Marks of a student were wrongly entered in computer as 83, actual marks of that student were 63. Due to this mistake average marks of whole class got increased by half (1/2). Find the total number of students in that class.

Answer: Option C

Explanation:

\begin{aligned}
= \frac {7(1+2+3+...+10)}{10}
\end{aligned}

\begin{aligned}
= \frac {7(10(10+1))}{10 \times 2}
\end{aligned}

\begin{aligned}
= \frac {7(110)}{10 \times 2} = 38.5
\end{aligned}

Answer: Option C

Explanation:

Let the average weight of the 59 students be A.
So the total weight of the 59 of them will be 59*A.

The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs

\begin{aligned}
\frac{59A + 45}{60} = A - 0.2
\end{aligned}

=> 59A + 45 = 60A - 12
=> 45 + 12 = 60A - 59A
=> A = 57

Answer: Option A

Explanation:

\begin{aligned}
Average = \frac{3(1+2+3+4+5)}{5} = \frac{45}{5} = 9
\end{aligned}

Answer: Option C

Explanation:

Sum of n natural numbers
\begin{aligned}
= \frac{n(n+1)}{2}
\end{aligned}

\begin{aligned}
= \frac{30(30+1)}{2} = \frac{30(31)}{2} = 465
\end{aligned}

Answer: Option B

Explanation:

\begin{aligned}
\frac{(55 \times 50) +(60 \times 55) +(45 \times 60) }{55 + 60 + 45}
\end{aligned}

\begin{aligned}
\frac{8750}{160} = 54.68
\end{aligned}