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# SNAP Ratio Questions PDF [Most Important]

The Ratio is an important topic in the Quant section of the SNAP Exam. You can also download this Free Ratio Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Ratio questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.

Question 1:Â Six years ago Jagannath was twice as old as Badri if the ratio of their present age is 9:5 respectively .What is the difference between their present ages?

a)Â 24

b)Â 30

c)Â 50

d)Â Cannot determined

e)Â None of these

Solution:

Let present age of Jagannath = $9x$ years

=> Badri’s present age = $5x$ years

According to ques, => $(9x-6)=2 \times (5x-6)$

=> $9x-6=10x-12$

=> $10x-9x=12-6$

=> $x=6$

$\therefore$ Difference between their present ages = $9x-5x=4x$

= $4 \times 6=24$

=> Ans – (A)

Question 2:Â An amount of Rs 1,25,000 is to be distributed among the Sudhir,Soni,Shakunthala in the respective ratio of 2 : 3 : 5.What will be the difference between Soni’s and Sudhir’s Â Share?

a)Â 25000

b)Â 12500

c)Â 18750

d)Â 2500

e)Â None of these

Solution:

Let amount received byÂ Sudhir,Soni and Shakunthala be $2x,3x$ and $5x$ respectively.

=> Total amount = $(2x+3x+5x)=125,000$

=> $10x=125,000$

=> $x=\frac{125,000}{10}=12500$

$\therefore$ Difference between Soni’s and Sudhir’s Â Share = $3x-2x=x = Rs.$ $12,500$

=> Ans – (B)

Question 3:Â A vessel contains a mixture of milk and water in the respective ratio of 14 : 3. 25.5 litres of the mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of pure milk is added to the mixture. If the resultant mixture contains 20% water, what was the initial quantity of mixture in the vessel before the replacement? (in litres)

a)Â 51

b)Â 102

c)Â 68

d)Â 85

e)Â 34

Solution:

Let the total quantity of mixture in the vessel initially = $17x$ litres

=> Quantity of milk = $\frac{14}{17} \times 17x = 14x$ litres

Quantity of water = $17x – 14x = 3x$ litres

Acc. to ques,

=> $\frac{14x – (\frac{14}{17} \times 25.5) + 5}{3x – (\frac{3}{17} \times 25.5) + 2.5} = \frac{80}{20}$

=> $\frac{14x – 21 + 5}{3x – 4.5 + 2.5} = \frac{4}{1}$

=> $\frac{14x – 16}{3x – 2} = \frac{4}{1}$

=> $14x – 16 = 12x – 8$

=> $14x – 12x = 16 – 8$

=> $x = \frac{8}{2} = 4$

$\therefore$ Initial quantity of mixture in the vessel before the replacement = $17 \times 4 = 68$ litres

Question 4:Â The perimeter of a rectangular field is 240 metre. The ratio between the length and breadth of the field is 8:7. Find the area of the field.

a)Â 3854 sq. m.

b)Â 3584 sq. m.

c)Â 3684 sq. m.

d)Â 3666 sq. m.

e)Â None of these

Solution:

Let the length and breadth of the rectangular field be $8x$ m and $7x$ m respectively.

Perimeter = $2 (8x + 7x) = 240$

=> $15x = \frac{240}{2} = 120$

=> $x = \frac{120}{15} = 8$

$\therefore$ Area of field = $8x \times 7x = 56 x^2$

= $56 \times (8)^2 = 3584 m^2$

Question 5:Â The respective ratio between the monthly salaries of Rene and Som is 5 : 3. Out of her monthly salary Rene gives ${1 \over 6}$th as rent, ${1 \over 5}$th to her mother, 30% as her education loan and keeps 25% aside for miscellaneous expenditure. Remaining Rs. 5000 she keeps as savings. What is Somâ€™s monthly salary?

a)Â Rs. 21000

b)Â Rs. 24000

c)Â Rs. 27000

d)Â Rs. 36000

e)Â Rs. 18000

Solution:

Let monthly salary of Rene = $Rs. 1500x$

=> Monthly salary of Som = $Rs. 900x$

Amount given as rent by Rene = $\frac{1}{6} \times 1500x = 250x$

Amount given by Rene to her mother = $\frac{1}{5} \times 1500x = 300x$

Amount for loan = $\frac{30}{100} \times 1500x = 450x$

Amount kept aside = $\frac{25}{100} \times 1500x = 375x$

=> Amount left = $1500x – (250x + 300x + 450x + 375x) = 5000$

=> $1500x – 1375x = 125x = 5000$

=> $x = \frac{5000}{125} = 40$

$\therefore$ Som’s salary = $900 \times 40 = Rs. 36,000$

TakeÂ  SNAP mock tests here

Question 6:Â A and B started a business with initial investments in the respective ratio of 18 : 7. After four months from the start of the business, A invested Rs. 2000 more and B invested Rs. 7000 more. At the end of one year, if the profit was distributed among them in the ratio of 2 : 1 respectively, what was the total initial investment with which A and B started the business?

