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Examveda

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:
Alligation mcq solution image
∴ Required ratio = $$\frac{2}{{91}}$$ : $$\frac{1}{{13}}$$ = 2 : 7

This Question Belongs to Arithmetic Ability >> Alligation

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Comments ( 3 )

  1. Shahriar Nayan
    Shahriar Nayan :
    3 years ago

    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
    138​x+75​y=139​
    135​x+72​y=134​
    Solving for x and y we get x=92​ and y=97​
    x:y=2:7

  2. Taibur Rahman
    Taibur Rahman :
    3 years ago

    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)

  3. Jyothi Pandu
    Jyothi Pandu :
    4 years ago

    Anyone can explain another method briefly

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