Mixture and Alligation

 


  Mixture and Alligation

M. Velrajan


1. In a mixture of 28 litres, the ratio of milk and water is 5 : 2. If 2 litres of water is added to the mixture, find the ratio of milk and water in the new mixture.  

(a) 2 : 1    (b) 1 : 2    (c) 2 : 3    (d) 1 : 3

Since 28/(5 + 2) = 4, the quantity of milk and water are 5 × 4 = 20, 2 × 4 = 8 

After adding 2 lit water quantity of milk and water become 20, 10 and 

the ratio of milk and water is 2 : 1

Aliter : the new mixture = 28 + 2 = 30. 

The sum of ratio is a divisor of 30 and milk is more than water, 

so by the answer choices only  2 : 1 satisfies these conditions


2. One litre of water is added to 5 litres of a 20% solution of alcohol in water. The strength of alcohol in the new solution is.           

(a) 16 23 %            (b) 15%          (c) 20%            (d) 16%

Alcohol in 5 litres solution = 5 × (20/100) = 1.

After adding 1 litre of water, alcohol in new 6 litre solution = 1 

Strength of alcohol in new solution = ⅙  × 100 = 16 ⅔ %


3. A milkman has a mixture of milk in which the ratio of milk and water is 7 : 4. He sells 55 litres of this mixture and then adds up 20 litres of pure water. Now the ratio of milk and water is 7 : 6. Find the new quantity of this mixture.

1. 70 litres      2. 195 litres      3. 130 litres     4. 210 litres

After selling 55 litres, the ratio of milk and water is same as 

the original ratio 7 : 4

After selling 55 litres, the quantity of milk in the mixture = 7x

and water = 4x, for some x

After adding 20 litres of water, water = 4x + 20. 

So, the new ratio is

7x/(4x + 20) = 7/6

42x = 28x + 140

14x = 140

x = 10

New qu antity = 7×10 + 4×10 + 20 = 130

Initial quantity = 70 + 40 + 55 = 165


A.  Suppose there are two components in a mixture and the ratio of the components

is a : b. And by adding quantity m and n of the first and the second components respectively to the mixture the ratio of the components  becomes c : d. Then the

ratio constant of the ratio a : b is given by x = (nc - md) / (ad - bc)

For,  If x is the ratio of constant of a : b then the original quantities of the two

components are ax and bx. After adding quantity m and n of the first and second

components the quantities of the components become

ax + m and bx + n.

Hence the new ratio is (ax + m) / (bx + n).

So, (ax + m) / (bx + n) = c/d

adx + md = bcx + nc

x = (nc - md) / (ad - bc) 


Aliter for 3 :

After selling 55 litres, the ratio of milk and water is same as 

the original ratio 7 : 4

By A., the ratio constant of the ratio a : b is given by x = (nc - md) / (ad - bc)

Here a = 7, b = 4, c = 7, d = 6, m = 0, n = 20

 So, x = (20×7 - 0×6) / (7×6 - 4×7) = 20 / (6 - 4) = 10

New quantity = 7×10 + 4×10 + 20 = 130


5. A milkman buys 20 litres of milk from a dairy at the cost of Rs 50 per litre.

He also purchases mineral water at the rate of Rs 20 per litre and adds it to the

milk. He then sells the adulterated milk at the rate of Rs 60 per litre and makes a

profit of Rs. 400. What is the amount of water he added to the milk? 

1. 10 litre      2. 8 litre          3. 5 litre        4. 4 litre

Suppose x litre of water is added. 

Then quantity of adulterated milk = 20 + x

Selling price = (20 + x)60 = 1200 + 60x

Cost = 20 × 50 + 20x = 1000 + 20x

Selling price - Cost = Profit 

1200 + 60x - (1000 + 20x) = 400

Without writing above steps we have to write the above equation 

40x = 400 - 200 = 200  

x = 5 litre      Ans. 3.


6. How much water (free of cost) should be added to 18 litres of milk worth Rs 70 per

litre so that the value of the mixture is Rs 60 per litre? 

1. 2 litres      2. 3 litres      3. 4 litres     4. 5 litres 

Suppose x litres of water is added.

Then the adultrated mixture = 18 + x litres

So, (18 + x)60 = 18 × 70

60x = 18(70 - 60) = 18 × 10 = 180

x = 180/60 = 3     Ans. 2.

Aliter :

The loss 18(70 - 60) by the less selling price for the milk has to be balanced by

water added × 60.

i.e. 18(70 - 60) = water added × 60

Water added = 18(70 - 60)/60 

                     = 180/60 = 3         Ans. 2.

 

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