Percentage Questions for MAH-CET

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Percentages Questions
Percentages Questions

Here you can download a free Percentage questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Pertcentage of 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: Paresh got 102 marks in Hindi, 118 marks in Science, 104 marks in Sanskrit, 114 marks in Maths and 96 marks in English. The maximum marks of each subject are 120. How much overall percentage of marks did paresh get ?

a) 89

b) 82

c) 77

d) 71

e) None of these

1) Answer (A)

Solution:

The total marks obtained by Paresh is 102+118+104+114+96 = 534
The total marks that could have been obtained is 120*5 = 600

So, the overall percentage obtained by Paresh is 534/600 * 100 = 89%

Question 2: The wheat sold by a grocer contained 10% low-quality wheat. What quantity of good-quality wheat should be added to 150 kg of wheat so that the percentage of low-quality wheat becomes 5% ?

a) 150 kg

b) 135 kg

c) 50 kg

d) 85 kg

e) None of these

2) Answer (A)

Solution:

150 Kg of weight contains 10% low quality weight that is 15 Kg low quality weight.
Let the quantity of good quality wheat to be added be X.
So, $15 = \frac{5}{100}*(150+X)$
Or, $300 = 150+X$

So X = 150 Kg

Question 3: Koel scored 49 marks in English, 37 marks in Science, 45 marks in Mathematics, 53 marks in Hindi and 55 marks in Social studies. The maximum marks a student can score in each subject is 70. How much approximate percentage did Koel get in this exam?

a) 53

b) 79

c) 68

d) 73

e) 88

3) Answer (C)

Solution:

Here the total subjects are 5 each subject has maximum 70 marks.

So total marks available are 70 × 5 = 350

Total marks of Koel is = 49 + 37 + 45 + 53 + 55 = 239

Percentage of Koel is = $\frac{239}{350}\times100$ = 68.28, which is equal to approximate 68 percent

Question 4: In a test, minimum passing percentage for reserved and unreserved category is 40% and 54% respectively. A candidate from unreserved category secured 300 marks and failed by 24 marks. What are the minimum passing marks of reserved category in that test ?

a) 280

b) 254

c) 230

d) 240

e) None of these

4) Answer (D)

Solution:

Let the Total marks is ‘T’ .

The unreserved guy got 300 but failed by 24 marks i.e. his pass mark is 300 + 24 = 324 .

Given that , the pass mark for unreserved guy is 54% of the total . i.e. $\frac{54}{100}$T = 324.
We get , T = 600 .
Also given that the pass mark for reserved guy is 40% of the total .i.e. {40/100}* 600 = 240.

Question 5: 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) 52

b) 54

c) 48

d) 53

e) None of these

5) Answer (A)

Solution:

So Sohail’s overall percentage marks-

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

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

=52

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Question 6: 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

6) 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 7: 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

7) 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 = $\frac{\frac{100}{9} – 10}{10} \times 100$

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

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

Instructions

<p “=””>Study the following information carefully to answer the questions that follow :  <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 8: 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) 35

b) 42

c) 46

d) 38

e) 31

8) Answer (B)

Solution:

Total passengers in train A = 700

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

Number of passengers In train A 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 class :

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 together = 140 + 241 = 381

Total number of passengers in Train-B = 910

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

= $41.86 \% \approx 42 \%$

Question 9: 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

9) 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 10: 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

10) 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

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