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Time require to fill the tank by pipe A = 45 MIN

Time require to empty the tank by pipe B = 1 hour 30 min = 60 + 30 = 90 min

Amount of tank FILLED in 30 min by pipe A,

⇒ Amount of tank filled = (1/45) × 30 = 2/3

⇒ Remaining part of tank = 1 - 2/3 = 1/3

For time require to fill the tank,

⇒ (1/45 - 1/90) × t = 1/3

⇒ [(2 - 1)/90] × t = 1/3

⇒ (1/90) × t = 1/3

⇒ t = (1/3) × (90)

⇒ t = 30 min

The ratio of EFFICIENCIES of P, Q and R is 4 : 3 : 8.

Let their efficiencies be 4T, 3T and 8T, respectively. Their total efficiency will be 15T.

We know, TIME to finish work is INVERSELY proportional to efficiency.

⇒ 4.5 × 3T = Time REQUIRED × 15T

⇒ Time required = 4.5/5 = 0.9 hours = 54 minutes

ASSUME the efficiency of a boy = B units/day

As a woman is TWICE as fast as a boy so the efficiency of a woman = 2B units/day

As a man is twice as fast as a woman so the efficiency of a man = 4B units/day

According to the question, all of them i.e. a man, a woman and a boy can finish a work together in 4 days.

In 4 days total work done by a man, a woman and a boy = 4 × (B + 2B + 4B) = 28B units.

∴ Total work = 28B units.

Time required for a boy to do the whole work alone = $(\FRAC{{Total\;work\;}}{{efficiency\;of\;a\;boy}}\; = \frac{{28B}}{B} = 28\;days)$

Quantity A:

Let total work be 120 units (LCM of 24, 30 & 40)

⇒ Work done by A in one day = 120/24 = 5 units

⇒ Work done by B in one day = 120/30 = 4 units

⇒ Work done by C in one day = 120/40 = 3 units

⇒ Work done by A and B in 1 day = (5 + 4) = 9 units

⇒ Work done by A and B in 6 days = 9 × 6 = 54 units

⇒ Remaining work = 120 – 54 = 66 units

⇒ Work done by A, B and C together in 1 day = 5 + 4 + 3 = 12 units

⇒ Time taken by A, B & C together to complete the remaining 66 units of work = 66/12 = 5.5 days

⇒ Total time taken = 6 + 5.5 = 11.5 days

Quantity B:

⇒ Time taken by B to complete the work = 20 days

⇒ Time taken by C to complete the work = 20/2 = 10 days

⇒ Time taken by A to complete the work alone = 10 + 2 = 12 days

Let total work be 60 units (LCM of 12, 20 & 10)

⇒ Work done by A in one day = 60/12 = 5

⇒ Work done by B in one day = 60/20 = 3

⇒ Work done by C in one day = 60/10 = 6

⇒ Work done by A, B and C together in 1 day = 5 + 3 + 6 = 14 units

⇒ Time taken by A, B and C together to complete the work = 60/14 = 30/7 days

∴ Quantity A > Quantity B

The total number of man-days will be same in both the cases.

⇒ 96 × 40 = 96 × 10 + (96 + N) × (34 – 10)

⇒ 3840 = 960 + 2304 + 24N

Number of days Kamal TAKES to complete a WORK = 20 days

∴ In 1 day Kamal can do = 1/20 part of work

Given: Bimal is 20% more EFFICIENT than Kamal

∴ In 1 day Bimal can do = 1.2/20 part of work

Given: SURESH is 25% more efficient than Bimal

∴ In 1 day Suresh can do = $(\frac{{1.2\; \times \;1.25}}{{20}})$ part of work

A works for D days and completed 1/3^rd of WORK

Work done by A in D days = 1/3

A alone can complete the work = 3D days

Work done by A in 1 day = (1/3) × (1/D) = 1/3D

Ratio of time taken by A and B = 100 : 125

Let’s assume B alone can complete the work in in x days.

