Mixture Problems MCQ Quiz - Objective Question with Answer for Mixture Problems - Download Free PDF

​Calculations:

Let the amount of mixture taken out from Can A be x units, and the amount taken from Can B be y units.

In Can A, the ratio of soda to water is 5:3. So, the amount of soda taken out from Can A is:

Soda taken from Can A = (5/8) × x, and water taken from Can A = (3/8) × x.

In Can B, the ratio of soda to water is 7 : 2. So, the amount of soda taken out from Can B is:

Soda taken from Can B = (7/9) × y, and water taken from Can B = (2/9) × y.

Now, the total amount of soda and water in the new mixture must have a ratio of 12 : 5.

The total amount of soda in the new mixture = (5/8) × x + (7/9) × y.

The total amount of water in the new mixture = (3/8) × x + (2/9) × y.

The ratio of soda to water in the new mixture is 12 : 5. So, we can write:

\(\dfrac{(5/8) \times x + (7/9) \times y}{(3/8) \times x + (2/9) \times y} = \dfrac{12}{5}\)

Cross-multiply:

(5/8) × x + (7/9) × y = (12/5) × ((3/8) × x + (2/9) × y)

Now, simplifying this equation, we multiply through by 40 (LCM of 8, 9, and 5):

40 × [(5/8) × x + (7/9) × y] = 40 × (12/5) × [(3/8) × x + (2/9) × y]

After simplification, solving this will give the ratio of P : Q.

Therefore, after solving the equation, you will get P : Q = 9 : 10.

Given:

The ratio of milk to water in 630 liters of mixture is 4:3.

140 liters of the mixture is taken out.

Calculations:

The total parts of the mixture = 4 + 3 = 7 parts.

Quantity of milk in the mixture = (4/7) × 630 = 360 liters of milk.

Quantity of water in the mixture = (3/7) × 630 = 270 liters of water.

When 140 liters of the mixture is taken out, the ratio of milk and water in the 140 liters will also be 4:3.

Milk taken out = (4/7) × 140 = 80 liters.

So, the remaining milk = 360 - 80 = 280 liters.

∴ The quantity of milk now is 280 liters.

Calculation

Initial ratio milk:water = 5:3
Let quantity = 8x ⇒ milk = 5x, water = 3x

New milk = 5x + 10  
New water = 3x + 7
New ratio:
5x + 10 / (3x + 7) = 8/5
25x  + 50 = 24x + 56
x = 6
Milk = 5x = 30 Liters

Given:

Total mixture = 60 litres

Ratio of milk to water = 2:1

New desired ratio of milk to water = 1:2

Formula Used:

Milk = (Total mixture × Milk's ratio) / (Sum of ratios)

Water = (Total mixture × Water's ratio) / (Sum of ratios)

New water required = (New water quantity - Original water quantity)

Calculation:

Original milk = (60 × 2) / (2 + 1)

⇒ Original milk = 120 / 3

⇒ Original milk = 40 litres

Original water = (60 × 1) / (2 + 1)

⇒ Original water = 60 / 3

⇒ Original water = 20 litres

In the new ratio (1:2), milk = water × (1 / 2)

Let the new water quantity be W.

Milk = W × (1 / 2) = 40 litres (Milk remains constant).

⇒ W × (1 / 2) = 40

⇒ W = 40 × 2

⇒ W = 80 litres

New water to be added = New water quantity - Original water quantity

⇒ New water to be added = 80 - 20

⇒ New water to be added = 60 litres

The quantity of water to be further added is 60 litres.

Given:

Initial ratio of A : B = 12 : 13

100 litres of mixture removed and replaced with 100 litres of B

New ratio becomes 7 : 8

Formula Used:

Let total initial quantity = x litres

Then A = (12/25)x, B = (13/25)x

After removing 100 litres, the remaining A and B will be in the same ratio:

A removed = 100 × (12/25), B removed = 100 × (13/25)

Calculation:

A left = (12/25)x - (100 × 12/25) = (12/25)(x - 100)

B left = (13/25)x - (100 × 13/25) + 100 = (13/25)(x - 100) + 100

Now the new ratio = 7 : 8

⇒ [(12/25)(x - 100)] / [(13/25)(x - 100) + 100] = 7 / 8

Cross multiply:

8 × (12/25)(x - 100) = 7 × [(13/25)(x - 100) + 100]

⇒ (96/25)(x - 100) = (91/25)(x - 100) + 700

⇒ (96/25)(x - 100) - (91/25)(x - 100) = 700

⇒ (5/25)(x - 100) = 700

⇒ (1/5)(x - 100) = 700

⇒ x - 100 = 3500

⇒ x = 3600

Initial quantity of liquid B = (13/25) × 3600 = 1872 litres

∴ The initial quantity of liquid B is 1872 litres.

