Percentage Questions for MAH-CET PDF

Here you can download a free Percentage questions PDF with answers for MAH MBA CET by Cracku. These are some tricky questions in the MAH MBA CET exam that you need to find Percentage answers for the given questions. These questions will help you to make practice and solve the Percentage questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Percentage MCQ PDF for MBA-CET 2022 for free.

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Question 1: The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kgs of wheat so that the percentage of low quality wheat becomes 5%

a) 150 kgs

b) 135 kgs

c) 50 kgs

d) 85 kgs

e) None of these

1) Answer (A)

Solution:

Let $x$ kg of good quality wheat should be added.

Acc to ques,

=> $\frac{10}{100} \times x = \frac{5}{100} \times (x + 150)$

=> $10x = 5x + 750$

=> $x = \frac{750}{5} = 150$ kg

Question 2: 42 per cent of first number is 56 per cent of the second number. What is the respective ratio of the first number to the second number ?

a) 4 : 5

b) 31 : 42

c) 4 : 3

d) Cannot be determined

e) None of these

2) Answer (C)

Solution:

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

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

Acc to ques,

=> $\frac{42}{100} * 100x = \frac{56}{100} * 100y$

=> $42x = 56y$

=> $\frac{x}{y} = \frac{56}{42} = \frac{4}{3}$

=> $x : y = 4 : 3$

Question 3: In a class of 55 students and 3 teachers, each student got sweets that are 20% of the total number of students and each teacher got sweets that are 60% of the total number of students. How many sweets were there?

a) 737

b) 671

c) 714

d) 638

e) None of these

3) Answer (E)

Solution:

There are 55 students and 3 teachers.

Sweets received by each student = $\frac{20}{100} * 55$ = 11

=> Total sweets received by all the students = 55 * 11 = 605

Sweets received by each teacher = $\frac{60}{100} * 55$ = 33

=> Total sweets received by both teachers = 3 * 33 = 99

$\therefore$ Total sweets distributed in class = 605 + 99 = 704

Question 4: One-seventh of a number is 39. What will be 56% of that number?

a) 164.66

b) 152.88

c) 178.22

d) 182.44

e) None of these

4) Answer (B)

Solution:

Let the number = $7x$

Acc to ques,

=> $\frac{1}{7} * 7x = 39$

=> $x = 39$

Now, 56% of the number = $\frac{56}{100} * 7 * 39$

= 152.88

Question 5: The difference between 73% of a number and 58% of the same number is 960. What is 22% of that number?

a) 1408

b) 1232

c) 1324

d) 1536

e) None of these

5) Answer (A)

Solution:

Let the number be $100x$

Acc to ques,

=> $(73 – 58) \%$ of $100x = 960$

=> $\frac{15}{100} \times 100x = 960$

=> $x = \frac{960}{15} = 64$

=> Number = 100 * 64 = 6400

$\therefore$ 22% of 6400 = $\frac{22}{100} * 6400$

= 1408

Question 6: 109. A plot of 715 sq. ft. is available at the rate of Rs. 3,850/- per sq.ft. If 40% of the total cost of the plot is to be paid for booking the plot, how much is the booking amount?

a) Rs. 11,10,000/-

b) Rs. 11,01,100/-

c) Rs. 11,01,000/-

d) Rs. 11,00,100/

e) None of these –

6) Answer (B)

Solution:

Total cost of the plot = 715 * 3850 = Rs. 27,52,750

40% of the total cost is to be paid for booking

=> Booking amount = $\frac{40}{100} * 2752750$

= Rs. 11,01,100

Question 7: The owner of a Gift shop charges his customer 28% more than the cost price: If a customer paid Rs. 1,408/- for some Soft toys, then what was the cost price of those Soft toys?

a) Rs. 1,300/-

b) Rs. 1,000/-

c) Rs. 1,200/-

d) Rs. 1,400/-

e) None of these

7) Answer (E)

Solution:

Let the cost price of the soft toys = Rs. $100x$

Selling price = $\frac{128}{100} * 100x = 128x$

Acc to ques,

=> $128x = 1408$

=> $x = \frac{1408}{128} = 11$

$\therefore$ Cost price of soft toys = 100 * 11 = Rs. 1,100

Question 8: Three-fourth of a number is equal to 60% of another number. What is the difference between the numbers?

a) 18

b) 32

c) 24

d) Cannot be determined

e) None of these

8) Answer (D)

Solution:

Let the numbers be $x$ and $y$

Acc to ques,

=> $\frac{3}{4} \times x = \frac{60}{100} \times y$

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

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

Clearly, no unique answer is possible.

