Time And Work — Part 1 MCQs with Explanations

1 st method: A and B complete a work in = 15 days One day's work of (A + B) = $$\frac{1}{{15}}$$ B complete the work in = 20 days; One day's work of B = $$\frac{1}{{20}}$$ Then, A's one day's work $$\eqalign{ & = \frac{1}{{15}} - \frac{1}{{20}} \cr & = \frac{{4 - 3}}{6} \cr & = \frac{1}{{60}} \cr} $$ Thus, A can complete the work in = 60 days. 2 nd method: (A + B)'s one day's % work = $$\frac{{100}}{{15}}$$ = 6.66% B's one day's % work = $$\frac{{100}}{{20}}$$ = 5% A's one day's % work = 6.66 - 5 = 1.66% Thus, A need = $$\frac{{100}}{{1.66}}$$ = 60 days to complete the work.

$$\eqalign{ & {\text{One}}\,{\text{day's}}\,{\text{work}}\,{\text{of}} \cr & \left( {A + B} \right) = \frac{1}{{18}}\,.......(1) \cr & {\text{One}}\,{\text{day's}}\,{\text{work}}\,{\text{of}} \cr & \left( {A + C} \right) = \frac{1}{{12}}\,.......\left( 2 \right) \cr & {\text{One}}\,{\text{day's}}\,{\text{work}}\,{\text{of}} \cr & \left( {B + C} \right) = \frac{1}{9}\,.......\left( 3 \right) \cr & {\text{Adding}}\,\left( {\text{1}} \right){\text{,}}\,\left( {\text{2}} \right)\,{\text{and}}\,\left( {\text{3}} \right) \cr & 2 \times \left( {A + B + C} \right) = { {\frac{1}{{18}}} + {\frac{1}{{12}}} + {\frac{1}{9}} } \cr & 2(A + B + C) = \frac{1}{4} \cr & {\text{One day's work of}} \cr & {A + B + C} = \frac{1}{8} \cr & B = {\frac{1}{8}} - \left( {A + C} \right) \cr & B = {\frac{1}{8}} - {\frac{1}{{12}}} \cr & {\text{One day's work of}} \cr & B = \frac{{ {3 - 2} }}{{24}} = \frac{1}{{24}} \cr & B\,{\text{need}}\,24\,{\text{days}} \cr} $$

1 st Method: (A+B)'s one day's work = $$\frac{1}{3}$$ part (A+B) works 2 days together = $$\frac{2}{3}$$ part Remaining work = $$1 - \frac{2}{3}$$ = $$\frac{1}{3}$$ part $$\frac{1}{3}$$ part of work is completed by A in two days Hence, one day's work of A = $$\frac{1}{6}$$ Then, one day's work of B = $$\frac{1}{3} - \frac{1}{6}$$ = $$\frac{1}{6}$$ So, B alone can complete the whole work in 6 days. 2 nd Method: (A+B)'s one day's % work = $$\frac{{100}}{3}$$ = 33.3% Work completed in 2 days = 66.6% Remaining work = 33.4% One day's % work of A = $$\frac{{33.4}}{2}$$ = 16.7% One day's work of B = 33.4 - 16.7 = 16.7% B alone can complete the work in, = $$\frac{{100}}{{16.7}}$$ = 5.98 days ≈ 6 days.

1 st Method: $$\eqalign{ & \left( {B + C} \right)\,2\,{\text{days}}\,{\text{work}} \cr & = 2 \times \left( {\frac{1}{{20}} + \frac{1}{{30}}} \right) \cr & = 2 \times {\frac{{3 + 2}}{{60}}} \cr & = \frac{1}{6}{\text{part}} \cr & {\text{Remaining}}\,{\text{work}} \cr & = 1 - \frac{1}{6} \cr & = \frac{5}{6}\text{part} \cr & {\text{A's}}\,{\text{one}}\,{\text{day's}}\,{\text{work}} \cr & = \frac{1}{{18}}{\text{part}} \cr & {\text{Time taken to complete the work}} \cr & = \frac{{ {\frac{5}{6}} }}{{ {\frac{1}{{18}}} }}\,{\text{days}} \cr & {\text{Hence,}} \cr & {\text{Time taken to complete the work}} \cr & = {\frac{5}{6}} \times 18 \cr & = 15\,{\text{days}} \cr} $$ 2 nd Method: % of work B completes in one day = $$\frac{{100}}{{20}}$$ = 5%; % of work C completes in one day = $$\frac{{100}}{{30}}$$ = 3.33%; % of work (A + B) completes together in one day = 5 + 3.33 = 8.33%; % work (A + B) completes together in 2 days = 8.66 × 2 = 16.66%; Remaining work = 100 - 16.66 = 83.34%; % of work A completes in 1 day = $$\frac{{100}}{{18}}$$ = 5.55% Time taken to complete the remaining work by A = $$\frac{{83.34}}{{5.55}}$$ = 15 days

1 st Method: Working 5 hours a day, A can complete the work in 8 days i.e. = 5 × 8 = 40 hours Working 6 hours a day, B can complete the work in 10 days i.e. = 6 × 10 = 60 hours (A + B)'s 1 hour's work, $$\eqalign{ & = \frac{1}{{40}} + \frac{1}{{60}} \cr & = \frac{{3 + 2}}{{120}} \cr & = \frac{5}{{120}} \cr & = \frac{1}{{24}} \cr} $$ Hence, A and B can complete the work in 24 hours i.e. they require 3 days to complete the work. 2 nd Method: % 1 hour's work of A = $$\frac{{100}}{{40}}$$ = 2.5% % 1 hour's work of B = $$\frac{{100}}{{60}}$$ = 1.66% (A + B) one hour's % work, = (2.5 + 1.66) = 4.16% Time to complete the work, = $$\frac{{100}}{{4.16}}$$ = 24 hours Then, $$\frac{{24}}{8}$$ = 3 days They need 3 days, working 8 hours a day to complete the work.

