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## Arithmetic : Level 2 Test -7

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Question 1 |

A train travels a distance of 600 km at a constant speed. If the speed of the train is increased by Â 5 km/h, the journey would take 4 h less. Find the speed of the train.

100 km/h | |

25 km/h | |

50 km/h | |

None of these |

Question 1 Explanation:

Distance = 600 km/h

Let the speed of the train = s km/h

Now increased speed = (s + 5) km/h

So 600/s = 600/(s+5) + 4

600/s - 600/(s+5) = 4

600[5/s(s+5)]= 4

s = 25 km/h

Let the speed of the train = s km/h

Now increased speed = (s + 5) km/h

So 600/s = 600/(s+5) + 4

600/s - 600/(s+5) = 4

600[5/s(s+5)]= 4

s = 25 km/h

Question 2 |

Three pipes A, B and C can fill a tank in 20 min, 10 min and 30 min respectively. When the tank is empty, all the three pipes are opened. Â A, Band C discharge chemical solutions x, y and z respectively. The proportion of solution y in the liquid in the tank after 3 min is

6/11 | |

7/11 | |

8/11 | |

5/11 |

Question 2 Explanation:

Work done by the pipes in one minutes = 1/20 + 1/10 +1/30

So the work done by the pipes in 3 minutes = 3/20 + 3/10 +3/30 = 11/20

The work done by the 2

The work done by the 2

So the required ratio will be = (3/10) / (11/20) = 6/11

So the work done by the pipes in 3 minutes = 3/20 + 3/10 +3/30 = 11/20

The work done by the 2

^{nd}pipe in one minute = 1/10The work done by the 2

^{nd}pipe in three minutes = 3/10So the required ratio will be = (3/10) / (11/20) = 6/11

Question 3 |

Three taps A, Band C can fill a tank in 12, 15 and 20 h respectively. If A is open all the time and B and C is open for one hour each alternatively; the tank will be filled in

6 h | |

7 h | |

5 h | |

None of these |

Question 3 Explanation:

Tap A and B can fill the tank in =1/12 + 1/15 = 3/20

Tap A and C can fill the tank in =1/12 + 1/20 = 2/15

Work done by all the taps in 2 hours = 3/20 + 2/15 = 17/60

Filling done in 6 hours = 17/60 x 3 = 51/60

So remaining work is = 1 â€“ 51/60 = 3/20

Now the taps which are opened are A and B and the work done by the taps in one hour is 3/20 as discussed earlier,

so the total time for the whole filling would be = 7 hours

Tap A and C can fill the tank in =1/12 + 1/20 = 2/15

Work done by all the taps in 2 hours = 3/20 + 2/15 = 17/60

Filling done in 6 hours = 17/60 x 3 = 51/60

So remaining work is = 1 â€“ 51/60 = 3/20

Now the taps which are opened are A and B and the work done by the taps in one hour is 3/20 as discussed earlier,

so the total time for the whole filling would be = 7 hours

Question 4 |

A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time it increases its speed by 250 km/h. What is its original speed?

1000 km/h | |

750 km/h | |

600 km/h | |

800 km/h |

Question 4 Explanation:

Let the actual speed of the plane = S km/h

Let the actual time taken = T h

Distance = 1500 km

Therefore T = 1500/Sâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦a

Now acc. to question

1500/(S + 250) = T â€“ 30/60 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..b

1500/ (s+ 250) = 1500/s â€“ 30/60

Solving the above equation S = 750 km/h

Let the actual time taken = T h

Distance = 1500 km

Therefore T = 1500/Sâ€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦a

Now acc. to question

1500/(S + 250) = T â€“ 30/60 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..b

1500/ (s+ 250) = 1500/s â€“ 30/60

Solving the above equation S = 750 km/h

Question 5 |

A computer can perform 30 identical tasks in 6 h. At that rate, what is the approximately minimum number of computers that should be assigned to complete 80 of the tasks within 3 h?

12 | |

6.66 | |

5.33 | |

16 |

Question 5 Explanation:

6 hour = 30 tasks

1 hour = 5 tasks

3 hour = 15 tasks

So 15 article are done by 1 computer in 3 hours

15N = 80

N = 80 / 15 = approx. 5.33 times

1 hour = 5 tasks

3 hour = 15 tasks

So 15 article are done by 1 computer in 3 hours

15N = 80

N = 80 / 15 = approx. 5.33 times

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