Preparing for the CAT exam can be a challenging task, especially when it comes to tackling the Quantitative Aptitude section. Among the various topics that are covered in this section, ratio and proportion play a crucial role in getting high score in Quant. To help students prepare for this topic, we have gathered the top 20 Ratio and Proportion Questions to practice along with the detailed solutions which are provided in this article. One can also download these questions in a PDF format for future reference. Click on the link given below to download the top 20 CAT questions on ratio and proportion PDF with detailed solutions.

Download Top 20 Ratio And Proportion Questions For CAT

**Question 1:Â **Three vessels, A, B and C, contain a soda and water solution. The ratio of soda and water in vessel A and vessel B are 2:3 and 1:4. After mixing B and C in the ratio of 1:2, the resulting mixture contains the soda and water in the ratio of 4:11. Find the percentage of water in a mix which has A, B and C combined in a ratio of 3:5:2

a)Â 60%

b)Â 62%

c)Â 70%

d)Â 72%

**1)Â AnswerÂ (D)**

**Solution:**

Let us assume the amount of soda per litre mixture of C = x

When B and C are mixed in the ratio of 1:2, the ratio of soda and water becomes = 4:11

Since B and C are mixed in a ratio of 1: 2, assuming the volume of B = 5 litres, then the volume of C = 10 litres

B contains soda and water in a ratio of 1:4 (Hence soda in 5 litres of B = 1 litre)

Soda in 10 litres of C will be = 10x

Hence total soda in the mixture of B and C = 10x+1

It will be equal to the soda in water ( 15 litres of mix B and C) =Â $\ \frac{4}{15}\times\ 15$ = 10x+1

Hence x = 0.3

Now A, B and C are combined in 3:5:2 ( Let us assume volume 30, 50 and 20 litres, respectively)

So the amount of water in A = $\ \frac{3}{5}\times\ 30$ = 18 litres

Similarly, the amount of water in B = $\ \frac{4}{5}\times\ 50$ = 40 litres

The amount of water in C = (1-0.3)20 = 14 Litres

Hence, the total amount of water in (30+50+20) = 100 litres mixture = 18+40+14 = 72 litres

Hence, the percentageÂ of water in the mix= 72%

**Question 2:Â **A contractor lends out a rock breaker machine for 1400 rupees per hour. The machine uses 10 litres of diesel per hour, costing 80 rupees per litre and has an operating cost of 200 rupees per hour. The contractor discovers that if he mixes kerosene, costing 40 rupees per litre, with diesel, he can get the same efficiency, but the operating cost will increase by 50 per cent. Find the ratio in which kerosene needs to be mixed with diesel if he intends to earn a profit per cent 70 per cent more than he was making before.

a)Â 7:3

b)Â 2:1

c)Â 4:3

d)Â 3:1

e)Â e

**2)Â AnswerÂ (B)**

**Solution:**

Total cost incurred initially = 80*10 + 200 = 1000 rupees, where he spends 800 on diesel and 200 on operations.

Profit made initially was 1400 – 1000 = 400 rupees.

Profit per cent =Â $\ \frac{\ 400}{1000}\times\ 100=40\%$

New profit per cent was 70%Â more than previous.

New profit % = 40*1.7 = 68%

The contractor will make 68% profit when he will mix kerosene with diesel.

New cost price = 1400/1.68 = 833.34

New operating cost will be 50 percent more than previous = 200*1.5= 300 rupees

Amount spent on mixture of kerosene and diesel = 833.34 -300 = 533.34 rupees.

Price of of diesel = 80 rupees per liter

Price of kerosene = 40 rupees per liter

Let quantityÂ of kerosene and diesel used be x and y respectively.

x + y = 10

40x + 80y = 533.34

On solving the above equations we can get x:y as 2:1

**Question 3:Â **The Municipal Authority of Nagaur district maintains the population record of the district, and every year it is updated. For the year 2020, females were 35% of the total population. In 2021, the Total population increased by 60%, and the population of males to that of females can be expressed in the ratio of a:b. The 2022 population was further increased by 50%, and the population of males to that of females can be expressed in the ratio of a:b. In 2022 the total population was 50,400. Which of the following cannot be possible?

