A) 8 hrs 45 min | B) 8 hrs 42 min |

C) 8 hrs | D) 8 hrs 34 min |

Explanation:

Number of pages typed by Adam in 1 hour = 36/6 = 6

Number of pages typed by Smith in 1 hour = 40/5 = 8

Number of pages typed by both in 1 hour = (6 + 8) = 14

Time taken by both to type 110 pages = (120 * 1/14) = $8\frac{4}{7}$ = 8 hrs 34 min.

A) 3 days | B) 2 days |

C) 6 days | D) 12 days |

Explanation:

From the given data,

12 children 16 days work,

One child’s one day work = 1/192.

8 adults 12 days work,

One adult’s one day’s work = 1/96.

Work done in 3 days = ((1/96) x 16 x 3) = 1/2

Remaining work = 1 – 1/2 = 1/2

(6 adults+ 4 children)’s 1 day’s work = 6/96 + 4/192 = 1/12

1/12 work is done by them in 1 day.

1/2 work is done by them in 12 x (1/2) = 6 days.

A) 24.5 days | B) 26.6 days |

C) 25 days | D) 20 days |

Explanation:

Arun has completed $\frac{1}{3}$rd of the work in 8 days

Then he can complete the total work in

$\frac{1}{3}$ ---- 8

1 ---- ?

= 24 days

But given Akhil is only 60% as efficient as Arun

Akhil =$\frac{1}{24}\times \frac{60}{100}=\frac{1}{40}$

Akhil can complete the total work in 40 days

Now, remaining 2/3rd of work can be completed in

1 ------ 40

$\frac{2}{3}$ ------ ?

= 26.66 days.

A) 50 days | B) 48 days |

C) 47.5 days | D) 49 days |

Explanation:

50 men can build a tank in 40 days

Assume 1 man does 1 unit of work in 1 day

Then the total work is 50×40 = 2000 units

50 men work in the first 10 days and completes 50×10 = 500 units of work

45 men work in the next 10 days and completes 45×10 = 450 units of work

40 men work in the next 10 days and completes 40×10 = 400 units of work

35 men work in the next 10 days and completes 35×10 = 350 units of work

So far 500 + 450 + 400 + 350 = 1700 units of work is completed and

Remaining work is 2000 - 1700 = 300 units

30 men work in the next 10 days. In each day, they does 30 units of work.

Therefore, additional days required = 300/30 =10

Thus, total 10+10+10+10+10 = 50 days required.

A) 10 days | B) 15 days |

C) 20 days | D) 5 days |

Explanation:

After 10 days, the remaining food would be sufficient for the 1000 students for 20 more days

-->If 1000 more students are added, it shall be sufficient for only 10 days (as the no. of students is doubled, the days are halved).

A) 4:5 | B) 3:4 |

C) 4:3 | D) 2:3 |

Explanation:

(20 x 18) men can complete the work in in one day.

one man's one day work = 1/360

(18 x 15) women can complete the work in 1 day

1 woman's one day work = 1/270

So, required ratio = $\frac{1}{270}:\frac{1}{360}$= 4:3

A) 5.5 days | B) 6 days |

C) 4.2 days | D) 5 days |

Explanation:

L's 10 days work = $\left(\frac{1}{15}\times 10\right)=\frac{2}{3}$

Remaining work = $\left(1-\frac{2}{3}\right)=\frac{1}{3}$

Now, work is done by K in one day = 1/18

1/3 work is done by K in $\left(18\times \frac{1}{3}\right)$ = 6 days

A) Option A | B) Option B |

C) Option C | D) Option D |

Explanation:

Raghu can complete the work in (12 x 9)hrs = 108 hrs.

Arun can complete the work in (8 x 11)hrs = 88 hrs.

Raghu's 1 hrs work = 1/108 and Arun's 1 hrs work = 1/88

(Raghu + Arun)'s 1 hrs work = $\left(\frac{1}{108}+\frac{1}{88}\right)=\frac{49}{2376}$

So, both Raghu and Arun will finish the work in $\frac{2376}{49}hrs$

Number of days of 12 hours each=$\left(\frac{2376}{49}\times \frac{1}{12}\right)$ = $\frac{198}{49}=4\frac{3}{49}days$

A) 3 am | B) 12 pm |

C) 1 pm | D) 3 pm |

Explanation:

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

$\Rightarrow \frac{1}{4}+\frac{1}{12}=\frac{1}{3}$

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours

Thus, work will be completed at 3pm.