⇒ 100 / 125 = 3D/x

⇒ x = 15D/4

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

2/3^rd of work completed by B in (D + 12) days.

⇒ 2/3 = (4/15D) × (D + 12)

⇒ 30D = 12D + 144

⇒ 18D = 144

⇒ D = 8

A can alone complete the work in 8 × 3 = 24 days

B alone can complete the work in = 15/4 × 8 = 30

If A and B work TOGETHER, work will be completed in = 1/[(1/24) + (1/30)] = 1/(9/120) = 120/9

NET part filled in 1 hour = (1/5 + 1/6 - 1/12) = 17/60

∴ TIME taken to fill the TANK = 1/(17/60) = 60/17 hours

Efficiency of Nikhil : Efficiency of Kanak = 3 : 1

Nikhil will TAKE 1/3 time as compared to Sunita

Let us ASSUME Kanak will take 3x DAYS to COMPLETE an assignment and Nikhil will take x days.

⇒ 3x - x = 6

⇒ x = 3

Let PIPES P and Q take p and q HOURS respectively to fill the cistern.

From the question, p = 2Q----(1)

They fill TOGETHER the tank in 28 hours.

$(\begin{ARRAY}{l} \therefore \frac{1}{p} + \frac{1}{q} = \frac{1}{{28}}\\ \Rightarrow \frac{1}{{2q}} + \frac{1}{q} = \frac{1}{{28}}\\ \Rightarrow \frac{{3}}{{2{q}}} = \frac{1}{{28}} \end{array})$

⇒ 2q = 84

∴ q = 42 hours

Let the distance to be travelled be ‘d’.

⇒ Anuj walks d distance in 15 days while resting 8 hours a DAY

1 day = 24 hours

Number of hours Anuj walks in a day = 24 – 8 = 16 hours

∴ Time TAKEN to walk d distance = 16 × 15 = 240 hours

∴ SPEED = d/240----(1)

New distance, d’ = 2D

New speed = 2 × speed = 2d/240 = d/120

Resting hours = 2 × 8 = 16 hours

⇒ Number of working hours in a day = 24 – 16 = 8 hours

Time taken to walk 2d distance = 2d/(d/120) = 240 hours

Rahi is 30% less EFFICIENT than Tanu.

Tanu TAKES 99 days to COMPLETE a work.

∴ NUMBER of days TAKEN by Rahi to finish a work = $(\frac{{99}}{{100 - 30}} \times 100 = \frac{{990}}{7}\;days)$

In one day, together they do = $(\frac{1}{{99}} + \frac{7}{{990}} = \frac{{17}}{{990}})$

∴ Time taken to do complete work = 990/17 ≈ 58 days

Ratio of time taken by A and B to complete a work = 2 ? 3

∴ Ratio of EFFICIENCY of A and B = 3 ? 2

Let, efficiency of A = 3X/day and efficiency of B = 2x/day

Let, quantity of total work = a

According to problem,

⇒ 5 × 3x + 10 × 2x = 0.7a

⇒ 0.7a = 35x

⇒ a = 50x

Let, they together do the work in = y days

⇒ y × (3x + 2x) = 50x

⇒ y = 50x/5x

⇒ y = 10

Let Jake can FINISH the piece of WORK alone in x days.

∴ ASHLEY can alone finish the work in x/2 days.

In 1 DAY Jake can do work = 1/x

In 1 day Ashley can do work = 1/(x/2) = 2/x

1/x + 2/x = 1/20

⇒ 3/x = 1/20

⇒ x = 60

∴ Ashley alone can finish the work in 60/2 i.e. 30 days

Shivi is 20% less efficient than Manik.

Manik takes 60 DAYS to complete a work.

