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# Average and Percentage Questions for CAT

Average and Percentage is one of the key topics in the CAT Quant Section. If you’re not very strong in this topic, it is essential that you know the basics of CAT Average and Percentage. Also, do check out all the Average and Percentage questions from the CAT Previous Papers with detailed video solutions. This article will look into some important Average and Percentage questions for the CAT Exam. If you want to practice these important Average and Percentage questions for CAT, you can download the PDF, which is completely Free.

Question 1:Â A fruit seller has a sale of â‚¹10,435, â‚¹9,927,â‚¹10,855 â‚¹10,230 and â‚¹9,562 for five consecutive months.How much sale (in â‚¹) must he have in the sixth month so that he gets an average sale of â‚¹10,500?

a)Â 9,231

b)Â 8,231

c)Â 8,991

d)Â 11,991

Solution:

As we know,

Average = $\frac{Sumofobservation}{total\ number\ of\ observation}$

Let the sale of 6th month be x

Average =Â $\frac{\left(10435+9927+10855+10230+9562+x\right)}{6}=10500$

=Â $51009\ +\ x=63000$

x = 11991

Hence, option D is correct.

Question 2:Â The mean of 100 items is 47.It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50, respectively. The correct mean is:

a)Â 48

b)Â 47

c)Â 50

d)Â 51

Solution:

Given,

Mean of 100 items = 47

Sum of observation =Â $mean\times\ number\ of\ observation$

i.e; $47\times\ 100\ =\ 4700$

Correct sum = 4700 + (60 + 70 + 80) – (40 + 20 +Â 50)

i.e; 4800

Correct mean =Â $\frac{Correct\ sum}{number\ of\ observation}$

i.eÂ $\frac{4800}{100}=\ 48$

Hence, Option A is correct

Question 3:Â The average weight of 6 people increased by 2.5 kg when a new person came in place of other person weighing 55 kg. what can be the weight of new person (in kg)?

a)Â 70

b)Â 62.5

c)Â 60

d)Â 75.5

Solution:

Total weight increased = number of personsÂ $\times\$ increased weight

i.e;$6\times\ 2.5=15$

Weight of new person = weight of replaced person + Total increased weight

i.e; 55 + 15 = 70 kg

Hence, Option A is correct.

Question 4:Â A girl spends 76% of her income. If her income increases by 18% and her expenditure increases by 25%,then what is the percentage increase or decrease in her savings (correct to one decimal place)?

a)Â 6.9%, decrease

b)Â 4.2%, decrease

c)Â 5.7%, increase

d)Â 8.4%, increase

Solution:

Let the income of girl is 100

Expenditure is 76% of income

i.e,Â $\frac{76}{100}\times\ 100$

= 76

Saving = Income – expenditure

100 – 76 = 24

According to question,

income is increased by 18%

Increased income =Â $\frac{18}{100}\times\ 100\ +\ 100=118$

Expenditure is increased by 25%

increased expenditure =Â $\frac{25}{100}\times\ 76\ +\ 76=95$

New saving = 118 – 95 = 23

% decrease in saving =Â $\frac{\left(24-23\right)}{24}\times\ 100$

i.e;Â $4.16\ \simeq\ 4.2\ \%$

Hence, Option B is correct.

Question 5:Â The average of eleven numbers is 56. The average of first three numbers is 52 and that of next five numbers is 60. The 9th and 10th number are 3 and 1 more than the 11th number respectively. What is the average of 9th and 11th numbers?

a)Â 53.5

b)Â 52

c)Â 52.5

d)Â 54

Solution:

The average of first three numbers is 52.

Sum of the first three numbers = 52 x 3 = 156

The average of next five numbers is 60.

Sum of the next five numbers = 60 x 5 = 300

Let the 11th number be ‘n’.

The average of eleven numbers is 56.

Sum of the eleven numbers = 56 x 11 = 616

156 + 300 + (n + 3) + (n + 1) + n = 616

3n + 460 = 616

3n = 156

n = 52

Average of 9th and 11th numbers = $\frac{(n+3)+n}{2}$

= $\frac{107}{2}$

= 53.5

Hence, the correct answer is Option A

Checkout: CAT Free Practice Questions and Videos

Question 6:Â The total number of students in a school is 1400, out of which 35% of the students are girls and the rest are boys. If 80% of the boys and 90% of the girls passed in an annual examination, then the percentage of the students who failed is:

a)Â 16.5

b)Â 21.5

c)Â 17.4

d)Â 15.8

Solution:

Total number of students in a school is 1400.

