Question Detail

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}

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

Answer: Option D

3. 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

4. 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}

5. 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}