# Mixtures and Alligations Questions For SSC CHSL PDF

SSC Stenographer Constable Mixtures and Alligations Question and Answers download PDF based on previous year question paper of SSC Stenographer exam. 20 Very important Mixtures and Alligations questions for Stenographer Constable.

MIXTURES AND ALLIGATIONS QUESTIONS FOR SSC CHSL PDF

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**Question 1: **Ram purchased three variety of barley at Rs. 100 per kg, Rs. 140 per kg and Rs. 180 per kg respectively. In what ratio he should mix these three varieties in that order so that the average cost of the mixture is Rs. 150 per kg?

a) 2 : 4 : 7

b) 1 : 5 : 10

c) 1 : 2 : 3

d) 2 : 5 : 5

e) None of the above

**Question 2: **A milkman mixed water in the pure milk in the ratio of 2 : 3. If water is available at one-sixth of the price of pure milk then find out the profit percentage made by the milkman when he sold the entire mixture at the cost price of pure milk?

a) 25 percent

b) 33.33 percent

c) 50 percent

d) 40 percent

e) 66.66 percent

**Question 3: **A solution containing alcohol and water in the ratio 7 : 2 is mixed with twice the amount of another solution containing alcohol and water in the ratio 5 : 1. What is the percentage of alcohol in the final mixture?

a) 78.91%

b) 90%

c) 75.12%

d) 79.38%

e) 81.48%

**Question 4: **Two varieties of rice A and B are mixed in the ratio 4 : 5 to form a mixture X and in the ratio 3 : 4 to form another mixture Y. When X and Y are mixed in equal quantities, a new mixture Z is formed. When 3 kg of rice A is mixed in Z, the ratio of both the varieties of rice in Z becomes equal. What was the quantity of X and Y mixed to form Z?

a) 11.8125 kg

b) 13.225 kg

c) 18 kg

d) 19.75 kg

e) 10.1125 kg

**Question 5: **A solution containing milk and water in the ratio 3 : 1 is mixed with twice the amount of another solution containing milk and water in the ratio 2 : 1. What is the percentage of milk in the final mixture?

a) 77.11%

b) 69.44%

c) 57.2%

d) 56%

e) 60.36%

**Question 6: **A solution containing milk and water in the ratio 5 : 1 is mixed with 20 litres of milk. After removing 10.7 litres of the new mixture, the ratio of water and milk in the remaining mixture is found to be 3 : 25. What was the initial quantity of mixture?

a) 23.4 litres

b) 36 litres

c) 30.6 litres

d) 24 litres

e) 21.6 litres

**Question 7: **A milkman has some quantity of milk with him. He sells 20% of the milk and replaces it water. He realized that the mixture is too dilute and hence, he adds 20L of pure milk to the mixture. He then again sells 40% of the mixture and replaces it water. He adds 48L of pure milk to the mixture. If he now has 300L of pure milk with him then what is the quantity of the milk with which he started?

a) 520L

b) 550L

c) 500L

d) 480L

e) 450L

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**Question 8: **The cost of two different varieties of rice A and B Rs. 40/kg and Rs. 60/kg respectively. If A and B are mixed together in the ratio of 5 : 7 what will be the total cost of the mixture?

a) Rs. $\frac{155}{3}$/kg

b) Rs. 50/kg

c) Rs. 80/kg

d) Rs. 53/kg

e) Cannot be determined

**Question 9: **The cost of three different varieties of milk A, B, and C is Rs. 20/litre, Rs. 25/litre and Rs. 30/litre respectively. If A, B and C are mixed together in the ratio of 2 : 1 : 3, what will be the total cost of the mixture?

a) Rs. $\frac{155}{6}$/litre

b) Rs. 25/litre

c) Rs. $\frac{155}{12}$/litre

d) Rs. 29/litre

e) Cannot be determined

**Question 10: **The cost of three different varieties of rice A, B, and C is Rs. 20/kg, Rs. 25/kg and Rs. 30/kg respectively. If A, B and C are mixed together in the ratio of 2 : 1 : 2, what will be the cost of the mixture per kg?

a) Rs. 75

b) Rs. 25

c) Rs. 30

d) Rs. 125

e) Rs. 100

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**Question 11: **A mixture contains water and milk in the ratio of 3 : 5. If 4 litres of water and 5 litres of milk are added to the mixture, the ratio of water and milk becomes 2 : 3. What is the quantity of water in the final mixture?

