+20y)/(x+y)
= 10 + {10y/(x+y)}

How did you get this (10 + {10y/(x+y)})? Please explain

regards,
hhk

hi HHk,
sorry for the late reply.
here is your explanation-
(10x+20y)/(x+y)
=(10x+10y +10y)/(x+y)
=[(10x+10y)/(x+y)] + [10y/(x+y)]
=[10(x+y)/(x+y)] + [10y/(x+y)]
=10 + {10y/(x+y)}

Hope this helps.
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Re: positive integer [#permalink] New post 01 Aug 2009, 16:24
guys u can also try dis way..its quite simple and short:

firstly let the ans be 'k'

therefore,

10x + 20y / x + y = k

10x + 20y = kx + ky

kx - 10x = 20y - ky

x(k - 10) = y(20 - k)

now since x<y...

therefore (k - 10) will have to be greater than (20 - k)

just plug in the options and the only one ur left wit is 16.

hence the ans must be C.

hope dis helps....
Re: positive integer   [#permalink] 01 Aug 2009, 16:24
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