Time And Work — Part 2 MCQs with Explanations

Time × Rate = total charges 100 × 100 = 10000 X × 125 = 110 [25% increase in rate, user can afford only 10% increase] X = $$\frac{{110}}{{125}} \times 100$$ = 88% Thus, decrease in time = 12%

Mango Packing rate of Ram Lal = $$\frac{{70}}{7}$$ = 10 per hour Guavas Packing rate of Ram Lal = $$\frac{{56}}{7}$$ = 8 per hour Mango Packing rate of Ram Lal's Wife = $$\frac{{30}}{6}$$ = 5 per hour Guavas Packing rate of Ram Lal's Wife = $$\frac{{24}}{6}$$ = 4 per hour On comparing, 8 Guavas = 10 Mangoes 1 Guavas = $$\frac{{10}}{8}$$ Mangoes 2400 Guavas = $$\frac{{10 \times 2400}}{8}$$ = 3000 Mangoes This means that they need to pack (6300 + 3000) Mangoes Combined Mango Packing rate of Both = 10 + 5 = 15 per hour We will take 2 days = 1 time unit. Working 10 hours a day they will pack 150 Mangoes in 1 time unit Total time taken to pack 6300 Mangoes = $$\frac{{6300}}{{150}}$$ = 42 time unit So, total time taken = 2 × 42 = 84 days

Let X be the initial number of men then, According to the question, 4X = X + (X - 20) + (X - 40) + (X - 60) + (X - 80) + (X - 100) + (X - 120) ⇒ 4X = 7X - 420 ⇒ 3X = 420 ⇒ X = $$\frac{{420}}{3}$$ ⇒ X = 140 men

Total efficiency of two persons = 50% [As they complete work in 2 days. ] First Person completes work = $$\frac{1}{3}$$ = 33.33% [In 2 days] Rest work will be completed by Second man = $$\frac{2}{3}$$ = 66.66% [In 2 days] So, efficiency of second person is greater. Efficiency of second person = $$\frac{{66.66}}{2}$$ = 33.33% per day Then, Second person will complete whole work in, = $$\frac{{100}}{{33.33}}$$ = 3 days.

Let initially X number of workers be there. Now, Using work equivalence method, X + (X - 1) + (X - 2) + . . . . . + 1 = X × 55% of X $$\frac{{{\text{X}} \times \left( {{\text{X}} + 1} \right)}}{2} = \frac{{5{\text{X}}}}{{100}}$$ [series is in AP. Sum of AP = {No. of terms (first term + last term)/2}] X = 10 workers.

Let X litres be the per day filling and L litres be the capacity of the reservoir, then 90X + L = 40000 × 90 ----------- (1) 60X + L = 32000 × 60 ----------- (2) Solving the equation, X = 56000 litres Thus, 56000 litres per day can be used without the failure of supply.

Ratio of efficiencies of A, B and C, = 5x : 4x : 6x Number of days required by A and B = $$\frac{{100}}{{9{\text{x}}}}$$ ------ (1) Number of days required by A, B and C = $$\frac{{100}}{{15{\text{x}}}}$$ ------ (2) $$\eqalign{ & \frac{{100}}{{9{\text{x}}}} - \frac{{100}}{{15{\text{x}}}} = \frac{8}{3} \cr & \Rightarrow {\text{x}} = \frac{5}{3} \cr} $$ Number of days required by A, B and C = $$\frac{{100}}{{15{\text{x}}}}$$ = $$\frac{{100}}{{15 \times \frac{5}{3}}}$$ = 4 days

Difference in times required by the first man (A) and second man (B) = 3 hours. Also, if t a and t b are the respective times, then t b - t a = 3 . . . . . . . . . ..(1) Also, B alone be take = (t a + 3) h According to the question, 2t b - t a = 8 2 × (t a + 3) - t a = 8 [Using equation (1)] t a = 2 hours. Now B and woman together take 2 hours and A also take 2 hours, so time required will be half when all 3 work together. So in 1 hour work would be completed.

Let E 1 and E 2 are the working efficiency of each student of IIT and NIT per hour respectively. Using work equivalence method, 4 × 10 × 60 × E 1 = 5 × 8 × 80 × E 2 $$\frac{{{{\text{E}}_1}}}{{{{\text{E}}_2}}} = \frac{4}{3}$$ As, 3E 1 = 4E % less efficient of NITians = $$\frac{{1 \times 100}}{4}$$ = 25% Thus, each engineer from NIT is 25% less efficient than that of IIT.

