# Ratio and Proportion Questions for TISSNET 2022 – Download PDF

Here you can download TISSNET 2022 Ratio and Proportion [PDF] by Cracku. Very Important Ratio and Proportion Questions for TISSNET 2022 based on asked questions in previous exam papers. These questions will help your TISSNET exam preparation. So kindly download the PDF for reference and do more practice.

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**Question 1:Â **An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is

a)Â 48

b)Â 84

c)Â 70

d)Â 96

**Question 2:Â **The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a)Â 3 : 10

b)Â 1 : 3

c)Â 1 : 4

d)Â 2 : 5

**Question 3:Â **Consider three mixtures â€” the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has

a)Â The same amount of water and liquid B

b)Â The same amount of liquids B and C

c)Â More water than liquid B

d)Â More water than liquid A

**Question 4:Â **Flights A and B are scheduled from an airport within the next one hour. All the booked passengers of the two flights are waiting in the boarding hall after check-in. The hall has a seating capacity of 200, out of which 10% remained vacant. 40% of the waiting passengers are ladies. When boarding announcement came, passengers of flight A left the hall and boarded the flight. Seating capacity of each flight is two-third of the passengers who waited in the waiting hall for both the flights put together. Half the passengers who boarded flight A are women. After boarding for flight A, 60% of the waiting hall seats became empty. For every twenty of those who are still waiting in the hall for flight B, there is one air hostess in flight A. What is the ratio of empty seats in flight B to the number of air hostesses in flight A?

a)Â 10 : 1

b)Â 5 : 1

c)Â 20 : 1

d)Â 1 : 1

**Question 5:Â **Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

a)Â 2 : 3

b)Â 4 : 3

c)Â 3 : 2

d)Â 3 : 4

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**Question 6:Â **A student gets an aggregate of 60% marks in five subjects in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?

a)Â 2

b)Â 3

c)Â 4

d)Â 5

**Question 7:Â **The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness’ be if the value of the first is four times that of the second?

a)Â 16 : 9

b)Â 9 : 4

c)Â 9 : 16

d)Â 4 : 9

**Question 8:Â **One bacterium splits into eight bacteria of the next generation. But due to environmental condition only 50% survives and remaining 50% dies after producing next generation. If the seventh generation number is 4,096 million, what is the number in first generation?

a)Â 1 million

b)Â 2 million

c)Â 4 million

d)Â 8 million

**Question 9:Â **I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.

a)Â 90

b)Â 85

c)Â 100

d)Â 105

**Instructions**

DIRECTIONS for the following two questions: The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

**Question 10:Â **Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1: 2: 3?

a)Â 1.3

b)Â 1

c)Â 0.6

d)Â 2.3

**Question 11:Â **Company ABC starts an educational program in collaboration with Institute XYZ. As per the agreement, ABC and XYZ will share profit in 60 : 40 ratio. The initial investment of Rs.100,000 on infrastructure is borne entirely by ABC whereas the running cost of Rs. 400 per student is borne by XYZ. If each student pays Rs. 2000 for the program find the minimum number of students required to make the program profitable, assuming ABC wants to recover its investment in the very first year and the program has no seat limits.

a)Â 63

b)Â 84

c)Â 105

d)Â 157

e)Â 167

**Question 12:Â **Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?

a)Â 328 units

b)Â 368 units

c)Â 392 units

d)Â 616 units

e)Â None of the above

**Question 13:Â **The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?

a)Â 65 percent

b)Â 60 percent

c)Â 70 percent

d)Â None of the above

**Question 14:Â **The ratio of â€˜metal 1â€™ and â€˜metal 2â€™ in alloy â€˜Aâ€™ is 3 :4. In alloy â€˜Bâ€™ same metals are mixed in the ratio 5:8. If 26 kg of alloy â€˜Bâ€™ and 14 kg of alloy â€˜Aâ€™ are mixed then find out the ratio of â€˜metal 1â€™ and â€˜metal 2â€™ in the new alloy.