a) 11 litres

b) 12 litres

c) 8 litres

d) 10 litres

e) 9 litres

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**Question 12: **A mixture contains water and alcohol in the ratio of 5 : 2. If 5 litres of water and 4 litres of alcohol are added to the mixture, the ratio of water and alcohol becomes 2 : 1. What is the quantity of alcohol in the final mixture?

a) 10 litres

b) 7 litres

c) 8 litres

d) 12 litres

e) Cannot be determined

**Question 13: **A mixture A contains milk and water in the ratio of 7 : 2. Another mixture B contains milk and water in the ratio of 2 : 1. If A and B are mixed in the ratio of 1 : 3, what is the ratio of milk and water in the final mixture?

a) 12 : 7

b) 23 : 17

c) 25 : 11

d) 29 : 11

e) 13 : 5

**Question 14: **A mixture P contains milk and water in the ratio of 5 : 1. Another mixture Q contains milk and water in the ratio of 3 : 2. If equal quantities of P and Q are mixed, what is the ratio of milk and water in the final mixture?

a) 19 : 7

b) 8 : 3

c) 43 : 17

d) 37 : 29

e) 1 : 1

**Question 15: **A mixture contains water and milk in the ratio 2 : 5. 2 litres of water is added to the mixture and the ratio of water and milk in the mixture becomes 1 : 2. How much more water must be added to the mixture so that the quantity of water and milk become equal in the mixture?

a) 10 litres

b) 16 litres

c) 3 litres

d) 6 litres

e) 12 litres

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**Question 16: **A mixture contains water and milk in the ratio 2 : 5. 2 litres of water is added to the mixture and the ratio of water and milk in the mixture becomes 1 : 2. What is the quantity of milk present in the mixture?

a) 20 litres

b) 10 litres

c) 15 litres

d) 25 litres

e) 5 litres

**Question 17: **A milkman has some milk with him. He sells 20% of it and mixes water to recover the milk he sold. He again sells 20% of the mixture and mixes water to recover the mixture he sold. Finally, he sells 50% of the mixture and mixes water to recover the mixture he sold. If he is now left with 96L of milk then what was the quantity of milk at the start?

a) 320

b) 300

c) 340

d) 280

e) 250

**Question 18: **A and B are mixed in 3:5 to obtain X. A and B are mixed in 7:4 to obtain Y. In what ratio should X and Y be mixed so that the resultant mixture has A and B in ratio 3:2?

a) 4:9

b) 2:9

c) 5:2

d) 3:4

e) 1:3

**Question 19: **Solutions A and B must be mixed in the ratio of 5:2 to make X. A, B and X must be mixed in a ratio of 3:5:2 to make Y. What is the ratio A:B in Y?

a) 30:37

b) 31:39

c) 15:17

d) 3:7

e) 4:5

**Question 20: **A milkman has 400L of milk. He sells x% of it and replaces it with water. He then sells 10% of the mixture and replaces it water. If the final mixture has 342L of milk and rest water then what is the value of x?

a) 9

b) 7

c) 5

d) 12

e) 10

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**Question 21: **A has 30% milk and rest water, B has 45% milk and rest water and C has 60% milk and rest water. They are mixed such that the final mixture has 50% milk. If they were mixed in A:B:C proportion then which of the following shows the correct relation between them?

a) 2C = B+4A

b) 3C = 2B+A

c) 2C+3A = 5B

d) 3A = 2C+B

e) C = A+B

**Question 22: **A shopkeeper buys some petrol and kerosene at 60 and 40 Rs respectively. He then mixes it and sells 1L of the mixture at the cost price of petrol and earns himself a profit of $\frac{40}{7}$ Rupees. In what ratio did he mix petrol and kerosene?

a) 5:2

b) 4:3

c) 2:1

d) 7:4

e) 3:2

**Question 23: **Solutions A and B must be mixed in the ratio 4:3 to obtain mixture X. Solutions A, B and mixture X must be mixed in the ratio 2:5:7 to obtain mixture Y. What should be the ratio of solutions A and B in mixture Y?

