Mixtures and Alligations

123

Q1: How much Pepsi at Rs. 6 a litre is added to 15 litre of ‘dew’ at Rs.10 a litre so that the price of the mixture be Rs. 9 a litre?

A. 5

B. 8

C. 10

D. None of these

Correct Answer: A

Solution:
Let $x$ be the amount of Pepsi added.
The total cost of 15 liters of dew is $15 \times 10 = \text{Rs.
} 150$.
The total cost of $x$ liters of Pepsi is $6x$.
The total volume of the mixture is $15 + x$ liters.
The total cost of the mixture is $150 + 6x$.
The price of the mixture is Rs.
9 per liter.
Therefore, $\frac{150 + 6x}{15 + x} = 9$ $$150 + 6x = 9(15 + x)$$ $$150 + 6x = 135 + 9x$$ $$150 - 135 = 9x - 6x$$ $$15 = 3x$$ $$x = \frac{15}{3} = 5 \text{ liters}$$ Therefore, 5 liters of Pepsi should be added.

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Q2: In a municipal parking there are some two wheelers and rest are 4 wheelers. If wheels are counted, there are total 520 wheels but the incharge of the parking told me that there are only 175 vehicles. If no vehicle has a stepney, then the no. of two wheelers is :

A. 75

B. 100

C. 90

D. 85

Correct Answer: C

Solution:
Let $x$ be the number of two-wheelers and $y$ be the number of four-wheelers.
We know that the total number of vehicles is 175, so $x + y = 175$.
The total number of wheels is 520, so $2x + 4y = 520$.
We can solve this system of linear equations.
From the first equation, we get $y = 175 - x$.
Substitute this into the second equation: $2x + 4(175 - x) = 520$ $2x + 700 - 4x = 520$ $-2x = -180$ $x = 90$ Therefore, there are 90 two-wheelers.

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Q3: In my pocket there are Rs. 25 consisting of only the denominations of 20 paise and 50 paise. Thus there are total 80 coins in my pocket.The no. of coins of the denomination of 50 paise is :

A. 30

B. 70

C. 50

D. 25

Correct Answer: A

Solution:
Let $x$ be the number of 20 paise coins and $y$ be the number of 50 paise coins.
We are given that the total number of coins is 80, so $x + y = 80$.
The total value of the coins is Rs.
25, which is equal to 2500 paise.
Therefore, $20x + 50y = 2500$.
We have a system of two linear equations: 1) $x + y = 80$ 2) $20x + 50y = 2500$ From equation (1), we can express $x$ as $x = 80 - y$.
Substitute this into equation (2): $20(80 - y) + 50y = 2500$ $1600 - 20y + 50y = 2500$ $30y = 900$ $y = 30$ Therefore, there are 30 coins of 50 paise denomination.
To verify, substitute $y = 30$ into $x = 80 - y$: $x = 80 - 30 = 50$.
So there are 50 coins of 20 paise and 30 coins of 50 paise.
Total coins: $50 + 30 = 80$ (Correct) Total value: $(50 \times 20) + (30 \times 50) = 1000 + 1500 = 2500$ paise = Rs.
25 (Correct)

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Q4: There are some shepherds and their sheep in a grazing field. The no. of total heads are 60 and total legs are 168 including both men and sheep. The no. of sheep is :

A. 18

B. 26

C. 24

D. 36

Correct Answer: C

Solution:
Let $x$ be the number of shepherds and $y$ be the number of sheep.
Each shepherd has 2 legs and each sheep has 4 legs.
We can set up a system of two linear equations based on the given information: $$ x + y = 60 $$ $$ 2x + 4y = 168 $$ We can solve this system of equations using substitution or elimination.
Let's use elimination.
Multiply Equation 1 by -2: $$ -2x - 2y = -120 $$ Now add this modified Equation 1 to Equation 2: $$ (-2x - 2y) + (2x + 4y) = -120 + 168 $$ $$ 2y = 48 $$ $$ y = 24 $$ Therefore, there are 24 sheep.
Substitute $y = 24$ back into Equation 1 to find $x$: $$ x + 24 = 60 $$ $$ x = 36 $$ There are 36 shepherds and 24 sheep.
The question asks for the number of sheep.

