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. 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}

2. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is

Answer: Option A

Explanation:

Number is (5*27) - (4*25) = 135-100 = 35

3. Find the average of all numbers between 6 and 34 which are divisible by 5

Answer: Option B

Explanation:

\begin{aligned}
Average = (\frac{10+15+20+25+30}{5}) = \frac{100}{5} =20
\end{aligned}

4. The average of four consecutive odd numbers is 24. Find the largest number.

Answer: Option B

Explanation:

Let the numbers are x, x+2, x+4, x+6, then
\begin{aligned}
=> \frac{x+(x+2)+(x+4)+(x+6)}{4} = 24
\end{aligned}

\begin{aligned}
=> \frac{4x+12)}{4} = 24
\end{aligned}

\begin{aligned}
=> x+3 = 24 => x = 21
\end{aligned}

So largest number is 21 + 6 = 27

5. 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}