To Make a Mixture from Two Mixtures MCQ Quiz - Objective Question with Answer for To Make a Mixture from Two Mixtures - Download Free PDF

Given:

Total volume of mixture = 108 liters

Ratio of milk to water = 5:4

Milk added = 20 liters

Water added = 36 liters

Formula used:

Milk in the mixture = (Total mixture) × (Ratio of milk / Total ratio)

Water in the mixture = (Total mixture) × (Ratio of water / Total ratio)

Calculations:

Total ratio of milk and water = 5 + 4 = 9

Milk in the mixture = (108 liters) × (5/9) = 60 liters

Water in the mixture = (108 liters) × (4/9) = 48 liters

After adding milk and water:

New amount of milk = 60 + 20 = 80 liters

New amount of water = 48 + 36 = 84 liters

Difference between milk and water in the mixture = 84 - 80 = 4 liters

Y = 4

7Y = 7 × 4 = 28

∴ The value of 7Y is 28.

​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 = 8 : 9.

Given:

Vessel 1: 31 litres, Milk : Water = 15 : 16

Vessel 2: 43 litres, Milk : Water = 26 : 17

Required: Add water to make final Milk : Water = 1 : 1

Formula used:

Milk = Total Quantity × (Milk Ratio / Total Ratio)

Water = Total Quantity × (Water Ratio / Total Ratio)

Calculations:

Milk in Vessel 1 = 31 × 15 / (15 + 16) = 31 × 15 / 31 = 15 L

Water in Vessel 1 = 31 - 15 = 16 L

Milk in Vessel 2 = 43 × 26 / (26 + 17) = 43 × 26 / 43 = 26 L

Water in Vessel 2 = 43 - 26 = 17 L

Total Milk = 15 + 26 = 41 L

Total Water = 16 + 17 = 33 L

Required: Milk = Water ⇒ 41 = 33 + x

⇒ x = 8 L

∴ Water to be added = 8 litres

Explanation:

Let the initial volume of liquids A and B be x and 2x.

Total volume of mixture = x + 2x + 20 = 3x + 20

Final volume of liquid A = x − 2

Final volume of liquid B = 2x + 7

Final volume of liquid C = 15

Therefore, 3(x-2)=(2x+7).

x = 13.

Volume of the mixture = 3x + 20 = 59 litres.

Given:

Cost of the first variety of pulses = Rs. 75 per kg

Cost of the second variety of pulses = Rs. 100 per kg

Cost of the mixture = Rs. 86.5 per kg

Formula Used:

Ratio of quantities = (Cost of second variety - Cost of mixture) : (Cost of mixture - Cost of first variety)

Calculation:

Let the ratio be x : y.

Cost of second variety - Cost of mixture = 100 - 86.5 = 13.5

Cost of mixture - Cost of first variety = 86.5 - 75 = 11.5

Ratio of quantities = 13.5 : 11.5

⇒ Ratio = 135 : 115

⇒ Ratio = 27 : 23

The grocer must mix the pulses in the ratio 27:23.

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:

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:

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 

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:

Price of X qualities of pulses = Rs.100/kg

Price of Y qualities of pulses = Rs.150/kg

Qunatity of X type of pulses : Qunatity of Y type of pulses = 7 : 20

Formula used:

Average price = (Total price of mixture of two quality of grain)/Total quantity

Calculation:

Let quantity of X type of pulses = 7x

Quantity of X type of pulses = 20x

Average price = (Total price of mixture of two quality of grain)/Total quantity

⇒ {(100 × 7x) + (150 × 20x)}/27x

⇒ (700x + 3000x)/27x

⇒ 3700x/27x = Rs.137.03 ≈ Rs.137

∴ The correct answer is Rs.137.

Shortcut Trick 

\(\frac{150-m}{m-100}= \frac{7}{20}\)

⇒ 3000 - 20m = 7m - 700

⇒ 3700 = 27m

⇒ m = 3700/27 

 ⇒ m =Rs.137.03 ≈ Rs.137

∴ The correct answer is Rs.137.

Given:

80 liters of a mixture contain milk and water in the ratio of 27 ∶ 5.

Calculation:

Milk = \(\dfrac{27}{32}\) × 80 = 67.5 L

Water = 80 - 67.5 = 12.5 L

Let, m = water to be added to get 3 : 1

⇒ \(\dfrac{67.5}{12.5 + m}\) = 3

⇒ 67.5 = 37.5 + 3m

⇒ m = 10

∴ The quantity of water to be added to get 3 : 1, is 10 litres.

Given:

A and B contain spirit and water in the ratio 3 : 8 and 6 : 5 respectively

Calculation:

∴ Required ratio = (1/11) : (2/11) = 1 : 2

Alternate Method

Let the total quantity in A be 11x and in B be 11y, which is to be mixed

According to the question:

(3x + 6y)/(8x + 5y) = 5/6

⇒ 18x + 36y = 40x + 25y

⇒ 22x = 11y

⇒ x/y = 1/2

∴ They should be mixed in the ratio 1 : 2

Total volume = 21 litres

2/3 of a milk-water mixture was milk,

⇒ Volume of milk = 2/3 × 21 = 14 litres

⇒ Volume of water = 21 – 14 = 7 litres

4 litres of water is added to it,

⇒ New volume of water = 7 + 4 = 11 litres.

⇒ Total volume of mixture = 14 + 11 = 25

Let the quantity of milk and water in the mixture be x and 56 – x

Now, 16ml of the solution is is removed

Quantity of solution replaced will be removed in the same ratio as the contents in the solution

∴ Quantity of milk left after removal = x – (16 × x/56) = 5x/7

Quantity of water left after removal = 56 – x – {16 × (56 – x)/56} = 40 – 5x/7

Ratio of milk and water after adding 5 ml of water is 5 ∶ 4

⇒ (5x/7)/(40 + 5 – 5x/7) = 5/4

⇒ 20x = 45 × 7 × 5 – 25x

⇒ x = 7 × 5 = 35

∴ Concentration of milk in the mixture initially was 35 ml

Alternate Method

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

Given:

Quantity of mixture is 18 litres

Percentage of soda is 30%

Percentage of wine is 70%

Calculation:

Quantity of soda = 30% of 18 = 5.4 litres

Quantity of wine = 70% of 18 = 12.6 litres

Let the quantity of soda added be x

Total quantity = 18 + x

According to the question

(5.4 + x)/(18 + x) = 40/100

⇒ (5.4 + x)/(18 + x) = 2/5

⇒ 27 + 5x = 36 + 2x

⇒ x = 3 litres

∴ The required quantity of soda is 3 litres.

Shortcut Trick

Ratio = 6 : 1

⇒ 6 = 18 litres

⇒ 1 = 3 litres 

∴ The required quantity of soda is 3 litres.