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Q5: In the 75 litres of mixture of milk and water, the ratio of milk and water is 4 :1. The quantity of water required to make the ratio of milk and water 3:1 is :

A. 1 litre

B. 3 litres

C. 4 litres

D. 5 litres

Correct Answer: D

Solution:
Let the quantity of milk in the mixture be $4x$ litres and the quantity of water be $x$ litres.
Total quantity of mixture = $4x + x = 5x$ litres.
Given that the total quantity of mixture is 75 litres, we have $5x = 75$.
Therefore, $x = 15$ litres.
Quantity of milk = $4x = 4 \times 15 = 60$ litres.
Quantity of water = $x = 15$ litres.
Let $y$ litres of water be added to make the ratio of milk and water 3:1.
The quantity of milk remains the same, which is 60 litres.
The quantity of water becomes $15 + y$ litres.
The new ratio of milk to water is $60 : (15 + y) = 3 : 1$.
We can set up a proportion: $$ \frac{60}{15 + y} = \frac{3}{1} $$ Cross-multiplying, we get: $60 = 3(15 + y)$ $60 = 45 + 3y$ $3y = 60 - 45$ $3y = 15$ $y = 5$ Therefore, 5 litres of water needs to be added.

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Q6: A car agency has 108 cars. He sold some cars at 9% profit and rest at 36% profit. Thus he gains 17% on the sale of all his cars. The no. of cars sold at 36% profit is :

A. 25

B. 32

C. 35

D. 75

Correct Answer: B

Solution:
Let $x$ be the number of cars sold at 9% profit and $(108-x)$ be the number of cars sold at 36% profit.
The total profit is 17% of 108 cars.
Let $CP$ be the cost price of each car.
The total cost price is $108CP$.
$$108CP \times 1.17 = 126.36CP$$ The total selling price is $126.36CP$.
The selling price from cars sold at 9% profit is $xCP \times 1.09 = 1.09xCP$.
The selling price from cars sold at 36% profit is $(108-x)CP \times 1.36 = 146.88CP - 1.36xCP$.
Therefore, $1.09xCP + 146.88CP - 1.36xCP = 126.36CP$ $$146.88CP - 0.27xCP = 126.36CP$$ $$0.27xCP = 20.52CP$$ $$x = \frac{20.52}{0.27} = 76$$ The number of cars sold at 36% profit is $108 - 76 = 32$.

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Q7: Rs. 69 were divided among 115 students so that each girl gets 50 paise less than a boy. Thus each boy received twice the paise as each girl received. The no. of girls in the class is :

A. 92

B. 42

C. 33

D. 23

Correct Answer: A

Solution:
Let the amount received by each girl be $x$ paise.
Then each boy received $2x$ paise.
Each girl received 50 paise less than a boy, so $x + 50 = 2x$.
Solving for $x$, we get $x = 50$ paise.
Therefore, each girl received 50 paise and each boy received 100 paise ($2 \times 50 = 100$).
Let the number of girls be $g$ and the number of boys be $b$.
The total number of students is 115, so $g + b = 115$.
The total amount distributed is Rs.
69, which is 6900 paise.
So, $50g + 100b = 6900$.
We have a system of two equations: $$g + b = 115$$ $$50g + 100b = 6900$$ From the first equation, $b = 115 - g$.
Substitute this into the second equation: $50g + 100(115 - g) = 6900$ $50g + 11500 - 100g = 6900$ $-50g = -4600$ $g = 92$ Therefore, there are 92 girls.

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Q8: A butler stole wine from a butt of sherry containing 50% of spirit, then he replenished it by different wine containing 20% spirit. Thus there was only 30% strength (spirit) in the new mixture. How much of the original wine did he steal?

