A) 30 | B) 40 |

C) 50 | D) 60 |

Explanation:

(A+B)'s 15 days work=

Remaining work =

Now, work is done by A in 10 days.

Whole work will bedone by A in days.

A) 25 | B) 24 |

C) 23 | D) 21 |

Explanation:

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work.

So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work-(B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.

A) 7 hrs | B) 6 hrs 10 min |

C) 5 hrs 25min | D) 8 hrs 15 min |

Explanation:

A can write in 1hour = 32/6 pages

similarly

B in 1 hour = 40/5 pages

Together (A+B) in 1 hour = 32/6 + 40/5 = 40/3

so,

A+B write 40/3 pages in 1 hour

A+B write 110 pages in (3/40) x 110 Hours = 8 hours 15 min.

A) 6 days | B) 12 days |

C) 4 days | D) 9 days |

Explanation:

K's 1 day's salary = 1/10

L's 1 day's salary = 1/15

Together their 1 day's salary = 1/10 + 1/15

= 3/30 + 2/30

= 5/30

= 1/6

So the money will be

enough for paying them both for 6 days.

A) 17 men | B) 14 men |

C) 13 men | D) 16 men |

Explanation:

M x T / W = Constant

where, M= Men (no. of men)

T= Time taken

W= Work load

So, here we apply

M1 x T1/ W1 = M2 x T2 / W2

Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?

Note that here, W1 = W2 = 1 road, ie. equal work load.

Clearly, substituting in the above equation we get, M2 = 14 men.

A) 45 metric tonnes | B) 47 metric tonnes |

C) 55 metric tonnes | D) 34 metric tonnes |

Explanation:

2 engines of former type for one hour consumes 2x24/(6x8) = 1 metric ton

i.e. 3 engines of latter type consumes 1 ton for one hour

hence 9 engines consumes 3 tons for one hour

for 15 hours it is 15 x 3 = 45 metric tonnes.