The average marks of a class of 35 children is 35. The marks of one of the students, who got 35, was incorrectly entered as 65. What is the correct average of the class?
(a) 33.76
(b) 34.14
(c) 35.24
(d) 36.50
(e) 34.49
Answer: (b)
Total correct marks of 35 children
= 35 × 35 + 35 – 65
= 1225 – 30 = 1195
Required average = \(\displaystyle \frac{{1195}}{{35}}=34.14\)
The average height of 30 boys out of a class of 50 is 160 cm. If the average height of the remaining boys is 165 cm, the average height of the whole class (in cm) is :
(a) 161
(b) 162
(c) 163
(d) 164
(e) 165
Answer: (b)
Average height of whole class
\(\displaystyle \frac{{30\times 160+20\times 165}}{{50}}cm\)
\(\displaystyle \frac{{4800+3300}}{{50}}cm\)
\(\displaystyle \frac{{8100}}{{50}}cm=162cm\)
The average of 7 consecutive numbers is 20. The largest of these numbers is :
(a) 24
(b) 23
(c) 22
(d) 20
(e) 19
Answer: (b)
Average of 7 consecutive numbers is 20.
Since the numbers are consecutive, they form an arithmetic series with common difference 1.
Since, 7 is odd, 20 must be the middle number.
We can write the series as below,
17, 18, 19, 20, 21, 22, 23
Therefore, The largest of these numbers is 23
The average of odd numbers upto 100 is
(a) 50.5
(b) 50
(c) 49.5
(d) 49
(e) None of these
Answer: (b)
Average of the first n natural odd numbers = \(\displaystyle \frac{n}{2}\)
Number of odd numbers upto 100 = \(\displaystyle \frac{{50}}{2}\)
Other Method:
Trick – The average of first ‘n’ consecutive odd natural numbers i.e. 1, 3, 5, ….. (2n – 1) = n
Odd numbers are 1, 3, 5, …………., 99
Total odd numbers are= 50
Therefore, Average = 50
The average of five consecutive positive integers is n. If the next two integers are also included, the average of all these integers will
(a) increase by 1.5
(b) increase by 1
(c) remain the same
(d) increase by 2
(e) None of these
Answer: (b)
Five consecutive integers are : x, x + 1, x + 2, x + 3 and x + 4
Their average \(\displaystyle \begin{array}{l}=\frac{{x\text{ }+\text{ }x\text{ }+\text{ }1\text{ }+\text{ }x\text{ }+\text{ }2\text{ }+\text{ }x\text{ }+\text{ }3\text{ }+\text{ }x\text{ }+\text{ }4}}{5}\\=\frac{{5x+10}}{5}=x+2\end{array}\)
New average \(\displaystyle \begin{array}{l}=\frac{{5x+10+x+5+x+6}}{7}\\=\frac{{7x+21}}{7}=x+3\end{array}\)
Difference = x + 3 – x – 2 = 1