A. 1 / 3

B. 2 / 3

C. 1 / 2

D. 1 / 4

Correct Answer: B

Solution:
Let $x$ be the amount of wine stolen.
The original mixture contained 50% spirit.
After $x$ amount is stolen, the remaining spirit is $0.5(1-x)$.
The butler then adds $x$ amount of 20% spirit wine.
The total volume remains 1.
The new mixture contains 30% spirit.
Therefore, we can set up the equation: $$0.5(1-x) + 0.2x = 0.3(1)$$ Solving for $x$: $0.5 - 0.5x + 0.2x = 0.3$.
This simplifies to $0.2 = 0.3x$.
Therefore, $x = \frac{0.2}{0.3} = \frac{2}{3}$.
The butler stole $\frac{2}{3}$ of the original wine.
To express this as a percentage, $\left(\frac{2}{3}\right) \times 100\% \approx 66.67\%$.
However, the question asks how much of the original wine, not percentage.
Therefore, the answer is $\frac{2}{3}$ or approximately 66.67% of the original wine.
Let's assume the butt contained 1 unit of wine.
Then the butler stole $\left(\frac{2}{3}\right)$ units of wine.

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Q9: In what proportion water be mixed with spirit to gain 12.5% by selling it at cost price?

A. 3 : 5

B. 1 : 8

C. 2 : 7

D. 1 : 9

Correct Answer: B

Solution:
Let the cost price of the spirit be Rs.
$100$ per liter.
To gain $12.5\%$, the selling price should be Rs.
$112.5$.
However, the spirit is sold at the cost price, implying that the seller is selling less spirit than a liter.
Let $x$ liters of spirit be sold for Rs.
$100$.
Then, the selling price per liter = $\frac{100}{x}$.
The gain is $12.5\%$, therefore, the selling price is $112.5\%$ of the cost price.
We have: $\frac{100}{x} = 1.125 \times 100$ $\frac{100}{x} = 112.5$ $x = \frac{100}{112.5} = \frac{1000}{1125} = \frac{8}{9}$ This means that $\frac{8}{9}$ liters of spirit are sold as 1 liter.
The quantity of water added is $1 - \frac{8}{9} = \frac{1}{9}$ liters.
The proportion of water to spirit is $\frac{1/9}{8/9} = \frac{1}{8}$.
Therefore, the proportion of water to spirit is $1:8$.

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Q10: Mr. Mittal purchased two steel factories, one in India and other one in Malaysia for total Rs. 72 crores. Later on he sold the Indian factory at 16% profit and Malasian factory at 24% profit. Thus he gained a total profit of 19%. The selling price of Indian factory is:

A. 45 crore

B. 52.2 crore

C. 8.55 crore

D. Cannot be determined

Correct Answer: B

Solution:
Let the cost price of the Indian factory be $x$ crores and the cost price of the Malaysian factory be $(72-x)$ crores.
The selling price of the Indian factory is $1.16x$ crores ($16\%$ profit).
The selling price of the Malaysian factory is $1.24(72-x)$ crores ($24\%$ profit).
Total selling price = $1.16x + 1.24(72-x) = 1.19 \times 72$ ($19\%$ total profit) $$1.16x + 89.28 - 1.24x = 85.68$$ $$-0.08x = 85.68 - 89.28$$ $$-0.08x = -3.6$$ $$x = 45$$ Selling price of Indian factory = $1.16x = 1.16 \times 45 = 52.2$ crores

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Q11: In a 25 litre mixture of milk and water, the water is only 20%. How many litres of water is required to increase the percentage of water to 90%?

A. 45 litre

B. 70 litre

C. 115 litre

D. 175 litre

Correct Answer: D

Solution:
Step 1: Calculate the initial amount of water: 20% of 25 litres = $\frac{20}{100} \times 25 = 5$ litres.
Step 2: Calculate the initial amount of milk: 25 litres - 5 litres = 20 litres.
Step 3: Let $x$ be the amount of water added.
The total volume becomes $25 + x$ litres.
Step 4: The amount of water becomes $5 + x$ litres.
Step 5: The new percentage of water is 90%, so we set up the equation: $\frac{5 + x}{25 + x} = \frac{90}{100}$.
Step 6: Simplify the equation: $100(5 + x) = 90(25 + x)$ Step 7: Expand: $500 + 100x = 2250 + 90x$ Step 8: Solve for $x$: $10x = 1750$ Step 9: $x = 175$ litres.
Therefore, 175 litres of water needs to be added.

