Let 25 persons can complete 5 units of work working 6 hours a day in x days.

⇒

⇒

⇒ x =

Hence, Required number of days = 16

Given:

Ratio of working efficiencies of a man and a woman = 3 : 2

Sum of ratios = 3 + 2 = 5

Let the total work be 330x units.

Then, combined efficiency of a man and a woman

= 330x/66 = 5x units/day

Therefore, efficiency of a man = (5x) ×  = 3x units/day

And efficiency of a woman = (5x) ×  = 2x units/day

Now, combined efficiency of 6 men and 2 women = (3x) × 6 + (2x) × 2

= 18x + 4x = 22x units/day

Time taken to complete the same work by 6 men and 2 women

= 330x/22x = 15 days

Hence, the correct answer is option A.

B is 25% more efficient than A.

As we know,

Also, Ratio of efficiency is opposite to ratio of time taken.

According to question:

5x = 15

⇒ x = 3

⇒ 4x = 12

Hence, B can do a work in 12 days.

Let total work = LCM (15,12) = 60 unit

Efficiency of A =

Efficiency of B =

Efficiency of A and B together = 4 + 5 = 9 unit/day

Both work together for 4 days only.

Work finished by them in 4 days =

Remaining work = 60 unit – 36 unit = 24 unit

C completed the remaining work in 8 days.

Efficiency of C =

Efficiency of A ,B and C together = 4 + 5 +3= 12 unit/day

Number of days taken by A,B and C to complete the whole work =

Let efficiency of one man =

Efficiency of one woman =

Two men and 7 women can complete a work in 28 days whereas 6 men and 16 women can do the same work in 11 days.

According to question:

⇒

⇒

⇒

⇒

⇒  …………..(1)

Let 7 men can do the same work in x days.

⇒

Using (1)

⇒

⇒

⇒

Hence, 7 men can do the same work in 22 days.

Given:

Efficiency of A = 2 × efficiency of C

A = 2C

Let the total work = LCM of 20 and 24 = 120 units

Efficiency of A and B =  = 6 units/day

A + B = 6 units/day

By putting value of A

 2C + B = 6 units/day           ……eq1

Efficiency of B and C =  = 5 units/day

B + C = 5units/day          .....eq2

By subtracting eq2 in eq1

 2C + B – B – C = 6 – 5

 C = 1 units/day

By putting value of C is eq2

 B + 1 = 5

 B = 4 units/day

Therefore, required time taken by B alone to complete 40% of work =  =  = 12 days

Hence, option B is correct.

Let the total work = 240 units

Efficiency of A =  = 4 units/hr

Efficiency of B =  =  = 5 units/hr

Work done by A and B in 15 days = 15 × (4 + 5) = 15 × 9 = 135 units

Remaining work = 240 – 135 = 105 units

Efficiency of C =  = 7.5 units/hr

Therefore, required time to complete th part of the work =  =  = 12 days

Hence, option D is correct.

Total work = 14 × 18 = 252 units

Work done in 5 days = 14 × 5 = 70 units

Work done on the last day = 14 × 1 = 14 units

Remaining work = 252 – 70 – 14 = 168

Required, time to complete remaining work =  =  = 21 days

Therefore, total time to complete the work = 5 + 21 + 1 = 27 days

Hence, option A is correct.

Total days of work of:

A = 24 days

B = 15 days

C = 12 days

Total work = 120 units

Efficiency of A = 120/24 = 5units

Efficiency of B = 120/15 = 8 units

Efficiency of C = 120/12 = 10 units

Now given that:

B and C started the work.

So, in 3 days B completes 8× 3 = 24 units

and C completes = 10 × 3 = 30 units

Total work completed by B and C in three days. = 24 + 30 = 54 units

Left work = 120 – 54 = 66 units

A completes 5 units in 1 day

⇒ 66 units = 66/5 =

Let total work =LCM(20,25) = 100 units

Efficiency of A = 5 unit per day

Efficiency of B = 4 unit per day

Work completed by B in 5 days = 20 units

Remaining work = 100 units – 20units = 80 units

Now this 80 units work is done by A.

Number of days taken by A to finish 80 unit work = 80/5 = 16 days

But for 5 days A and B worked together.

Hence, Number of more days taken by A to finish the remaining work = 16-5 = 11 days.

16M × 8 = (8M + 9W)10

64M = 40M + 45W

24 M = 45W

M/W = 45/24 = 15/8

Efficiency of Men = 15 unit

Women = 8 unit

Total work = 16 × 15 × 8

Times taken by 20 women = (16×15×8)/(20×8)

= 12 days.