Question 1: The average age of 20 students in a class is 12 years. If the age of the teachers is also added then the average becomes 13 years. Find out the age of the teacher?
Solution:
Method 1: solving by using the formula.
Let the age of the teacher be X years.
Then the formula for averages becomes
13 = (20 x 12 + X)/21 = (240 + X)/21
Multiplying both the sides by 21 to get
=> 273 = 240 + X
on solving for X
X = 33 years
So the age of the teacher is 33 years
Method 2 :
Adding the teachers age to the group increases the age of the students by 1 full year.
Since there are 20 students the total years added by the inclusion of the teacher = 20 years
And since the teacher is new addition to the group having an average of 13, 13 years more years are added.
Therefore total years added to the group due to the inclusion of the teacher = 13 + 20 = 33 years.
Solution:
Method 1: solving by using the formula.
Let the age of the teacher be X years.
Then the formula for averages becomes
13 = (20 x 12 + X)/21 = (240 + X)/21
Multiplying both the sides by 21 to get
=> 273 = 240 + X
on solving for X
X = 33 years
So the age of the teacher is 33 years
Method 2 :
Adding the teachers age to the group increases the age of the students by 1 full year.
Since there are 20 students the total years added by the inclusion of the teacher = 20 years
And since the teacher is new addition to the group having an average of 13, 13 years more years are added.
Therefore total years added to the group due to the inclusion of the teacher = 13 + 20 = 33 years.
Question 2: Rakesh bought 10 shirts for 250 each, 15 pants for 350 each and 20 handkerchiefs for 50 each. Find the average cost of each item.
Solution:
Here simply use the average formula :
Average = (10 x 250 + 15 x 350 + 20x50)/(45) = (2500 + 5250 + 1000)/45
= 194.44
Question 3: The average contribution of an office towards the health fund is 120 month. If the officer level employees are contributing Rs. 460 and non-officers staff is paying Rs 110. Then, if there are 15 officers then find the number of non-officer staff.
Solution:
Let the non-officer staff be X.
Then total staff would be (15 + X)
Then total contribution by all the staff = 120 (15 + X )
Using the formula for the average calculation
120 (15 + X ) = 6900 + 110 X
Solving for X
1800 + 120X = 6900 + 110X
120X - 110X = 6900 - 1800 = 5100
10X= 5100
X = 510
There fore non-Officer staff is 510
Question 4: A certain college while taking admission in the math course is assigning following weight age to the subjects.
Math - 4
Science - 3
English - 2
Hindi - 1
If a student has secured following marks in these subjects then find out arithmetic average and the weighted average of the marks.
Math - 70
Science - 60
English - 70
Hindi - 80
Solution:
Using the Formula for arithmetic averages we have,
Arithmetic average = (70 + 60 + 70 + 80) / 4 = 70
Using the formula for the weighted average we have
weighted average = ( 4 x 70 + 3x60 + 2x70 + 1x80)/(4 + 3 + 2 + 1)
= (280 + 180 + 140 + 80)/10
= 68
Difference between the arithmetic and the weighted average = 2
and it is due to the different weight age to various subjects in calculating the weighted average.
Question 5: One-third of the journey is covered at the rate of 25 km/hr , one-fourth at the rate of 30 km/hr and the rest at 50 km/hr. Find the average speed for the whole journey.
Solution:
For calculating speed we have to first find the total distance covered and total time taken.
Let the total distance be X km.
Then according to the question
1. (X/3) distance is covered at 25 Km/hr and time taken = X/75 hr
2. (1/4)X distance is covered at 30 Km.hr and time taken = X/120
3. Rest of the journey ( X - ( X/3 + X/4)) = 5X/12 at the rate 50 Km/hr and time taken = X/120
Therefore calculating the average time taken
Average Speed = Total Distance Covered/Total Time Taken
= X/ ( X/75 + X/120 + x/120)
= 1/(3/100) = 100/3
= 33(1/3) Km/Hr
Question 6: The average weight of 8 persons is increased by 2 Kg when one of them who weighs 56 kg is replaced by another person. The weight of the new person is :
Solution:
The new person added 2 kg to the average weight of 8 people. Then total new weight added to the group is 8 X 2 = 16 kg. In addition to this weight , The new person also replaced a member with 56 kg of weight. Then, then total weight of the new person = 56 + 16 = 72 Kg.
Question 7: In an examination, one section of class X, having 60 students scored an average marks of 55 and the second section, having 40 students scored an average of 45 marks.What is the overall average.
Solution:
Average = Total Marks scored by all the students / Total numbers of students
1. Total Marks scored by 60 students of first section = 55x60 = 3300
2. Total marks scored by 40 students of second section = 40 X 45 = 1800
Therefore total marks scored by all the students = 5100
and total students = 60 + 40 = 100
Using the formula for average mentioned above,
we have ,
average marks = 5100/100 = 51
Question 8: The average marks of a student in four subjects is 65. If the student scored 75 marks in the fifth subject then the average of the total marks is ?