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Q12: In a class of 30 students, the average weight of boys is 20 kg and the average weight of the girls is 25 kg. The fraction of boys out of the total students of the class is :

A. 4 / 5

B. 5 / 6

C. 3 / 4

D. Data Inadequate

Correct Answer: D

Solution:
Let $b$ be the number of boys and $g$ be the number of girls.
We know that $b + g = 30$.
The total weight of boys is $20b$ kg, and the total weight of girls is $25g$ kg.
The average weight of the class is a weighted average of the boys' and girls' average weights.
Let's assume the average weight of the class is $x$.
We don't need to calculate $x$ directly to solve the problem.
Using the concept of alligation, we can set up a ratio: $$ \frac{25 - x}{x - 20} = \frac{b}{g} $$ We also know that $\frac{b}{30}$ is the fraction of boys.
We can use another approach.
Let the total weight of the class be $W$.
Then: $$ W = 20b + 25g $$ The average weight of the class is $\frac{W}{30}$.
We don't need this value directly.
Let's use the weighted average concept to find the fraction of boys.
Let $f$ be the fraction of boys.
Then $(1-f)$ is the fraction of girls.
The average weight of the class can be expressed as: Average = $f \times 20 + (1-f) \times 25$ We don't need to solve for the average.
We are given that the total number of students is 30.
Therefore: $f = \frac{b}{30}$ and $(1-f) = \frac{g}{30}$ We can't solve for $f$ directly using this method without more information.
However, we can use the given information to find the ratio.
Let's reconsider the alligation approach.
We don't have the overall average.
Instead, we are looking for $\frac{b}{30}$.
Let's use another approach: Suppose we assume an overall average weight.
Let’s say 22 kg.
Using the alligation method: $$ \frac{25-22}{22-20} = \frac{3}{2} $$ This represents the ratio of girls to boys.
So, the ratio is $g:b = 3:2$.
This means $g = 3k$ and $b = 2k$ for some integer $k$.
Since $g + b = 30$, we have $5k = 30$, which means $k = 6$.
Therefore, $b = 12$ and $g = 18$.
The fraction of boys is $\frac{b}{b+g} = \frac{12}{30} = \frac{2}{5}$

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Q13: The price of petrol is Rs. 60 per litre and the price of spirit is Rs. 40 per litre. In what ratio the petrol and spirit be mixed such that the profit after selling the mixture at Rs. 75 per litre be 25%?

A. 1 : 1

B. 3 : 2

C. 5 : 1

D. Such a mixture is not possible

Correct Answer: D

Solution:
Let the ratio of petrol to spirit be $x:y$.
Cost price of petrol = $60x$ Cost price of spirit = $40y$ Total cost price = $60x + 40y$ Selling price of mixture = $75(x+y)$ Profit = Selling Price - Cost Price = $75(x+y) - (60x + 40y) = 15x + 35y$ Profit percentage = $\frac{\text{Profit}}{\text{Cost Price}} \times 100 = 25\%$ Therefore, $\frac{15x + 35y}{60x + 40y} = 0.25$ Solving the equation: $15x + 35y = 0.25(60x + 40y)$ $15x + 35y = 15x + 10y$ $25y = 0$ This equation implies that there's an error in the question's formulation, as it leads to $y=0$, which is not a valid ratio.
The problem likely lies in the profit percentage calculation or the given selling price.
Let's assume the profit is 25% of the cost price.
Let's re-examine.
Let the ratio be $x:y$.
Cost price of mixture = $\frac{60x + 40y}{x+y}$ Selling price = 75 Profit = $75 - \frac{60x + 40y}{x+y}$ Profit percentage = 25% $\frac{75 - \frac{60x + 40y}{x+y}}{\frac{60x + 40y}{x+y}} = 0.25$ $75(x+y) - (60x + 40y) = 0.25(60x + 40y)$ $75x + 75y - 60x - 40y = 15x + 10y$ $15x + 35y = 15x + 10y$ $25y = 0$ This still leads to an invalid solution.
There must be a mistake in the question's parameters.
A valid solution cannot be obtained with the given information.

