Averages Part 8

Problem 1: The average marks secured by 36 students was 52. But it was discovered that a term 64 was misread as 46. What is the correct mean of marks?

Solution:

64 – 46 = 18

52+\frac{18}{36}=52.50

Problem 2: The average age of the four players is 20 years. If the age of the coach is also included, the average age is increased by 20%. The age of the coach is?

Solution:

20\times \frac{1}{5}=4

20+\frac{5\times 4}{20}=40

Problem 3: The age of the captain of a cricket team of 11 members is 26 years and the wicketkeeper of the team is 3 years older than the captain, if the ages of these two excluded then the average age of the team is one year less than the average age of the whole team. What is the average age of the team?

Solution:

Captain =26

Wicket keeper = 29

11-2=9

2a+9*1=26+29

2a = 55-9 = 46

a = 23

Averages Part 7

Problem 1: A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is?

Solution:

Average for 11 tests =100

Average = 5

Average+number*increment = Total

a+11*5=100

a=100-55=45

Shortcut:

a=100-11*5=45

N=a+5=45+5=50

Problem 2: The average age of 20 boys is 14 years. A boy who is 16 years old left them and one new boy replaces him. Then the average becomes 13.8 years. Find the age of the new boy?

Solution:

20*14 – 16+a= 13.8*20

280-16+a=276

a = 276-264 = 12

Shortcut:

16-0.2*20=12

Problem 3: The average temperature of Monday, Tuesday and Wednesday was 30^{\circ}C and that of Tuesday, Wednesday and Thursday was 33^{\circ}C. If the temperature on Monday was 32^{\circ}C, then the temperature on Thursday was?

Solution:

T+W+TH = 33*3

M+T+W=30*3

TH-32=9

TH = 32+9 =41

Averages Part 6

Problem 1: The average weight of A, B, and C is 50kg. If the average weight of A and B be 48 kg and that of B and C be 45 kg, then find the weight of B?

Solution:

A+B+C = 50*3 =150

A+B = 48*2 =96

B+C = 45*2 = 90

A+B+B+C = 186

A+B+C=150

B=36

Problem 2: The average age of eleven players in a cricket team decreases by 2 months when two new players are included in the team replacing two players of ages 17 years and 20 years. The average age of new players is?

Solution:

\frac{17+20}{2}=\frac{37}{2}

\frac{2\times 11}{2}=11

18years 6 months – 11

17 years 7 months

Problem 3: The average wages of a worker for 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7days was Rs.92/day. What was his wage on the 8th day?

Solution:

15*90 – 7*87 – 7*92

1350 – 609 -644

1350 – 643 = 97

Shortcut:

90+(3-2)*7=97

Averages Part 5

Problem 1: Average of n numbers is‘a’. The first number is increased by 2, the second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is?

Solution:

Average of n numbers is a

Total =an+2+4+8+16

a=2, r=2, n=n

S_{n}=\frac{a\times (r^{n}-1)}{r-1} , r>1

\frac{a^{n}+\frac{2\times (2^{n}-1)}{2-1}}{n}

a+2\times \frac{2^{n-1}}{n}

Problem 2: In a class of 50 boys, the average height of 30 boys is 160 cm. If the average height of remaining boys is 100 cm, then what will be the average height of boys(in cm)?

Solution:

\frac{30\times 160+20\times 100}{50}

\frac{30\times 160}{50}+\frac{20\times 100}{50}=136

Problem 3: The average of the first three numbers is thrice the fourth number. If the average of all the four numbers is 10, then find the fourth?

Solution:

\frac{a+b+c}{3}=3d

a+b+c=9d

\frac{a+b+c+d}{4}=10

9d+d=4\times 10

10d=40

d=4

Problem 4: The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number, the average becomes 17.5. Then the numbers of natural numbers are?

Solution:

15n+30-5 = 17.5n

25 = 2.5n

n = 10

Shortcut:

\frac{30-5}{2.5}=\frac{25}{2.5}=\frac{250}{25}=10

 

Averages Part 4

Problem 1: The average age of class A and Class B is 11 years, the average age of class A is 8 years and the average age of class B is 15 years. If the number of students in class A is 52. Then find the number of students in class B?

Solution:

8*52 + 15*B =(52+B)*11

8*52 +15B = 52*11 +11B

4B=52(11-8)

B=39

Problem 2: The average of 20 numbers is 25. If one of the numbers is considered as 35 instead of 75, what will be the new average?

Solution:

25+\frac{35-75}{20}

25-\frac{40}{20}=23

Problem 3: The Average age of A and B is 50 years and that of B and C is 38 years. What is the difference between the age of A and C?

Solution:

A+B = 50*2 =100

B+C = 38*2 = 76

A-C = 24

Problem 4: The average of 30 numbers is 20. The average of the first 12 numbers is 26. Find the average of remaining numbers?

Solution:

20-\frac{6\times 12}{18}=16