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positive integer

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Manager
Joined: 24 Apr 2009
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positive integer [#permalink]  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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Joined: 18 Jun 2009
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Kudos [?]: 64 [1] , given: 15

Re: positive integer [#permalink]  21 Jul 2009, 12:48
1
KUDOS
(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.

Manager
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Kudos [?]: 57 [0], given: 13

Re: positive integer [#permalink]  21 Jul 2009, 14:06
Good explanation gmanjesh, I did the same method.
Manager
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Kudos [?]: 20 [1] , given: 11

Re: positive integer [#permalink]  21 Jul 2009, 19:55
1
KUDOS
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.
Current Student
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Kudos [?]: 64 [0], given: 15

Re: positive integer [#permalink]  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
Manager
Joined: 27 Jun 2008
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Kudos [?]: 20 [0], given: 11

Re: positive integer [#permalink]  21 Jul 2009, 22:37
Duh! I read * as ^. I should call it a day. Thanks gmanjesh for your patient elaboration.
Manager
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Kudos [?]: 8 [0], given: 2

Re: positive integer [#permalink]  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..
Current Student
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Kudos [?]: 64 [0], given: 15

Re: positive integer [#permalink]  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.
Manager
Joined: 27 Jun 2008
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Kudos [?]: 20 [0], given: 11

Re: positive integer [#permalink]  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.
Director
Joined: 23 May 2008
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Kudos [?]: 37 [0], given: 0

Re: positive integer [#permalink]  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.
Manager
Joined: 27 Jun 2008
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Kudos [?]: 20 [0], given: 11

Re: positive integer [#permalink]  22 Jul 2009, 15:43
Can you pls provide the steps
Manager
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Kudos [?]: 8 [0], given: 2

Re: positive integer [#permalink]  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..
Manager
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Kudos [?]: 8 [0], given: 2

Re: positive integer [#permalink]  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..
Manager
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Kudos [?]: 10 [1] , given: 1

Re: positive integer [#permalink]  29 Jul 2009, 21:36
1
KUDOS
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.
Intern
Joined: 13 Jul 2009
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Re: positive integer [#permalink]  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
Senior Manager
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Kudos [?]: 54 [0], given: 23

Re: positive integer [#permalink]  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.
Intern
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Re: positive integer [#permalink]  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
Senior Manager
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Re: positive integer [#permalink]  31 Jul 2009, 17:02
nice work sudiptak.
Manager
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Kudos [?]: 10 [1] , given: 1

Re: positive integer [#permalink]  01 Aug 2009, 07:20
1
KUDOS
@sudiptak
(10x+20y)/(x+y)
= 10 + {10y/(x+y)}

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

regards,
hhk

hi HHk,
(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.
Manager
Joined: 20 May 2008
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Re: positive integer [#permalink]  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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