Question 9: Sophia invests 25% of her monthly salary in insurance policies. She spends 15% of her monthly salary in shopping and 35% of her salary on household expenses. She saves the remaining amount of Rs. 9,050.What is Sophia’s annual income ?

a) Rs. 84,500

b) Rs. 5, 30,000

c) Rs. 3, 25,200

d) Rs. 4, 34,400

e) None of these

9) Answer (D)

Solution:

Let Sophia’s monthly salary be Rs. $100x$

Sophia’s % monthly expenditure = (25 + 15 + 35)% = 75%

=> Savings % = 100 – 75 = 25%

Acc to ques,

=> $\frac{25}{100} * 100x = 9050$

=> $x = \frac{9050}{25} = 362$

=> Sophia’s annual income = $12 * 100x$ = Rs. $1200x$

= 1200 * 362 = Rs. 4,34,400

Question 10: The number of employees in Companies A, B and C are in a ratio of 4 : 5 : 6 respectively. If the number of employees in the three Companies is increased by 25%, 30% and 50% respectively, what will be the new ratio of employees working in Companies A, B and C respectively?

a) 13 : 10 : 18

b) 10 : 13 : 17

c) 13 : 15 : 18

d) Cannot be determined

e) None of these

10) Answer (E)

Solution:

Let the no. of employees in companies A, B and C in respectively be $4x , 5x$ and $6x$

After respective increase in the number of employees :

A = $\frac{125}{100} \times 4x = 5x$

B = $\frac{130}{100} \times 5x = 6.5x$

C = $\frac{150}{100} \times 6x = 9x$

=> Required ratio = 5 : 6.5 : 9

= 10 : 13 : 18

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Question 11: 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) 76

b) 82

c) 68

d) 78

e) None of these

11) Answer (D)

Solution:

Total marks obtained by Matthew

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

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

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

Question 12: 855 candidates applied for a job, out of which 80% of the candidates were rejected. How many candidates were selected for the job ?

a) 684

b) 151

c) 676

d) 179

e) None of these

12) Answer (E)

Solution:

% of candidates selected = (100 – 80)% = 20 %

Number of candidates selected = $\frac{20}{100} \times 855$

= $\frac{855}{5} = 171$

Question 13: A student was awarded certain marks in an examination. However, after reevaluation, his marks were reduced by 40% of the marks that were originally awarded to him so that the new score now became 96. How many marks did the student lose after re-evaluation ?

a) 58

b) 68

c) 63

d) 56

e) 64

13) Answer (E)

Solution:

Let the original marks of the student = $100x$

Marks after re-evaluation = $\frac{60}{100} \times 100x = 60x$

=> $60x = 96$

=> $x = \frac{96}{60} = 1.6$

Thus, original marks = 100 * 1.6 = 160

$\therefore$ Marks lost after re-evaluation = 160 – 96 = 64

Question 14: 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) 35

b) 40

c) 45

d) 36

e) 48

14) Answer (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 15: 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) 650

b) 900

c) 850

d) 950

e) None of these

15) Answer (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 16: 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) 854

b) 848

c) 798

d) 782

e) None of these

16) Answer (A)

Solution:

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

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

= $7 \times 122$ = 854

Question 17: The salary of Sarthak is 40% of that of Sarvagya. Harish’s salary is 60% of that of Sarthak. By what per cent is Sarvagya’s salary more than that of Harish ?

a) 317

b) 217

c) 228

d) 281

e) None of these

17) Answer (A)

Solution:

Let Sarvagya’s salary = Rs. 100

=> Sarthak’s salary = $\frac{40}{100} * 100$ = Rs. 40

=> Harish’s salary = $\frac{60}{100} * 40$ = Rs. 24

=> Required % = $\frac{100 – 24}{24} * 100$

= 317%

Question 18: Sushant spent 18% of his monthly salary on buying electronic goods and 32% of the monthly salary on repair work in his house: Out of the remaining amount he invested 42% in fixed deposits. If he was left with Rs.12,325, how much is his annual salary?

a) Rs. 5,18,000

b) Rs. 5,02,600

c) Rs. 5,15,600

d) Rs. 5,00,200

e) Rs. 5,10,000

18) Answer (E)

Solution:

Let monthly salary = Rs. $100x$

Amount spent on buying electronic goods = $\frac{18}{100} \times 100x = 18x$

Amount spent on repair work = $\frac{32}{100} \times 100x = 32x$

Amount left = $100x – 18x – 32x = 50x$

Amount invested in fixed deposits = $\frac{42}{100} \times 50x = 21x$

=> Amount left = $50x – 21x = 29x$

Acc to ques,

=> $29x = 12325$

=> $x = \frac{12325}{29} = 425$

=> Monthly salary = 100 * 425 = Rs. 42,500

$\therefore$ Annual salary = 42500 * 12 = Rs. 5,10,000

Question 19: In a village, three people contested for the post of Village Pradhan. Due to their own interest, all the voters voted and no one vote was invalid. The losing candidate got 30% votes. What could be the minimum absolute margin of votes by which the winning candidate led by the nearest rival, if each candidate got an integral percent of votes ?

a) 4

b) 2

c) 1

d) 3

e) None of these

19) Answer (B)

Solution:

Let total number of voters in the village = 100

No. of votes received by losing candidate = $\frac{30}{100} \times 100 = 30$

No. of votes received by remaining candidate = 100 – 30 = 70

Since, the winner will get more votes than the loser => Winner will get more than 35 votes out of the 70.

For the margin of votes to be minimum, votes received by winner = 36

and votes received by loser = 34

$\therefore$ Absolute margin of votes = 36 – 34 = 2

Question 20: If tax on a commodity is reduced by 10%, total revenue remains unchanged. What is the percentage increase in its consumption?

a) $11\frac{1}{9}$

b) $20$%

c) $10$%

d) $9\frac{1}{11}$

e) None of these

20) Answer (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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