Let X be the number of women required to finish the work in 21 days. Now, using Work Equivalence Method: 42 × 18 = X × 21 X = 36. Number of women required = 36

Given, A is thrice good workman as B. Means, A = 3B Let B can finish work in X days, then A will finish same work in (X - 40) days alone Now, BX = 3B × (X - 40) X = 60 days B can finish work in 60 days, then A can finish the work in 20 days. One day work of B = $$\frac{1}{{60}}$$ One day work of A = $$\frac{1}{{20}}$$ One day work of (A+B) = $$\frac{1}{{60}} + \frac{1}{{20}} \Rightarrow \frac{{1 + 3}}{{60}} \Rightarrow \frac{1}{{15}}$$ So, they can finish work together in 15 days

A and B can complete the work in 18 days, work rate = $$\frac{{100}}{{18}}$$ = 5.55% per day They together can complete the work in 12 days = 5.55 × 12 = 66.60% Now, A leaves and B takes another 15 days to complete the whole work, Work rate of B = $$\frac{{33.30}}{{15}}$$ = 2.22% per day B work for (12 + 15) = 27 days So, Work done by B in 27 days = 2.22 × 27 ≈ 60% And So 40% work is done by A. so there share should be 60% and 40% ratio. A's share = 40% of 1500 = Rs. 600

3 men or 4 women can plough the field in 43 days 3 men = 4 women 1 man = $$\frac{4}{3}$$ women 7 man = $$\frac{{28}}{3}$$ women 7 men and 5 women = $$5 + \frac{{28}}{3}$$ = $$\frac{{43}}{3}$$ women 4 women can plough field in 43 days So, 1 women can plough in = 43 × 4 days Hence, $$\frac{{43}}{3}$$ women can plough = $$\frac{{43 \times 4 \times 3}}{{43}}$$ = 12 days

Given (A+B)'s one day's work = $$\frac{1}{8}$$ (B + C)'s one day's work = $$\frac{1}{{12}}$$ (A + B + C) 's 1 day's work = $$\frac{1}{6}$$ Work done by A, alone= (A + B + C) 's 1 day's work - (B + C)'s one day's work $$ = \frac{1}{6} - \frac{1}{{12}} = \frac{{2 - 1}}{{12}} = \frac{1}{{12}}$$ Work done by C, alone = (A + B + C) 's 1 day's work - (A + B)'s one day’s work $$ = \frac{1}{6} - \frac{1}{8} = \frac{{4 - 3}}{{24}} = \frac{1}{{24}}$$ ⇒ (A + C)’s one day’s work $$\eqalign{ & = \frac{1}{{12}} + \frac{1}{{24}} \cr & = \frac{{2 + 1}}{{24}} \cr & = \frac{3}{{24}} = \frac{1}{8} \cr} $$ ⇒ (A + C) will take 8 days to complete the work together

Work done by all the tanks working together in 1 hour, $$ \Rightarrow \frac{1}{{10}} + \frac{1}{{12}} - \frac{1}{{20}} = \frac{2}{{15}}$$ Hence, tank will be filled in $$\frac{{15}}{2}$$ = 7.5 hour.

A, B and C together can fill 100% empty tank in 10 minutes Work rate of (A + B + C) = $$\frac{{100}}{{10}}$$ = 10% per minute A alone can fill the tank in 30 minutes Work rate of A = $$\frac{{100}}{{30}}$$ = 3.33% per minute B alone can fill the tank in 40 minutes Work rate of B = $$\frac{{100}}{{40}}$$ = 2.5% Work rate of (A + B) = 3.33 + 2.5 = 5.83% per minute Work rate of C, = Work rate of (A + B + C) - (A + B) = 10 - 5.83 = 4.17% per minute So, C takes = $$\frac{{100}}{{4.17}}$$ ≈ 24 minutes to fill the tank

Work rate of A = $$\frac{{100}}{9}$$ = 11.11% work per day Work rate of B = $$\frac{{100}}{{15}}$$ = 6.66% work per day They together can do (A + B) = 11.11 + 6.66 ≈ 18% work per day They are working in alternate day, so we take 2 days = 1 unit of day Therefore, in one unit of day they can complete 18% of work (A + B) can complete 90% of work in 5 units of days. i.e. (5 × 18) And the rest 10% work will be completed by A in Next day So, total number of day, = 5 Unit of days + 1 day of A = 2 × 5 + 1 = 11 days

