A mixture contain x litre milk and y litre water. if 90 litre of mixture is taken out and replaced with water such that ratio of milk and water becomes 5:7 but if 180 litre of mixture is taken out and replaced with water then ratio of milk and water becomes 5:13 . find x?
Answers
Answer:
x = 200
Step-by-step explanation:
x - milk
y - water
total liquid = x + y
1) 90l mixture taken out, and added 90l water
90l mixture = 90x/(x + y) milk + 90y/(x + y) water
new mixture
milk = x - 90x/(x + y)
water = y - 90y/(x +y) + 90
(x^2 + xy - 90x)/(x + y) : (y^2 + xy - 90y + 90x + 90y)/(x + y) = 5:7
=> 7(x^2 + xy - 90x) = 5(y^2 + xy + 90x) -----(1)
2) 180l mixture taken out, and added 180l water
180l mixture = 180x/(x + y) milk + 180y/(x + y) water
new mixture
milk = x - 180x/(x + y)
water = y - 180y/(x +y) + 180
(x^2 + xy - 180x)/(x + y) : (y^2 + xy - 180y + 180x + 180y)/(x + y) = 5:13
=> 13(x^2 + xy - 180x) = 5(y^2 + xy + 180x)
=> 13(x^2 + xy - 180x) = 5(y^2 + xy + 90x + 90x)
=> 13(x^2 + xy - 180x) = 5(y^2 + xy + 90x) + 450x
=> 13(x^2 + xy - 180x) = 7(x^2 + xy - 90x) + 450x ---- from (1)
=> 13x^2 + 13xy -2340x = 7x^2 + 7xy - 630x + 450x
=> 6x^2 + 6xy - 2160x = 0
=> x^2 + xy - 360x = 0
=> x(x + y -360) = 0
=> x = 0 or x + y = 360
=> y = 360 - x
putting x = 0 in (1)
7(x^2 + xy - 90x) = 5(y^2 + xy + 90x) -----(1)
=> 7(0^2 + y*0 - 90*0) = 5(y^2 + y*0 + 90*0)
=> y = 0
putting y = 360 - x in (1)
7(x^2 + xy - 90x) = 5(y^2 + xy + 90x) -----(1)
=> 7(x^2 + x*(360 - x) - 90x) = 5((360 - x)^2 + x*(360 - x) + 90x)
=> 7(x^2 + 360x - x^2 -90x) = 5( 360^2 + x^2 - 720x + 360x - x^2 + 90x)
=> 7(x^2 + 360x - x^2 -90x) = 5( 360^2 + x^2 - 720x + 360x - x^2 + 90x)
=> 7 * 270x = 5 (360^2 -270x)
=> 12 * 270x = 5 * 360 * 360
=> x = 360 * 360 * 5 / (270 * 12)
=> x = 200
Answer:
Step-by-step explanation:
200