value of K

The question is incomplete.

Please correct it.

If x, y, and k are positive numbers such that (x/x+y)10+(y/x+y)(20) = k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30

Take x=1 and y=4 to get the answer

howcm u arrived at x=1 and y=4.pls explain

Soln
= (x/x+y)10+(y/x+y)(20)
= 10(x/x+y)+10(y/x+y)+10(y/x+y)
= 10(x+y)/(x+y) + 10(y/x+y)
= 10 + 10 (y/x+y)


since x and y are positive so
the answer will certainly be greater than 10
A is out


since y > x

so y/(x+y) will always be greater than 1/2
had y been equal to x, the value could have been equal to 1/2

so 12 and 15 are also out


Now we are left with 18 and 30

y/(x+y) will always be less than 1
it can never be greater
so 30 is out

we are left with 18, the answer


You will not require this much effort evrytime, once you are through with the concept, by looking at statement you will be able to judge
that result is between 15 and 20 [non inclusive]


Also,
No offenses but I personally dont like number substitution.

If you want to do with substitution, you can do so

choose any set of numbers and solve
than take another set and solve

you will find the result to be between 15 - 20
and
if u r fortunate you may get 1 and 4 as your first choice and 18 as answer.


Hope this helps!!!

vivek.kapoor83 wrote:howcm u arrived at x=1 and y=4.pls explain

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

how did you go from the first line to the second. I dont understand.

Good explanation Stop at 800.


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


10(x/x+y)+10(y/x+y) is same as 10(x+y)/x+y

I dont see how.

10 (1/y) + 10(1/x) is the only thing i see with combining those too.

OA is 18

Ian Stewart wrote:I gave a short, albeit very conceptual, solution here:

www.beatthegmat.com/number-system-sets-t14684.html

Ian, thanks for the great explanation in your other post.

On a general note, could you please let us know how to recognize a problem to be weighted averages and apply these techniques to different situations. Especially, when not to use weighted averages to solving problems.

Thanks for all your help.