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Question Detail
A motorist travel to a place 150 km away at an avearge speed of 50 km/hr and returns ar 30 km/hr.His average speed for the whole jouney in km/hr is
- 36.5 km/hr
- 37.5 km/hr
- 35.5 km/hr
- 34.5 km/hr
Answer: Option B
Explanation:
Average speed will be
\begin{aligned}
\frac{2xy}{x+y} km/hr
= \frac{2(50)(30)}{50+30} km/hr
\end{aligned}
= 37.5 km/hr
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
- 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
2. A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17th match
- 36
- 37
- 38
- 39
Answer: Option D
Explanation:
Let the average after 17th match is x
then the average before 17th match is x-3
so 16(x-3) + 87 = 17x
=> x = 87 - 48 = 39
3. 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}
4. Average of 10 numbers is zero. At most how many numbers may be greater than zero
- 0
- 1
- 5
- 9
Answer: Option D
5. 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
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