a) 4:3

b) 5:2

c) 2:5

d) 3:4

e) None of these

**Question 24: **Satish is coconut water seller. He mixes water and coconut water in the ratio 2:7. Then, he mixes water with this mixture in the ratio 2:5. He sells the mixture at 25 % more than the cost price of coconut water. What is the profit percentage of Satish? (Water is available free of cost)

a) 100%

b) 150%

c) 125%

d) 80%

e) None of the above

**Question 25: **A dishonest milkman mix water with milk, which costs Rs.35 per litre, to make a profit of 30%. If he sells the resulting mixture at Rs. 39 per litre find out what is the ratio of water and milk in the mixture sold by him ?(Water is available without any cost)

a) 1:3

b) 1:5

c) 1:4

d) 1:2

e) 1:6

**Question 26: **A dishonest shopkeeper mixes brick powder to chilli powder. If the cost price of the chilli powder is 10 times the cost price of the brick powder, the ratio in which chilli powder must be mixed with brick powder to realise a profit of 50% by marking up the price by 25% is

a) 4:1

b) 22:5

c) 23:6

d) 19:5

e) 21:4

**Question 27: **The ratio in which 2 metals A and B must be mixed to obtain alloy C is 7:3. The ratio in which the same 2 metals must be mixed to obtain alloy D is 4:5. In what ratio must the 2 alloys be mixed if the final mixture must contain the 2 metals in the ratio 3:2 respectively?

a) 3:20

b) 3:16

c) 20:7

d) 3:10

e) 14:9

**Question 28: **A milkman sets out to sell milk in a vessel that is filled to its brim. Once he sells 10% of the total volume of milk with him, he adds water to the mixture and fills the vessel. After he sells 20% of the remaining mixture, he again adds water and fills the vessel. What is the ratio of milk to water in the final mixture?

a) 25:8

b) 21:4

c) 13:7

d) 18:7

e) 11:5

**Question 29: **A shopkeeper buys some coffee and chicory. The cost price of coffee is 1.5 times the cost price of chicory. He mixes coffee and chicory in some ratio. He makes a profit of 20% by selling the mixture at the cost price of coffee. What is the ratio in which chicory and coffee was mixed by the shop keeper?

a) 2:1

b) 1:2

c) 2:3

d) 3:2

e) 1:1

**Question 30: **Solutions A and B must be mixed in the ratio 4:5 to obtain mixture X. Solutions A and B must be mixed in the ratio 7:2 to obtain mixture Y. In what ratio must mixtures X and Y be mixed if the final mixture must contain A and B in the ratio 26:19?

a) 3:2

b) 4:5

c) 5:2

d) 5:3

e) 3:1

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**Answers & Solutions:**

**1) Answer (D)**

Let us check the options.

Option (A): Average cost price of the mixture = $\dfrac{200+560+1260}{2+4+7}$ = Rs. 155.38

Option (B): Average cost price of the mixture = $\dfrac{100+700+1800}{1+5+10}$ = Rs. 162.50

Option (C): Average cost price of the mixture = $\dfrac{100+280+540}{1+2+3}$ = Rs. 153.33

Option (D): Average cost price of the mixture = $\dfrac{200+700+900}{2+5+5}$ = Rs. 150

Therefore, option D is the correct answer.

**2) Answer (C)**

Let ‘$\text{P}$’ be the cost price of pure milk per liter. Therefore, cost price of water = $\dfrac{\text{P}}{6}$.

Let us assume that milkman mixed $\text{2L}$ and $\text{3L}$ volume of water and pure milk.

Hence, total cost incurred by the milkman = $2\text{L}*\dfrac{\text{P}}{6} + 3\text{L*P}$ = $\dfrac{10\text{LP}}{3}$

Profit made by the milkman in the transaction = $5\text{L*P} – \dfrac{10\text{LP}}{3}$ = $\dfrac{5\text{LP}}{3}$

Therefore, the profit percentage made by the milkman = $\dfrac{\dfrac{5\text{LP}}{3}}{\dfrac{10\text{LP}}{3}}\times 100$ = 50 percent.

Therefore, option C is the correct answer.

**3) Answer (E)**

Let 9 litres of the first solution be mixed with 18 litres of the second solution.

Therefore, total volume of the solution = (9 + 18) litres = 27 litres

Quantity of alcohol in the first solution = $\frac{7}{9} * 9$ litres = 7 litres

Quantity of alcohol in the second solution = $\frac{5}{6} * 18$ litres = 15 litres

Total quantity of alcohol = (7 + 15) litres = 22 litres

% of milk = $\frac{22}{27} * 100$% = 81.48%

Hence, option E is the correct answer.