∴ Number of days taken by Shivi to finish a work = $(\FRAC{{60}}{{100 - 20}} \times 100 = 75\;days)$

By Unitary method, Work done by Shivi in ONE day = 1/75

Similarly, work done by Manik in one day = 1/60

Work done by Shivi on first two days = 2/75 = 2/75

Work done by Manik on third day =$(\;\frac{1}{{60}})$

∴ TOTAL work done in a span of three days = $(\frac{2}{{75}} + \frac{1}{{60}} = \frac{{8 + 5}}{{300}} = \frac{13}{{300}})$

∴ Total number of complete spans = Integer part of $(\frac{{300}}{13} = 23)$

Remaining work = $(1 - \left( {23 \times \frac{13}{{300}}} \right) = \frac{1}{{300}})$

After completion of three day cycle, once again Shivi will work on first two days.

∴ Time taken by Shivi to complete $(\frac{1}{300})$work = $(\frac{1}{{300}} \times 75 = 1/4\;days)$ which is less than 2 days.

∴ Approximate total number of days taken to complete the work = (23 × 3 + (1/4))

= 69.25 days ≈ 70 days

Concept: If M₁ Persons can do a work W₁ in H₁ hours and M₂ persons can do a work W₂ in H₂ hours then we have a general formula in this case.

$(\frac{{{M_1}{H_1}}}{{{W_1}}} = \frac{{{M_2}{H_2}}}{{{W_2}}})$

If 12 furnaces consume 13 tonnes of coal in $(6\frac{1}{2})$ hours, then for 7 furnaces for consuming 14 tonnes

We have, M₁ = 12 furnaces

W₁= 13 tones

H₁ = $(6\frac{1}{2} = \frac{{13}}{2}\;hours)$

M₂ = 7 furnaces

W₂ = 14 tones

Let H₂ = H hours

By putting all these values in the above general formula

$(\begin{array}{l} \frac{{12 \times \frac{{13}}{2}}}{{13}} = \frac{{7 \times H}}{{14}}\\ \RIGHTARROW \;H\; = \frac{{12 \times 13 \times 14}}{{2 \times 7 \times 13}} = 12hours\; \end{array})$

Work done by Imran in ONE day = 1/12

Work done by Hashim in one day = 1/36

Work done by Rajat in one day = 1/18

Work done by Imran and Hashim TOGETHER in one day = (1/12) + (1/36) = 1/9

Work done by Imran and Rajat together in one day = (1/12) + (1/18) = 5/36

Work done in 2 DAYS = (1/9) + (5/36) = 1/4

Time taken to COMPLETE the work = 2/(1/4) = 2×4 = 8 days

A can do a PIECE of WORK in 80 days.

$(\begin{array}{l} {\rm{Work\;done\;by\;A\;in\;}}1{\rm{\;days}} = \frac{1}{{80}}\\ {\rm{Work\;done\;by\;A\;in\;}}10{\rm{\;days}} = \frac{1}{{80}}{\rm{\;}} \times 10 = \frac{1}{8}\\ \THEREFORE {\rm{Remaining\;work\;to\;be\;done\;by\;B}} = 1 - \frac{1}{8} = \frac{7}{8}\\ \frac{7}{8}{\rm{\;of\;the\;work\;is\;done\;by\;B\;in\;}}42{\rm{\;days\;}}\\ \therefore {\rm{Complete\;work\;is\;done\;by\;B\;in}} = 42{\rm{\;}} \times \frac{8}{7} = 6 \times 8 = 48{\rm{\;days\;}}\\ {\rm{Work\;done\;by\;A\;in\;}}1{\rm{\;day}} = \frac{1}{{80}}{\rm{\;and\;work\;done\;by\;B\;in\;}}1{\rm{\;day}} = \frac{1}{{48}}\\ \therefore {\rm{A\;and\;B's\;}}1{\rm{\;day's\;work}} = \frac{1}{{80}} + \frac{1}{{48}} = \frac{8}{{240}} = \frac{1}{{30}} \end{array})$

∴ Both will finish the work in 30 days.