a)Â 3:2

b)Â 2:5

c)Â 2:3

d)Â None of the above

**Question 15:Â **X and Y are the two alloys which were made by mixing Zinc and Copper in the ratio 6 : 9 and 7 : 11 respectively. If 40 grams of alloy X and 60 grams of alloy Y are melted and mixed to form another alloy Z, what is the ratio of Zinc and Copper in the new alloy Z?

a)Â 6 : 9

b)Â 59 : 91

c)Â 5 : 9

d)Â 59 : 90

**Question 16:Â **Ravindra and Rekha got married 10 years ago, their ages were in the ratio of 5 : 4. Today Ravindraâ€™s age is one sixth more than Rekhaâ€™s age. After marriage, they had 6 children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of 3 : 2 : 1. What is the largest possible value of the present total age of the family?

a)Â 79

b)Â 93

c)Â 101

d)Â 107

**Question 17:Â **There are two alloys P and Q made up of silver, copper and aluminium. Alloy P contains 45% silver and rest aluminum. Alloy Q contains 30% silver, 35% copper and rest aluminium. Alloys P and Q are mixed in the ratio of 1 : 4 . 5. The approximate percentages of silver and copper in the newly formed alloy is:

a)Â 33% and 29%

b)Â 29% and 26%

c)Â 35% and 30%

d)Â None of the above

**Question 18:Â **A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

a)Â 1 : 1

b)Â 8 : 7

c)Â 4 : 3

d)Â 6 : 5

**Question 19:Â **Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a)Â 27:14

b)Â 27:13

c)Â 27:16

d)Â 27:18

**Question 20:Â **If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

a)Â 201

b)Â 205

c)Â 207

d)Â 210

**Answers & Solutions:**

**1)Â AnswerÂ (B)**

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.’. 12xy+8xy+42xy=130 => 65xy=130 => xy=2.

.’.`The weight, in kg. of the metal C is 42xy=84.

**2)Â AnswerÂ (B)**

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given thatÂ three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$Â $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ =Â $\dfrac{90 – 4c + 7c}{9}$ =Â $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ =Â $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

**3)Â AnswerÂ (C)**

The proportion of water in the first mixture is $\frac{1}{3}$

The proportion of Liquid A in the first mixture is $\frac{2}{3}$

The proportion of water in the second mixture is $\frac{1}{4}$

The proportion of Liquid B in the second mixture is $\frac{3}{4}$

The proportion of water in the third mixture is $\frac{1}{5}$

The proportion of Liquid C in the third mixture is $\frac{4}{5}$

As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$

The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$

The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$

The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$

Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C Â is correct.

**4)Â AnswerÂ (A)**

Out of 200 of the seating capacity, 180 seats are filled out of which 108 are males and 72 are females. Remaining 20 seats are vacant. According to given condition seating capacity in both the planes is 120 . Considering flight A – we can find that 100 passenger in waiting hall will be taking fight A . So 80 people remain in in the waiting hall who will be taking flight B . Now for every 20 people taking flight B we have a air hostess in flight A . So in total there are 4 air hostess in flight A. Flight B having 120 as seating capacity, 40 remain vacant. So required ratio 40:4 = 10:1 .

**5)Â AnswerÂ (D)**

Fraction of A in contained 1 = $\frac{5}{6}$

Fraction of A in contained 2Â = $\frac{1}{4}$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($\frac{5}{6}$) + (1-x)($\frac{1}{4}$) =Â $\frac{1}{2}$

$\frac{7x}{12}$ = $\frac{1}{4}$

=> x = $\frac{3}{7}$

=> Ratio = 3:4

**6)Â AnswerÂ (C)**

Let’s say he scored marks as $10x,9x,8x,7x,6x$ or total of $40x$ which is 60% of total maximum marks(T).