**4) Answer (A)**

Let there be 4k kg of rice A and 5k kg of rice B in mixture X and

let there be 3m kg of rice A and 4m kg of rice B in mixture Y

Total quantity of X = (4k + 5k) kg = 9k kg

Total quantity of Y = (3m + 4m) kg = 7m kg

When X and Y are mixed in equal quantities to form Z,

Quantity of rice A in Z = (4k + 3m) kg

Quantity of rice B in Z = (5k + 4m) kg

It is given that, 4k + 3m + 3 = 5k + 4m

or, k + m = 3……………(i)

Also, Quantity of X = Quantity of Y

=> 9k = 7m…………….(ii)

On solving (i) and (ii), we get

k = $\dfrac{21}{16}$ and m = $\dfrac{27}{16}$

Quantity of X and Y mixed = 9k or 7m = 11.8125 kg

Hence, option A is the correct answer.

**5) Answer (B)**

Let 12 litres of the first solution be mixed with 24 litres of the second solution.

Therefore, total volume of the solution = (12 + 24) litres = 36 litres

Quantity of milk in the first solution = $\frac{3}{4} * 12$ litres = 9 litres

Quantity of milk in the second solution = $\frac{2}{3} * 24$ litres = 16 litres

Total quantity of milk = (9 + 16) litres = 25 litres

% of milk = $\frac{25}{36} * 100$% = 69.44%

**6) Answer (B)**

Let the milk and water in the original mixture be 5$k$ and $k$.

After adding 20 litres of milk, the ratio of milk and water changed to 25 : 3

(Removing 10.7 litres of the mixture is irrelevant as it will not change the ratio of milk and water in the mixture)

So, $\frac{5k + 20}{k} = \frac{25}{3}$

On solving, we get $k$ = 6

Total volume of the initial mixture = $5k + k$ = 36 litres

Hence, option B is the correct answer.

**7) Answer (C)**

Let the quantity of milk with the milkman be xL. He sells 20% of it and thus he is left with 0.8x.

He adds 20L of pure milk and thus. he now has 0.8x+20 L of milk with him.

Out of this he sells 40% of the milk.

He is now left with (0.8x+20)*0.6 = 0.48x+12 L of milk.

He adds 48L of pure milk and he now has 0.48x+12+48 = 0.48x+60 L of milk.

Given, 0.48x+60 = 300

Thus, x = 240/0.48 = 500L

Hence, option C is the correct answer.

**8) Answer (E)**

As the total quantity of the mixture is not known, we cannot calculate the total cost of the mixture.

Hence, option E is the correct answer.

**9) Answer (E)**

As the total quantity of the mixture is not known, we cannot calculate the total cost of the mixture.

Hence, option E is the correct answer.

**10) Answer (B)**

Let 2 kg of A, 1 kg of B and 2 kg of C are mixed together.

Total rice = (2 + 1 + 2) kg = 5 kg

Cost of A in the mixture = Rs. (2 * 20) = Rs. 40

Cost of B in the mixture = Rs. (1 * 25) = Rs. 25

Cost of C in the mixture = Rs. (2 * 30) = Rs. 60

Total cost of the mixture = Rs. (40 + 25 + 60) = Rs. 125

Cost per kg = Rs. $\frac{125}{5}$ = Rs. 25

Hence, option B is the correct answer.

**11) Answer (D)**

Let the quantity of water and milk be $3x$ and $5x$

It is given that, $\frac{3x + 4}{5x + 5} = \frac{2}{3}$

On solving, we get $x$ = 2

Therefore, the quantity of water in the final mixture = (3 * 2 + 4) litres = 10 litres.

Hence, option D is the correct answer.

**12) Answer (A)**

Let the quantity of water and alcohol be $5x$ and $2x$

It is given that, $\frac{5x + 5}{2x + 4} = \frac{2}{1}$

On solving, we get $x$ = 3

Therefore, the quantity of alcohol in the final mixture = (3 * 2 + 4) litres = 10 litres.

Hence, option A is the correct answer.

**13) Answer (C)**

As A and B are mixed in the ratio of 1 : 3, let us assume that 9 litres of A and 27 litres of B are mixed.

In 9 litres of A, milk is 7 litres, and water is 2 litres.

In 27 litres of B, milk is 18 litres, and water is 9 litres.

When A and B are mixed, quantity of milk = (7 + 18) litres = 25 litres

When A and B are mixed, quantity of water = (2 + 9) litres = 11 litres

Required ratio is 25 : 11

Hence, option C is the correct answer.

**14) Answer (C)**

Let us assume that 30 litres each of P and Q are mixed.

In 30 litres of P, milk is 25 litres and water is 5 litres.

In 30 litres of Q, milk is 18 litres and water is 12 litres.