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Q14: A trader sells total 315 TV sets. He sells black and white TV sets at a loss of 6% and colour TV sets at a profit of 15%.Thus he gains 9% on the whole. The no. of B/ W TV sets, which he has sold, is :

A. 126

B. 216

C. 135

D. 90

Correct Answer: D

Solution:
Let $x$ be the number of black and white TV sets and $y$ be the number of color TV sets.
Total TV sets sold = $x + y = 315$ (1) Let $CP_{bw}$ be the cost price of one black and white TV set and $CP_c$ be the cost price of one color TV set.
We assume that $CP_{bw} = CP_c = CP$ (for simplification, this doesn't affect the final answer because we're only dealing with ratios).
The selling price of black and white TV sets = $CP \times (1 - 0.06) = 0.94 CP$ The selling price of color TV sets = $CP \times (1 + 0.15) = 1.15 CP$ Total cost price = $x \times CP + y \times CP = (x+y) \times CP = 315 \times CP$ Total selling price = $0.94x \times CP + 1.15y \times CP$ Overall profit is 9%, so total selling price = $315 \times CP \times 1.09 = 343.35 \times CP$ Therefore, $0.94x \times CP + 1.15y \times CP = 343.35 \times CP$ Dividing by $CP$, we get $0.94x + 1.15y = 343.35$ (2) Substitute (1) into (2): $y = 315 - x$ $0.94x + 1.15(315 - x) = 343.35$ $0.94x + 362.25 - 1.15x = 343.35$ $-0.21x = -18.9$ $x = \frac{18.9}{0.21} = 90$ Therefore, the number of black and white TV sets sold is 90.

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Q15: The average age of boys in class is 16.66, while the average age of girls is 18.75. Thus the average age of all the 40 students of the class is 17.5. If the difference between the no. of boys and girls is 8, then the no. of girls in the class is :

A. 12

B. 16

C. 18

D. Data Inadequate

Correct Answer: B

Solution:
Let $b$ be the number of boys and $g$ be the number of girls.
Total number of students = $b + g = 40$ (1) Total age of boys = $16.66b$ Total age of girls = $18.75g$ Total age of all students = $16.66b + 18.75g$ Average age of all students = $\frac{16.66b + 18.75g}{40} = 17.5$ $16.66b + 18.75g = 17.5 \times 40 = 700$ (2) The difference between the number of boys and girls is 8.
$b - g = 8$ or $g - b = 8$ Case 1: $b - g = 8 \Rightarrow b = g + 8$.
Substitute in (1): $g + 8 + g = 40 \Rightarrow 2g = 32 \Rightarrow g = 16, b = 24$ Substitute in (2): $16.66 \times 24 + 18.75 \times 16 = 400 + 300 = 700$.
This is correct.
Case 2: $g - b = 8 \Rightarrow g = b + 8$.
Substitute in (1): $b + b + 8 = 40 \Rightarrow 2b = 32 \Rightarrow b = 16, g = 24$ Substitute in (2): $16.66 \times 16 + 18.75 \times 24 = 266.56 + 450 = 716.56$.
This is incorrect.
Therefore, the number of girls is 16.

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Q16: The ratio of water and alcohol in two different containers is 2: 3 and 4: 5 In what ratio we are required to mix the mixtures of two containers in order to get the new mixture in which the ratio of alcohol and water be 7 : 5?

A. 7 : 3

B. 5 : 3

C. 8 : 5

D. 2 : 7

Correct Answer: B

Solution:
Let the ratio of water and alcohol in container 1 be $2:3$.
This means for every 5 units of mixture, there are 2 units of water and 3 units of alcohol.
The fraction of alcohol is $\frac{3}{5}$.
Let the ratio of water and alcohol in container 2 be $4:5$.
This means for every 9 units of mixture, there are 4 units of water and 5 units of alcohol.
The fraction of alcohol is $\frac{5}{9}$.
Let $x$ be the ratio in which the two mixtures are mixed.
Then the ratio of alcohol in the new mixture is given by: $$ \frac{\frac{3}{5}x + \frac{5}{9}}{x + 1} $$ We want the ratio of alcohol to water in the new mixture to be $7:5$, which implies the fraction of alcohol in the new mixture is $\frac{7}{12}$.
Therefore we set up the equation: $$ \frac{\frac{3}{5}x + \frac{5}{9}}{x + 1} = \frac{7}{12} $$ Cross-multiplying: $$ 12\left(\frac{3}{5}x + \frac{5}{9}\right) = 7(x + 1) $$ $$ 12\left(\frac{27x}{45} + \frac{25}{45}\right) = 7x + 7 $$ $$ 12\left(\frac{27x + 25}{45}\right) = 7x + 7 $$ $$ \frac{4(27x + 25)}{15} = 7x + 7 $$ $$ 4(27x + 25) = 15(7x + 7) $$ $$ 108x + 100 = 105x + 105 $$ $$ 3x = 5 $$ $$ x = \frac{5}{3} $$ Therefore, the ratio in which the mixtures should be mixed is $5:3$.

