Question Detail

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}

1. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students

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

2. 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}

3. 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}

4. Reeya obtained 65, 67, 76, 82 and 85 out of 100 in different subjects, What will be the average.

Answer: Option B

Explanation:

\begin{aligned}
( \frac {65+67+76+82+85}{5} ) = 75
\end{aligned}

5. Average of all prime numbers between 30 to 50

37
37.8
39
39.8

Answer: Option D

Explanation:

Prime numbers between 30 and 50 are:
31, 37, 41, 43, 47

Average of prime numbers between 30 to 50 will be
\begin{aligned}
(\frac{31+37+41+43+47}{5}) = \frac{199}{5} = 39.8
\end{aligned}