$\frac{T \times 60}{100}=40x$

So T (total maximum marks)=$\frac{400x}{6}$

Or Individual max. marks = $\frac{T}{5}=\frac{80x}{6}$

Passing marks =50% of individual max. marks =$\frac{40x}{6}=6.66x$

Hence he scored more than passing marks in four subjects as $10x,9x,8x$ and $7x$

and failed in one subject as scoring $6x$ marks which is less than passing marks of $6.66x$

**7)Â AnswerÂ (B)**

Value of coin = $k (2r)^2 t$ (where k is proportionality constant, 2r is diameter and t is thickness)

So (value of first coin) = 4 (value of second coin)

$k (2r_1)^2 t_1 = 4 \times (k(2r_2)^2 t_2)$

or $\frac{t_1}{t_2} = \frac{9}{4}$ Â (As ratio of diameters 2r will be 9:4)

**8)Â AnswerÂ (A)**

let’s say x is the initial number of bacterias :

So in 2nd generation no. of bacterias = \frac{8x}{2} = 4x

In 3rd generation, it will be = 16x

4th gen. = 64x

5th gen. = 256x

6th gen. = 1024x

7th gen. = 4096x

Hence x = 1 million

**9)Â AnswerÂ (D)**

Let’s say number of coins are 2.5x , 3x and 4x

So total amount will be = 2.5x + 3x(0.5) + 4x(0.25) = 210

So x = 42

And number of 1 rs. coins = 2.5x = 105

**10)Â AnswerÂ (A)**

The relative sweetness of the mixture is (1*0.74 + 2*1 + 3*1.7) / (1+2+3) = 7.84/6 = 1.30

Option a) is the correct answer.

**11)Â AnswerÂ (A)**

For program to be profitable both companies must recover costs before they can start making profits.

Since, ABC wants to recoup investment in the first year and there is no limit of number of students, profits can only be shared after both companies can reach a situation of minimum profits (zero profit), which would be:

Let x be minimum number of students required to reach a situation of minimum profits (in this case 0).

Comparing total costs and revenuesÂ :

=> $400x + 100000 = 2000x.$

=> $x = \frac{100000}{(2000 – 400)} = 62.5 = 63$

**12)Â AnswerÂ (B)**

Let the quantities of the chemicals X and Y, mixed to produce product M be $5c$ and $4c$ respectively.

X is prepared by mixing A and B in the ratio = 1 : 3

=> Quantity of B in X = $\frac{3}{4} \times 5c = \frac{15 c}{4}$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $\frac{2}{3} \times 4c = \frac{8 c}{3}$

Quantity of B in M = $\frac{15 c}{4} + \frac{8 c}{3} = \frac{77 c}{12}$

Now, 864 units of M was mixed with water to prepare the final mixture.

=> Total quantity of M = $9c = 864$ => $c = \frac{864}{9} = 96$

Concentration of raw material B in the final mixture is 50 %

=> Quantity of final mixture = $\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$

$\therefore$ Quantity of water added to M = $1232 – 864 = 368$ units

**13)Â AnswerÂ (A)**

The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?

Let ‘9x’ be the number of total journalists in the office. Then, we can say that the number of male and female journalists are ‘5x’ and ‘4x’ respectively.

It is given that 30 percent of the male journalists and 40 percent of the female journalists are covering political news. Hence, total number of journalists who are covering political news = 0.3*5x + 0.4*4x = 3.1x

Therefore, the total numberÂ journalists who are covering sports news = 9x – 3.1x = 5.9x.

Hence, theÂ percentage of the journalists in the newspaper is currently involved in sports reporting = $\dfrac{5.9x}{9x}\times 100$ $\approx$ 65 percent. Therefore, option A is the correct answer.

**14)Â AnswerÂ (C)**

The ratio of â€˜metal 1â€™ and â€˜metal 2â€™ in alloy â€˜Aâ€™ is 3 :4.Therefore, we can say that 14 kg of alloy ‘A’ will contain $\dfrac{3}{7} 14$ = 6 kg of ‘metal 1’ andÂ $\dfrac{4}{7} 14$ = 8 kg of ‘metal 2’.