a)Â Males(2021) = 12,600 and Females(2022) = 31,500

b)Â Males(2021) = 13440 and Females(2022) = 30,240

c)Â Males(2021) = 21,000 and Females(2022) = 18,900

d)Â Males(2021) = 11,200 and Females(2022) = 33,500

**3)Â AnswerÂ (D)**

**Solution:**

Let the total population in 2020 be 100x

Population in 2021(increase of 60%) = 160x

Population in 2022 (an increase of 50%) = 240x

It is given that 240x = 50400

Therefore x= 210 So, the Population in 2021 is = 33,600 and Population in 2022 = 50,400

The ratio of males to females in 2021 and 2022 is the same = a:b

From this, we calculate the possible values of a:b from the information given in option

Option(A) – Males(2021) = 12,600 and Females(2022) = 31,500

Males 2021 = 12600, then females (2021) = 21,000

So a:b (2021) = 3:5

Females (2022) =31,500 then males (2022) = 18900

So a:b(2022) = 3:5

It is possible

Similarly, we check for options B, C and D

Option D cannot be possible

=> Males(2021) = 11,200 and Females(2022) = 33,500

Males(2021) = 11,200 then females(2021) = 22,400

So a:b (2021) = 1:2

Females(2022) = 33,500 (This is incorrect as for males : females = 1:2 in 2022 number of females should be 33,600)

Thus, Option D is not possible.

**Question 4:Â **The ratio of the monthly incomes of two people is 7:3, while their monthly expenses are in a ratio of 5:2. If both of them save 2000 rupees monthly, find their total monthly income.

**4)Â Answer:Â 60000**

**Solution:**

Let their incomes be 7x and 3x and expenses be 5y and 2y.

7x – 5y = 2000

3x – 2y = 2000

7x – 5y = 3x – 2y

4x = 3y

2y =Â $\frac{8x}{3}$

3x – $\frac{8x}{3}$ = 2000

x = 6000

Total income = 7x + 3x = 10x = Rs 60000

The answer is 60000.

**Question 5:Â **Chetan and Ajay are two employess of ABC Pvt Ltd. They are working on the same type of project. If Chetan’s work efficiency is, two-thirds of Ajay and Ajay’s working hours are three-fifths more than Chetan’s. ABC Pvt Ltd. develops a tool to identify the productivity of employees, where the productivity is given as

Productivity =Â $k\ \times\ \left(working\ hours\right)^2\times\ \left(workÂ efficiency\right)$

What is the ratio of the productivity of Ajay to that of Chetan?

a)Â 50:192

b)Â 96:25

c)Â 81:49

d)Â 49:81

**5)Â AnswerÂ (B)**

**Solution:**

Let the Ajay’s work efficiency be 3x, Chetan’s work efficiency will be 2x

Working hours of Chetan be 5t, Ajay’s working hours will be 8t

Productivity =Â $k\ \times\ \left(working\ hours\right)^2\times\ \left(work efficiency\right)$

=> Ajay’s productivity =Â $k\ \times\ \left(8t\right)^2\times\ \left(3x\right)$

=> 192ktx

=> Chetan’s productivity = $k\ \times\ \left(5t\right)^2\times\ \left(2x\right)$

=> 50ktx

Ratio of productivity of Ajay and Chetan = 192:50

=> 96:25

**Question 6:Â **The cost of fish varies directly as the cube of its weight. If the fish is cut into three pieces with weights in the ratio of 2:2:1, then the fishmonger will get 43200 rupees less after selling all the pieces. Find the original price of the whole fish.

**6)Â Answer:Â 50000**

**Solution:**

Let the weight of the pieces of fish be 2x,2x, and x kg, respectively.

The weight of the whole fish is (2x+2x+x) = 5x kg.

Let theÂ costÂ of the whole fish be C.

It is given,Â the cost of fish varies directly as the cube of its weight.

Therefore, the cost of the whole fish $C=k\left(5x\right)^3\ =>\ 125kx^3$

Similarly,

The cost of the first pieceÂ $\left(C_1\right)$ =Â $k\left(2x\right)^3\ = 8kx^3$

The cost of the second pieceÂ $\left(C_2\right)$ =Â $8kx^3$

The cost of the third pieceÂ $\left(C_3\right)$ =Â $kx^3$

The cost of the whole fish after cutting it into pieces will be =Â $8kx^3+8kx^3+kx^3\ =17kx^3$

It is given that,

$125kx^3-17kx^3=43200$

$108kx^3$ = 43200

Therefore,Â $kx^3$ = 400

The original cost of whole fish =Â $125kx^3$ = 125*400 = 50000

The answer is 50000.