TIME-taken by PIPE P to fill the tub = 6 mins

∴ Part of tub filled by Pipe P in one minute = 1/6

Time-taken by pipe Q to fill the tub = 18 min.

∴ Part filled by Pipe Q in one minute = 1/18

The part of tank filled in a span of 2 minutes $(= \;\frac{1}{6} + \frac{1}{{18}} = \frac{4}{{18}} = \frac{2}{9})$

∴ part of tub filled in 8 minutes $(= \;4\; \times \frac{2}{9}\; = \frac{8}{9})$

⇒ Part of tub left to be filled = 1 – (8/9) = 1/9

After 8 minutes, it’s pipe A’s turn to fill the tub and 1/9^th part of the tub is empty.

Pipe A fills the entire tub in 6 minutes, so the time taken by pipe A to fill 1/9^th of tub = 6/9 min.

⇒ Total time taken $(= 8 + \frac{6}{9} = 8\frac{2}{3}minutes)$

Amount EMPTIED by pipe R in one hour = 1

As pipe P fills the entire cistern in 2 hours,

PART of cistern FILLED by pipe P in one hour = ½

As pipe Q fills the entire cistern in 2.5 hours,

Amount filled by pipe Q in one hour = 1/2.5

For the first half an hour only pipe P is open.

∴ Part of cistern filled by pipe P in starting ½ hour = ½ × ½ = ¼

For the next half an hour, pipe P and pipe Q both are open.

∴ Part of cistern filled by pipe P and pipe Q in next half an hour $(= \frac{1}{2}\left( {\frac{1}{2} + \frac{2}{5}} \right) = \frac{1}{4} + \frac{1}{5} = \frac{9}{{20}})$

∴ Total pat of cistern filled after one hour $(= \frac{1}{4} + \frac{9}{{20}} = \frac{7}{{10}})$

After 1 hour all three PIPES are open.

Let x be the number of hours for which all 3 pipes are open

∴ At the end of x hours, the cistern will be completely empty

∴ $(\frac{7}{{10}} + x\left( {\frac{1}{2} + \frac{1}{{2.5}} - \frac{1}{1}} \right) = 0)$

$(\Rightarrow \frac{7}{{10}} - \frac{x}{{10}} = 0)$

⇒ x = 7

∴ All three tanks are open for 7 hours.

⇒ Tank will get empty 7 hours after 7:00 am.

Total capacity of Cistern = LCM of 25, 20 and 10 = 100 units

Efficiency of pipe 1 = 100/25 = 4 units/min

Efficiency of pipe 2 = 100/20 = 5 units/min

Efficiency of pipe 3 = - 100/10 = -10 units/min

Efficiency of pipe 1, pipe 2 and pipe 3 together = 4 + 5 - 10 = -1 units/min

∴ TIME taken to empty the cistern = 100/1 = 100 min

Given,

Time to FILL the CISTERN without leak = 4 hrs

∴ Part FILLED without leak in 1 hour = 1/4

Time TAKEN to fill the cistern with leak = 6 hrs

∴ Part filled with leak in 1 hour = 1/6

∴ Part emptied by leak in 1 hour = 1/4 - 1/6 = 1/12

(A + B)’s 2 day’s work $( = \FRAC{1}{{15}} + \frac{1}{{20}} = \frac{7}{{60}})$

Work done in 8 pairs of DAYS $( = \left\{ {\frac{7}{{60}} \times 8} \RIGHT\} = \frac{{14}}{{15}})$

REMAINING work $( = \left\{ {1 - \frac{{14}}{{15}}} \right\} = \frac{1}{{15}})$

Work done by A on 17^th day $( = \frac{1}{{15}})$

Total time taken = 17 days.

LET Baman takes X DAYS to complete the work then AMAN will TAKE 3X days to complete it.

∴ 1/X + 1/3X = 1/12

The ratio of efficiencies of Gary, Sunny and Allan is 3 : 2 : 7. Suppose Gary, Sunny and Allan finish WORK alone in T/3, T/2 and T/7 DAYS, respectively.

