Mathematically, it is defined as the ratio of summation of all the data to the number of units present in the list.
Average = $\mathbf{\frac{X_{1}+X_{2}+X_{3}+X_{4}+…..X_{n}}{n}}n$
OR
Average = $\frac{Sum \;of\; Observations\;}{Total\; Number\; of\; Observations\;}$
Average Speed : Average speed is calculated using the below formula
Average Speed = $\frac{Total\; Distance\;}{Total\; Time\;}$
Also , Formula to calculate Average speed when $X$ travels at speed $a$ and $b$ for the same amount of time is $\frac{a+b}{2}$
Average Velocity : We calculate Average Velocity using the below formula : $\frac{Displacement}{Total\; Time\;}$
1: The average age of 39 boys and a girl is are 10 years. If the age of the girl is excluded the average age of the group is reduced by 1. What is the age of the girl ?
Options:
A. 45 years
B. 46 years
C. 50 years
D. 49 years
Solution:
The average of 39 boys + 1 girl = 10
Sum of ages of 39 boys + 1 girl = 10 x (39+1)= 400 ……………… Eqn (i)
When the girl is excluded from the average age the new average = 10 - 1= 9
The sum of the ages of 39 boys= 39 x 9 = 351 ………………Eqn(ii)
Girl’s age= (i)- (ii) = 400 - 351 = 49
Alternatively :When the girl leaves the group, she takes with her 1 year (i.e the change in the new average) from each of the 39 students along with the 10 years of her average age.
So the age of the girl will (39 x 1)+ 10 that is 49 years.
Correct option: D
2. The age of Ritesh at the time of his wedding was 27 years, while the age of his wife at that time was 25 years. Four years after their marriage the average age of Ritesh, his wife, and his son is 21 years. Find Age of Ritesh’s son.
Options:
A. 3 years
B. 2 years
C. 5 years
D. 4 years
Solution:
Let the current age son be x
Ages of Ritesh and his wife after 4 years of their marriage = 27+4 = 31 and 25+4 =29 years respectively.
Average age of the family =21
$\implies \frac{(31+29+x)}{3} = 21$
$\implies x + 60 = 63$
$\implies x = 63 - 60$
$\implies x = 3$ year’s
Therefore, the current age Ritesh's son is 3 years.
Correct option: A
1: The batting average of Sachin in 15 innings is 50. The difference between the runs of his best and worst innings is 40. Excluding the best and the worst innings the average of 13 innings played by Sachin is 50. Calculate Sachin’s best score.
Options:
A. 80 runs
B. 70 runs
C. 100 runs
D. 98 runs
Solution:
Total score of Sachin in 15 innings = 50 x 15 = 750
Total score of Sachin in 13 innings (excluding his best and worst inning’s scores) = 13 x 50 = 650
Sum of Sachin’s score in his best and worst innings = 750 - 650 = 100
The difference of Sachin’s score in his best and worst innings = 40
B+W = 100…(1)
B-W = 40…(2)
adding equation 1 and 2; (1) + (2)
2B = 140
B = 70
Therefore, Sachin’s score in his best innings was 70 runs.
Correct option: B
2. The average marks of a class gets increased by 5 when Raj's marks are wrongly entered as 75 instead of 55. Find the number of students in the class.
Options:
A. 3 students
B. 4 students
C. 5 students
D. 7 students
Solution:
Let the total strength of class be n.
Let sum of scores of rest of students is x
Let the average when marks of Raj are 75 = A1= $\frac{75+x}{n}$
The average of class when Raj's marks is 55 = A2 = $\frac{55+x}{n}$
A1-A2 = 5
$\frac{75+x}{n} - \frac{55+x}{n} = 5n$
20 = 5n
n = 4
Therefore, the total number of students in the class is 4.
Correct option: B
3. A team of 5 players participated in a competition. The best player of the team scored 50 points. Had he scored 80 points, the average score of the team would have been 75. Calculate the total points scored by the team.
Options:
A. 356
B. 390
C. 345
D. 333
Solution:
Let the sum of of points scored by 4 players of the team other then the best player be x.
Actual average = $\frac{x+50}{5}$
Required average = 75 = $\frac{x+80}{5}$
$=> \frac{(31+29+x)}{3} = 21$
x = (75×5)−80(75×5)-80
x = 375-80
x = 295
Total points scored by the team = 295 + 50 = 345
Correct option: C
1: Simon travels from his home to office at the speed of 50 Kmph. While returning from office to home his speed increases by 20%. Calculate Simon’s average speed
Options:
A. 54.54 Kmph
B. 56.45 Kmph
C. 54.45 Kmph
D. 56.54 Kmph
Solution:
Simon’s speed while returning home = $120$ x $\frac{50}{100} = 60km$
Average speed = $\frac{2ab}{a+b}$
That is, $\frac{(2×50×60)}{(50+60)} = \frac{6000}{110}$ = 54.54 Kmph
Therefore, the average speed of Simon will be 54.54 Kmph.
Correct option: A
2. Mikel travels at speed 40 km/hr for half the time and speed 30 km/hr for other half of the time. Then, what will be average speed of Mikel?
Options:
A. 45 km/h
B. 35 km/h
C. 67 km/h
D. 23 km/h
Solution:
Average speed of Mikel = $\frac{a+b}{2}$ = $\frac{40+30}{2}$ = 35 km/h
Correct option: B
3. Lilesh traveled a distance of 500 km partially on a motorbike and partially in car. The speed of the motorbike was 50 Kmph and distance covered was 200 Km. The total time taken in the journey was 6 hours. Find the average speed for the entire journey.
Options:
A. 87.54 Kmph
B. 78 Kmph
C. 93.45 Kmph
D. 83.33 Kmph
Solution:
Average Speed = $\frac{Total\; Distance\;}{Total\; Time\;}$
Average speed = $\frac{500}{6}$
= 83.33 Kmph
Correct option: D
1:The average of 5 numbers is 50. The average of the first and the second number is 40. Similarly, the average of the fourth and the fifth number is 25. Find the third term of the series.
Options:
A. 128
B. 220
C. 120
D. 112
Solution:
Let the third term be x
Average of 5 numbers = 50
Therefore, the sum total of all 5 numbers = 50 x 5 = 250
Sum of first two numbers = 40 x 2 = 80
Sum of fourth and fifth term = 25 x 2 = 50
80+x+50 = 250
x = 250-130
x = 120
Correct option: C
2.The average of 5 consecutive numbers is 50. Find the numbers.
Options:
A. 48, 49 , 50, 51, 52
B. 47, 48 , 49, 50, 51
C. 49, 50 , 51, 52, 53
D. 46, 47 , 48, 49, 50
Solution:
Let the numbers be (n-2), (n-1), (n), (n+1), (n+2)
Sum of all numbers = n-2+n-1+n+n+1+n+2 = 50 x 5
5n = 250
n = 50
Numbers = 48, 49 , 50, 51, 52
Correct option: A
3.Calculate the average of 5 consecutive odd numbers greater than 15.
Options:
A. 22
B. 19
C. 21
D. 20
Solution:
Odd numbers greater than 15 = 17, 19, 21, 23, 25
Average = $\frac{(17+19+21+23+25)}{5} = 21$
Correct option: C