Pipe A can fill empty tank in 36 min. Pipe A can fill the tank = $$\frac{{100}}{{36}}$$ = 2.77% per minute Pipe B can fill empty tank in 45 min. Pipe B can fill the tank = $$\frac{{100}}{{45}}$$ = 2.22% per min. A and B can together fill the tank = (2.77 + 2.22) ≈ 5% per minute So, A and B can fill the tank in 7 min. = 7 × 5 = 35% of the tank Rest tank to be filled = 100 - 35 = 65% C can empty the full tank in 30 min. C can empty the tank = $$\frac{{100}}{{30}}$$ = 3.33% per min. C is doing negative work i.e. emptying the tank A, B and C can together fill the tank, = 2.77% + 2.22% - 3.33% = 1.67% tank per minute So, A, B and C will take time to fill 65% empty tank, = $$\frac{{65}}{{1.67}}$$ = 39 min. (Approx)

Time taken by A =x hours. Therefore taken by A, B and C together = (x - 6) Time taken by B = (x - 5) Time taken by C = 2(x - 6) Now, rate of work of A + Rate of work of B + Rate of work of C = Rate of work of ABC. $$ \Rightarrow \frac{1}{x} + \frac{1}{{x - 5}} + \frac{1}{{2\left( {x - 6} \right)}} = \frac{1}{{x - 6}}$$ On solving above equation, we get x = 3, $$\frac{{40}}{6}$$ When x = 3, the expression (x - 6) becomes negative, thus it's not possible. $$ \Rightarrow x = \frac{{40}}{6}$$ Time taken by A & B together = $$\frac{1}{{\frac{3}{{20}} + \frac{3}{5}}}$$ = $$\frac{4}{3}$$ hours

To do half of the work in one sixth of the time means A is 3 times as efficient as B. To understand this point just think if same work would have been done in one sixth of the time then A would have been six times as efficient as B. Now they together complete the work in 10 days. B is working with someone who is 3 times as efficient as himself. That means 4 people of B's efficiency finished the work in 10 days. So B alone would have done it in 40 days. A alone would have taken one third of this time. Alternatively, Given, $${\text{A}} \times \frac{1}{6} = {\text{B}} \times \frac{1}{2}$$ So, A = 3B Now, Given they together complete the work in 10 days So, One Day''s work of, (A + B) = $$\frac{{100}}{{10}}$$ = 10% (3B + B) = 10% 4B = 10% So, one day work of B = $$\frac{{10}}{4}$$ = 2.5% So, B can complete 100% work in = $$\frac{100}{2.5}$$ = 40 days

His Per day pay = Rs. 26 Total pay employee got = Rs. 829 Total pay he gets if he did not remain idle a single day, = 26 × 56 = Rs.1456 He Forfeits or fined = 1456 - 829 = Rs. 627 Per day he Forfeits Rs. 7 Means per idle day he loses = 26 + 7 = Rs. 33 So, Total idle days = $$\frac{{627}}{{33}}$$ = 19 days Alternatively, Let he works for X days and remain idle for (56 - X) Now, According to question, X × 26 - 7 × (56 - X) = 829 26X + 7X - 392 = 829 33X = 829 + 392 = 1221 X = 37 days Number of days remain idle = 56 -37 =19 days