Work done in one day by Gary, Sunny and Allan will be 3/T, 2/T and 7/T, respectively.

Gary and Sunny work for a week. After that, Gary leaves and Allan JOINS Sunny. After one more week, the work GETS finished

⇒ Work done in FIRST week = 7 × (3/T + 2/T) = 35/T

⇒ Work done in second week = 7 × (3/T + 7/T) = 70/T

⇒ 35/T + 70/T = 1

⇒ T = 105

Let the total units of WORK be 600

A, B, and C started the work and they work for 2 days

⇒ Number of units PRODUCED in TWO days by A, B and C = (30 + 60 + 20) × 2 = 220 units

Then A and B left the work and D and C work another 3 days together

⇒ Number of units produced in 3 days by D and C = (15 + 20) × 3 = 35 × 3 = 105 units

⇒ Work left for D and E = 600 – 220 – 105 = 275 units

B’s 1 day work = 1/15

Hence, B’s 5 DAYS work $( = \frac{5}{{15}} = \frac{1}{3})$

Work remaining $(= \left\{ {1 - \frac{1}{3}} \right\} = \frac{2}{3})$

2/3 of the work can be DONE by A in $(\left\{ {\frac{2}{3} \times 24} \right\} = 16)$

Let the filling capacity of the CONTAINER = x m³ 

Emptying capacity = (x + 15) m³ 

⇒ 3000/x – 3000/(x + 15) = 10 

⇒ 3000x + 45000 – 3000x = 10x(x + 15) 

⇒ 45000 = 10x² + 150X 

⇒ x² + 15x – 4500 = 0 

⇒ x = 60 , -75 (It can be neglected)

∴ Emptying capacity = x + 15 = 75

PART of the work done by KHALEESI = (1/15 × 6) = 2/5

Part of the work done by John Snow = (1/12 × 6) = 1/2

Part of the work done by Cersi = 1 – (1/2 + 2/5) = 1/10

∴, (Khaleesi’s share) : ( John’s share) : (Cersi’s share) = 2/5 : ½ : 1/10 = 4 : 5 : 1

∴ Khaleesi’s share = (4/10 × 2400) = Rs.960

John’s share = (5/10 × 2400) = Rs.1200

Cersi’s share = (1/10 × 2400) = Rs.240

Now, Khaleesi’s daily WAGES = Rs. (960/6) = Rs.160

John’s daily wages = Rs. (1200/6) = Rs.200

Cersi’s daily wages = Rs. (240/4) = Rs.60

∴ Daily wages of john and cersi are (60 + 200) Rs.260.

Quantity A:$

Let Sammy can complete a piece of work in ‘x’ hours

TIME TAKEN by Billy to complete the work = (2x) hours

⇒ (x + 6) = 2x

⇒ x = 6 hours

Work done by Sammy in ONE hour = 1/x = 1/6

Work done by Billy in one hour = 1/2x = 1/12

Work done by Sammy and Billy together in one hour = (1/6) + (1/12) = 1/4

Time taken by both Sammy and Billy to complete the work together = 1/(1/4) = 4 hours

∴ Time taken by both Sammy and Billy to complete the work together = 4 hours

Quantity B: 5

Quantity B > Quantity A

Work DONE by Ojha in ONE day = 1/18

Work done by Ojha and BALI together in one day = 1/10

Work done by Bali in one day = (1/10) – (1/18) = 2/45

Work done by Bali in one day if he works for only half day = 2/(2×45) = 1/45

Work done by Ojha and Bali together if Bali works only for half a day daily in one day

= (1/18) + (1/45)

= 7/90

Time taken by Ojha and Bali to complete the work together when Bali works only for half a day

Ratio of their efficiency = 3 : 6 : 2

Days of completion ratio = 1/3 : 1/6 : 1/2 = 2 : 1 : 3

3 ⇒ 30

1 ⇒ 10

2 ⇒ 20

Part of work done by A and B in 6 days = (6/10 + 6/20) = 18/20 = 9/10

Remaining work = (1 – 9/10) = 1/10

∴ 1/10 of work will be completed by C in 3 days.