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Q17: The diluted wine contains only 8 litres of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many litres of mixture shall be replaced with pure wine if there was initially 32 litres of water in the mixture ?

A. 4

B. 5

C. 8

D. None of these

Correct Answer: B

Solution:
Initially, the mixture contains 8 liters of wine and 32 liters of water, totaling 40 liters.
The wine concentration is $8/40 = 20\%$.
Let $x$ liters of the mixture be replaced with pure wine.
This means $x$ liters of the 20% mixture is removed, and $x$ liters of pure wine (100% wine) is added.
The amount of wine removed is $0.20x$.
The amount of water removed is $0.80x$.
The remaining wine is $8 - 0.20x$.
The remaining water is $32 - 0.80x$.
After adding $x$ liters of pure wine, the new amount of wine is $(8 - 0.20x) + x = 8 + 0.80x$.
The total volume remains 40 liters.
The new concentration is 30%, so we have the equation: $$ \frac{8 + 0.80x}{40} = 0.30 $$ $$ 8 + 0.80x = 12 $$ $$ 0.80x = 4 $$ $$ x = \frac{4}{0.80} $$ $$ x = 5 $$ Therefore, 5 liters of the mixture must be replaced with pure wine.

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Q18: The shopkeeper mixed 40 kg refined oil with vegetable oil worth Rs. 60 per kg. Thus he gains Rs. 10 after selling the mixture of the two oils. The price of the first oil is :

A. 20

B. 25

C. 45

D. Cannot be determined

Correct Answer: D

Solution:
Let the price of refined oil be $x$ Rs/kg.
The shopkeeper mixed 40 kg of refined oil with vegetable oil.
Let's assume $y$ kg of vegetable oil was mixed.
Total cost price of the mixture = $40x + 60y$ Total quantity of the mixture = $40 + y$ kg Let the selling price of the mixture be $S$ Rs/kg.
Total selling price = $S(40+y)$ Profit = Total selling price - Total cost price = $S(40+y) - (40x + 60y) = 10$ We need more information to solve this problem.
The question is incomplete; it lacks information about the selling price ($S$) or the quantity of vegetable oil ($y$) mixed.
Therefore, a definitive answer cannot be calculated.
The question as stated is unsolvable.

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Q19: 4 kg of a metal contains 1/5 th copper and rest is zinc. Another 5 kg of metal contains 1/6th copper and rest is zinc. The ratio of copper and zinc into the mixture of these 2 metals is:

A. 49 : 221

B. 39 : 231

C. 94 : 181

D. None of these

Correct Answer: A

Solution:
Step 1: Calculate the amount of copper in 4 kg of metal: $\frac{1}{5} \times 4 \text{ kg} = 0.8 \text{ kg}$ copper.
The remaining amount is zinc: $4 \text{ kg} - 0.8 \text{ kg} = 3.2 \text{ kg}$ zinc.
Step 2: Calculate the amount of copper in 5 kg of metal: $\frac{1}{6} \times 5 \text{ kg} = \frac{5}{6} \text{ kg}$ copper.
The remaining amount is zinc: $5 \text{ kg} - \frac{5}{6} \text{ kg} = \frac{25}{6} \text{ kg}$ zinc.
Step 3: Calculate the total amount of copper in the mixture: $0.8 \text{ kg} + \frac{5}{6} \text{ kg} = \frac{4.8 + 5}{6} \text{ kg} = \frac{9.8}{6} \text{ kg} = \frac{49}{30} \text{ kg}$ Step 4: Calculate the total amount of zinc in the mixture: $3.2 \text{ kg} + \frac{25}{6} \text{ kg} = \frac{19.2 + 25}{6} \text{ kg} = \frac{44.2}{6} \text{ kg} = \frac{221}{30} \text{ kg}$ Step 5: Calculate the ratio of copper to zinc in the mixture: $\frac{49}{30} : \frac{221}{30} = 49 : 221$

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Q20: 450 litres of a mixture of milk and water contain the milk and water in the ratio 9 :1. How much water should be added to get a new mixture containing milkand water in the ratio 3:1?