Given:

Cost of 1 kg tea of box A (cheaper) = Rs.300

Cost of 1 kg tea of box B (dearer) = Rs.400

Formula used:

Rule of alligation 

Calculation:

Let the mean price be Rs. X

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

According to question,

Given ratio = 5/6

So, 5/6 = (400 - X)/(X- 300)

⇒ 11x = 3,900

⇒ x = 354.54 ≈ 355

∴ The price of mixture  of 1 kg tea is Rs.355

Shortcut TrickAlloy A = 5 : 2 --sum--> 7]  × 4

Alloy B = 3 : 4 --sum--> 7] × 5

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Since the sum of quantity is same so multiplication by 4 and 5 just because the amount of A and B are taken in the ratio 4 : 5

Alloy A = 20 : 8

Alloy B = 15 : 20

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Alloy C = 35 : 28 = 5 : 4

Total quantity = 5 + 4 = 9 

Required % = (5/9) × 100% =  \(55\frac{5}{9}\)

∴ The required percentage of x in alloy C is \(55\frac{5}{9}\).

Alternate Method 

Given:

The mixture of x and y in Alloy A = 5 : 2

The mixture of x and y in Alloy B = 3 : 4

The ratio of A and B in alloy C = 4 : 5

Calculation:

Let the quantity of metal x in alloy C be x

Quantity of metal x in alloy A = \(\frac{5}{{7}}\)

Quantity of metal y in alloy A = \(\frac{2}{{7}}\)

Quantity of metal x in alloy B = \(\frac{3}{{7}}\)

Quantity of metal y in alloy B = \(\frac{4}{{7}}\)

According to the question

The ratio of x and y in alloy C = [(\(\frac{5}{{7}}\) × 4) + (\(\frac{3}{{7}}\) × 5)]/[(\(\frac{2}{{7}}\) × 4) + (\(\frac{4}{{7}}\) × 5)]

⇒ (\(\frac{20}{{7}}\) + \(\frac{15}{{7}}\))/(\(\frac{8}{{7}}\) + \(\frac{20}{{7}}\))

⇒ (\(\frac{35}{{7}}\))/(\(\frac{28}{{7}}\))

⇒ (\(\frac{35}{{7}}\) × \(\frac{7}{{28}}\)

⇒ \(\frac{5}{{4}}\)

Now,

Quantity of x in alloy C = \(\frac{5}{{(5 + 4)}}\)

⇒ \(\frac{5}{{9}}\)

Percentage of x in alloy C = (\(\frac{5}{{9}}\) × 100)

⇒ \(\frac{500}{{9}}\)

⇒ \(55\frac{5}{9}\)

∴ The required percentage of x in alloy C is \(55\frac{5}{9}\)

Shortcut Trick

Solution quantity = 400ml

Let the quantity of pure alcohol to be added in 400ml be A ml.

alcohol in 400ml solution = 16 × 400/100 = 64ml.

Then,

⇒ 400 × 16/100 + A = (400 + A) × 40/100

⇒ 64 + A = 160 + 2A/5

⇒ 3A/5 = 96

⇒ A = 96 × 5/3

⇒ A = 160

Alternate Method

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

5 units → 400 ml

Then, 2 units → 400/5 × 2 = 160 ml

Given:

A container contains 25 litre of milk. From this container, 5 litre of milk is taken out and replaced by water. 

Concept used:

Left Quantity = Initial Quantity(1 - [Fraction removed])^N (Where N = Number of times the process is carried out)

Calculation:

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

Now, the quantity of milk left in the container

⇒ 25(1 - 1/5)³

⇒ 25 × (4/5)³

⇒ 25 × 64/125

⇒ 12.8 litre

∴ There is 12.8 litre of milk left in the container.

Shortcut Trick 

Given:

Ratio of milk and water is 2 : 3

60 liters of mixture is taken out

Then ratio of milk and water is 1 : 2

Calculation:

Let the milk and water in the total mixture be 2x and 3x.

⇒ Milk in the total mixture = 2x/5x

⇒ Milk in the total mixture = 2/5

⇒ Water in the total mixture = 3x/5x

⇒ Water in the total mixture = 3/5

In 60 liters of mixture 

⇒ Milk = 2/5 × 60

⇒ Milk = 24 liters

⇒ Water = 3/5 × 60

⇒ Water = 36 liters 

When 60 liters of mixture is taken out, 

Replaced with 60 liters of water.

Then the ratio of Milk and Water is 1 : 2

⇒ (2x – 24) : (3x – 36 + 60) = 1 : 2

⇒ (2x – 24)/(3x + 24) = 1 : 2

⇒ 2(2x – 24) = 1(3x + 24)

⇒ 4x – 48 = 3x + 24

⇒ 4x – 3x = 24 + 48

⇒ x = 72

⇒ Total capacity of container = 2x + 3x

⇒ Total capacity of container = 5x

⇒ Total capacity of container = 5 × 72

⇒ Total capacity of container =  360 liters

∴ Total capacity of the container is 360 liters.