35% of the students are girls and the rest are boys.

Number of girls =Â $\frac{35}{100}\times1400$ = 490

Number of boys = 1400 – 490 = 910

80% of the boys and 90% of the girls passed in an annual examination.

Number of students passed in the examination =Â $\frac{80}{100}\times910+\frac{90}{100}\times490$ =Â $728+441$ = 1169

Number of students failed in the examination = 1400 – 1169 = 231

Percentage of the students failed in the examination =Â $\frac{231}{1400}\times100$

= 16.5%

Hence, the correct answer is Option A

Question 7:Â The average of 23 numbers is 51. The average of first 12 numbers is 49 and the average of last 12 numbers is 54. If the twelfth number is removed, then the average of the remaining numbers (correct to two decimal places) is:

a)Â 53:25

b)Â 50.45

c)Â 51.75

d)Â 52.65

Solution:

The average of 23 numbers is 51.

Sum of the 23 numbers = 51 x 23 = 1173…….(1)

The average of first 12 numbers is 49.

Sum of the first 12 numbers = 49 x 12 = 588

Sum of the first 11 numbers + 12th number = 588…..(2)

The average of last 12 numbers is 54.

Sum of the last 12 numbers = 54 x 12 = 648

12th number +Â Sum of the last 11 numbers = 648…..(3)

Sum of the first 11 numbers + 12th number +Â 12th number + Sum of the last 11 numbers = 588 + 648

(Sum of the first 11 numbers + 12th number +Â Sum of the last 11 numbers) +Â 12th number = 1236

Sum of the 23 numbers +Â 12th number = 1236

1173 +Â 12th number = 1236

12th number = 63

Required average =Â $\frac{1173-63}{22}$

= 50.4545

Hence, the correct answer is Option B

Question 8:Â The average weight of a certain number of students in a class is 55.5 kg. If 4 students with average weight 60 kg join the class, then the average weight of all students in the class increases by 360 g. The number of students in the class, initially, is:

a)Â 46

b)Â 31

c)Â 41

d)Â 36

Solution:

Let the initial number of students = n

Average of the weight of ‘n’ students = 55.55

Sum of the weight of ‘n’ students = 55.55n

Sum of the weight of 4 students = 60 x 4 = 240 kg

According to the problem,

$\frac{55.55n+240}{n+4}=55.55+\frac{360}{1000}$

$\frac{55.55n+240}{n+4}=55.55+0.36$

$\frac{55.55n+240}{n+4}=55.86$

55.55n + 240 = 55.86n + 223.44

0.36n = 16.56

n = 46

The initial number of students = n = 46

Hence, the correct answer is Option A

Question 9:Â The average monthly salary of 60 employees of a factory is â‚¹29900. If two officers are getting â‚¹90000 each and the average salary of 8 supervisors is â‚¹65000, then what is the average salary (in â‚¹) of the remaining employees?

a)Â 22680

b)Â 29080

c)Â 21080

d)Â 21880

Solution:

The average monthly salary of 60 employees of the factory is â‚¹29900.

Total monthly salary of 60 employees of the factory = 29900 x 60 = â‚¹1794000

Two officers are getting â‚¹90000 each.

Sum of the salary of two officers = 2 x 90000 = â‚¹180000

The average salary of 8 supervisors is â‚¹65000.

Total salary of 8 supervisors = 65000 x 8 =Â â‚¹520000

Total salary of remaining 50 employees of the factory =Â 1794000 –Â 180000 –Â 520000 =Â â‚¹1094000

Average of remaining 50 employees of the factory =Â $\frac{1094000}{50}$ =Â â‚¹21880

Hence, the correct answer is Option D

Question 10:Â Weight of A is 20% more than weight of B, whose weight is 30% more than weight of C. By how much percent weight of A is more than weight of C?

a)Â 44

b)Â 56

c)Â 69

d)Â 35.89

Solution:

Weight of B is 30% more thanÂ weight of C.

B =Â $\frac{130}{100}\times$C

Weight of A is 20% more than weight of B.

A =Â $\frac{120}{100}\times$B =Â $\frac{120}{100}\times\frac{130}{100}\times$C =Â $\frac{156}{100}$C

Required percentage =Â $\frac{\frac{156}{100}C-C}{C}\times100$

=Â $\frac{56}{100}\times100$

= 56%

Hence, the correct answer is Option B