**Question 7:Â **An alloy of the steelÂ contains 3 different grades of iron. The first alloy has the 3 grades in the ratio 5:3:6, andÂ the second alloy has them in the ratio 2:3:1. If two alloys are mixed and the final ratio of 3 grades of iron is x:3:5. Find x, if the weight of the first alloy is 280 kgs.

a)Â $\dfrac{22}{3}$

b)Â $\dfrac{25}{3}$

c)Â $\dfrac{22}{5}$

d)Â $\dfrac{25}{4}$

**7)Â AnswerÂ (C)**

**Solution:**

Let the weight of the second alloy be ‘y’.

The weight of the first alloy is = 280kgs.

The composition of different grades =Â $\dfrac{5}{14}\times280=100,\ \dfrac{3}{14}\times\ 280=60,\ \dfrac{6}{14}\times\ 280=120$

Let the composition of different grades in second alloy be 2x, 3x and x

Two alloys are mixed and the resultant composition of different grades is 100+2x, 60+3x, 120+x

It is given,

$\ \dfrac{\ 60+3x}{120+x}=\dfrac{3}{5}$

300 + 15x = 360 + 3x

12x = 60

x = 5

Two alloys are mixed and the resultant composition of different grades is 110, 75, 125

Ratio = 110:75:125 =Â $\frac{110}{25}$ : 3 : 5

$x$ =Â $\dfrac{110}{25}$ =Â $\dfrac{22}{5}$

The answer is option C.

**Question 8:Â **There are 4 stations on a rectangular railway track P, Q, R and S. 2 trains, A and B, run round trips on this track. Train A is thrice as fast as B. On a particular day, train A starts from station P in a clockwise direction, and train B starts from station R in an anti-clockwise direction. After completing 6 round trips, train A arrives at station Q at the same time as train B arrives there after completing 2 round trip. Find the ratio of length and breadth of the railway track.

a)Â 3:1

b)Â 5:2

c)Â 4:3

d)Â 7:2

**8)Â AnswerÂ (A)**

**Solution:**

Let us assume Length and breadth of the rectangle to be L and B respectively.

Let the speed of Trains A and B be 3S and S respectively.

Since Both the trains have arrived at the same time, we can say that distance covered by them upon their respectiveÂ speed should be equal.

Distance covered by A = 12 *(L +B) + L

Distance covered by B = 4(L+B) + B

$\ \ \frac{\ 13L+12B\ \ }{3S}$ =Â $\ \ \frac{\ 4L+5B\ \ }{S}$

13L + 12B = 12L + 15B

L= 3B

Hence, Option A is the correct answer.

**Question 9:Â **Manish has a 150-litreÂ alcohol andÂ waterÂ solution in a ratio of $2\ :\ x$, respectively, where x is a positive integer. It is mixed with y litres of a solution having 44.44% alcohol concentration to get a solution having 40% alcohol concentration. IfÂ $4<\ x<12$, andÂ the amount ofÂ water present in bothÂ initial quantities is a whole number, find the value of xy.

a)Â 4050

b)Â 5400

c)Â 5670

d)Â 5832

**9)Â AnswerÂ (B)**

**Solution:**

It is given that the amount of water inÂ bothÂ initial quantities is a whole number.

Hence $\frac{150x}{2+x}$ is a whole number, and the range of x is (4, 12).

Only one value is possible for x.(i.e. x = 8).

If x = 8

$\frac{150x}{2+x}=\frac{150\times\ 8}{2+8}=120$.

Only one value is possible for x.(i.e. x = 8).

150 L acid solution is mixed with y L to get 40% alcohol concentration.

$\frac{150\times\ \frac{2}{10}+\frac{4y}{9}}{150+y}=0.4$

$30+\frac{4y}{9}=60+0.4y$

$\frac{0.4y}{9}=30$

$y=675\ L$

The initial volume of water in y litres =Â $\frac{5}{9}\times\ 675=375$.

Then, $xy=8\times\ 675=5400.$

The answer is Option (B).

