A farmers contains two beaker X and Y having mixture of milk and water in the ratio 3 : 2 and 3: 5 respectively. 40% of mixture is taken out from beaker X and poured into Y by this concentration milk in beaker Y is increased by 80%. After this he poured of mixture from beaker Y to beaker X. If total quantity of milk is reduced by 31 in beaker X then find the total mixture farmer have?
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Solution :-
Let
- In X , milk = 3x and water = 2x .
- In y , milk = 3y and water = 5y .
now, in 40% of mixture of X we have,
- Milk = 3x * (2/5) = (6x/5)
- Water = 2x * (2/5) = (4x/5)
when this mixture is poured in Y ,
→ Total milk in Y = 3y + (6x/5)
A/q,
→ 3y + (6x/5) = (180/100) * 3y
→ (15y + 6x)/5 = 9/5(3y)
→ 15y + 6x = 27y
→ 6x = 27y - 15y
→ 6x = 12y
→ x/y = 2/1 .
then,
→ Mixture X = 5x = 5 * 2 = 10
→ Mixture Y = 8y = 8 * 1 = 8
therefore,
→ Total mixture farmar has = 10 + 8 = 18 (Ans.)
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