Two vessels A and B contain milk and water mixed in the ratio 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing $$69\frac{3}{{13}}\% $$ milk, is-
A. 2 : 7
B. 3 : 5
C. 5 : 2
D. 5 : 7
Answer: Option A
Solution(By Examveda Team)
Let cost of 1 litre milk be Rs. 1 Milk in 1 litre mixture in A = $$\frac{8}{13}$$ litre;Cost price of 1 litre mixture in A = Rs. $$\frac{8}{13}$$
Milk in 1 litre mixture in B = $$\frac{5}{7}$$ litre;
Cost price of 1 litre mixture in B = Rs. $$\frac{5}{7}$$
Milk in 1 litre of final mixture
$$\eqalign{ & = {\frac{{900}}{{13}} \times \frac{1}{{100}} \times 1} \cr & = \frac{9}{{13}}{\text{ litre}}{\text{}} \cr} $$
Mean price = Rs. $$\frac{9}{{13}}$$
By the rule of alligation, we have:
∴ Required ratio = $$\frac{2}{{91}}$$ : $$\frac{1}{{13}}$$ = 2 : 7
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Comments ( 3 )
Related Questions on Alligation
A. $$\frac{{1}}{{2}}$$ kg
B. $$\frac{{1}}{{8}}$$ kg
C. $$\frac{{3}}{{14}}$$ kg
D. $$\frac{{7}}{{9}}$$ kg
A. 81 litres
B. 71 litres
C. 56 litres
D. 50 litres
A→8:5
138 milk and 135 water
B→5:2
75 milk and 72 water
69133%=1313×69+3=13900×1001=139
139 milk and 134 water
Let x:y be the ratio in which they are mixed
138x+75y=139
135x+72y=134
Solving for x and y we get x=92 and y=97
x:y=2:7
Let, From A =x part is taken
From B=y part is taken
So, milk taken from A= 8x/13
milk taken from B= 5y/7
ATQ,
(8x/13)+(5y/7)= (69*3/13)% of (x+y)
Or, (56x+65y)/91=(900/13×100) of (x+y)
Or, (56x+65y/7=9x+9y
Or, 56x+65y=63x+63y
Or, 7x=2y
So, x:y=2:7 (Ans)
Anyone can explain another method briefly