# Ratio Questions for SNAP Exam: Important Questions With Solutions PDF

The SNAP 2022 exam is going to be conducted on the 10th, 18th & 23rd of December 2022. So, the aspirants are advised to achieve a strong grasp of the quant basics, which is very essential to score high in SNAP 2022. The ratio is one of the most important topics in the Quant section of the SNAP Exam. You can also download this Free Ratio Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Ratio questions in the SNAP exam. Utilize this **PDF practice set, **which is one of the best sources for practising.

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**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

**1)Â AnswerÂ (A)**

**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

**2)Â AnswerÂ (B)**

**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 **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

**3)Â AnswerÂ (D)**

**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 = $900 \times 40 = Rs. 36,000$

**Question 4: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

**4)Â AnswerÂ (A)**

**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 5: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${1 \over 2}$

**5)Â AnswerÂ (B)**

**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 6: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

**6)Â AnswerÂ (B)**

**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

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**Question 7: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

**7)Â AnswerÂ (A)**

**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 8: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

**8)Â AnswerÂ (E)**

**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 9: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

**9)Â AnswerÂ (E)**

**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 10: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

**10)Â AnswerÂ (B)**

**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 11: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

**11)Â AnswerÂ (D)**

**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 12: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

**12)Â AnswerÂ (C)**

**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 13: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

**13)Â AnswerÂ (D)**

**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 14: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

**14)Â AnswerÂ (E)**

**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 15: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

**15)Â AnswerÂ (D)**

**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 16:Â **Some chocolates were distributed among four friends A, B, C and D such that the respective ratio between the number of chocolates received by A and C be 7:9. B got 29 chocolates more than that of A and D got 33 chocolates more than that of C. If B got 15 chocolates more than that of C, how many chocolates did D receive?

a)Â 84

b)Â 96

c)Â 72

d)Â 99

e)Â 87

**16)Â AnswerÂ (B)**

**Solution:**

Chocolates got byÂ A = $7x$

Chocolates got by C = $9x$

Chocolates got by B = $7x + 29$

Chocolates got by D = $9x + 33$

Acc to ques,

=> $(7x + 29) – (9x) = 15$

=> $2x = 29 – 15 = 14$

=> $x = 7$

$\therefore$ Chocolates received by D

= $9 \times 7 + 33 = 63 + 33 = 96$

**Question 17:Â **The angles in a triangle are in a ratio of 19 : 10 : 7. What is the sum of thrice the smallest angle and the twice the largest angle ?

a)Â 275Â°

b)Â 295Â°

c)Â 280Â°

d)Â 273Â°

e)Â None of these

**17)Â AnswerÂ (B)**

**Solution:**

Let the angles of the triangle be $19x , 10x , 7x$

=> Sum of angles = $19x + 10x + 7x = 180^{\circ}$

=> $36x = 180^{\circ}$

=> $x = \frac{180}{36} = 5^{\circ}$

=> Angles are = $95^{\circ} , 50^{\circ} , 35^{\circ}$

$\therefore$ Sum of thrice the smallest angle and the twice the largest angle

= $(3 \times 35^{\circ}) + (2 \times 95^{\circ})$

= $105 + 190 = 295^{\circ}$

**Question 18:Â **Five years ago, the ratio between ages of Meena and Sita was 3 : 4. After five years from now the ratio between their ages will be 5 : 6. Find the present age of Meena ?

a)Â 22 years

b)Â 25 years

c)Â 15 years

d)Â 20 years

e)Â None of these

**18)Â AnswerÂ (D)**

**Solution:**

Let the ages of Meena and Sita 5 years ago be $3x$ and $4x$ years

=> Present age of Meena = $(3x + 5)$ years

Present age of SitaÂ = $(4x + 5)$ years

Acc. to ques,

=> $\frac{3x + 10}{4x + 10} = \frac{5}{6}$

=> $18x + 60 = 20x + 50$

=> $2x = 60 – 50 = 10$

=> $x = \frac{10}{2} = 5$

$\therefore$ Present age of Meena = $(3 \times 5 + 5) = 20$ years

**Question 19:Â **The respective ratio between the present ages of father, mother and daughter is 7 : 6 : 2. The difference between motherâ€™s and the daughterâ€™s age is 24 years. What is the fatherâ€™s age at present ?

a)Â 43 years

b)Â 42 years

c)Â 39 years

d)Â 38 years

e)Â None of these

**19)Â AnswerÂ (B)**

**Solution:**

LetÂ the present ages of father, mother and daughter respectively = $7x , 6x , 2x$

Acc to ques,

=> $6x – 2x = 24$

=> $x = \frac{24}{4} = 6$

$\therefore$ Father’s Â present age = 7 * 6 = 42 years

**Question 20:Â **Kajal spends 55% of her monthly income on grocery, clothes and education in the ratio of 4 : 2 : 5 respectively. If the amount spent on clothes is Rs. 5,540, what is Kajalâ€™s monthly income

a)Â Rs. 55,400

b)Â Rs. 54,500

c)Â Rs. 55,450

d)Â Rs.55.650

e)Â None of these

**20)Â AnswerÂ (A)**

**Solution:**

Let Kajal’s monthly salary = Rs. $100x$

Amount spent by Kajal = $\frac{55}{100} \times 100x = $Rs. $55x$

=> Amount spent on clothes = $\frac{2}{(4 + 2 + 5)} \times 55x = 10x$

Acc to ques,

=> $10x = 5540$

=> $x = 554$

$\therefore$ Kajal’s monthly salary = $100 \times 554$ = Rs. 55,400