# SNAP Percentage Questions PDF [Most Important]

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

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**Question 1:Â **Matthew scored 42 marks in biology, 51 marks in chemistry, 58 marks in mathematics, 35 marks in physics and 48 marks in English. The maximum marks a student can score in each subject are 60. How much overall percentage did Matthew get in this exam?

a)Â 76

b)Â 82

c)Â 68

d)Â 78

e)Â None of these

**1)Â AnswerÂ (D)**

**Solution:**

Total marks obtained by Matthew

= 42 + 51 + 58 + 35 + 48 = 234

Maximum marks of the five subjects = 5 * 60 = 300

=> Required % = $\frac{234}{3} \times 100$ = 78Â %

**Question 2:Â **Aryan got 350 marks and Vidya scored 76 percent marks in the same test. If Vidya scored 296 marks more than Aryan, what were the maximum marks of the test ?

a)Â 650

b)Â 900

c)Â 850

d)Â 950

e)Â None of these

**2)Â AnswerÂ (C)**

**Solution:**

Marks scored by Vidya = 350 + 296 = 646

Let the maximum marks be $x$

=> $\frac{76}{100} x = 646$

=> $x = \frac{646}{0.76} = 850$

**Question 3:Â **Last year there were 610 boys in a school. The number decreased by 20 percent this year. How many girls are there in the school if the number of girls is 175 percent of the total number of boys in the school this year ?

a)Â 854

b)Â 848

c)Â 798

d)Â 782

e)Â None of these

**3)Â AnswerÂ (A)**

**Solution:**

Present population of boys = $\frac{80}{100} \times 610$ = 488

=> Number of girls = $\frac{175}{100} \times 488$

= $7 \times 122$ = 854

**Instructions**

<p “=””>Study the following information carefully to answer the questions that follow :Â <p “=””>There are two Trains, Train-A and Train-B. Both Trains have four different types of Coaches viz. General Coaches, Sleeper Coaches, First Class Coaches and AC Coaches. In Train A there are total 700 passengers. Train-B has thirty percent more passengers than Train A. Twenty percent of the passengers of Train-A are in General Coaches. One-fourth of the total number of passengers of Train-A are in AC coaches. Twenty three percent of the passengers of Train-A are in Sleeper Class Coaches. Remaining passengers of Train-A are in first class coaches. Total number of passengers in AC coaches in both the trains together is 480. Thirty percent of the number of passengers of Train-B is in Sleeper Class Coaches. Ten percent of the total passengers of Train-B are in first class coaches. Remaining passengers of Train-B are in general class coaches.

**Question 4:Â **Total number of passengers in General Class coaches in both the Trains together is approximately what percentage of total number of passengers in Train-B ?

a)Â 35

b)Â 42

c)Â 46

d)Â 38

e)Â 31

**4)Â AnswerÂ (B)**

**Solution:**

Total passengers in train A = 700

=> Total passengers in train B = $\frac{130}{100} \times 700 = 910$

Number of passengers In train AÂ in the class :

General = $\frac{20}{100} \times 700 = 140$

AC = $\frac{1}{4} \times 700 = 175$

Sleeper = $\frac{23}{100} \times 700 = 161$

First = $700 – 140 – 175 – 161 = 224$

Number of passengers in train B in the classÂ :

AC = $480 – 175 = 305$

Sleeper = $\frac{30}{100} \times 910 = 273$

First = $\frac{10}{100} \times 910 = 91$

General = $910 – 305 – 273 – 91 = 241$

Total number of passengers in General Class coaches in both the Trains togetherÂ = 140 + 241 = 381

Total number of passengers in Train-BÂ = 910

=> Required % = $\frac{381}{910} \times 100$

= $41.86 \% \approx 42 \%$

**Question 5:Â **If tax on a commodity is reduced by 10%, total revenue remains unchanged. What is the percentage increase in its consumption?

