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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 of four consecutive odd numbers is 24. Find the largest number.
- 25
- 27
- 29
- 31
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
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. 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
- 50
- 55
- 60
- 65
Answer: Option C
Explanation:
Total weight increased is 1.5 * 10 = 15.
So weight of new person is 45+15 = 60
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.
- 25
- 30
- 35
- 40
- 45
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. 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}
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