Shortcut Trick 

Given:

A dairy farmer's can contains 6 litres of milk.

His wife adds some water to it such that milk and water are in the ratio 4 ∶ 1.

Calculation:

Milk : Water = 4 : 1

Let the quantity of milk and water be 4x and x.

Quantity of Milk = 4x = 6 litres

⇒ x = 1.5 litres

Quantity of Water = x = 1.5 litres

According to the question,

\(\dfrac{6+x}{1.5}\) = \(\dfrac{5}{1}\)

⇒ 6 + x = 7.5

⇒ x = 7.5 - 6 = 1.5 litres

Alternate Method 

Given:

Initial ratio of acid and base = 17 : 3

Final mixture of acid and base = 1 : 1

Calculation:

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)

According to the question, ratio of acid and base in resultant mixture is 1:1

Thus, equating Equations (1) and (2)

(340x - 17y)/20 = (60x + 17y)/20

⇒ 340x - 17y = 60x + 17y 

⇒ 34y = 280x

⇒ y/x = 280/34

⇒ 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

∴ 7/17 fraction of the mixture must be drawn off and substituted by the base so that the ratio of acid and base in the resultant mixture in the solution becomes  1:1
 Shortcut Trick

Let us remove some quantity of the mixture from the solution.

After that

                   Acid         Base

Initial ratio     17    :       3

Final ratio       1    :       1

We are adding the Base so the quantity of Acid will remain the same.

So multiply the second ratio by 17.

                   Acid         Base

Initial ratio     17    :       3

Final ratio     17    :      17

So Base added = 17 - 3 = 14 units

Here note that the initial quantity of mixture = Final quantity of mixture

So

Initial quantity of mixture = 17 + 17 = 34 units

the required ratio = 14/34 = 7/17

Given:

The ratio of milk and water in the first vessel = 8 : 7

The ratio of milk and water in the second vessel = 7 : 9

The ratio of water and milk in the resultant mixture = 9 : 8

Calculation:

Let x litre of the first mixture and y litre of the second mixture are mixed.

Quantity of milk in x litre of the first mixture = 8x/15

Quantity of milk in y litre of the second mixture = 7y/16

Total quantity of the resultant mixture = (x + y)

Quantity of milk in (x + y) litre of the resultant mixture = 8(x + y)/17

8x/15 + 7y/16 = 8(x + y)/17

⇒ 8x/15 + 7y/16 = 8x/17 + 8y/17

⇒ 8x/15 – 8x/17 = 8y/17 – 7y/16

⇒ (136x – 120x)/15 × 17 = (128y – 119y)/17 × 16

⇒ 16x/15 = 9y/16

⇒ 256x = 135y

⇒ x/y = 135/256

∴ The required ratio is 135 : 256

Alternative Method:

The concentration of milk in the first mixture = 8/15

The concentration of milk in the second mixture = 7/16

The concentration of milk in the resultant mixture = 8/17

By the rule of Allegation,

⇒ 9/272 : 16/255

⇒ 9 × 255 : 16 × 272

⇒ 9 × 15 : 16 × 16

⇒ 135 : 256     

∴ The required ratio is 135 : 256.

Given: 

The cost price of wine = Rs. 60 per litre

The cost price of water = Rs. 0 per litre

The cost price of mixture = Rs. 40 per litre

Calculations:

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

∴ The ratio in which water and wine should be mixed is 1 : 2.

Alternate Method

Given: 

The cost price of wine = Rs. 60 per litre

The cost price of water = Rs. 0 per litre

The cost price of mixture = Rs. 40 per litre

Concept used:

If two indigents are mixed, then

Calculation:

Using alligation, 

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

∴ The ratio in which water and wine should be mixed is 1 : 2.

Important Points

Given:

The ratio of sugar to water in solution A = 1 ∶ 4

The ratio of salt to water in solution B = 1 ∶ 26

Calculation:

First, make the quantity of solution A and solution B same.

Total unit of sugar and water in solution A = 1 + 4 = 5 units

Total unit of salt and water in solution B = 1 + 26 = 27 units

Now, multiply the ratio of solution A by 27 and multiply the ratio of solution B by 5.

The ratio of sugar to water in solution A = 1 × 27 ∶ 4 × 27 = 27 : 108

The ratio of salt to water in solution B = 1 × 5 ∶ 26 × 5 = 5 : 130

Now, mix the solution 2 : 3.

Therefore, multiply the new ratio of solution A by 2 and multiply the new ratio of the solution B by 3.

The new required ratio of solution A = 54 : 216

The new required ratio of solution B = 15 : 390

The ratio of sugar, salt and water in ORS = 54 : 15 : 606

The ratio of sugar and salt = 54 : 15 = 18 : 5

Therefore, "18 : 5" is the required answer.

Shortcut Trick