Mixture and Alligation Questions with Solution (Advanced)

Solution:

Ratio = (32 - 30) : (40 - 32)

Ratio = 2 : 8

Ratio = 1 : 4

Solution:

Initial milk volume = 40 × 17 / (17 + 3)

Initial milk volume = 34 litres

Initial water volume = 6 litres

Let the added milk be x litres.

Final milk volume = Initial milk volume + x

Final water volume = Initial water volume

New ratio (milk to water) = (34 + x) / 6 = 7 / 1

34 + x = 42

x = 8 litres

Solution:

Profit = 20% of CP

⇒ SP = CP + 0.20 × CP

⇒ 320 = 1.20 × CP

⇒ CP = 320 / 1.20

⇒ CP = ₹266.67/kg

Using the formula for the cost price of the mixture:

CP of mixture = (C1 × Quantity1 + C2 × Quantity2) / (Quantity1 +  Quantity2)

266.67 = (180 × Quantity1 + 280 × Quantity2) / (Quantity1 + Quantity2)

Let Quantity1 = x and Quantity2 = y

266.67 = (180x + 280y) / (x + y)

x / y = 13.33 / 86.67

x / y = 2 / 13

Ratio = (21 - 17)/(17 - 11)

= 2/3

Solution:

The quantity of mixture in vessel C before mixing 10 litres of milk 70 - 10 =  60 litres

The quantity of milk in vessel C = 60 × 3/4 = 45

The quantity of water in vessel C = (60 - 45) = 15

Let the quantity of mixture in vessel A and B be X and Y respectively. 60% of milk from vessel A is poured in vessel C

0.6X = 45

X = 75 litres

50% of the water from vessel B is poured into vessel C

0.50Y = 15

Y = 30 litres

Required ratio X: Y = 75: 30 =5: 2

Let the mean price be Rs. X

So, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) = (400 - X) : (X  - 300)

5/6 = (400 - X)/(X- 300)

x = 354.54 ≈ 355

Solution:

By alligation

The ratio of solution to pure alcohol = 60 ∶ 24 = 5 ∶ 2

5 units → 400 ml

Then, 2 units → 400/5 × 2 = 160 ml

Solution:

Fraction of milk removed = 5/25 = 1/5

Now, the quantity of milk left in the container

25(1 - 1/5)³

25 × 64/125

12.8 litre

Solution:

Let the acid and base be 17x litres and 3x litres resp

Total mixture = 20x

Let the drawn part of the mixture be 'y' liters

Acid in (20 - y) litres mixture

(20x - y) × (17/20) = (340x - 17y)/20 ----(i)

Now adding 'y' litres of base to the mixture

Base in the resultant mixture

(3/20) × (20x - y) + y = (60x + 17y)/20 ----(ii) (340x - 17y)/20 = (60x + 17y)/20

y/x = 140/17

Total mixture = 20x = (20 × 17) liters

Mixture to be removed and replaced = y = 140 liters Required fraction = (140)/(20 × 17) = 7/17

By alligation :

Ratio of wine and water = 40 : 20 = 2 :1

x = 5

The original amount of kerosene = 5x = 25 litres

Solution:

Selling price of mixture is Rs. 408 per kg

Profit% = (selling price/Cost price – 1) × 100

⇒ 20 = {(408/Cost price) – 1} × 100

⇒ 0.2 + 1 = 408/Cost price

⇒ Cost price of mixture = 408/1.2 = Rs.340 per kg

∴ Ratio of Assam Tea to the Darjeeling Tea = (400 - 340) : (340 - 300)  = 60 : 40 = 3 : 2

Solution:

Quantity of Solution = 56 ml (Given)

16 ml of the solution replaced with 5 ml of water

then the solution quantity will be = (56 – 16) + 5 = 45 ml

Milk : Water = 5 : 4

M + W = 45 (will have to be distributed in the ratio of 5 : 4)  In the Final Mixture,

Milk = 25 ml, Water = 20 ml

In 20 ml Water, 5 litre was added after replacement.

∴ Before Replacement, Water = 20 - 5 = 15 ml

Ratio of Milk and Water (Initially) = 25 : 15 = 5 : 3

16 ml of solution was replaced (will be calculated in the ratio of 5 : 3) Replaced Quantity of Milk and Water = 10 ml and 6 ml

Hence, Initial Quantity of Milk = 25 ml + 10 ml = 35 ml

Solution:

Silver = 40%; Copper = 30%; Nickel = 30%

Let, the total weight of alloy = 100 units

So, the silver = 40 units

If we add 20 units silver then, alloy = (100 + 20) = 120 units

also, silver = 40 + 20 = 60 units

Now, silver is 50% of the alloy [∵ silver = 60 and alloy = 120] So, 100 units = 25 kg

⇒ 20 units = (25/100) × 20 = 5 kg

Solution:

Let the quantity of wine and water added in final mixture be x and y respectively. According to the question:

60 × x + 0y = (x + y) × 40

⇒ 60x = 40x + 40y

⇒ 60x - 40x = 40y

⇒ 20x = 40y

⇒ x : y = 2 : 1