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## Arithmetic : Level 1 Test -10

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*Arithmetic : Level 1 Test -10*. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%% Your answers are highlighted below.

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 = 2

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 = 2

^{1}/_{6}d = 5 kmQuestion 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:

9 ^{1}/_{7} | |

3 ^{3}/_{11} min | |

7 ^{3}/_{11}min | |

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 = 9

Given that the first tap is closed 3 minutes before Therefore (d – 3 ) /15 + d/18 = 1

d = 108/11

So d = 9

^{9}/_{11}min Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

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