PART filled by PIPE A in 1 minute = 1/10

Part filled by pipe B in 1 minute = 1/40

(A + B)’s 10 minute work = 5 × (1/10 + 1/40) = 5/8

REMAINING work = 1 – 5/8 = 3/8

⇒ 1/40 : 3/8 = 1 : x

⇒ 1/15 = 1/x

⇒ x = 15

Suppose TOTAL work = 72 UNITS (LCM of 24, 36 and 3)

⇒ Efficiency of Sumit = 72/3 = 24

⇒ Efficiency of the 1^st SON = 72/24 = 3

⇒ Efficiency of the 2^nd son = 72/36 = 2

Suppose the efficiency of the 3^rd son = x

Since Sumit can do TWICE the work as much as done by all the 3 sons together

⇒ 24 = 2 × (3 + 2 + x)

⇒ x = 7

⇒ Ratio in which the amount will be distributed among the 3 sons = 3 ? 2 ? 7

∴ Share of the 3^rd son = 600 × 7/12 = RS. 350

If m₁MEN and w₁woman working together can do a piece of work in d₁ days, and m₂men and w₂ women working together can do it in d₂ days men are p times efficient than women then

⇒ d₁/d₂ = (pm₂ + w₂)/(pm₁ + w₁)

⇒ 8/2 = (26P + 48)/(8p + 9)

⇒ 32p + 36 = 26p + 48

⇒ 6p = 12

⇒ P = 2

WORK DONE by a tap in one HOUR = 1/5

Work done by leak in one hour = 1/30

Total work done in 1 hour = {(1/5) – (1/30)} = 1/6

Let the efficiency Q be T.

P works twice as FAST as Q.

⇒ Efficiency of P = 2T

Also, Q and R together can finish a work together in one third of the time in which Q finishes the work alone.

⇒ Efficiency of Q + Efficiency of R = 3 × Efficiency of Q

⇒ Efficiency of R = 2 × Efficiency of Q = 2T

We know, share of a person is PROPORTIONAL to his efficiency.

⇒ Share of Q = (Efficiency of Q/Sum of efficiencies of P, Q and R) × Rs. 30000

⇒ Share of Q = T/(2T + T + 2T) × Rs. 30000 = Rs. 6000

We know that WORK = M × D × H

Here,

M = no. of men

D = no of DAYS

H = hours

So,

M₁ × D₁ × H_{1 =} M₂ × D₂ × H₂

9 × 20 × 7 = 7 × D₂ × 10

? when using 2.25 liter bottle, PER bottle fill 1/8 of BUCKET.

∴ when using 1.2 liter bottle, per bottle fill (1.2/2.25) × 1/8 = 1/15 of the bucket.

let Nikhil can COMPLETE the project in X DAYS.

∴ Rohit will complete it in x/3 days and Sanjeev will do it in x/6 days.

? In one day Nikhil, Rohit and Sanjeev will complete 1/x, 3/x and 6/x of the project.

∴ combined one day work of all partner = 1/x + 3/x + 6/x = 10/x = 1/30 (given)

GIVEN,

A can complete the WORK in hours = 5 × 8 = 40 hours

B can complete the work in hours = 6 × 10 = 60 hours

A's 1 HOUR's work = 1/40

B's 1 hour's work = 1/60

(A + B)’s 1 hour's work = (1/40) + (1/60) = 1/24

Both will finish work in 24 DAYS.

Number of days of 8 hours a day = 24 × 1/8 = 3

Let, T worked for ‘y’ DAYS.

⇒ R worked for (y - 8) days and S worked for (y - 12) days.

R completes the work in 36 days.

⇒ R’s 1 day’s work = 1/36

⇒ R’s (y - 8) day’s work = (y - 8)/36

S completes the work in 54 days.