Given, A is 60% more efficient of B Means, $$\eqalign{ & {\text{A}} = {\text{B}} + 60\% \,{\text{of B}} \cr & {\text{A}} = {\text{B}} + \frac{{60{\text{B}}}}{{100}} \cr & {\text{A}} = \frac{{100{\text{B}} + 60{\text{B}}}}{{100}} \cr & {\text{A}} = \frac{{160{\text{B}}}}{{100}} \cr & {\text{A}} = \frac{{8{\text{B}}}}{5} \cr} $$ A can complete whole work in 15 days. So, One day work of A = $$\frac{1}{{15}}$$ One day work of A = $$\frac{{8{\text{B}}}}{5}$$ = $$\frac{1}{{15}}$$ One day work of B = $$\frac{5}{{120}}$$ = $$\frac{1}{{24}}$$ One day work, (A + B) = $$ \frac{1}{{15}} + \frac{1}{{24}}$$ One day work, (A + B) = $$\frac{{24 + 15}}{{360}}$$ = $$\frac{{39}}{{360}}$$ So, Time taken to finish the work by A and B together = $$\frac{{360}}{{39}}$$ = $$\frac{{120}}{{13}}$$ days Alternatively We can solve it through percentage method Given, A = $$\frac{{8{\text{B}}}}{5}$$ A can complete whole work in 15 days Work rate of A = $$\frac{{100}}{{15}}$$ = 6.66% per day Work rate of $$\frac{{8{\text{B}}}}{5}$$ = 6.66% Work rate of B = 4.16% per day Work rate of (A + B) = 6.66 + 4.16 = 10.82% per day So, A and B can complete 100% work in $$\frac{{120}}{{13}}$$ days

Pipe A can fill the empty tank in = 0.9 hours So work rate of the Pipe A = $$\frac{{100}}{{0.9}}$$ % per hour Pipe B can empty the tank in = 0.7 hours Negative Work rate of B = $$\frac{{100}}{{0.7}}$$ % per hour. (B is removing water, so, taken as negative work) Tank fill per hour = $$\frac{{100}}{{0.7}} - \frac{{100}}{{0.9}}$$ = 31.75% per hour Time Taken to empty the tank = $$\frac{{100}}{{31.75}}$$ ≈ 3.15 hours So, Capacity of the tank = 3.15 × 180 = 567 liters

$$\eqalign{ & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{15}}; \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{20}}; \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1day's}}\,{\text{work}} \cr & = {\frac{1}{{15}} + \frac{1}{{20}}} = \frac{7}{{60}} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{4}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{7}{{60}} \times 4} = \frac{7}{{15}} \cr & \therefore {\text{Remaining}}\,{\text{work}}\, = {1 - \frac{7}{{15}}} = \frac{8}{{15}} \cr} $$

$$\eqalign{ & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{4} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{16}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{12}} \cr & \therefore {\text{C's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \frac{1}{4} - \left( {\frac{1}{{16}} + \frac{1}{{12}}} \right) = {\frac{1}{4} - \frac{7}{{48}}} = \frac{5}{{48}} \cr & {\text{So,}}\,{\text{C}}\,\,{\text{alone}}\,{\text{can}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}} \cr & \frac{{48}}{5} = 9\frac{3}{5}\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{A's}}\,{\text{2}}\,{\text{day's}}\,{\text{work}} = {\frac{1}{{20}} \times 2} = \frac{1}{{10}} \cr & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{60}}} = \frac{6}{{60}} = \frac{1}{{10}} \cr & {\text{Work}}\,{\text{done}}\,{\text{in}}\,{\text{3}}\,{\text{days}} = {\frac{1}{{10}} + \frac{1}{{10}}} = \frac{1}{5} \cr & {\text{Now}},\,\frac{1}{5}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{in}}\,{\text{3}}\,{\text{days}} \cr & \therefore {\text{Whole}}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{in}}\, \cr & {3 \times 5} = 15\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{Ratio}}\,{\text{of}}\,{\text{times}}\,{\text{taken}}\,{\text{by}}\,{\text{A}}\,{\text{and}}\,{\text{B = 1:3}} \cr & {\text{The}}\,{\text{time}}\,{\text{difference}}\,{\text{is}}\,\left( {{\text{3 - 1}}} \right)\,{\text{2}}\,{\text{days}} \cr & {\text{while}}\,{\text{B}}\,{\text{take}}\,{\text{3}}\,{\text{days}}\,{\text{and}}\,{\text{A}}\,{\text{takes}}\,{\text{1}}\,{\text{day}}{\text{.}} \cr & {\text{If}}\,{\text{difference}}\,{\text{of}}\,{\text{time}}\,{\text{is}}\,{\text{2}}\,{\text{days,}}\,{\text{B}}\,{\text{takes}}\,{\text{3}}\,{\text{days}}{\text{.}} \cr & {\text{If}}\,{\text{difference}}\,{\text{of}}\,{\text{time}}\,{\text{is}}\,{\text{60}}\,{\text{days,}} \cr & {\text{B}}\,{\text{takes}}\,\left( {\frac{3}{2} \times 60} \right) = 90\,{\text{days}} \cr & {\text{So,}}\,{\text{A}}\,{\text{takes}}\,{\text{30}}\,{\text{days}}\,{\text{to}}\,{\text{do}}\,{\text{the}}\,{\text{work}}{\text{.}} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{30}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{90}} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{30}} + \frac{1}{{90}}} = \frac{4}{{90}} = \frac{2}{{45}} \cr & \therefore {\text{A}}\,{\text{and}}\,{\text{B}}\,{\text{together}}\,{\text{can}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}}\, \cr & \frac{{45}}{2} = 22\frac{1}{2}\,{\text{days}}\, \cr} $$