**Question 10:Â **A, B and C are three containers of pure milk coffee with volumes 100L, 200L and 300L. From container A, 20L of the solution is replaced with water. From container B, 20L of the solution is replaced with water twice. From container C, 30L of the solution is replaced with water thrice. After all these operations, if the solutions present in containers A, B and C are mixed in the ratio 5:3:2, what is the percent of pure milk coffee in final mixture?

a)Â 75.26%

b)Â 76.82%

c)Â 77.44%

d)Â 78.88%

**10)Â AnswerÂ (D)**

**Solution:**

Final milk coffee quantity in container A = 100-20 = 80L

Milk coffe percentage in container A = 80%

Final milk coffee quantity in container B = $200(0.9)^2$ = 162L

Milk coffee percentage in container B = 81%

Final milk coffee quantity in containder C = $300(0.9)^3$ = 218.7L

Milk coffee percentage in container C = 72.9%

Contents in containers A, B and C are mixed in the ratio 5:3:2

% of milk coffee in final solution =Â $\ \frac{\ 5\left(80\right)+3\left(81\right)+2\left(72.9\right)}{5+3+2}=\ \frac{\ 400+243+145.8}{10}=78.88\%$

The answer is option D.

**Question 11:Â **In Emerald society, there areÂ ‘m’ number of people.Â TheÂ ratio of the number of males to the number of people who don’t read paper is 7:5, and the ratio of the number of females to people who read paper is 4:5. If there are 2 males who read paper for every female who doesn’t read. Find the square of minimum possible value of m.

a)Â 225

b)Â 400

c)Â 144

d)Â 256

**11)Â AnswerÂ (A)**

**Solution:**

Let

The number of males who read = a

The number of males who don’t read = b

The number of females who read = c

The number of females who don’t read = d

From the given information,

$\frac{a+b}{b+d}=\frac{7}{5}$

$5a=2b+7d$…..(1)

$\frac{c+d}{a+c}=\frac{4}{5}$

$c+5d=4a$…..(2)

$\frac{a}{d}=\frac{2}{1}$

$a=2d$…..(3)

From (3) and (1)

We get, 10d = 2b + 7d

3d = 2b

From (2) and (3), we get

8d = 5d + c

c = 3d.

Let d = k.

Then, $d=\frac{c}{3}=\frac{2b}{3}=\frac{a}{2}=k$

$m=a+b+c+d=2k+\frac{3k}{2}+3k+k=\frac{15k}{2}$

$\frac{15k}{2}$ should be a natural number; for that, k has to be a multiple of 2.

Hence min value of k is 2.

The minimum value of m = 15.

$m^2=15^2=225$

The answer is Option (A).

**Question 12:Â **Product Chlorobenzene is produced by mixing Benzene and Chlorine in a ratio of 7:5. Benzene is prepared by mixing two raw materials, Phenol and Nacl, in a ratio of 2:3.Â Chlorine is prepared by mixing two raw materials, Nacl and Kcl, in the ratio of 1:4. Then, the final mixture of Chlorobenzene and Ethanol is prepared by mixing 900 units of Chlorobenzene with Ethanol. Suppose the concentration of the raw material Nacl in the final mixture is 30%. How much Ethanol had been added to Chlorobenzene?

a)Â 360 units

b)Â 420 units

c)Â 450 units

d)Â 400 units

**12)Â AnswerÂ (D)**

**Solution:**

First, Chlorobenzene is produced by mixing Benzene and Chlorine in 7:5. Hence, 7/12th of

product is Benzene, and 5/12th of the product is Chlorine.

Now, Benzene has Phenol and Nacl in the ratio of 2:3, which implies that 3/5th of Benzene is

Nacl. Therefore, a fraction of Nacl in the product of Chlorobenzene from Benzene

= (7/12Ã—3/5) = 21/60

Similarly, the chemical Chlorine has Nacl and Kcl in the ratio of 1:4, which implies 1/5th of Chlorine is

Nacl. Therefore, a fraction of Nacl in the product of Chlorobenzene from Chlorine = (5/12Ã—1/5) = 5/60

So, the total fraction of Nacl in Chlorobenzene = (21/60+5/60) = 26/60

Since there is no Nacl in Ethanol, that means all Nacl in the final mixture is obtained by the Chlorobenzene mixture.

The amount of NaCl in final mixture = 900Ã—26/60 = 390 units

It is also given that the concentration of Nacl in the final mixture is 30% which implies:

30% of the Final mixture = 390 units

30/100 (Final mixture)=390 units Final mixture=390Ã—100/30 = 1300 units.

Ethanol added in the final mixture = (1300-900) = 400 units.