⇒ S’s 1 day’s work = 1/54

⇒ S’s (y - 12) day’s work = (y - 12)/54

T can complete the work in 72 days.

⇒ T’s 1 day’s work = 1/72

⇒ T’s y day’s work = y/72

According to the QUESTION,

⇒ $(\FRAC{{y - 8}}{{36}} + \frac{{y - 12}}{{54}} + \frac{y}{{72}} = 1 \Rightarrow \frac{{6\left( {y - 8} \right) + 4\left( {y - 12} \right) + 3y}}{{216}} = 1)$

⇒ 13y - 96 = 216

⇒ 13y = 312

⇒ y = 24 days

∴ The number of days for which T worked = 24 days

Time TAKEN by the inlet PIPE to fill the cistern = 6 hours

So, part of cistern filled by inlet pipe in 1 hour = 1/6

Time taken to fill the cistern with leak = 6 hours + 4 hours = 10 hours

part of cistern filled by inlet pipe with leak in 1 hour = 1/10

Let the leak take x hours to empty the filled tank.

If an inlet pipe can fill the tank in x hours, then the portion filled in 1 hour = 1/x

$(\BEGIN{array}{l}\frac{1}{6}\; - \;\frac{1}{x}\; = \frac{1}{{10}}\\ \Rightarrow \;\frac{1}{6}\; - \frac{1}{{10}}\; = \frac{1}{x}\\ \Rightarrow \;\frac{{10\; - \;6}}{{60}}\; = \;\;\frac{1}{x}\\ \Rightarrow \;\frac{4}{{60}}\; = \;\frac{1}{x}\end{array})$ 

⇒ 4x = 60

⇒ x = 15 hours

Let the capacity of the tank = X ltr

Time TAKEN by A and B to fill the tank = 3 hr

Tank filled by A and B in 1 hr = x/3

Time taken by B to fill the tank = 2 hr

Tank filled by B in 1 hr = x/2

Tank filled by A in 1 hr = x/2 - x/3 = x/6

Time taken by A to completely empty the tank = 6 hr

Time taken by A empty the 25% (1/4 )tank = 6 × (1/4) = 1.5 hr

Mike and Mindy’s 1 day work = 1/6

Mindy and JOSH’s 1 day work = 1/10

Mike, Mindy and Josh’s 1 day work = 1/4

So, Mike and Josh’s 1 day work = 2 × (Mike, Mindy and Josh’s 1 day work) - (Mike and Mindy’s 1 day work) - (Mindy and Josh’s 1 day work)

⇒ 2 × (1/4) - (1/6) - (1/10)

⇒ (1/2) - (4/15) = 7/30

C drains out one UNIT of water in one hour

Capacity of the tank = 180

Half the capacity of the tank = 90

∴ TIME required to EMPTY the half-filled tank = 90/1 = 90 hours

Work done by the MAN in 1 DAY = 1/10

Work done by the FATHER in 1 day = 1/15

Work done by the son in 1 day = 1/20

Work done by them together in 1 day = 1/10 + 1/15 + 1/20 = (6 + 4 + 3) /60 = 13/60

Time taken by them to complete the job = 60/13 DAYS

Total provision (in men days) = 500 × 27 = 13500

After 3 days, provision left (in men days) = 13500 – 3 × 500 = 13500 – 1500 = 12000

Now, total new number of men = 500 + 300 = 800

X can complete the work in = 60 days

∴ X’s 1 day’s work = 1/60

Y can complete the work in = 90 days

∴ Y’s 1 day’s work = 1/90

Z can complete the work in = 180 days

∴ Z’s 1 day’s work = 1/180

X completes 1/30 work in 2 days.

∴ (X + Y + Z)’s 3^rd day work,

⇒ 1/60 + 1/90 + 1/180

⇒ 6/180

⇒ 1/30

(X + Y + Z) can do 1/30 + 1/30 work in 3 days = 1/15