$$\eqalign{ & {\text{C's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \frac{1}{3} - \left( {\frac{1}{6} + \frac{1}{8}} \right) \cr & = \frac{1}{3} - \frac{7}{{24}} \cr & = \frac{1}{{24}} \cr & {\text{A's}}\,{\text{wages:B's}}\,{\text{wages:C's}}\,{\text{wages}} \cr & = \frac{1}{6}:\frac{1}{8}:\frac{1}{{24}} \cr & = 4:3:1 \cr & \therefore {\text{C's}}\,{\text{share}}\,{\text{(for}}\,{\text{3}}\,{\text{days}}\,{\text{)}} \cr & = Rs.\,\left( {3 \times \frac{1}{{24}} \times 3200} \right) \cr & = Rs.\,400 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{1}}\,{\text{man's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = x\,{\text{and}} \cr & 1\,{\text{boy's}}\,{\text{1day's}}\,{\text{work}} = y \cr & {\text{Then,}}\,6x + 8y = \frac{1}{{10}}\,{\text{and}}\, \cr & 26x + 48y = \frac{1}{2} \cr & {\text{Solving}}\,{\text{these}}\,{\text{two}}\,{\text{equations,}}\,{\text{we}}\,{\text{get}} \cr & x = \frac{1}{{100}}\,and\,y = \frac{1}{{200}} \cr & {\text{(15}}\,{\text{men}}\,{\text{ + 20}}\,{\text{boy)'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{{15}}{{100}} + \frac{{20}}{{200}}} = \frac{1}{4} \cr & \therefore {\text{15}}\,{\text{men}}\,{\text{and}}\,{\text{20}}\,{\text{boys}}\,{\text{can}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}}\,{\text{4}}\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{A's}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{4}; \cr & {\text{(B + C)'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{3}; \cr & {\text{(A + C)'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{2}. \cr & {\text{(A + B + C)'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} \cr & = {\frac{1}{4} + \frac{1}{3}} = \frac{7}{{12}} \cr & {\text{B's}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} \cr & = {\frac{7}{{12}} - \frac{1}{2}} = \frac{1}{{12}} \cr & \therefore {\text{B}}\,{\text{along}}\,{\text{will}}\,{\text{take}}\,{\text{12}}\,{\text{hours}}\,{\text{to}}\,{\text{do}}\,{\text{the}}\,{\text{work}} \cr} $$

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{10}} \cr & {\text{C's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{50}} \cr & \left( {{\text{A + B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{10}} + \frac{1}{{50}}} = \frac{6}{{50}} = \frac{3}{{25}}........\left( {\text{i}} \right) \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = \left( {{\text{B + C}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}\,........\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right){\text{,we}}\,{\text{get}}:2 \times \left( {{\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}} \right) \cr & = \frac{3}{{25}} \cr & \Rightarrow {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{3}{{50}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}}\left( {\frac{1}{{10}} - \frac{3}{{50}}} \right) \cr & = \frac{2}{{50}} = \frac{1}{{25}} \cr & {\text{So,}}\,{\text{B}}\,\,{\text{alone}}\,{\text{could}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}}\,{\text{25}}\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{Whole}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & = {20 \times \frac{5}{4}} = 25\,{\text{days}} \cr & {\text{Now}},\,\left( {1 - \frac{4}{5}} \right)\,\,i.e., \cr & \frac{1}{5}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{and}}\,{\text{B}}\,{\text{in}}\,{\text{3}}\,{\text{days}} \cr & {\text{Whole}}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{and}}\,{\text{B}}\,{\text{in}} \cr & = \left( {3 \times 5} \right) = 15\,{\text{days}} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{25}}, \cr & \left( {{\text{A + B}}} \right)\,{\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{15}} \cr & \therefore {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{15}} - \frac{1}{{25}}} = \frac{4}{{150}} = \frac{2}{{75}} \cr & {\text{So,}}\,{\text{B}}\,\,{\text{alone}}\,{\text{would}}\,{\text{do}}\,{\text{the}}\,{\text{work}}\,{\text{in}} \cr & \frac{{75}}{2} = 37\frac{1}{2}\,{\text{days}} \cr} $$

