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. The average age of the mother and her six children is 12 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother

Answer: Option C

2. The average of six numbers is X and the average of three of these is Y.If the average of the remaining three is z, then

x = y + z
2x = y + z
x = 2y + z
x = y + 2z

Answer: Option B

Explanation:

X =((3y+3z)/6)
or
2X= y + z

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. 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}

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}