**Question 13:Â **Two samples A and B consist of Ethylene & Propane in the ratio of 7:9 for sample A & 9:7 for sample B. The two samples A and B are mixed in the ratio of 2:3 to create a new sample C.Â FromÂ sample C, some part of ethylene is replaced with propane, resulting in a 1:1 Ethylene to Propane ratio. If the replaced part is x% of the total solution, what is the value of 100x?

**13)Â Answer:Â 125**

**Solution:**

Consider sample A- Ratio of Ethylene & Propane = 7:9

Proportion of Ethylene=$\ \frac{\ 7}{16}$ , Proportion of Propane =Â $\ \frac{\ 9}{16}$

Mixed in ratio of 2:3 so by taking 2 parts , Ethylene = 14/16 & Propane = 18/16

Consider sample B- Ratio of Ethylene & Propane = 9:7

Proportion of Ethylene=$\ \frac{\ 9}{16}$ , Proportion of Propane = $\ \frac{\ 7}{16}$

Mixed in ratio of 2:3 so by taking 3 parts , Ethylene = 27/16 & Propane = 21/16

Adding Ethylene =Â $\frac{\ \left\{\ \frac{14}{16}+\ \frac{27}{16}\right\}\ }{5}$ = 41/80

Similarly, Propane = 39/80

There are 41 units of Ethylene and 39 units of Propane in the solution.

1 unit from Ethylene should be replaced with propane to get 1:1 ratio.

$\frac{x}{100}\times80=1$

x = 1.25

100x = 125

The answer is 125.

**Question 14:Â **The amount of money with Radhey and Shyam is in the ratio of 5:6. Both of themÂ invested some amount in the stock market. The amount invested by Radhey was 150% of the amount invested by Shyam. They both earned a profit of 25% on their respective investments. If the final amount of money with Radhey is equal to the initial amount with Shyam, what is the final ratio of the money with them?

a)Â 1:1

b)Â 9:10

c)Â 4:5

d)Â 7:8

**14)Â AnswerÂ (B)**

**Solution:**

Let the amount of money with Radhey and Shyam be 15x and 18x,

respectively. Also, let the amount invested by Shyam be 4y. Thus, the

amount invested by Radhey will be 6y.

Now, both got a return of 25% on their investments.

Thus, the final amount received by Radhey will be $15x-6y+\frac{5}{4}\left(6y\right)\ =\ 15x-\frac{3}{2}y$

Also, this amount is equal to the initial investment by Shyam.

Thus, $\ 15x\ +\frac{3y}{2}=\ 18x$

=> $\ \frac{3}{2}y=\ 3x$

=> $y=2x$

Thus, amount invested by Radhey and Shyam will be 12x and 8x, respectively.

Also, total amount which was not invested will be 3x and 10x.

Now, after 25% profit, the total amount with Shyam = $\frac{5}{4}\left(8x\right)+10x\ =\ 20x$

The required ratio = 18x : 20x = 9:10

The correct option is B.

**Question 15:Â **There are two cans A and B with a capacity of 40 litres and 30 litres, respectively. The can A have 60% alcohol, and the can B have 46.67% alcohol. If 15 litres from can AÂ and 12 litres from can B are exchanged between both the cans, what isÂ the final ratio ofÂ alcohol content present in can A to can B?

a)Â $\frac{103}{87}$

b)Â $\frac{124}{83}$

c)Â $\frac{78}{63}$

d)Â $\frac{110}{81}$

**15)Â AnswerÂ (A)**

**Solution:**

Can A has 60% alcohol content, andÂ 15 litres have been removed.

Amount of alcohol removed from can A = $15\times\ 0.6=9\ litres$

Amount of water removed from can A = 15 – 9 = 6 litres.

Can B has 46.67% alcohol content, and 12 litres have been removed.

Amount of alcohol removed from can B=Â $12\times\ \frac{7}{15}=5.6\ litres$.

Amount of water removed from can B = 12 – 5.6 = 6.4 litres.

Removed quantities from both cans are exchanged.

The final alcohol content in can A =Â $40\times\ 0.6-9+5.6=20.6\ litres$

The final alcohol content in can B =Â $30\times\ \frac{7}{15}-5.6+9=17.4\ litres$

The ratio of alcohol content of can A to can BÂ =Â $\frac{20.6}{17.4}=\frac{103}{87}$

The answer is option A.

**Question 16:Â **Â A has 20 red balls and 30 green balls. If B also has red and green coloured balls butÂ twice the total number of balls as A and when he gives a certain number of his red balls to A the ratio of red balls to green balls with A reverses. If the ratio of final number of red balls with B to his initial number of red balls is 2 : 3. How many green balls does B have after giving his red balls to A ?

