Question Detail

If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is

54.48
54.68
54.60
54.58

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}

1. Find the sum of first 30 natural numbers

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}

2. Average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then weight of the new man is

Answer: Option C

Explanation:

Total weight increased is 1.5 * 10 = 15.
So weight of new person is 45+15 = 60

3. Average of 10 numbers is zero. At most how many numbers may be greater than zero

Answer: Option D

4. Find the average of first 10 multiples of 7

35.5
37.5
38.5
40.5

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}