A. 54

B. 90

C. 45

D. 63

Correct Answer: B

Solution:
Let the quantity of milk in the mixture be $9x$ liters and the quantity of water be $x$ liters.
Total quantity = $9x + x = 10x = 450$ liters Therefore, $x = 45$ liters.
Quantity of milk = $9x = 9 \times 45 = 405$ liters Quantity of water = $x = 45$ liters Let $y$ liters of water be added to the mixture.
The new ratio of milk to water is 3:1.
The quantity of milk remains the same (405 liters).
The quantity of water becomes $45 + y$ liters.
The ratio is given by: $$ \frac{405}{45 + y} = \frac{3}{1} $$ $$ 405 = 3(45 + y) $$ $$ 405 = 135 + 3y $$ $$ 3y = 405 - 135 $$ $$ 3y = 270 $$ $$ y = \frac{270}{3} $$ $$ y = 90 \text{ liters} $$ Therefore, 90 liters of water should be added.

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Q21: The ratio of petrol and kerosene in the container is 3: 2 when 10 litres of the mixture is taken out and is replaced by the kerosene, the ratio becomes 2: 3. The total quantity of the mixture in the container is :

A. 25

B. 30

C. 45

D. Cannot be determined

Correct Answer: B

Solution:
Let the quantity of petrol be $3x$ and kerosene be $2x$.
Total quantity = $5x$ When 10 liters of mixture is taken out, the amount of petrol removed is $\frac{3}{5} \times 10 = 6$ liters and kerosene removed is $\frac{2}{5} \times 10 = 4$ liters.
Remaining petrol = $3x - 6$ Remaining kerosene = $2x - 4$ 10 liters of kerosene is added.
New quantity of kerosene = $2x - 4 + 10 = 2x + 6$ New ratio of petrol to kerosene = $\frac{3x - 6}{2x + 6} = \frac{2}{3}$ Cross-multiplying: $3(3x - 6) = 2(2x + 6)$ $9x - 18 = 4x + 12$ $5x = 30$ $x = 6$ Total quantity = $5x = 5 \times 6 = 30$ liters

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Q22: From a container, 6 litres milk was drawn out and was replaced by water. Again 6 litres of mixture was drawn out & was replaced by the water. Thus the quantity of milk and water in the container after these two operations is 9 : 16. The quantity of mixture is :

A. 15

B. 16

C. 25

D. 31

Correct Answer: A

Solution:
Let the initial quantity of milk in the container be $x$ litres.
After the first operation, the quantity of milk remaining is $x - 6$ litres.
The total quantity remains $x$ litres.
The fraction of milk remaining is $\frac{x-6}{x}$.
After the second operation, 6 litres of the mixture is drawn out.
The amount of milk drawn out is $6 \times \frac{x-6}{x} = \frac{6(x-6)}{x}$ litres.
The amount of milk remaining after the second operation is $(x-6) - \frac{6(x-6)}{x} = (x-6)(1 - \frac{6}{x}) = (x-6)\frac{(x-6)}{x} = \frac{(x-6)^2}{x}$ litres.
The ratio of milk to water after two operations is 9:16.
The total quantity is still $x$ litres.
Therefore, the amount of milk is $\frac{9x}{9+16} = \frac{9x}{25}$ litres.
Equating the amount of milk remaining after two operations: $$ \frac{(x-6)^2}{x} = \frac{9x}{25} $$ $$ 25(x-6)^2 = 9x^2 $$ $$ 25(x^2 - 12x + 36) = 9x^2 $$ $$ 25x^2 - 300x + 900 = 9x^2 $$ $$ 16x^2 - 300x + 900 = 0 $$ Dividing by 4: $$ 4x^2 - 75x + 225 = 0 $$ This is a quadratic equation.
We can solve it using the quadratic formula or factorization.
Factorization yields: $$ (x-15)(4x-15) = 0 $$ Therefore, $x = 15$ or $x = \frac{15}{4}$.
Since the initial quantity must be greater than 12 (to allow for two 6-litre extractions), $x = 15$ litres.
The quantity of the mixture is $x = 15$ litres.

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