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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. 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
2. 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
- 55
- 56
- 57
- 58
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
3. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is
- 45
- 50
- 55
- 60
Answer: Option A
4. Find the sum of first 30 natural numbers
- 470
- 468
- 465
- 463
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}
5. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.
- 70
- 76
- 78
- 80
Answer: Option B
Explanation:
Average after 11 innings should be 36
So, Required score = (11 * 36) - (10 * 32)
= 396 - 320 = 76
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