Correct Answers: | |
Wrong Answers: | |
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Question 1
The speed of a steamer in still water, the upstream speed of the steamer and the downstream speed of the steamer form an A.P.(Arithmetic Progression) whose common difference is $1{1/6}$. What is the speed of the river?
$1{1/6}$ units.
$2{3/5}$ units.
$2{3/8}$ units.
$3{1/8}$ units.
Ans: a
Let the speed of the steamer in still water be u and the speed of the river be v. The three speeds are u - v, u and u + v. They are in an A.P. with a common difference equal to v. Since v is also the speed of the river, the required answer is: $1{1/6}$.
Question 2
The difference of the squares of the downstream and upstream speeds of a boat in a stream is 12. What is the product of the speed of the stream and that of the boat in still water?
3.
4.
2.
6.
Ans: a
Let the speed of the boat in still water be u, and let v be the speed of the stream. The difference of squares of the speeds is $(u + v)^2 - (u - v)^2$, which is $u^2 + v^2 + 2uv - (u^2 + v^2 - 2uv)$ = 4uv = 12. We get uv = $12/4$ = 3.
Question 3
A steamer boat can travel 200 km downstream in 20 hours. If it covers the same distance upstream in 50 hours, what is the speed of the boat in still water?
7 kmph.
8 kmph.
6 kmph.
10 kmph.
Ans: a
The downstream speed = $200/20$ = 10 km/h, and the upstream speed = $200/50$ = 4 km/h. By the shortcut method, the speed of the steamer boat in still water is average of its downstream and upstream speeds. So the required speed = ${10 + 4}/2$ = 7 km/h.
Question 4
The speed of a boat in still water is 16 km/h and rate of flow of the stream is 6 km/h. If it travels upstream for 7 hours, what is the distance travelled by the boat during the journey?
Question 5
A man can row a boat at the rate of 12 km/h in still water, and at the rate of 8 km/h against the stream. At what rate can he row down the stream?
This Blog Post/Article "Boats and Streams Quiz Set 005" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2020-02-07. Published on: 2016-05-06