A) 15 | B) 30 |

C) 25 | D) 10 |

Explanation:

Combined efficiency of all the three boats = 60 passenger/trip

Now, consider option(a)

15 trips and 150 passengers means efficiency of B1 =

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 =

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

A) [7 + (215/345)] days | B) [6 + (11/215)] days |

C) [6 + (261/315)] days | D) [5 + (112/351)] days |

Explanation:

After day 1, A finishes 1/9 of the work.

After day 2, B finishes 1/7 more of the total work. (1/9) + (1/7) is finished.

After day 3, C finishes 1/5 more of total work. Total finished is 143/315.

So, after day 6, total work finished is 286/315.

Now remaining work = 315 - 286 = 29 /315

On day 7, A will work again

Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work.

Since he takes 9 days to finish the total task, he will need 261/315 of the day.

Total days required is 6 + (261/315) days.

A) 44 | B) 32 |

C) 56 | D) 49 |

Explanation:

Girl eats 112 chocolates in 30 sec

so she can eat in 12 sec is 12 x 112/30 = 44.8 chocolates.

Her boy friend can eat one-half of 112 in twice of 30 sec

so he can eat 56 in 60 sec

Then he can eat in 12 sec is 56 x 12/60 = 11.2 chocolates.

Hence, together they can eat

=> 44.8 + 11.2

56 chocolates in 12 seconds.

A) 24 $ | B) 22 $ |

C) 16 $ | D) 14 $ |

Explanation:

Let the 1 day work of a boy=b and a girl=g, then

2b + g = 1/5 ---(i) and

b + 2g = 1/6 ---(ii)

On solving (i) & (ii), b=7/90, g=2/45

As payment of work will be in proportion to capacity of work and a boy is paid $ 28/week,

so a girl will be paid 28x = 16 $.

A) 3 days | B) 26/3 days |

C) 13/4 days | D) 24/5 days |

Explanation:

we know that **m1 x d1 = m2 x d2**

=> 4 x 6 = 5 x d

where d = no. of days taken by 5 men to paint

d = 24/5 days.

A) 2 days | B) 3 days |

C) 4 days | D) 5 days |

Explanation:

One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get

15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.