When P and Q are mixed, quantity of milk = (25 + 18) litres = 43 litres

When P and Q are mixed, quantity of water = (5 + 12) litres = 17 litres

Required ratio is 43 : 17

Hence, option C is the correct answer.

**15) Answer (A)**

Let the quantity of water and milk be $2x$ and $5x$

It is given that, $\frac{2x + 2}{5x} = \frac{1}{2}$

On solving, we get $x$ = 4

Therefore, the quantity of milk = 5 * 4 = 20 litres

and the quantity of water = 2 * 4 = 8 litres

Quantity of water required to be added = Difference between the quantities of milk and water = (20 – 8) litres = 12 litres

But, 2 litres of water is already added.

So, 10 litres more water needs to be added.

Hence, option A is the correct answer.

**16) Answer (A)**

Let the quantity of water and milk be $2x$ and $5x$

It is given that, $\frac{2x + 2}{5x} = \frac{1}{2}$

On solving, we get $x$ = 4

Therefore, the quantity of milk = 5*4 = 20 litres.

Hence, option A is the correct answer.

**17) Answer (B)**

Let the quantity of milk with the milkman be X.

He sells 20% of it and mixes water to recover the milk he sold thus, the quantity of milk with him now is = 0.8X

He again sells 20% of the mixture and mixes water to recover the mixture he sold thus, the quantity of milk with him now is = 0.8*0.8X=0.64X

Finally, he sells 50% of the mixture and mixes water to recover the mixture he sold thus, the quantity of milk with him now is = 0.64*0.5X = 0.32X

Given 0.32X = 96

Hence, X = 300L.

Hence, option B is the correct answer.

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**18) Answer (B)**

let quantity of X mixed be 8x and quantity of Y mixed be 11y.

Therefore amount of A and B in X and Y is 3x, 5x and 7y, 4y respectively.

If A:B = 3:2 then,

$ \dfrac{3x+7y}{5x+4y} = \dfrac{3}{2} $

$ 6x+14y = 15x+12y $

$ i.e. 2y = 9x $

Therefore, $ x:y = 2:9 $

Hence, option B is the correct answer.

**19) Answer (B)**

Let’s consider there is 14L of X. Out of that there will be 10L of A and 4L of B. In the final mixture if X is 2 parts out of 10 then 14L is 2 parts out of 10 and hence total mixture is 70L. Quantity of A in final mixture added directly is 3*70/10 = 21L, Quantity of B in final mixture added directly is 5*70/10 = 35L. Therefore, total quantity of A and B in final mixture is 21+10= 31L and 35+4 = 39L.

Therefore, ratio of A:B = 31:39.

Hence, option B is the correct answer.

**20) Answer (C)**

When he sells x% of mile he is selling 400*x/100 amount of milk = 4x i.e. after selling 4x milk he is left with 400-4x milk. He then sells 10% of the mixture i.e. now he has 90% of mixture in which milk is (400-4x)*0.9. This is the amount of milk in the final mixture. Therefore, (400-4x)*0.9 = 342

i.e. 400-4x = 342/0.9

i.e. 400-4x = 380

i.e. 4x = 20 and x = 5

Hence, option C is the correct answer.

**21) Answer (A)**

If they are mixed in A:B:C then,

$\dfrac{0.3A+0.45B+0.6C}{A+B+C} = \dfrac{1}{2} $

$ 0.6A+0.9B+1.2C = A+B+C $

$ 2C = B+4A $

Hence, option A is the correct answer.

**22) Answer (A)**

Let the ratio of petrol to kerosene mixed be a:b. Let be a litre the quantity of the petrol mixed and hence, b litres will be the quantity of the kerosene mixed. Cost price of a+b litres of the solution will be $60a+40b$. He is selling the solution at the cost of the petrol, i.e. the cost price of a+b litres of the solution will be $60a+60b$. He is earning a profit of Rs $\frac{40}{7}$ i.e. $60a+60b$+$60a+40b$=$\frac{40}{7}$

Therefore, b = $\frac{2}{7}$ and hence a will be $\frac{5}{7}$.

Therefore, $a:b$ = $5:2$.

Therefore, our answer is option ‘A’.

**23) Answer (D)**

Let us assume the volume of mixture X = 7a ml.

Solution A in mixture X = $\frac{4}{3+4}$*7a = 4a ml …(1)

Solution B in mixture X = $\frac{3}{3+4}$*7a = 3a ml …(2)

Let us assume the volume of mixture Y = 14a ml.

