Averages - competitive exams

1. A v e r a g e s Dr. S. Murali Professor & Head Dept. of Mechanical Engineering

2. Definition The average of any set of quantities can be found by adding all the numbers together then dividing by the number of quantities in the set average = total of quantities / number of quantities

3. Math Formula Formula

4. In a Crude way- Simple Initial Avg. Final Avg. Diff Final Avg. =Initial + Diff.

5. In a Crude way- Simple Excess Deficiency Excess loss = Deficiency gain I Set Avg. I I Set Avg. Final Avg.

6. 1. Find the average of 4,6,9 and 11 Sol: given numbers are small apply conventional formula Sum of the numbers is = 30 Average = Sum/No. of Numbers =30/4 =7.5

7. Sol: Numbers are larger: avoid using conventional method Method 1 Sum 36 This excess is shared among 6 numbers =36/5=7.2 Average of 6numbers is =Assumed avg. + avg. of excess 146+7.2= 153.2 146 152 149 162 157 2 . Find the average of 146,152, 149, 157 and 162 +6 +3 +16 11 assume average 146 and excess available at other numbers are

8. 146,152, 149, 157 and 162 Method 2 arranging the numbers in ascending order Sum 6 This excess is shared among 6 numbers =6/5=1.2 Average of 5 numbers is =Assumed avg. + avg. of excess 152+1.2= 153.2 146 149 152 157 162 -6 -3 5 10 assume average 152 and find excess/less available at other numbers are

9. 3 . Find the average of 16,24,32,40 and 48 Sol: Hence middle number 32 is the average 16 24 32 40 48 Series in with odd numbers(5) 8 8 8 8 Arithmetic Progression

10. 4. Find the average of 16,24,32,40,48 and 56 Sol: Average of middle numbers is the average =(32+40)/2=36 16 24 32 40 48 56 Series in with even numbers(6) Arithmetic Progression 8 8 8 8 8

11. 5.The avg. of 7 consecutive numbers is 33. the largest of these numbers is? Sol: 7 consecutive numbers 7 consecutive even numbers 7 consecutive odd numbers All these combinations are in AP and as 7 being odd number, the middle number equals to the average The largest of numbers is x=33+3=36 X-3 X-2 X-1 X+1 X+2 X+3 X-6 X-4 X-2 X+2 X+4 X+6 X-5 X-3 X-1 X+1 X+3 X+5 X X X

12. 6. The avg. of 30 students in a class is 12 years. The avg. age of a group of 5 of the students is 10 years and that of another group of 5 of them is 14 years. What is the avg. of the remaining students? Excess of 5 students =less required by 5 students No change in remaining average, hence average of remaining is 12 20 5 5 12 10 10 14 10

13. 7. The average of eight numbers is 14. the average of six of these numbers is 16.What is the average of remaining two numbers. 14 16 Gain by 6 = loss by 2 12 = x  2 Average loss=6 Average of remaining two = 14-6=8 8

14. 8 . If the average mark of a student in four examinations is 40. If he got 80 marks in 5 th Examination. What is the New average ? Sol: Assume Avg. of 5 books is 40 Excess available at 5 th book is 40 Increase in average =40/5=8 New average =40+8 =48 40 8 80-40=40

15. 9 . If the average mark of a student in four examinations is 40. If he got 20 marks in 5 th Examination. What is the New average ? Sol: Assume Avg. of 5 books is 40 less shared by 5 th book is 20 Decrease in average =-20/5=-4 New average =40-4 =36 40 2 20-40=-20

16. 10. If the avg. age of 30 boys in a class is 15 yrs. When the age of class teacher is included the avg. is 16 years. What is the age of the class teacher? Sol: Un contributed age of teacher =Avg. of students= 15 Avg increase due to teacher addition is =New avg – initial avg = 16-15=1 Distributed age of teacher = X =1×31=31  Teacher age=assumed average + surplus =15+ X =15+31=46 15 16 X 15

17. 11. The average of 50 numbers is 38. What is the average of the numbers, if two numbers namely 45 and 55 are discarded. Excess at 45 and 55 is = 7 + 17 =24 38 55 45 17 7 Less available at 48 numbers is =24/48=0.5 The average of 48 numbers is =38-0.5 =37.5

18. 12. In a group the Avg. income of 6 men is Rs. 500 And that of 5 women is Rs. 280. What is the average income of the group? Sol: Assume average of 11 members as 280 Then excess average available with 6 men= 500-280=220 Total excess income available with 6 men=6×220 Increase in average of 11 members =6×220/11=120  New average =280+ 120=400 6 5 500 280 500-280=220 280

19. 13. A class has 2 sections, in one of which there are 40 students with an average of 14.5 years. The average of the class is 14.2 years . If there are be 32 students in the other section, its average is ? Excess available at 40 students =0.3 × 40=12 Increase in average 32 students =12/32=0.375 Average of 32 students x = 14.2-0.375 x = 13.825 40 32 14.5 x 14.2 0.375 Excess

20. 14. The average of 11 observations is 60. if the average of first five observations is 58 and that of last five is 56, then the sixth observations is Excess supplied by sixth observation to first five and last five is = 4*5+2*5 =30 Sixth observation=60 + 30=90 60 /11 56 /5 58 /5 X /1 Excess

21. 15. Mukhesh has twice money as sohan and sohan has 50% more money than pankaj has. If the average money of with them is Rs. 110, then muksh has Total money available =110*3 3x+1.5X+x =110*3 5.5 x = 110 *3 x = 110/5.5*3 x =60 Mukhesh money= 1.5 x 60=90. 110 x 1.5x 3x

22. 16. Three years ago, the average age of a family of 5 members is 17 years. A baby having been born. the average age of the family is the same today. What is the present age of the baby 17 20 Excess supplied 5 family members to the baby is = 3*5 =15 Age of the baby is=17 - 15=2 Excess 2 15

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24. Puzzle 1. A series from 1-9 is arranged in the boxes as shown such that the sum along vertical and horizontal direction is 13. find the number at G block? Sol: A B C D E F G H I 1 2 3 4 5 6 7 8 9

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