a)Â $11\frac{1}{9}$

b)Â $20$%

c)Â $10$%

d)Â $9\frac{1}{11}$

e)Â None of these

**5)Â AnswerÂ (A)**

**Solution:**

Revenue = consumption $\times$ tax amount

Let consumption = 10 and tax = 10

=> Revenue = $10 \times 10 = 100$

Now, after tax is reduced by 10 %, new tax = $10 – \frac{10}{100} \times 10$

= $10 – 1 = 9$

Total revenue remains unchanged

=> New consumption = $\frac{100}{9}$

$\therefore$ % increase in consumption = $\frac{\frac{100}{9} – 10}{10} \times 100$

= $\frac{10}{9} \times 10 = \frac{100}{9}$

= $11\frac{1}{9} \%$

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**Question 6:Â **Sujata scored 2240 marks in an examination that is 128 marks more than the minimum passing percentage of 64%. What is the percentage of marks obtained by Meena if she scores 907 marks less than Sujata?

a)Â 35

b)Â 40

c)Â 45

d)Â 36

e)Â 48

**6)Â AnswerÂ (B)**

**Solution:**

Let maximum marks in the examination = $100x$

Minimum passing marks = $\frac{64}{100} \times 100x = 64x$

Marks scored by Sujata = $2240$

Acc to ques,

=> $64x + 128 = 2240$

=> $64x = 2240 – 128 = 2112$

=> $x = \frac{2112}{64} = 33$

=> Maximum marks = $100 \times 33 = 3300$

Marks scored by Meena = $2240 – 907 = 1333$

$\therefore$ % marks obtained by Meena = $\frac{1333}{3300} \times 100$

= $40.39 \% \approx 40 \%$

**Question 7:Â **The difference between the 5/6 th of a number and 35 percent of the same is 1392 What will be 55% of that number â€˜?

a)Â 2880

b)Â 1584

c)Â 1854

d)Â 1485

e)Â None of these

**7)Â AnswerÂ (B)**

**Solution:**

Let the number be ‘n’.

According to question,

$\frac{5}{6}$n-35% of n = 1392.

$\frac{5}{6}n-\frac{35\times{n}}{100}=1392$.

$\frac{29\times{n}}{60}=1392$.

$n=2880$.

55 % of n = $\frac{55\times2880}{100}$.

=1584.

Hence, Option B is correct.

**Question 8:Â **Sohail scored 33 marks in English, 37 marks in Science, 28 marks in Mathematics, 26 marks in Hindi and 32 marks in Social studies. The maximum marks a student can score in each subject in 60. How much percentage did Sohail get in this exam ?

a)Â 52

b)Â 54

c)Â 48

d)Â 53

e)Â None of these

**8)Â AnswerÂ (A)**

**Solution:**

So Sohail’s overall percentage marks-

=$\frac{33+37+28+26+32}{300}$ $\times$100

=$\frac{156}{300}$ $\times$100

=52

**Question 9:Â **In an examination the maximum aggregate marks that a student can get are 1020. In order to pass the exam a student is required to get 663 marks out of the aggregate marks. Shreya got 612 marks. By what percent did Shreva fail in the exam ?

a)Â 5

b)Â 8

c)Â 7

d)Â Cannot he determined

e)Â None of these

**9)Â AnswerÂ (A)**

**Solution:**

Passing percentage=663/1020*100=65%

Shreya percentage=612/1020*100=60%

Therefore she failed by 5%

**Question 10:Â **Sushil got 103 marks in Hindi, 111 marks in Science, 98 marks in Sanskrit. 110 marks in Maths and 88 marks in English. If the maximum marks of each subject are equal and if Sushil scored 85 percent marks in all the subjects together, what is the maximum mark of each subject ?

a)Â 110

b)Â 120

c)Â 115

d)Â 100

e)Â None of these

**10)Â AnswerÂ (B)**

**Solution:**

Total marks scored by sushil=103+111+98+110+88=510

Let total max. Marks be x.

Now, it is given that, 85% of x=total marks scored by sushil

I.e, 0.85*x=510

Therefore, x=510*100/85

x=600

This is for 5 subjects altogether.

Therefore, max. Marks for one subject = 600/5 =120 marks.

Option B is the right answer.