Mixtures and Alligations Questions for CAT with Answers PDF

Total distance travelled = 285 km
Total time taken = 6 hours
Average speed of journey = 285/6 km/h
By alligation method,

Therefore, the ratio of time spent travelling at 40 km/h to that travelling at 55 km/h =
Total time of journey = 6 hours
Therefore,
Time of travel at 55 km/h = 3 hours
Therefore, distance travelled at this speed = 55 × 3 = 165 km
This is our required solution.

Required ratio = 61 : 29

Applying the rule of alligation,

Gold : Copper = 6 : 4 = 3 : 2

20 lbs of the first alloy contains 20 × 5/8 of A (in lbs) and 20 × 3/8 of B (in lbs).
65 lbs of the second alloy contains 65 × 6/13 of A (in lbs) and 65 × 7/13 of B (in lbs).
Suppose X lbs of pure A is required.
⇒ (X + 65 + 20) × 
⇒ X = 21.25 lbs

Gold in the first alloy = 20%
Gold in the second alloy = 40%
By alligation:

Ratio of the two alloys = 2 : 3
Quantity of the second alloy = 100 × 3/2 = 150 gm

Cost price of the mixture = (10 × 40 + 30 × 20) = 1000
Selling price of the mixture = Cost price + (Cost price × 15%)
Selling price of the mixture = 1000 + 
Per kg selling price = = = 28.75

Let, the volume of each container = V
Now according to the question:
In container 1 W : A = V - 3x : 3x
In container 2 W : A = V - x : x
Now

V - 2x = 3x
V = 5x
So, In container 1 W : A = 2x : 3x = 2 : 3
In container 2 W : A = 4x : x = 4 : 1

Selling price of the mixture = Rs. 125
Profit = 25%


Now, apply the rule of alligation:

Total amount of milk = 40 litres
Amount of milk replaced by water = 10 litres
Now, ratio of milk to the total mixture = 
Water content in the mixture = 64 - 27 = 37
Hence, ratio of milk to water = 27 : 37

Sugar : Water = 36 : 13
Sugar syrup in the container = of the total volume of the container.
After repeating the process twice, we have

Suppose we mix 1 L, 2 L and 3 L of A, B and C, respectively.
So, total milk in the solution would be = 1/4 + 4/5 + 9/5 = (5 + 16 + 36)/20 = 57/20
Total water in the solution would be= 3/4 + 6/5 + 6/5 = (15 + 24 + 24)/20 = 63/20
So, milk and water ratio in the new solution would be= 57 : 63 = 19 : 21

Let x kg of metal X be added.
Weight of metal X in the first alloy = 6/11 × 11 kg = 6 kg
Weight of metal X in the second alloy = 7/20 × 20 kg = 7 kg
Total weight of new alloy = (11 + 20 + x) kg
Total weight of metal X in the new alloy = (6 + 7 + x) kg.

130 + 10x = 186 + 6x
4x = 56
x = 14 kg

Let the volume of initial solution be x litres.
So, the amounts of liquids A and B are 3x/5 and 2x/5, respectively.
When 10 litre of solution is taken out, the remaining liquid A in the solution would be

Quantity of liquid B in the new solution would be

So, the ratio of liquids A and B in the new solution would be

2(3x - 30) = 2x + 30
6x - 2x = 30 + 60
4x = 90
x = 22.5 litres

10 litres of mixture from the 1st container is taken.
Thus, amount of milk in 1st mixture = 8 litres
16 litres of mixture from the 2nd container is taken.
Thus, amount of milk in 2nd mixture = 4 litres
Total milk after mixing two mixtures = 12 litres
Total amount of mixture = 26 litres
Fraction of milk in new mixture = 
Fraction of milk in 3rd container containing some pure milk = 1
Fraction of milk in the final mixture = 
By Alligation method:

Therefore, the ratio of two mixtures to pure milk will be
= 26 : 9
Therefore, the quantity of milk in the third container is 9 litres.
Total volume of the new mixture = 26 + 9 = 35 litres
This is our required solution.