a)Â Rs. 50,000

b)Â Rs. 25,000

c)Â Rs. 1,50,000

d)Â Rs. 75,000

e)Â Rs. 1,25,000

Solution:

Let amount invested by A = $Rs. 18x$

=> Amount invested by B = $Rs. 7x$

After four months from the start of the business, A invested Rs. 2000 more and B invested Rs. 7000 more

Thus, ratio of profit received by A : B

= $[(18x \times 4) + (18x + 2000) \times 8] : [(7x \times 4) + (7x + 7000) \times 8]$

= $(72x + 144x + 16000) : (28x + 56x + 56000)$

= $(216x + 16000) : (84x + 56000) = (54x + 4000) : (21x + 14000)$

Acc. to ques, => $\frac{54x + 4000}{21x + 14000} = \frac{2}{1}$

=> $54x + 4000 = 42x + 28000$

=> $54x – 42x = 12x = 28000 – 4000 = 24000$

=> $x = \frac{24000}{12} = 2000$

$\therefore$ Total initial investment = $18x + 7x = 25x$

= $25 \times 2000 = Rs. 50,000$

Question 7:Â A vessel contains 100 litres mixture of milk and water in the respective ratio of 22 : 3. 40 litres of the mixture is taken out from the vessel and 4.8 litres of pure milk and pure water each is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?

a)Â 78${1 \over 2}$

b)Â 79${1 \over 6}$

c)Â 72${5 \over 6}$

d)Â 76

e)Â 77${1 \over 2}$

Solution:

Quantity of milk in vessel = $\frac{22}{25} \times 100 = 88$ litres

=> Quantity of water = $100 – 88 = 12$ litres

40 litres of the mixture is taken out, i.e., $\frac{40}{100} = (\frac{2}{5})^{th}$

=> Milk left = $88 – \frac{2}{5} \times 88 = 52.8$ litres

Water left =Â $12 – \frac{2}{5} \times 12 = 7.2$ litres

Now, 4.8 lires of milk and water are added.

=> Quantity of milk in the vessel = 52.8 + 4.8 = 57.6 litres

Quantity of water in the vessel = 7.2 + 4.8 = 12 litres

$\therefore$ Required % = $\frac{57.6 – 12}{57.6} \times 100$

= $\frac{475}{6} = 79 \frac{1}{6} \%$

Question 8:Â Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1. Jar B which had 40 litres of mixture of milk and water was emptied into jar A, as a result in jar A, the respective ratio of milk and water became 13 : 7. What was the quantity of water in jar B?

a)Â 8 litres

b)Â 15 litres

c)Â 22 litres

d)Â 7 litres

e)Â 1 litre

Solution:

Jar A has 60 litres of mixture of milk and water in the respective ratio of 2 : 1

=> Quantity of milk in Jar A = $\frac{2}{3} \times 60 = 40$ litres

Quantity of water in Jar A = $60 – 40 = 20$ itres

Let quantity of water in Jar B = $x$ litres

=> Quantity of milk in Jar B = $(40 – x)$ litres

Acc. to ques, => $\frac{40 + (40 – x)}{20 + x} = \frac{13}{7}$

=> $560 – 7x = 260 + 13x$

=> $13x + 7x = 560 – 260$

=> $20x = 300$

=> $x = \frac{300}{20} = 15$ litres

Question 9:Â Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4. Jar B which had 20 litres of mixture of milk and water, was emptied into jar A, and as a result in jar A, the respective ratio of milk and water becomes 5: 3. What was the quantity of water in jar B?

a)Â 5 litres

b)Â 3 litres

c)Â 8 litres

d)Â 2 litres

e)Â 1 litre

Solution:

Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4

=> Quantity of milk in Jar A = $\frac{5}{9} \times 36 = 20$ litres

Quantity of water in Jar A = $36 – 20 = 16$ itres

Let quantity of water in Jar B = $x$ litres

=> Quantity of milk in Jar B = $(20 – x)$ litres

Acc. to ques, => $\frac{20 + (20 – x)}{16 + x} = \frac{5}{3}$

=> $120 – 3x = 80 + 5x$

=> $5x + 3x = 120 – 80$

=> $8x = 40$

=> $x = \frac{40}{8} = 5$ litres

Question 10:Â A starts a small business with Rs. 3600. At the end of few months from the start of business, B joined the business with Rs. 4000. If the annual profit between A and B was divided between them in the respective ratio of 6 : 5, then B joined the business after how many months from the start of the business?

a)Â Four

b)Â Two

c)Â Six

d)Â Five

e)Â Three

Solution:

Let B remained in the business for $x$ months.

