Average Maths Questions Methods and Tricks

AVERAGE

Example 1: The average weight of five persons sitting in a boat is 38 kg. The average weight of boat and the persons sitting in the boat is 52 Kg. What is the weight of the boat?

Solution:  Method 1:  Avg. weight of 5 persons = 38 Kg.
  ⇒ Sum of weight of persons = 38 x 5 = 190
Avg. weight of boat and persons = 52 kg.
  ⇒ Sum of weight of boat and persons = 6 x 52= 312
  ⇒ Weight of boat = 312 – 190 = 122 Kg

 Method 2:  Quickest & Less calculative:
Weight of new entrant = n x new avg. + No. of old members x change in avg.
  ⇒ 1 x 52 + 5 x 14 = 122 Kg.

Example 2: The average of 100 numbers is 44. The average of these 100 numbers and 4 other new numbers is 50. The average of the four new numbers will be?

Solution: Sum of 4 new numbers = n x new avg. + original numbers x change in avg.
  ⇒ 4 x 50 + 100 x 6 = 800
  ⇒ Avg. of 4 new numbers = 800/4 = 200

Example 3: The average of all the numbers between 6 and 50 which are divisible by 5 will be?

Solution: Sum of multiples of 5 between 6 & 50 = [Sum of multiples of 5 below 50 – Sum of multiples of 5 below 6]
   ⇒ 5 + 10 +….+45 – 5
   ⇒ 5(1+2+…+9) – 5
   ⇒ 5(9 x 10)/2 – 5   {Since Sum of n numbers = n(n+1)/2}
   ⇒ 220
   ⇒ Avg. of multiples of 5 between 5 & 50 = 220/8 = 27.5

Example 4: The average of 15 numbers is 7. If the average of the first 8 numbers be 6.5 and the average of last 8 numbers be 9.5, then the middle number is?

Solution:  Method 1:  Sum of 15 numbers = 15 x 7 = 105
Sum of first 8 numbers = 8 x 6.5 = 52
Sum of last 8 numbers = 8 x 9.5 = 76
  ⇒ Middle 8th number = 52 + 76 - 105 = 23

 Method 2:  Direct Formula:
Avg. of all numbers = A
Avg. of first few numbers = x
Avg. of last few numbers = y
Desired number = A + Middle no. x [(y - A) + (x - A)]
  ⇒ Middle 8th number = 7 + 8 x [(9.5 - 7) + (6.5 - 7)]
  ⇒ 7 + 16 = 23

Example 5: The average of 7 consecutive numbers is 20. The largest of these numbers is?

Solution:  Method 1:  Basic: Let the numbers be x, x+1, x+2, x+3, x+4, x+5, x+6
  ⇒ Sum of numbers = 7x + 21
  ⇒ Avg. of numbers = (7x + 21)/7 = 20
  ⇒ x = 119/7 = 17
  ⇒ Largest number = 17 + 6 = 23

 Method 2:  Quick and conceptual: Since the numbers are consecutive, they form an arithmetic series with common difference.
As 7 is odd number, so 20 must be the middle number.
  ⇒ Series will be = 17, 18, 19, 20, 21, 22, 23
Hence largest number = 23

Example 6: If the average of 6 consecutive even numbers is 25, the difference between the largest and the smallest number is?

Solution: As numbers are 6 consecutive even, so there must be three numbers before 25 and three after 25.
Series will be = 20, 22, 24, 26, 28, 30
Difference between largest and smallest = 30 - 20 = 10

Example 7: The average of nine consecutive numbers is n. If the next two numbers are also included the new average will be increased by?

Solution: Since there are 9 consecutive numbers with average n, then middle number must be ‘n’.
  ⇒ Tenth and eleventh number will be = (n+5) & (n+6)
  ⇒ New avg. = (9n + n+5 + n+6)/11 = n + 1
Hence new avg. will increase by 1

Example 8: The average salary, per head, of all the workers of an institution is Rs 60. The average salary of 12 officers is Rs 400; the average salary, per head, of the rest is Rs 56. The total number of workers in the institution are?

