Solutions
Problem 1 :
A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 5, 3, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What should be the number in place of x ?
Solution :
Given : Average of 3, 11, 7, 9, 5, 3, 8, 19, 17, 21, 14 and x is 12.
That is,
Sum of all the given numbers / number of numbers = 12
(3 + 11 + 7 + 9 + 5 + 3 + 8 + 19 + 17 + 21 + 14 + x) / 12 = 12
(117 + x) / 12 = 12
Multiply both sides by 12.
(117 + x) = 12 ⋅ 12
117 + x = 144
Subtract 117 from both sides.
x = 27
So, the number should be in the place of 'x' is 27.
Problem 2 :
The average age of 30 kids is 9 years. If the teacher's age is included, the average age becomes 10 years. Find the teacher's age.
Solution :
Given : The average age of 30 kids is 9 years.
That is,
Total age of 30 kids / 30 = 9
Multiply both sides by 30.
Total age of 30 kids = 9 ⋅ 30
Total age of 30 kids = 270
Given : If the teacher's age is included, the average age becomes 10 years
That is,
(Total age of 30 kids + Age of the teacher) / 31 = 10
(270 + Age of the teacher) / 31 = 10
Multiply both sides by 31.
270 + Age of the teacher = 10 ⋅ 31
270 + Age of the teacher = 310
Subtract 270 from both sides.
Age of the teacher = 40 years
Problem 3 :
The average of 6 numbers is 8. What is the 7th number , so that the average becomes 10 ?
Solution :
Given : The average 6 numbers is 8.
That is,
Sum of 6 numbers / 6 = 8
Multiply both sides by 6.
Sum of 6 numbers = 8 ⋅ 6
Sum of 6 numbers = 48
Given : If the 7th number is included, the average becomes 10.
That is,
(Sum of 6 numbers + 7th number) / 7 = 10
(48 + 7th number) / 7 = 10
Multiply both sides by 7.
48 + 7th number = 10 ⋅ 7
48 + 7th number = 70
Subtract 48 from both sides.
7th number = 22
Problem 4 :
David's average score in the last 9 tests is 80. What should be his score in his next test, so that his average score will be 82 ?
Solution :
Given : The average score of 9 tests is 80.
That is,
Sum of scores in 9 tests / 9 = 80
Multiply both sides by 9.
Sum of scores in 9 tests = 80 ⋅ 9
Sum of scores in 9 tests = 720
Let "x" be his score in his next test.
Given : Average score of 10 tests is 82.
Then, we have,
(Sum of scores in 9 tests + x) / 10 = 82
(720 + x) / 10 = 82
Multiply both sides by 10.
720 + x = 82 ⋅ 10
720 + x = 820
Subtract 720 from both sides.
x = 100
So, David score in the next test should be 100.
Problem 5 :
In Kevin's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kevin and he thinks that Kevin's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight is less than 68 kg. If all of them are correct in their estimation, what is the average of different probable weights of Kevin ?
Solution :
Let Kevin's weight be "x" kg.
According to Kevin, we have
65 < x < 72
According to Kevin's brother, we have
60 < x < 70
According to Kevin's mother, we have
x < 68
The values of 'x' which satisfy all the above inequalities are 66 and 67.
So, the different probable weights of Kevin are 66 kg and 67 kg.
Average of 66 and 67 = (66 + 67) / 2
Average of 66 and 67 = 133 / 2
Average of 66 and 67 = 66.5
So, the average of different probable weights of Kevin is 66.5 kg.
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