Amount invested by A = Rs. 3600

Amount invested by B = Rs. 4000

Ratio of share in profit received by A and B

=> $\frac{3600 \times 12}{4000 \times x} = \frac{6}{5}$

=> $\frac{9 \times 2}{10 \times x} = \frac{1}{5}$

=> $x = \frac{18 \times 5}{10} = 9$

$\therefore$Â B joined the business after = $12 – 9 = 3$ months from the start of the business.

Question 11:Â In a class, the respective ratio between the number of boys and the number of girls is 3:1. A test was conducted, wherein the average score of the boys was 73, while that of the entire class was 71. What was the average score of the girls?

a)Â 68

b)Â 71

c)Â 67

d)Â 65

e)Â 63

Solution:

Let number of boys = $3x$

=> Number of girls = $x$

Let average score of girls = $y$

Acc. to ques,

=> $\frac{(73 \times 3x) + (y \times x)}{3x + x} = 71$

=> $\frac{x (219 + y)}{4x} = 71$

=> $219 + y = 71 \times 4 = 284$

=> $y = 284 – 219 = 65$

Question 12:Â Jar A contains 78 litres of milk and water in the respective ratio of 6 : 7. 26 litres of the mixture was taken out from Jar A. What quantity of milk should be added to jarA, so that water constitutes 40% of the resultant mixture in jar A?

a)Â 8 litres

b)Â 36 litres

c)Â 12 litres

d)Â 14 litres

e)Â 18 litres

Solution:

Jar A has 78 litres of mixture of milk and water in the respective ratio of 6 : 7

=> Quantity of milk in Jar A = $\frac{6}{13} \times 78 = 36$ litres

Quantity of water in Jar A = $78 – 36 = 42$ litres

26 litres of the mixture was taken out from Jar A, i.e., $\frac{26}{78} = (\frac{1}{3})^{rd}$

=> Milk left = $36 – \frac{1}{3} \times 36 = 24$

Water leftÂ = $42 – \frac{1}{3} \times 42 = 28$

Let milk added to jar A = $x$ litres

Acc. to ques, => $\frac{24 + x}{28} = \frac{60}{40}$

=> $\frac{24 + x}{28} = \frac{3}{2}$

=> $48 + 2x = 84$

=> $2x = 84 – 48 = 36$

=> $x = \frac{36}{2} = 18$ litres

Instructions

Read the following information carefully to answer these questions.
These statistical data were collected in the year 2005.Hamirpur in Gujarat is a small township with a population of 75000. 40% of the population belongs to the above 35 age category. The ratio of males to females is 1 : 1.5. Past records indicate that the population in Hamirpur grows at an annual rate of 7%. The total cultivable area in Hamirpur is 2 lakh acres. Paddy is the major crop of Hamirpur and has shown average productivity levels of 2.5 tonnes per acre. Hamirpur receives about 10 inches of rainfall in a normal monsoon year.

Question 13:Â If the ratio of males to females in 2006 remains the same as that in 2005, then the number of males in 2006 would be

a)Â 30600

b)Â 32100

c)Â 31500

d)Â 32700

e)Â 33000

Solution:

Population of Hamirpur in 2005 = 75000

% increase every year = 7%

=> population in 2006 = 75000 * 1.07 = 80250

MalesÂ : Females = 1Â : 1.5

=> No. of males in 2006 = $\frac{1}{2.5} * 80250$

= 32100

Question 14:Â Of the two numbers, 48 per cent of first number is 60 per cent of the second number. What is the respective ratio of the first number to the second number ?

a)Â 4 : 7

b)Â 3 : 4

c)Â 5 : 4

d)Â Cannot be determined

e)Â None of these

Solution:

Let the numbers be $100x$ and $100y$

We need to find = $\frac{100x}{100y} = \frac{x}{y} = ?$

Acc to ques,

=> $\frac{48}{100} * 100x = \frac{60}{100} * 100y$

=> $48x = 60y$

=> $\frac{x}{y} = \frac{60}{48} = \frac{5}{4}$

=> $x:y = 5:4$

Question 15:Â A sum of money is divided among A, B, C and D in the ratio of 4 : 5 : 7 : 11 respectively. If the share of C is Rs. 1,351/- then what is the total amount of money of A and D together?

