Q.1. Find the average of numbers 87, 84, 86, 90, 82, 88, 78.

A. 85

B. 84

C. 83

D. 82

Sol : Option A
The sum of all the observations here is 87 + 84 + 86 + 90 +82 + 88 +78 = 595
Number of observations = 7
So, Average = 595/7 = 85

Q.2. The average of 4 terms is 20 and the 1st term is 1/3 of the remaining terms. What will be the first number?

A. 30

B. 20

C. 60

D. 80

Sol : Option B
Average of 4 terms = 20
Hence, the total sum of 4 terms = 80
Let terms be A,B,C,D
So, the sum will be A+B+C+D =80
Given, 3A = B+C+D
So, 4A = 80,
A = 20

Q.3. The average age of A, B and C was 25 years and that of B and C was 25 years. A’s present age is:

A. 30 years

B. 25 years

C. 40 years

D. 42 years

Sol : Option B
Average of A,B,C is 25
So, sum of their ages =75
Now, the sum of B and C will be 50 (because their average is 25)
So age of A =75 - 50 = 25 years

Q.4. The average of 7 consecutive numbers is n. If the next two numbers are included, the average will

A. increased by 2

B. remains the same

C. increased by 1

D. increased by 2

Sol : Option C
The average of 7 consecutive numbers is n implies that the 4th term is equal to n.
Now if we include next two terms then the average of 9 terms will be the 5th term. Now as the terms are consecutive, so the 5th term will be n + 1.

Q.5. For 9 innings, Boman has an average of 75 runs. In the tenth inning, he scores 100 runs, thus increasing his average . His new average is

A. Rs. 75

B. Rs. 100

C. Rs. 72

D. Rs. 77.5

Sol : Option D
Total score for 9 innings is 75×9 = 675
Total score after 10th innings = 675 + 100 = 775
So, average = 775 / 10 = 77.5

Q.6. For 9 innings, Roman has an average of 65 runs. In the tenth inning, he scores 200 runs, thus increasing his average . His average increased by

A. 78.5

B. 72

C. 13.5

D. 77.5

Sol : Option C
Total score for 65 innings = 65×9 = 585
Total score after 10th innings = 585 + 200 = 785
So, the new average is 785/10 = 78.5
So, the increment is of 13.5

Q.7. In a family of 8, the men eat on average 72kg of food and women eat on an average 50kg of food. The men and women are equal in number. A hungry woman named Neetu joined the family for dinner and the average consumption became 67.How much did Neetu eat (in kgs)?

A. Rs. 115

B. Rs. 80

C. Rs. 90

D. Rs. 85

Sol : Option A
As men and women are equal so , there are 4 women and 4 men so, total consumption will be 72×4 = 288(by men) and 50×4 = 200(by women)
Total consumption = 488.
But after including Neetu the average consumption for 9 people is given to be 67.So the total consumption will be 67×9 = 603. So, Neetu’s consumption will be = 603 – 488 = 115

Q8. In a hotel, the tariff for every odd dates is Rs.1000 and for even dates is Rs. 2000. If the man paid total of 30000 in all. For how many days did he stay in the hotel given that the first day is 5th date of the month?

A. 50

B. 20

C. 40

D. 60

Sol : Option B
Total tariff = 30000
So, for odd dates (5th , 7th , and so on) = 1000
And for even dates (6th , 8th and so on ) = 2000
So, the average amount of money for 2 days is Rs. 1500.
So, total amount paid = 30000
So , number of days he stayed in the hotel = 30000/1500 = 20.

Q9. The average of 5 terms is 50. If the first 4 terms are 45, 42, 119, and 84, what will be the last term?

A. 56

B. -20

C. -40

D. -50

Sol : Option C
Sum of all the terms = 250
Sum of first four terms = 45+42+119+84 = 290
So, the 5th term should be 250 – 290 = - 40.

Q10. If the average number of 8 terms is given to be 40 and the average of first 6 terms is given to be 35.What is the average of the remaining 2 terms?

A. 30

B. 55

C. 40

D. 42

Sol : Option B
Sum of all the 8 terms = 320.
The sum of first 6 terms = 210
Now , the sum of remaining terms = 320 – 210 = 110
So , the average of 2 terms would be = 110/2 = 55