A) 6 | B) 4 |

C) 2 | D) 3 |

Explanation:

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.

A) 10 5/2 days | B) 11 days |

C) 11 2/5 days | D) 12 1/2 days |

Explanation:

Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

⇒ m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

A) 980 | B) 1232 |

C) 810 | D) 1121 |

Explanation:

From the given data,

Day Capacity

A -> 12 2

B -> 8 3 2

A + B + C -> 2 12

=> Capacity of C = 12 - 5 = 7

Ratio of capacity of A : B : C = 2 : 3 : 7

Difference of wages of C & B = 4/5 x 1540

= 4 x 308 = Rs. 1232

A) 16 days | B) 18 days |

C) 19 days | D) 21 days |

Explanation:

(16M + 12W) x 20 = 18W x 40

=> 2M = 3W

Then,

Convert all men into women

12M + 27W = 27 + 12 x 3/2W = 45W

Let number of days required be 'D'

=> 18 x 40 = 45 x D

=> D = 16 days.

A) 6 days | B) 8 days |

C) 11 days | D) 13 days |

Explanation:

M = 10 days

The ratio of efficiency of M & N are 3 : 2

Hence, the time rquired for N alone = 15 days

=> Required time taken t=by both to complete the work = M x N / M + N

= 10 x 15/ 10 + 15

= 150/25

= 6 days.

A) 20 hrs | B) 18 hrs |

C) 22 hrs | D) 20.5 hrs |

Explanation:

Time taken by both Meghana and Ganesh to work together is given by =

$\sqrt{{t}_{1}x{t}_{2}}$

Where

${t}_{1}=32hrs\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{t}_{2}=12\frac{1}{2}hrs=\frac{25}{2}hrs$

Therefore, time took by both to work together =

$\sqrt{32x\frac{25}{2}}=\sqrt{16x25}=4x5=20hrs$

A) 96 days | B) 48 days |

C) 24 days | D) 12 days |

Explanation:

Let number of days Charan can do the same work alone is 'd' days.

According to the given data,

$\frac{\mathbf{5}}{\mathbf{12}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{7}}{\mathbf{16}}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{14}}{\mathbf{d}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{14}{\mathrm{d}}=1-\frac{\left(20+21\right)}{48}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{14}{\mathrm{d}}=\frac{48-41}{48}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{d}=\frac{14\mathrm{x}48}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{d}\mathbf{}\mathbf{=}\mathbf{}\mathbf{96}\mathbf{}\mathbf{days}$

Therefore, Charan alone can complete the work in **96 days.**

A) 3 hrs | B) 3.5 hrs |

C) 2.5 hrs | D) 2 hrs |

Explanation:

Given Prabhas is twice as good a workman as Rana.

Prabhas finishes the work in 3 hrs

=> Rana finishes the work in 6 hrs.

Number of hours required together they could finish the work

**= 3 x 6 / 3 + 6 **

**= 18/9 **

**= 2 hrs.**

A) 2 hrs 24 min | B) 2 hrs 44 min |

C) 1 hrs 24 min | D) 1 hrs 24 min |

Explanation:

Given that three athletes can complete one round around a circular field in 16, 24 and 36 min respectively.

Now, required time after which they met for the first time = LCM of (16, 24 & 36) min

Now, LCM of 16, 24, 36 = 144 minutes = 2 hrs 24 min.