A) 27 days | B) 12 days |

C) 25 days | D) 18 days |

Explanation:

$\frac{3}{4}\times (x-2)x=(x+7)(x-10)$

$\Rightarrow {x}^{2}-6x-280=0$

=> x= 20 and x=-14

so, the acceptable values is x=20

Therefore, Total work =(x-2)x = 18 x 20 =360 unit

Now 360 = 30 x k

=> k=12 days

A) 1/5 | B) 1/6 |

C) 1/7 | D) 1/8 |

Explanation:

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 =** 1/6.**

A) 215 days | B) 225 days |

C) 235 days | D) 240 days |

Explanation:

Given that

**(10M + 15W) x 6 days = 1M x 100 days**

=> 60M + 90W = 100M

=> 40M = 90W

=> **4M = 9W.**

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in **25 x 9 = 225 days.**

Hence, 1 woman can do the same work in** 225 days.**

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work **= 2 + 7.6 + 2 = 11.6 = 11 3/5 days.**

A) 4 | B) 5 |

C) 3 | D) 6 |

Explanation:

Let M, N and O worked together for x days.

From the given data,

M alone worked for 8 days

M,N,O worked for x days

N, O worked for 1 day

But given that

M alone can complete the work in 18 days

N alone can complete the work in 36 days

O alone can complete the work in 54 days

The total work can be the LCM of 18, 6, 54 = 108 units

M's 1 day work = 108/18 = 6 units

N's 1 day work = 108/36 = 3 units

O's 1 day work = 108/54 = 2 units

Now, the equation is

8 x 6 + 11x + 5 x 1 = 108

48 + 11x + 5 = 108

11x = 103 - 48

11x = 55

x = 5 days.

Hence, all M,N and O together worked for 5 days.

A) 36 days | B) 30 days |

C) 24 days | D) 22 days |

Explanation:

Given the ratio of efficiencies of P, Q & R are **3 : 8 : 5**

Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively

They can do work for 12 days.

=> Total work = **12 x 16x = 192x**

Now, the required time taken by Q to complete the job alone = $\frac{\mathbf{162}\mathbf{x}}{\mathbf{8}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{24}\mathbf{}$days.

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) 200, 600, 400 | B) 400, 600, 200 |

C) 600, 200, 400 | D) 400, 200, 600 |

Explanation:

Ratio of efficiencies of P, Q and R = 2 : 3 : 4

From the given data,

Number of working days of P, Q, R = 5 : 10 : 5

Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20

Amounts of P, Q, R = 200, 600 and 400.

A) 16 days | B) 72/5 days |

C) 15 days | D) 96/5 days |

Explanation:

Work done by Shyam and Rahim in 8 days **= 8/32 = 1/4**

Remaining work to be done by Shyam and Ram **= 1 - 1/4 = 3/4**

Given efficieny of Ram is half of Rahim i.e, as Rahim can do the work in 48 days, Ram can do the work in 24 days.

One day work of Ram and Shyam **= (1/32 - 1/48) + 1/24 = 5/96**

Hence, the total work can be done by Shyam and Ram together in **96/5 days.**

Therefore, remaining work 3/4 can be done by them in **3/4 x 96/5 = 72/5 = 14.4 days.**