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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. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is
- 35
- 45
- 55
- 65
Answer: Option A
Explanation:
Number is (5*27) - (4*25) = 135-100 = 35
2. 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}
3. 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
4. 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
5. 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}
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