Solution:
Marks in the four subjects = 65 x 4 = 260
Marks in the fifth subject = 75
Total marks in five subjects = 335
Therefore average marks in five subjects = (335)/5 = 67
Question 9: The average of 13 results is 50. If the average of the first 6 results is 49 and that of the last 6 is 53, then find the 7th results.
Solution:
Given that the averages of the first 6 results and the last six result are 49 and 53 respectively.
Then the total value of all the 12 result ( except the 7 th one) is = 49x6 + 53x6 = 612
And the total value of all the 13 result = 650
Therefore the seventh value would be = 650 -612 = 38
Question 10: A driver drives to the office at 50 km/her and returns at 30km/hr. Find the average speed.
Solution:
The key here is that the driver drives the same distance to and fro. Let the distance be D in one direction.
Then the total distance travelled wound be 2D.
Now, time taken while driving at 50 km/hr = D/ 50 hr
And time taken to while driving at 30 km/hr = D/30 hr
Now,
Average Speed = Total Distance Traveled/ Total Time Taken = 2D/(D/ 50 + D/30)
= 2/8/150 = 150/4 = 37.50 km/hr
Q 11. Out of 30 teachers of a school, a teacher of age 60 years retired. In his place another teacher of 30 years was appointed. As a result , the mean age of the teachers will
(a) decrease by 1 year (b) remain same
(c) decrease by 2 years (d) decrease by 6 month.
Answer:
Solution:
First Method:
Net result of the retirement and appointment of the teachers result in a net decrease of 30 years. These 30 years are distributed across 30 teachers. Thus, the mean decreases by 1 years.
Second method:
Let the mean of the 29 teachers that remain the same be x years, then
Final Mean - Initial mean = (29x +30 )/30 - (29x + 60)/30 = (29x + 30 -29x - 60)/30
= - 30/30 = - 1
Thus, the final mean is less than the initial mean by 1 year.
Q12. Average age of A, B and C is 84 years.When D joins them the average age becomes 80 years. A new person E, whose age is 4 years more than D replaces A and the average of B, C , D and E becomes 78 years. What is the age of A?
(a) 70 years (b) 80 years
(c) 50 years (d) 60 years
Answer: (b) 80 years
Solution:
Since Addition of D to the group reduces the average age of the group to 80 years. This means addition of D, reduces 4 x 4 = 16 years from the total age. Thus, the age of D is 84 - 16 = 68 years.
Again, given that the age of E is 4 years less than D then the age of E is 68 + 4 = 72 years.
Addition of E and exit of A reduces the group average to 78 years. This means A total of 4x2 = 8 years have been taken from the group total. Thus, age of A is 72 + 8 = 80 years.
Q13. The average of six numbers is 32. If each of the first three numbers is increased by 2 and each of the remaining three numbers is decreased by 4, then the new average is
(a) 35 (b) 34
(c) 31 (d) 30
Answer:(c) 31
Solution:
Each of the first three numbers is increased by 2 then the increases in total
= 3 x 2 = +6
And each of the 3 remaining three numbers are decreased by 4, then total decreased is 4 x 3 = - 12
Therefore, net increase/decrease = + 6 - 12 = -6
And the decrease in the mean = -6/6 = -1. And new mean = 32- 1 = 31
14. Five years ago, the average age of P and Q was 25. The average age of P, Q and R today is 25. Age of R after 5 years will be
(a) 15 (b) 20
(c) 40 (d) 35
Answer: (b) 20
Solution:
Sum of ages of P and Q, five years ago, is {(P - 5) + (Q - 5 ) = 50 years} And their present ages =P + Q = 60 years
Again, When R is added to the group then sum = (P + Q + R) = 60 + R
and average = (60 + R)/3 = 25 (given)
=> 60 + R = 75
=> R =15 years
And age of R after 5 years = 15+ 5 = 20 years
15. The average marks scored by 36 students was 52. But it was discovered that an item 64 misread as 46.What is the correct mean of marks?
(a) 54 (b) 53.5
(c) 53 (d) 52.5
Answer:(d) 52.5
Solution:
First method:
Total marks of the student with the misread item = 36 x 52 = 1872
Total marks of the student with the correct item = 1872 - 46 + 64 = 1890
Therefore, average = 1840/36 = 52.5
Faster Method:
Due to the error in reading, the total reduced by 64-46= 18. And the loss of 18 was distributed among 36 members. That is average reduced by 19/36 = 0.5. Therefore correct average is 52+ 0.5 = 52.5
Click Here for
Formula for arithmetic average and weighted averages
How to find Compound Interest
How to calculate Simple Interest
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