Important for :

1

C

Daily wages = hourly wages × work hours.

Let the original hourly wages and work hours be Rs. x and y hours respectively.

Since he used to earn Rs. 120 earlier, xy = 120

New hourly wages = Rs. (1.25x) and new working hours = 0.84y

Hence, option C is correct.

2

C

Let total work = 150 unitsSince Radhe does 70% of the work (i.e. 105 units) in 15 days,

Radhe = | 105 | = 7 units per day |

15 |

Work left = 150 − 105 = 45 units

Let Shyam do x units of work per day. Shyam and Radhe finish the pending work in 4 days.

Hence, option C is correct.

3

A

A can do 10% work in a day. A has worked 6 days in total. And so has BA completed 60% work in 6 days and B did 40% in 6 days.

Efficiency of A and B = 6 : 4

B's share = | 4 | × 54000 = 21600 |

10 |

Hence, option A is correct.

4

D

Let the speed of doing work of the three persons be 1x, 3x and 5x respectivelyTime taken by each person = amount of work done/speed of doing work

Let the amount of work for each person = y (∵ work done is same)

The time taken by the first person = | y |

x |

The time taken by the second person = | y |

3x |

The time taken by the third person = | y |

5x |

The ratio of time taken = | y | : | y | : | y |

x | 3x | 5x |

⇒ 1 : | 1 | : | 1 | = 15 : 5 : 3 |

3 | 5 |

Hence, option D is correct.

5

C

6 workers complete some work in 10 hours.

Let the total work be equivalent to 60 man-hours.

From 9.00 am to 3.00 pm, all 6 workers work together for 6 hours.

From 3.00 pm to 4.00 pm, 7 man-hours of work will be done.

From 4.00 pm to 5.00 pm, 8 man-hours of work will be done.

From 5.00 pm to 6.00 pm, 9 man-hours of work will be done.

Thus, total amount of work done up to 6.00 pm is 36 + 7 + 8 + 9 = 60 man-hours

Thus, the computer will be assembled at 6.00 pm.

Hence, option C is correct.