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Arithmetic : Level 1 Test -10
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Question 1 |
A boatman goes 2 km against the current of the stream in 1 h and goes 1 km along the current in 10 min. How long will he take to travel 5 km in stationary water?
1 h 30 min | |
1 h 15 min | |
1 h | |
40 min |
Question 1 Explanation:
Speed of the boatman in upstream = 2 km/h
Speed of the boatman in downstream = 1/10 x 60 = 6 km/h
Therefore the speed in the stationary water = (2+6)/ 2 = 4 km/h
Therefore time will be = 5/4 = 1h and 15 min.
Speed of the boatman in downstream = 1/10 x 60 = 6 km/h
Therefore the speed in the stationary water = (2+6)/ 2 = 4 km/h
Therefore time will be = 5/4 = 1h and 15 min.
Question 2 |
A man rowed against a stream flowing 1.5 km/h to a certain point and then turned back, stopping 2 km short of the place from where he originally started. If the whole time occupied in rowing was 2 h 10 min and his uniform speed in still water is 4.5 km/h, the man went up the stream a distance of:
4 km | |
8 km | |
7 km | |
5 km |
Question 2 Explanation:
Speed of the man = 1.5km/h
Let d be the distance travelled by him in upstream
Therefore his return distance will be = (d – 2)
Speed of man in still water = 4.5 km/h
So therefore {d/ (4.5 – 1.5)} – {(d – 2)/(4.5 + 1.5)} = 2 h 10m
(2d + d – 2)/6 = 21/6 d = 5 km
Let d be the distance travelled by him in upstream
Therefore his return distance will be = (d – 2)
Speed of man in still water = 4.5 km/h
So therefore {d/ (4.5 – 1.5)} – {(d – 2)/(4.5 + 1.5)} = 2 h 10m
(2d + d – 2)/6 = 21/6 d = 5 km
Question 3 |
A, Band C can do a work in 8, 16 and 24 days respectively. They all begin together. A continues to work till it is finished, C left after 2 days and B one day before its completion. In what time is the work finished?
7 days | |
5 days | |
6 days | |
Cannot be determined |
Question 3 Explanation:
Let us suppose the work will finish in D days
So , d/8 + (d-1)/16 + 2/24 = 1
d = 5 Hence, the work will be done in 5 days
So , d/8 + (d-1)/16 + 2/24 = 1
d = 5 Hence, the work will be done in 5 days
Question 4 |
Two pipes can fill a tank in 10 h and 15 h respectively. However, leakage at the bottom of the tank delays the filling of the tank by 3 h when both the pipes are open simultaneously. How much time would the leak take to empty the full cistern?
22 h | |
18 h | |
12 h | |
21 h |
Question 4 Explanation:
Both the pipes will fill the tank in 1/10 +1/15 = 1/6 = 6 h
And due to the leakage the tank will fill in 6+3 hours i.e. 9 hours
Since we know that if pipe a fill the tank in x hours and pipe b empty the tank
in y hours then the time taken to fill the tank = 1/x -1/y
Therefore 1/6 – 1/9 = 1/18 = 18 hours
And due to the leakage the tank will fill in 6+3 hours i.e. 9 hours
Since we know that if pipe a fill the tank in x hours and pipe b empty the tank
in y hours then the time taken to fill the tank = 1/x -1/y
Therefore 1/6 – 1/9 = 1/18 = 18 hours
Question 5 |
Two pipes can fill a cistern in 15 min and 18 min respectively. Both the pipes are operating together, but 3 min before the cistern is full, the first pipe is closed. The cistern will be filled now in:
91/7 | |
33/11 min | |
73/11min | |
None of these |
Question 5 Explanation:
Let d be the time which is taken by both the pipes to fill the tank
Given that the first tap is closed 3 minutes before Therefore (d – 3 ) /15 + d/18 = 1
d = 108/11
So d = 99/11 min
Given that the first tap is closed 3 minutes before Therefore (d – 3 ) /15 + d/18 = 1
d = 108/11
So d = 99/11 min
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