**16)Â Answer:Â 25**

**Solution:**

Given that initially A has 20 red balls and 30 green balls.

The ratio of red to green is 2:3

When he gets certain number of red balls from B the ratio of red to green reverses.

3:2

Since the number of green balls does not change so he still has 30 balls which is 2/5th of his total balls hence A must have 45 balls after taking B’s balls.

Since B was having a total of 100 balls of which let x be red and 100 – x be green.

As he has given 25 balls toÂ A he must have x-25 red balls and 100-x green balls.

The ratio of final red balls with A to his initial is 2/3.

Hence x-25/x = 2/3

3x-75 = 2x.

x = 75.

Hence 100-x = number of green balls is 25.

**Question 17:Â **A, B, C and D are natural numbers. A is 500% of the sum of C and D. A:B:C is 21:30:2. If B:D = x:y, where x and y are co-prime numbers, find the value of |x-y|.

**17)Â Answer:Â 139**

**Solution:**

A = 5(C+D)

Let A,B,C be 21X, 30X and 2X.

21X = 5(2X + D)

D = 2.2X

B:D = 30X:2.2X = 150:11.

x-y = 139.

**Question 18:Â **A, B, C and D are natural numbers. A is 500% of the sum of C and D. A:B:C is 21:30:2. If B:D = x:y, where x and y are co-prime numbers, find the value of |x-y|.

a)Â 139

b)Â 49

c)Â 61

d)Â 95

**18)Â AnswerÂ (A)**

**Solution:**

A = 5(C+D)

Let A,B,C be 21X, 30X and 2X.

21X = 5(2X + D)

D = 2.2X

B:D = 30X:2.2X = 150:11.

x-y = 139.

**Question 19:Â **A can contains mixture of two liquids P and Q in the ratio 7:5. If 24 liters of the mixture is taken out and replaced by 24 liters of Q the ratio of P and Q in the can changes to 7:11. Find the initial quantity (in liters) of P in the can.

a)Â 44

b)Â 30

c)Â 49

d)Â 42

**19)Â AnswerÂ (D)**

**Solution:**

We have ratio of P and Q as 7:5

Now let the quantity be x liters

now (x-24) liters of mixture will also have ratio of P and Q as 7:5

When (x-24)liters of mixture is mixed with 24liters of Q the ratio becomes 7:11

so by alligation :

$\frac{Qc}{Qd}=\frac{\left(b-m\right)}{m-a}$

$\frac{24}{x-24}=\frac{\left(\frac{7}{12}-\frac{7}{18}\right)}{\left(\frac{7}{18}-0\right)}$

solving we get x=72liters

Therefore initial quantity of P will be :$\frac{7}{12}\times\ 72=42l$

**Question 20:Â **If a Jar A contains water and juice in the ratio 2: 3 with a total volume of 1 litre which is completely filled and Jar B which is of 2 litres and completely filled has water and juice in the ratio 3 : 4. If A and B are completely completely poured in Jar C and 1 litre of this solution is taken and mixed in a Jar of volume 1 litre which contains water and juice in the ratio 1: 4. What is the ratio of water and juice in the final mixtureÂ ?

a)Â 14 : 29

b)Â 13 : 29

c)Â 1 : 2

d)Â 12 : 31

**20)Â AnswerÂ (B)**

**Solution:**

Initially in Jar A the water and Juice are in the ratio 2 : 3.

The total volume of Jar A is 1 litre and hence the volume of the water and Juice are 2/5 litre and 3/5 litre.

Similarly for Jar B the water and Juice are in the ratio 3 : 4 and a total volume of 2 litres and hence :

6/7 litre water and 8/7 litre juice.

Now if they are mixed the new solution contains :

Water = 2/5+ 6/7 litres of water = 44/35 litres of water.

Juice = 3/5 + 8/7 = 61/35 litres of Juice.

Of which 1 litre is taken out = Hence this contains :

44/105 litres of water and 61/105 litres of juice.

This is mixed in jar of volume 1 litre which contains water and Juice in the ratio 1 : 4.

Hence 1/5 litre water and 4/5 litre juice this can be written as 21/105 litre water and 84/105 litre Juice .

Hence the final mixture has :

(44+21)/105 litre water and (61+84)/105 litre Juice.

65/105 : 145/105

13: 29