The ratio of â€˜metal 1â€™ and â€˜metal 2â€™ in alloy â€˜Bâ€™ is 5 :8.Therefore, we can say that 26 kg of alloy ‘B’ will contain $\dfrac{5}{13} 26$ = 10 kg of ‘metal 1’ andÂ $\dfrac{8}{13} 26$ = 16 kg of ‘metal 2’.

Hence, total weight of ‘metal 1’ in the new alloy = 6 + 10 = 16 kg

Total weight of ‘metal 2’ in the new alloy = 8 + 16 = 24 kg

Therefore,Â the ratio of â€˜metal 1â€™ and â€˜metal 2â€™ in the new alloy. = 16 : 24 = 2 :3. Hence, option C is the correct answer.

**15)Â AnswerÂ (B)**

Alloy X has zinc and copper in ratio 6:9

If 40 grams is taken then weights of zinc and copper are:

Zinc = $\frac{6}{15}$ * 40 = 16 grams

Copper = $\frac{9}{15}$ * 40 = 24 grams

Alloy Y has zinc and copper in ratio 7:11

If 60 grams is taken then weights of zinc and copper are:

Zinc = $\frac{7}{18}$ * 60 = $\frac{70}{3}$ grams

Copper = 60 –Â $\frac{70}{3}$ = $\frac{110}{3}$ grams

When mixed together, the weights of Zinc and Copper are:

Zinc = 16 +Â $\frac{70}{3}$ =Â $\frac{118}{3}$

Copper = 24 + $\frac{110}{3}$ =Â $\frac{182}{3}$

Ratio = 118/3:182/3 = 59:91

**16)Â AnswerÂ (D)**

10 years ago, Let age of Ravindra be 5x and Rekha be 4x

At present, Ravindra is 7/6 times of Rekha’s age.

5x + 10 = $\frac{7}{6}$ (4x + 10)

Solving, x =5

Ravindra was 25 years (10 years ago) and Rekha was 20 years (10 years ago)

Now, ages of their children is 3:2:1

Maximum possible ages of children is 9,6,3 years.

Total age of family is: 35 + 30 + 9*3 + 6*2 + 3 = 107 years.

**17)Â AnswerÂ (A)**

Composition of alloy P

Silver:Copper:Aluminium = 45:0:55

Composition of alloy Q

Silver:Copper:Aluminium = 30:35:35

They are mixed in ratio of 1: 4.5

Let us consider alloy P is taken 200 grams and alloy Q is taken 900 grams.

Then for alloy P :-

Silver:Copper:Aluminium = 90:0:110

For alloy Q:

Silver:Copper:Aluminium = 270:315:315

Total weight of P and Q combined is 1100 grams.

When P and Q are mixed, the new combined ratio of

Silver:Copper:Aluminium = 360:315:425

Percentage of Silver in mixture = $\frac{360}{1100}$ x 100 $\cong$ 33%

Percentage of Copper in mixture = $\frac{315}{1100}$ x 100 $\cong$ 29%

**18)Â AnswerÂ (A)**

The ratio of L, S, J for popcorn = 7 : 17 : 16

Let them be 7$x$, 17$x$ and 16$x$

The ratio of L, S, J for chips = 6 : 15 : 14

Let them 6$y$, 15$y$ and 14$y$

Given, 40$x$ = 35$y$, $x = \frac{7y}{8}$

Jumbo popcor = 16$x$ = 16 *Â $\frac{7y}{8}$= 14$y$

Hence, the ratio of jumbo popcorn and jumbo chips = 1 : 1

**19)Â AnswerÂ (B)**

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4

Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$

The total volume of water is $\frac{2X}{9}+\frac{4}{13}$

They are in the ratio 3:1

Hence,Â $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$

Therefore, $91X+81=78X+108$

Therefore $X = \frac{27}{13}$

**20)Â AnswerÂ (C)**

a : b = 3:4 and b : c = 2:1 => a:b:c = 3:4:2

=> a = 3x, b = 4x , c = 2x

=> a + b + c = 9x

=> a + b + c is a multiple of 9.

From the given options only, option C is a multiple of 9