a milkman mixed two solutions A and B of milk and water in the ratio of 4:5 respectively, then the new solution obtained in which the concentration of milk was 75%. in a 5 litre solution A, the quantity of water was 2 litre, then in 15 litre solution B , what was the quantity (in ml) of milk?
Answers
Answer:r 4*Q/15 = 18/5 or Q = 54/4 = 13.5 (L) [Ans]
Step-by-step explanation:
- Statement of the given problem,
(i) A milkman mixed two solutions A and B of milk and water in the ratio of 4:5 respectively.
(ii) A new solution was obtained in which the concentration of milk was 75%.
(iii) In 5L solution A, the quantity of water was 2L.
(iv) In 15L solution B, let Q denotes the quantity (in L) of milk present.
- Basic assumptions,
(v) Let (MA, WA) & (MB, WB) denote (milk, water) in solutions A & B respectively.
(vi) Let MN & WN denote milk & water respectively in new solution.
- From (i), (v) & (vi) we get following relations,
(MA + WA)/(MB + WB) = 4/5 …… (1a)
MN = MA + MB …… (1b)
WN = WA + WB …… (1c)
- From (ii) & (vi) we get,
MN/WN = 75%/25% = 3/1 …… (1d)
- From (iii), (iv) & (v) we get following relations,
MA /WA = (5 - 2)/2 …… (1e)
MB /WB = Q/(15 - Q) …… (1f)
- From (1b), (1c) & (1d) we get,
(MA + MB)/(WA + WB) = 3/1
or MA + MB = 3*(WA + WB) …… (1g)
- From (1e) we get,
MA = 3/5 …. (2a) and WA = 2/5 …. (2b)
From (1f) we get,
MB = Q/15 …. (3a) and WB = (15 - Q)/15 …. (3b)
- Therefore from (1g) and (2a) & (2b) and (3a) & (3b) we get,
3/5 + Q/15 = 3*[2/5 + (15 - Q)/15] = 6/5 + 3 - Q/5
or Q/15 + Q/5 = 6/5 + 3 - 3/5
or 4*Q/15 = 18/5 or Q = 54/4 = 13.5 (L) [Ans]