$$\eqalign{ & \left( {{\text{P + Q + R}}} \right){\text{'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} \cr & = {\frac{1}{8} + \frac{1}{{10}} + \frac{1}{{12}}} = \frac{{37}}{{120}} \cr & {\text{Work}}\,{\text{done}}\,{\text{by}}\,{\text{P,}}\,{\text{Q}}\,{\text{and}}\,{\text{R}}\,{\text{in}}\,{\text{2}}\,{\text{hours}} \cr & = {\frac{{37}}{{120}} \times 2} = \frac{{37}}{{60}} \cr & {\text{Remaining}}\,{\text{work}} = {1 - \frac{{37}}{{60}}} = \frac{{23}}{{60}} \cr & \left( {{\text{Q + R}}} \right){\text{'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} \cr & = {\frac{1}{{10}} + \frac{1}{{12}}} = \frac{{11}}{{60}} \cr & {\text{Now}},\frac{{11}}{{60}}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{Q}}\,\,{\text{and}}\,{\text{R}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & {\text{So}},\frac{{23}}{{60}}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{Q}}\,{\text{and}}\,{\text{R}}\,{\text{in}} \cr & = {\frac{{60}}{{11}} \times \frac{{23}}{{60}}} = \frac{{23}}{{11}}\,{\text{hours}} \approx 2\,{\text{hours}} \cr & {\text{So,}}{\text{the}}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{finished}}\,{\text{approximately}} \cr & {\text{2}}\,{\text{hours}}\,{\text{after}}\,{\text{11}}\,{\text{A}}{\text{.M}}{\text{.,}}\,{\text{i}}{\text{.e}}{\text{.,}}\,{\text{around}}\,{\text{1}}\,{\text{P}}{\text{.M}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{B's}}\,{\text{10}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{15}} \times 10} = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{work}} \cr & = {1 - \frac{2}{3}} = \frac{1}{3} \cr & {\text{Now}},\frac{1}{{18}}{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{1}}\,{\text{day}} \cr & \therefore \frac{1}{3}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & {18 \times \frac{1}{3}} = 6\,{\text{days}} \cr} $$

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y Then, 4x + 6y = $$\frac{1}{8}$$ and 3x + 7y = $$\frac{1}{{10}}$$ Solving the two equations, we get $$x = \frac{{11}}{{400}},\,\,\,y = \frac{1}{{400}}$$ ∴ 1 woman's 1 day's work = $$\frac{1}{{400}}$$ ⇒ 10 women's 1 day's work = $$ {\frac{1}{{400}} \times 10} $$ = $$\frac{1}{{40}}$$ Hence, 10 women will complete the work in 40 days

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{20}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{30}} \times 20} = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{work}} \cr & = {1 - \frac{2}{3}} = \frac{1}{3} \cr & {\text{Now}},\frac{1}{3}\,{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}}\,{\text{20}}\,{\text{days}} \cr & \therefore {\text{The}}\,{\text{whole}}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{A}}\,{\text{in}} \cr & {20 \times 3} = 60\,days \cr} $$