a)Â Rs. 2,123/-

b)Â Rs. 2,316/-

c)Â Rs. 2,565/-

d)Â Rs. 2,895/-

e)Â None of these

Solution:

Let the total amount to be divided = $27x$

A : B : C : D = 4Â : 5 : 7 : 11

=> Share of C = $\frac{7}{27} * 27x = 1351$

=> $x = \frac{1351}{7} = 193$

Now, total amount of money with A & D together

= $\frac{4 + 11}{27} * 27x$

= $15 * 193 =$Rs. $2,895$

Question 16:Â 38 per cent of first number is 52 per cent of the second number. What is the respective ratio of the first number to the second number?

a)Â 5 : 4

b)Â 16 : 9

c)Â 26 : 19

d)Â Cannot be determined

e)Â None of these

Solution:

Let the numbers be $100x$ and $100y$

We need to find = $\frac{100x}{100y} = \frac{x}{y} = ?$

Acc to ques,

=> $\frac{38}{100} * 100x = \frac{52}{100} * 100y$

=> $38x = 52y$

=> $\frac{x}{y} = \frac{52}{38} = \frac{26}{19}$

=> $x : y = 26 : 19$

Question 17:Â A. B, and C divide an amount of Rs. 4,200 amongst themselves in the ratio of 7 : 8 : 6 respectively. If an amount of Rs. 200 is added to each of their shares, what will be the new respective ratio of their shares of amount?

a)Â 8 : 9 : 6

b)Â 7 : 9 : 5

c)Â 7 : 8 : 6

d)Â 8 : 9 : 7

e)Â None of these

Solution:

Ratio of amounts received by A, B and C = 7Â : 8Â : 6

Sum of ratios = 7 + 8 + 6 = 21

=> Sum received byÂ :

A = $\frac{7}{21} \times 4200$ = Rs. 1400

BÂ = $\frac{8}{21} \times 4200$ = Rs.Â 1600

CÂ = $\frac{6}{21} \times 4200$ = Rs.Â 1200

On adding Rs. 200 to the share of each one, the required ratio

= 1600Â : 1800Â : 1400

= 8Â : 9Â : 7

Question 18:Â The number of employees in companies A, B and C are in a ratio of 3 : 2 : 4 respectively. If the number of employees in the three companies is increased by 20%, 30% and 15% respectively, what will be the new ratio of employees working in companies A, B and C respectively ?

a)Â 18 : 13 : 24

b)Â 13 : 18 : 23

c)Â 17 : 3 : 23

d)Â 18 : 11 : 23

e)Â None of these

Solution:

Let the no. of employees in companies A, B and C in respectively be $3x , 2x$ and $4x$

After respective increase in the number of employees :

A = $\frac{120}{100} \times 3x = 3.6x$

BÂ = $\frac{130}{100} \times 2x = 2.6x$

C = $\frac{115}{100} \times 4x = 4.6x$

=> Required ratio = 3.6Â : 2.6Â : 4.6

= 18Â : 13Â : 23

Question 19:Â The ratio of the ages of A and B seven years ago was 3 : 4 respectively. The ratio of their ages nine years from now will be 7 : 8 respectively. What is Bâ€™s age at present?

a)Â 16 years

b)Â 19 years

c)Â 28 years

d)Â 23 years

e)Â None of these

Solution:

Let present age of A = $x$ years and present age of B = $y$ years

Seven years ago,

=> $\frac{x – 7}{y – 7} = \frac{3}{4}$

=> $4x – 28 = 3y – 21$

=> $4x – 3y = 7$ ————-Eqn(1)

After 9 years,

=> $\frac{x + 9}{y + 9} = \frac{7}{8}$

=> $8x + 72 = 7y + 63$

=> $8x – 7y = -9$ ————–Eqn(2)

Solving equations (1) and (2), we getÂ :

$y = 23$

$\therefore$ Present age of B = 23 years

Question 20:Â 28% members of a certain group are married: What is the respective ratio between the number of married members to the number of unmarried members?

a)Â 7 : 17

b)Â 5 : 18

c)Â 7 : 18

d)Â Cannot be deterimed

e)Â None of these

Let the total members in the group = $100x$
Married members = $\frac{28}{100} \times 100x = 28x$
Unmarried members = $100x – 28x = 72x$
=> Required ratio = $28x : 72x$
= $7 : 18$