Solution:  Method 1:  Basic: Let no. of other workers except officers = x
Total salary of all the workers = 60 x (12 + x)
Total salary of 12 officers = 12 x 400 = Rs 4800
Total salary of the rest except officers = 56x
  ⇒ 4800 + 56x = 60 x (12 + x)
  ⇒ x = 1020
  ⇒ Total no. of workers = 1020 + 12 = 1032

 Method 2:  By Alligation:

  ⇒ Total No. of Officer : Total No. of other workers = 4:340 = 1:85   
It means for 1 officer there are 85 other workers
  ⇒ For 12 officer, total no. of other workers = 12 x 85 = 1020
  ⇒ Total no. of workers = 1020 + 12 = 1032

Example 9: The average of marks of 14 students was calculated as 71. But it was later found that the marks of one student had been wrongly entered as 42 instead of 56 and of another as 74 instead of 32. The correct average is?

Solution:  Method 1:  Sum of marks of 14 candidates = 14 x 71 = 994
New sum after correcting error = 994 + (56 - 42) + (32 - 74) = 966
  ⇒ Correct Average = 966/14 = 69

 Method 2:  Difference of marks = (42 + 74) - (56 + 32) = 28
It means 28 marks were extra during calculation of average.
In term of average, 28 marks = 28/14 = 2 marks
So average was 2 marks more.
  ⇒ Correct average will be 2 marks less = 71 – 2 = 69

Example 10: A student finds the average of ten 2-digit numbers. While copying numbers, by mistake, he writes one number with its digits interchanged. As a result his answer changed. As a result his answer is 1.8 less than the correct answer. The difference of the digits of the number, in which he made mistake, is?

Solution: Average is 1.8 marks less.
In term of sum of numbers, avg. of 1.8 marks for 10 numbers = 1.8 x 10 = 18 marks
Hence, Difference between the number and the number formed by interchanging digits = 18
So, number must be 35. Since 53 – 35 = 18
Difference of digits = 5 – 3 = 2

Example 11: The average marks obtained by 40 students of a class is 86. If the highest marks are removed, the average reduces by one mark. The average marks of the top 5 students is?

Solution:  Method 1:  Basic:
Sum of marks of 40 students = 40 x 86 = 3440
Sum of marks of 35 students = 35 x 85 = 2975
Sum of marks of 5 students = 3440 – 2975 = 465
Average of marks of 5 students = 465/5 = 93

 Method 2:  Formula based:
Marks of removed candidates = n x New avg. + Old number of candidates x Change in avg.
   ⇒ 5 x 85 + 40 x 1 = 465 marks
Average of 5 students = 465/5 = 93

Example 12: A cricket batsman had a certain average of runs for his 11 innings. In the 12th innings, he made a score of 90 runs and thereby his average of runs was decreased by 5. His average of runs after 12th innings is?

Solution:  Method 1:  Basic: Let the batsman’s average in 11 innings be x runs.
  ⇒ (11x + 90)/12 = x – 5
  ⇒ x = 150
Required Avg. = 150 – 5 = 145 Runs

 Method 2:  Formula Based:
Runs in last innings = |n x New average - No. of old innings x (change in average)|
Let new average be x.
   ⇒ 90 = x – 11 x 5
   ⇒ x = 145 Runs

Example 13: The average age of 8 men is increased by 2 years when two of them whose age are 21 and 23 years replaced by two new men. The average age of the two new men is?

Solution:  Method 1:  Basic: Let the average age was x.
Sum of ages of 8 men = 8x
New average = x + 2
   ⇒ 8x + weight of new men – 21 – 23 = 8(x + 2)
Weight of new men = 60 years
   ⇒ Avg. = 60/2 = 30 years

 Method 2:  Direct Formula:
Weight of new persons = |Weight of removed persons + No. of persons x change in avg.|
   ⇒ Weight of new persons = 21 + 23 + 8 x 2 = 60 years
   ⇒ Avg. = 60/2 = 30 year