Author 
Message 
Manager
Joined: 24 Apr 2009
Posts: 93

positive integer [#permalink]
Show Tags
21 Jul 2009, 12:36
1
This post was BOOKMARKED
Question Stats:
0% (00:00) correct
0% (00:00) wrong based on 0 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum.
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



Current Student
Joined: 18 Jun 2009
Posts: 356
Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD

Re: positive integer [#permalink]
Show Tags
21 Jul 2009, 13:48
1
This post received 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.
Please confirm the answer



Manager
Joined: 03 Jul 2009
Posts: 106
Location: Brazil

Re: positive integer [#permalink]
Show Tags
21 Jul 2009, 15:06
Good explanation gmanjesh, I did the same method.



Manager
Joined: 27 Jun 2008
Posts: 157

Re: positive integer [#permalink]
Show Tags
21 Jul 2009, 20:55
1
This post received 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
Joined: 18 Jun 2009
Posts: 356
Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD

Re: positive integer [#permalink]
Show Tags
21 Jul 2009, 21:08
(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
Posts: 157

Re: positive integer [#permalink]
Show Tags
21 Jul 2009, 23:37
Duh! I read * as ^. I should call it a day. Thanks gmanjesh for your patient elaboration.



Manager
Joined: 24 Apr 2009
Posts: 93

Re: positive integer [#permalink]
Show Tags
22 Jul 2009, 02:18
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
Joined: 18 Jun 2009
Posts: 356
Location: San Francisco
Schools: Duke,Oxford,IMD,INSEAD

Re: positive integer [#permalink]
Show Tags
22 Jul 2009, 06:13
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
Posts: 157

Re: positive integer [#permalink]
Show Tags
22 Jul 2009, 09:15
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
Posts: 806

Re: positive integer [#permalink]
Show Tags
22 Jul 2009, 10:50
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
Posts: 157

Re: positive integer [#permalink]
Show Tags
22 Jul 2009, 16:43
Can you pls provide the steps



Manager
Joined: 24 Apr 2009
Posts: 93

Re: positive integer [#permalink]
Show Tags
23 Jul 2009, 05:58
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
Joined: 24 Apr 2009
Posts: 93

Re: positive integer [#permalink]
Show Tags
25 Jul 2009, 12:49
guys, i confirmed this question with the experts and the answer to this question is definitely C.. oa is incorrect..



Manager
Joined: 22 Mar 2008
Posts: 53

Re: positive integer [#permalink]
Show Tags
29 Jul 2009, 22:36
1
This post received 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
Posts: 20

Re: positive integer [#permalink]
Show Tags
30 Jul 2009, 13:57
@sudiptak (10x+20y)/(x+y) = 10 + {10y/(x+y)}
How did you get this (10 + {10y/(x+y)})? Please explain
regards, hhk



Manager
Joined: 21 Jul 2009
Posts: 249
Location: New York, NY

Re: positive integer [#permalink]
Show Tags
31 Jul 2009, 09:33
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
Joined: 13 Jul 2009
Posts: 20

Re: positive integer [#permalink]
Show Tags
31 Jul 2009, 17:06
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
Joined: 26 Jul 2009
Posts: 347

Re: positive integer [#permalink]
Show Tags
31 Jul 2009, 18:02
nice work sudiptak.



Manager
Joined: 22 Mar 2008
Posts: 53

Re: positive integer [#permalink]
Show Tags
01 Aug 2009, 08:20
1
This post received 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, 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.



Manager
Joined: 20 May 2008
Posts: 56

Re: positive integer [#permalink]
Show Tags
01 Aug 2009, 17:24
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, 17:24






