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Question Detail
Average of first five multiples of 3 is
- 9
- 11
- 13
- 15
Answer: Option A
Explanation:
\begin{aligned}
Average = \frac{3(1+2+3+4+5)}{5} = \frac{45}{5} = 9
\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. 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}
3. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is
- 40
- 35
- 45
- 55
Answer: Option A
Explanation:
Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (20 * 2 + 5 * 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years
4. 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
5. Average age of boys in a class is 16 years and average age of girls is 15 years, what is the average age of all
- 15.5
- 15
- 16
- Cant be computed
Answer: Option D
Explanation:
As number of girls and boys is not given so result cant be computed
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