Mixture and Alligation Questions with Solution (Basic)

Solution:

The Required ratio = 4:1

Solution:

Ratio of water to milk = 1:5

Solution:

Ratio of Boys to Girls = 2 : 3

Girls = 3/5 x 1000 = 600

Solution:

Let all are boys then total sum = 100 x 100 = 10000 P. Let all are girls then total sum = 100 x 50 = 5000 P.

Ratio of Boys to Girls = 7 : 3. Number of girls = 3/10 x 100 = 30

Solution

(40-x) / (x-35) = (8/7)

15x = 560

x = 37.33

Profit % = 37.50 - 37.33 / 37.33 x 100 ≅ .455%

Solution:

Ratio of the quantity of tea of the first kind to that of the second kind in the  mixture should be 1 : 4

Solution:

Ratio = 1 : 1.

Solution:

Ratio = 3:1

Solution:

Required ratio will be

75 - 60: 90 – 75 i.e.

15 : 15

⇒ 1 : 1.

Solution:

Ratio = 13:7

Solution:

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 - 8 / x)⁴ Liters  x ( 1 - 8 / x)⁴ / x = (2 / 3)⁴

⇒�� = 24

Solution:

By the rule of alligation:

ratio of 1st and 2nd parts = 4 : 6 = 2 : 3.

Therefore, Quantity of 2nd kind

= (3/5 x 1000)kg

= 600 kg.

Solution:

Since first second varieties are mixed in equal proportions, so their average  price = Rs. (126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50  per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

x-153/22.50 = 1

=> x - 153 = 22.50

=> x=175.50

Hence, price of the third variety = Rs.175.50 per kg

Solution:

Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind. Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.

Solution:

Let the price of the mixed variety be Rs. x per kg. By the rule of alligation,  we have :

∴(20-x)/(x-15) = 2/3

60 - 3x = 2x – 30

x = 18.

Solution:

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

Quantity of A in mixture left = 7x – 63/12 = 7x – 21/4

Quantity of B in mixture left = 5x – 15/4

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

X = 3

So, the can contained 21 litres of A.

Solution:

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 = 1000/120 = 25/3

Therefore, Ratio of water and spirit = 1 : 5

Solution:

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)

we get P = 10 liters

Solution:

The 20 litre mixture contains milk and water in the ratio of 3 : 2.  Therefore, there will be 12 litres of milk in the mixture and 8 litres of water  in the mixture.

Step 1. When 10 litres of the mixture is removed, 6 litres of milk is removed  and 4 litres of water is removed.

Therefore, there will be 6 litres of milk and 4 litres of water left in the container. It is then replaced with pure milk of 10 litres. Now the container  will have 16 litres of milk and 4 litres of water.

Step 2. When 10 litres of the new mixture is removed, 8 litres of milk and 2  litres of water is removed. The container will have 8 litres of milk and 2  litres of water in it.

Now 10 litres of pure milk is added.

Therefore, the container will have 18 litres of milk and 2 litres of water in  it at the end of the second step.

Therefore, the ratio of milk and water is 18 : 2 or 9 : 1.

Solution:

Cost of 1 litre of water = Rs.0 = cheaper quantity.

Cost of 1 litre of juice = Rs.24 = dearer quantity.

And, the mean price = m = Rs.20 Therefore, (Cheaper quantity) : (Dearer  quantity) = (d - m) : (m - c) = 4:20 = 1:5

Hence, the required answer is 1:5.