Question 1

The average age of A, B , C, D five years ago was 45 years . by including P, the present average age of all the five is 49 years . the present age of P is?

Accepted Answer

Present age of P = [(49 x 5 ) - (4 x 45 + 4 x 5 ) ] years = 45 years.

Question 2

Out of three number, the first is twice the second and is half of the third. If the average of the three number is 56, the three number on order are?

Accepted Answer

Let the numbers be 2y, y and 4y ∵ Average = (2y + y + 4y)/3 ⇒ 7y/3 = 56 ∴ y = (3 x 56) / 7 = 24 Hence , the number in order are 48, 24 and 96

Question 3

The sum of seven numbers is 235. The average of the first three is 23 and that of the last three is 42. The fourth number is?

Accepted Answer

Sum of 7 numbers = sum of first 3 numbers + N4 + sum of last 3 numbers. (23 x 3) + N4 + (42 x 3) = 235 ⇒ N4 = 40 ∴ 4th number = 40

Question 4

Find the average of the following set of scores 253, 124, 255, 534, 836, 375, 101, 443, 760

Accepted Answer

Required average = (Sum of all scores) / (Number of scores) = (235 + 124 + 255 + 534 + 836 + 375 + 101 + 433 + 760)/9 = 3681/9= 409

Question 5

Find the average of the following set of scores. 124, 856, 331, 227, 963, 338, 259, and 662

Accepted Answer

Required average = ( Sum of scores) / (No of scores) = (124 + 856 + 331 + 227 + 963 + 338 + 259 + 662)/8 = 3760/8 = 470

Question 6

Find the average of all prime numbers between 60 and 90.

Accepted Answer

Average of all prime numbers between 60 and 90 = (61 + 67 + 71 + 73 + 79 + 83 + 89)/7 = 523/7 = 74.7

Question 7

The average of nine numbers is 50. The average of the first five numbers is 54 and that of the last three numbers is 52. Then the sixth number is

Accepted Answer

Required number = (Sum of all numbers) - [(Sum of first 5 numbers) + (Sum of last 3 numbers) = 50 x 9 - ( 54 x 5 + 3 x 52) = 450 - (270 + 156) = 450 - 426 ≈ 24

Question 8

The Average of six observation is 12. The average decreases by 1 when a new observation is included. What is the seventh observation?

Accepted Answer

Total of 6 observation = 6 x 12 = 72 Total of 7 observation (including the 7th) = 7 x (12 - 1) = 77 ∴ 7th observation = 77 - 72 = 5

Question 9

Average to ten number is 7 . If every number is multiplied by 12, What will be the average of new numbers?

Accepted Answer

Here initial average = 7 As we know that, if all the numbers are multiple by a certain number, then their average must be a multiple of that number. ∴ New average = 7 x 12 = 84

Question 10

The average weight of 60 students in a class was calculated to be 40 kg. Later it was found out that the weight of one of the students was calculated as 36 kg, whereas his actual weight was 33 kg. What is the actual average weight of the students in…

Accepted Answer

Total correct weight = (40 x 60) - 36 + 33 = 2400 - 3 = 2397 kg ∴ Required average weight = 2397/60 = 39.95 kg

Question 11

A batsman has a certain average of runs for 14th innings. He scores 72 runs in the 15th innings, thus his average increased by 2. Find the average after 15th innings.

Accepted Answer

Let the average of 14th innings = x ∴ Average of 15th innings = x+2 Then, 14x+72 = 15(x+2) ⇒ 14x+72 = 15x +30 ∴ x = 42 ∴ Average for 15th innings = x+2 = 42 + 2 = 44

Question 12

The Average of 5 observation is 15. If three new observation 16.5, 18 and 14.5 be added. Then find the new average.

Accepted Answer

Total of 5 observations = 5 x 15 = 75 New sum = 75 + 16.5 + 18 + 14.5 = 124 New Average = 124 ⁄ 8 = 15.5

Question 13

The average of x 1, x 2 and x 3 is 14. Twice the sum of x 2 and x 3 is 30. What is the value of x 1?

Accepted Answer

x1 + x2 + x3 = 3 x 14 = 42 ....(i) and (x1 + x2) x 2 = 30 ⇒ x1 + x2 =15 ....(ii) From Eqs. (i) and (ii), we get x1 + 15 = 42 ∴ x1 = 42 - 15 = 27

Question 14

The average age of 45 students in a class in 13 yr, 15 new students with an average age of 18 yr join the class. What is the average age of all the students in the class?

Accepted Answer

Sum of the age of 45 students = 45 x 13 = 585 yr Sum of the age of 15 students = 15 x18 = 270 yr Sum of the age of all students = sum of the age of 45 students + sum of the age of 15 students = 585 yr + 270 yr =855 yr ∴ Average age of all students = 855 ⁄ 60 = 14.25 yr

Question 15

The average salary of the entire staff in an office is ₹ 200 per day. The average salary of officers is ₹ 550 and that of non officers is ₹ 120. If the number of officers is 16, then find the number of non officers in the office.

Accepted Answer

Let the number of non-officers be y. Then number of entire staff = 16 + y Sum of total salary = 200(16 + y) =16 x 550 + 120y ⇒ 120y + 550 x 16 = 200x16 + 200y ⇒ 3y +220 = 80 + 5y ⇒ 2y = 140 ∴ y = 140/2 = 70

Question 16

Average value of property of three persons A, B, C is ₹ 11111 crore. The property of C is as less as the property of A is greater than the average property of both C and A. The value of property of B is.

Accepted Answer

( A + B + C ) / 3 = 11111 ....(i) Also, ( A+C ) / 2 = 11111 ....(ii) From Eqs. (i) & (ii), we get B = ₹ 11111 crore

Question 17

The average age of 25 girls in a class decreases by 6 months when a new girl takes the place of 20 year old girl. Find out the age of the new girl.

Accepted Answer

Total decrease in the age = 25 x 1/2 =12.5 yr ∴ Age of new girl = 20 - 12.5 = 7.5 yr

Question 18

The average age of all the members of a family is 25 yr. The average age of males is 30 yr while the average age of females is 20 yr. If the number of females in the family is 15, then find out the number of males.

Accepted Answer

Let number of males = y Then, total number of family members = 15 + y Sum of age of family members = sum of age of males + sum of age of females 25(15 + y) = 30y + 15x20 ⇒ 375 + 25y = 30y + 300 ⇒ 5y = 75 ∴ y = 15

Question 19

Out of three annual exams, each of maximum marks of 600, Lalit secured average marks of 55% and 60% in the first two exams. How many marks should Lalit score in 3rd test exam to make his overall average equal to 70%?

Accepted Answer

Total required marks in the three examinations = (3 x 600 x 70)/100 =1260 Marks secured in first examination = 55% of 600 = ( 600 x 55)/100 =330 Marks secured in second examination = 60% of 600 = ( 600 x 60)/100 =360 Thus, the required marks in third examination = 1260 - (330 + 360) = 1260 - 690 = 570

Question 20

The mean marks of 30 students in a class is 58.5. Later on, it was found that 75 was wrongly recorded as 57. Find the correct mean.

Accepted Answer

Correct mean = ( sum of total marks - 57 + 75 ) / 30 = ( 30 x 58.5 - 57 + 75 ) / 30 = (1755 + 18 ) / 30 = 59.1

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