Net quantity of solution A in mixture Y = Part that’s directly added + quantity of solution A from mixture X

$\Rightarrow$ $\frac{2}{2+5+7}$*14 + 4a = 2a + 4a = 6a ml

Net quantity of solution B in mixture Y = Part that’s directly added + quantity of solution B from mixture X

$\Rightarrow$ $\frac{5}{2+5+7}$*14 + 3a = 5a + 3a = 8a ml

Ratio of solution A and B in mixture Y = 6a : 8a = 3:4

**24) Answer (C)**

Let assume Satish had 700 ml of coconut water with himself for which he paid 70 rupees.

cost price incurred = 700/70 =10 ml/rupee

First, he will mix water in the ratio of 2: 7 hence he will mix 200 ml of water. So total quantity of mixture will be = 700+200 = 900 ml

Now he mixes water in this mixture in the ratio of = 2 : 5

So the quantity of water mixed in this mixture = $\frac{2}{5}*900$ = 360 ml

So net quantity of mixture that’s to be sold at 25% profit = 900+360 = 1260 ml

Selling price = $\frac{100}{100+25} * 10$ = 8 ml / rupee

So revenue generated on selling entire mixture = 1260 / 8 = 157.5 rupees

Profit percentage = $\frac{157.5-70}{70} *100$ = 125 %

**25) Answer (E)**

Selling price of mixture = 39 per litre

To gain 30% on this mixture this mixture should cost him = $\frac{100}{100+30} * 39$ = 30 per litre

By the rule of alligation

Hence the milkman should mix water and milk in the ratio of 1 : 6 in order to gain 30% profit.

**26) Answer (B)**

Let the cost price of chilli powder be ‘c’.

The cost price of brick powder = 0.1c.

Let us assume that chilli powder and brick powder are mixed in the ratio x:1.

Total amount of mixture = x + 1 kg.

Total cost incurred by the shopkeeper = xc + 0.1c

Selling price of the mixture = (x+1)*1.25*c

Profit realised = 50%

=> 1.5*(xc + 0.1c) = (x+1)*1.25c

1.5xc + 0.15c = 1.25xc + 1.25c

0.25xc = 1.1c

0.25x = 1.1

=> x = 4.4

Therefore, the ratio in which chilli powder and brick powder must be mixed = 4.4:1 = 22:5.

Therefore, option B is the right answer.

**27) Answer (E)**

Concentration of metal A in alloy C = 7/(3+7) = 7/10.

Concentration of metal A in alloy D = 4/(4+5) = 4/9.

Let the ratio in which alloys C and D are mixed be x:y.

(7x/10 + 4y/9)/(x+y) = 3/5

63x + 40y = 54x + 54y

9x = 14y

x/y = 14/9.

Therefore, option E is the right answer.

**28) Answer (D)**

After 10% of the milk is sold, the quantity of milk remaining will be 90 litres.

After 20% of the mixture is sold, the quantity of milk remaining in the vessel will be 0.8*90 = 72 litres.

Ratio of milk to water = 72:28 = 18:7.

Therefore, option D is the right answer.

**29) Answer (E)**

Let us assume the price of chicory to be P.

Price of coffee = 1.5p

Let us assume that coffee and chicory are mixed in the ratio x:y.

We know that he sells the mixture at the cost price of coffee and realises a profit of 20%.

=> Profit = 1.5p(x+y) – 1.5px – py = 0.5py

Cost = 1.5px + py

Profit %age = 0.5py/(1.5px+py)

0.5y/(1.5x + y) = 0.2

0.5y = 0.3x + 0.2y

0.3x = 0.3y

x/y = 1:1

Therefore, option E is the right answer.

**30) Answer (A)**

Let the quantity of mixture X be 9x.

Amount of A in mixture X = 4x

Amount of B in mixture X = 5x

Concentration of A = 4/9

Let the quantity of mixture Y be 9x.

Amount of A in mixture Y = 7x

Amount of B in mixture Y = 2x

Concentration of A = 7/9

Let us assume that the 2 mixtures are mixed in the ratio a:b to obtain a mixture which has A and B in the ratio 26:19.

=> Concentration of A = 26/(26+19) = 26/45.

Let us assume that the 2 mixtures are mixed in the ratio a:b.

(4a/9 + 7b/9)/(a+b) = 26/45

(4a+7b)/(9(a+b)) = 26/45

20a + 35b = 26a + 26b

6a = 9b

=> a/b = 3:2.

Therefore, option A is the right answer.

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