OG12-PS-148 qn

[kishokbabu]

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

OG explanation for this qn is very big, is there any easy approach or is there any number plugging options for this qn?

[GMATGuruNY]

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

OG explanation for this qn is very big, is there any easy approach or is there any number plugging options for this qn?

Check my explanation here:

https://www.beatthegmat.com/tough-algebr ... tml#292229.

[pemdas]

restated q.

10x/(x+y)+20y/(x+y) = K and x < y
find k?

(10x+20y)/(x+y) = K, (10(x+y)+10y)/(x+y), 10 + 10y/(x+y)
since x<y, we deduce y/(x+y) could be min. 1/2 and max 1
x<y, say x=0,.000001 or some other small value, x is +ve and less than y, y(x+y) gives close to 1 then OR say x=9.99999999 or some other value, x <y and y/(x+y) gives close to 1/2 then

Hence 10y/(x+y) is min. 5 and max. 10

answer choice d suits here only

(A) 10
(B) 12
(C) 15
(D) 18
(E) 30

OG explanation for this qn is very big, is there any easy approach or is there any number plugging options for this qn?[/quote]

[LalaB]

below is my very strange approach -

((x/(X+Y))(10) +(y/(X+Y))(20)) = K

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

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

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

y/(x+y)=k/10-1

y/(x+y)=(k-10)/10 or

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

it means

x+y=10
y=k-10

now we see that our k cant be 10(then y=0) and 30 (then y=20 and x<0).so A and E are out

answ choice C is also out, since then x=y

answ choice B is also out, since then x>y

answ choice D k=18 .then y=18-10=8 x=10-8=2 bingo!