Mixture and Alligation Quiz – 1

Let the quantity of the wine in the cask originally be x litres

 

Then, quantity of wine left in cask after 4 operations = 

x(1-\frac{8}{x})^{4}

=> (1-\frac{8}{x})^{4} = \frac{2}{3}^{4}

X = 24  

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of A in mixture left = (7x- 7/12 * 9) = (7x - 21/4)

 Quantity of B in mixture left = (5x - 5/12 *9) = (5x - 15/4)

 (7x - 21/4) /  (5x - 15/4) = 7 / 9

28x-21 / 20x+21 = 7/9

=> x = 3

So, the can contained 21 litres of A

Let the C.P of spirit be = Rs.10 per litre

Then S.P l litre of the mixture = Rs. 11

Profit = 20%

Therefore, C.P of the mixture = Rs. [100 / 120 ×10]

100 / 120×10 = Rs. 25/3 per litre

Therefore, Ratio of water and spirit = 1 : 5

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

 Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

 Thus, (30 + P) = 25% of (150 + P)

 Solving, we get P = 10 liters

Video solution:

pool : kerosene 

   3  :  2(initially) 

   2  :  3(after replacement)

 

Remaining Quantity  / Initial Quantity = (1 - Replaced Quantity / Total Quantity)

For petrol  (2/3) = (1- 10/k)

=> K = 30

Therefore the total Quantity of the mixture in the container is 30 litres

Video solution:

Given that container has 50 litres of milk.

 After replacing 8 litres of milk with water for three times, milk contained in the container is:

=>(50×42 / 50×42 / 50×42 / 50)

⇒50×42 / 50×42 / 50×42 / 50  = 29.63 litres.

 Wine                         Water

            8L                              32L

            1                :               4

           20 %                           80% (original ratio)

           30 %                           70% (required ratio)

 

In  this case, the percentage of water being reduced when the mixture is being replaced with wine.

 so the ratio of left quantity to the initial quantity is 7:8

Therefore , 7/8 = [1 - k/8]

=> k = 5 LIT

Video solution:

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

 Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

 Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

 i.e (35,70), (42,63), (45,60)

 So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

But in the question asked the ratio of water to milk = 193 : 122

Let the capacity of the pot be 'P' litres.

Quantity of milk in the mixture before adding milk = 4/9 (P - 8)

After adding milk, quantity of milk in the mixture = 6/11 P.

6P/11 - 8 = 4/9(P - 8)

10P = 792 - 352 => P = 44.

The capacity of the pot is 44 liters.

Let the percentage of benzene = X

(30 - X)/(X- 25) = 6/4 = 3/2

=> 5X = 135  

X = 27

So, required percentage of benzene = 27 %

Video solution: