Question Detail

Average of first five multiples of 3 is

Answer: Option A

Explanation:

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

1. 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 D

Explanation:

Suppose total number of students are = X
\begin{aligned}
Total increase = x*\frac{1}{2} = \frac{x}{2} \\
=> \frac{x}{2} = 83-63 = 20 \\
=> x = 40
\end{aligned}

2. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.

Answer: Option B

Explanation:

Average after 11 innings should be 36

So, Required score = (11 * 36) - (10 * 32)
= 396 - 320 = 76

3. 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}

4. 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

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

Answer: Option D