A vessel contains 50 litres of a solution of water and acid in the ratio 4:1. How many litres of water must be added to the solution to reverse the ratio?
Answers
Answered by
0
40 litrs of water must be added to the solution to reverse the ratio.
Answered by
0
w: A= 4:1 in 5o litre of solution
water = 4 X 50 / 4+1= 200/5= 40 litre
Acid = 1 X 50/ 5 = 50/5 = 10 litre
let X litre's of water to be added to solution
acid becomes 10 litres
water becomes. (40 + X) litre's
40+x : 10 =1:4
(40 + X) X 4 = 10 X 1
160+ 4x = 10
4 (40+ X ) = 10
40 + X = 10/ 4
X = 10/ 4 - 40
X = - 150 / 4
X = - 37.5 litre's
I think we have to remove the water instead of adding so. 50 - 37.5 =12.5 litre's makes the ratio of 1:4. I. e reverse ratio of mixture .......may b I m not sure abt this u may check the answer ......if not then comment plz .....thnku
water = 4 X 50 / 4+1= 200/5= 40 litre
Acid = 1 X 50/ 5 = 50/5 = 10 litre
let X litre's of water to be added to solution
acid becomes 10 litres
water becomes. (40 + X) litre's
40+x : 10 =1:4
(40 + X) X 4 = 10 X 1
160+ 4x = 10
4 (40+ X ) = 10
40 + X = 10/ 4
X = 10/ 4 - 40
X = - 150 / 4
X = - 37.5 litre's
I think we have to remove the water instead of adding so. 50 - 37.5 =12.5 litre's makes the ratio of 1:4. I. e reverse ratio of mixture .......may b I m not sure abt this u may check the answer ......if not then comment plz .....thnku
Similar questions