A) 10 % | B) 14 2/7 % |

C) 20 % | D) Can't be determined |

Explanation:

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x

=> D= 25 days

Now , the work done in 25 days = 25x

Total work = 175x

therefore, workdone before increasing the no of workers = %

A) 9 3/5 days | B) 7 days |

C) 6 3/7 days | D) 8 days |

Explanation:

Let the total work be 'W'

As per the given information,

P can complete the work 'W' in 24 days.

=> one day work of P = W/24

And also given that,

The time taken by P to complete one-third of work is equal to time taken by Q to complete half of the work

=> PW/3 = QW/2

=> P's

1 day = W/24

? = W/3

=> ? = 24W/3W = 8

=> QW/2 = 8 days

=> Q alone can complete the work W in 16 days

=> P + Q can complete the work in

1/24 + 1/16 = 5/48

=> 48/5 days = 9 3/5 days.

A) 24 days | B) 25 days |

C) 20 days | D) 19 days |

Explanation:

Let Q complete that work in 'L' days

=>

=>

L = 25 days.

A) 17 days | B) 11 days |

C) 15 days | D) 16 days |

Explanation:

We know that Time is inversely proportional to Efficiency

Here given time ratio of P & P+Q as

P/P+Q = 150/100 = 3:2

Efficiency ratio of P & Q as

P:Q = 2:1 ....(1)

Given Efficiency ratio of Q & R as

Q:R = 3:1 ....(2)

From (1) & (2), we get

P:Q:R = 6:3:1

P alone can finish the work in

22.5(3+1)/6 = 15 days.

A) 8 | B) 6 |

C) 4 | D) 2 |

Explanation:

Total Water reqduired = 5000 × 150 lit

= 750,000 litres = 750 cu.m.

Volume of tank = 20 × 15 × 5 = 1500 Cu.m.

Number of days required =1500/750 = 2 days.

A) 8 days | B) 12 days |

C) 14 days | D) 10 days |

Explanation:

Ratio of efficiencies of Priya and Sai is

Sai : Priya = 160 : 100 = 8 : 5

Given Priya completes the work in 16 days

Let number of days Sai completes the work be 'd'

=> 5×16 = 8×d

d = 10 days.

Therefore, number of days Sai completes the work is 10 days.