$$\eqalign{ & {\text{P}}\,{\text{can}}\,{\text{complete}}\,{\text{the}}\,{\text{work}} \cr & = \,\left( {12 \times 8} \right){\text{hrs}}{\text{.}} = 96\,{\text{hrs}}{\text{.}} \cr & {\text{Q}}\,{\text{can}}\,{\text{complete}}\,{\text{the}}\,{\text{work}} \cr & = \left( {8 \times 10} \right){\text{hrs}}{\text{.}} = 80\,{\text{hrs}}{\text{.}} \cr & \therefore {\text{P's}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{{96}}\,{\text{and}} \cr & \therefore {\text{Q's}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} = \frac{1}{{80}} \cr & \left( {{\text{P + Q}}} \right){\text{'s}}\,{\text{1}}\,{\text{hour's}}\,{\text{work}} \cr & = {\frac{1}{{96}} + \frac{1}{{80}}} = \frac{{11}}{{480}} \cr & {\text{So,}}\,{\text{both}}\,{\text{P}}\,{\text{and}}\,{\text{Q}}\,{\text{will}}\,{\text{finish}}\,{\text{the}}\,{\text{work}} \cr & = {\frac{{480}}{{11}}} {\text{ hrs}}{\text{.}} \cr & \therefore {\text{Number}}\,{\text{of}}\,{\text{days}}\,{\text{of}}\,{\text{8}}\,{\text{hours}}\,{\text{each}} \cr & {\frac{{480}}{{11}} \times \frac{1}{8}} = \frac{{60}}{{11}}{\text{days}} = 5\frac{5}{{11}}{\text{days}} \cr} $$

$$\eqalign{ & 1\,woman's\,1\,day's\,work = \frac{1}{{70}} \cr & {\text{1}}\,{\text{child's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{140}} \cr & \left( {{\text{5}}\,{\text{women + 10}}\,{\text{children}}} \right){\text{'s}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{5}{{70}} + \frac{{10}}{{140}}} = {\frac{1}{{14}} + \frac{1}{{14}}} = \frac{1}{7} \cr & \therefore {\text{5}}\,{\text{women}}\,{\text{and}}\,{\text{10}}\,{\text{chidren}}\,{\text{will}}\,{\text{complete}} \cr & {\text{the}}\,{\text{work}}\,{\text{in}}\,{\text{7}}\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{work}}\,{\text{done}}\,{\text{by}}\,{\text{X}}\,{\text{in}}\,{\text{4}}\,{\text{days}} \cr & = {\frac{1}{{20}} \times 4} = \frac{1}{5} \cr & {\text{Remaining}}\,{\text{work}} \cr & = {1 - \frac{1}{5}} = \frac{4}{5} \cr & \left( {{\text{X + Y}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{20}} + \frac{1}{{12}}} = \frac{8}{{60}} = \frac{2}{{15}} \cr & {\text{Now}},\frac{2}{{15}}{\text{work}}\,{\text{is}}\,{\text{done}}\,{\text{by}}\,{\text{X}}\,{\text{and}}\,{\text{Y}}\,{\text{in}}\,{\text{1}}\,{\text{day}}. \cr & {\text{So}},\,\frac{4}{5}\,{\text{work}}\,{\text{will}}\,{\text{be}}\,{\text{done}}\,{\text{by}}\,{\text{X}}\,{\text{and}}\,{\text{Y}}\,{\text{in}} \cr & {\frac{{15}}{2} \times \frac{4}{5}} = 6\,{\text{days}} \cr & {\text{Hence,}}\,{\text{total}}\,{\text{time}}\,{\text{taken}} \cr & = \left( {6 + 4} \right)\,{\text{days}} \cr & = 10\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{Ratio}}\,{\text{of}}\,{\text{times}}\,{\text{taken}}\,{\text{by}}\,{\text{A}}\,{\text{and}}\,{\text{B}} \cr & = 100:130 = 10:13 \cr & {\text{Suppose}}\,{\text{B}}\,{\text{takes}}\,x\,{\text{days}}\,{\text{to}}\,{\text{do}}\,{\text{the}}\,{\text{work}} \cr & {\text{Then}},10:13::23:x \cr & \Rightarrow x = {\frac{{23 \times 13}}{{10}}} \cr & \Rightarrow x = \frac{{299}}{{10}} \cr & {\text{A's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{1}{{23}} \cr & {\text{B's}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} = \frac{{10}}{{299}} \cr & \left( {{\text{A + B}}} \right){\text{'s}}\,{\text{1}}\,{\text{day's}}\,{\text{work}} \cr & = {\frac{1}{{23}} + \frac{{10}}{{299}}} \cr & = \frac{{23}}{{299}} \cr & = \frac{1}{{13}} \cr & \therefore A\,{\text{and}}\,{\text{B}}\,{\text{together}}\,{\text{can}}\,{\text{complete}} \cr & \,{\text{the}}\,{\text{work}}\,{\text{in}}\,{\text{13}}\,{\text{days}}{\text{.}} \cr} $$