According to question, Ganga begins at 9 am and she does 3 units/hours Saraswati begins at 10 am and she does 2 units/hours So by 11 am they complete 5 units Time $$ = \frac{{{\text{T}}{\text{.W}}.}}{{3 + 2}} = \frac{{24}}{5}$$ (4 cycle of 2 hrs each + 4 units left) And now ganga will complete 3 unit out of 4 units in 1 hr Now, rest 1 unit work done by = $$\frac{1}{2}$$ hr Total time = 8 + 1 + $$\frac{1}{2}$$ = 9$$\frac{1}{2}$$ hr Hence, Work finished at = 9 am + 9$$\frac{1}{2}$$ hr = 6:30 PM Alternate Solution: Work done by Ganga in 1 hour = $$\frac{1}{8}$$ Work done by Saraswati in 1 hour = $$\frac{1}{12}$$ They are working alternatively with Ganga beginning the job. Work done in every two hours = $$\frac{1}{8}$$ + $$\frac{1}{12}$$ = $$\frac{5}{24}$$ Work done in 4 × 2 = 8 hours = $$\frac{5\times4}{24}$$ = $$\frac{5}{6}$$ Remaining work = 1 - $$\frac{5}{6}$$ = $$\frac{1}{6}$$ In 9th hour, Ganga starts the work and does $$\frac{1}{8}$$ of the work Work remaining = $$\frac{1}{6}$$ - $$\frac{1}{8}$$ = $$\frac{1}{24}$$ In 10th hour, Saraswati starts the work Time needed to finish the remaining work $$\eqalign{ & = \frac{{\frac{1}{{24}}}}{{\frac{1}{{12}}}} \cr & = \frac{1}{{24}} \times 12 \cr} $$ $$=$$ 0.5 hours $$=$$ 30 minutes i.e., work will be completed in 9 hour 30 minutes, after 9 AM i.e., at 6:30 PM

Here, we use work equivalence method; 10 × 12 = 12 × x Or, x = 10 days To understand the work equivalence method, we use a graphic as follows: Men Days 10 ↓ 12 12 ↑ x (let) Here, the two arrows, downward (↓) and upward (↑) show variation between men and days. [If downward arrows show decrements then upward arrows show increments and vice-verse.] Thus, $$\eqalign{ & \frac{{10}}{{12}} = \frac{{\text{x}}}{{12}} \cr & {\text{or, x}} = \frac{{10 \times 12}}{{12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 10\,{\text{days}} \cr} $$

$$\eqalign{ & {\text{We have}} \cr & {\text{B}} = \frac{3}{2} \times {\text{A}} \cr & \to {\text{A}} = \frac{2}{3} \times {\text{B}} \cr & {\text{One day's work,}} \cr & \Rightarrow {\text{A}} + {\text{B}} = \frac{1}{{18}} \cr & \Rightarrow \frac{2}{3} \times {\text{B}} + {\text{B}} = \frac{1}{{18}} \cr & \Rightarrow \frac{5}{3} \times {\text{B}} = \frac{1}{{18}} \cr & {\text{One}}\,{\text{day's}}\,{\text{work}}\,{\text{of}}\,{\text{B}} \cr & = \frac{3}{{90}} \cr} $$ B alone can complete the work in $$\eqalign{ & = \frac{{90}}{3} \cr & = 30\,{\text{days}} \cr} $$

10 men = 20 boys →1 men = 2 boys 8 men = 2 × 8 boys = 16 boys Then, (16 boys + 4 boys) = 20 boys can make 260 mats in 20 days Now, It can be calculated by work equivalence method: 20 × 260 × 20 = x × 20 × 20 x = 260 mats

$$\frac{7}{{10}}$$ part of work has been completed by A in 15 days. Then, Rest work = 1 - $$\frac{7}{{10}}$$ = $$\frac{3}{{10}}$$ part Given, That $$\frac{3}{{10}}$$ part of the work is completed by A and B together in 4 days. Means, (A + B) completed the $$\frac{3}{{10}}$$ of work in 4 days So, (A + B)'s 1 day's work = $$\frac{3}{{10 \times 4}}$$ = $$\frac{3}{{40}}$$ Hence, (A + B) can complete the work in $$\frac{{40}}{3}$$ = $$13\frac{1}{3}$$ days

Let the work be completed in x days. C work for x days then A works for (x - 8) days and B works for (x - 12) days. According to the question, $$\eqalign{ & {\frac{{ {x - 8} }}{{36}} + \frac{{ {x - 12} }}{{54}} + \frac{x}{{72}}} = 1 \cr & {\frac{{ {6x - 48 + 4x - 48 + 3x} }}{{216}}} = 1 \cr & 13x - 96 = 216 \cr & 13x = 216 + 96 = 312 \cr & x = \frac{{312}}{{13}} \cr & x = 24\,{\text{days}} \cr} $$

A can complete the work in 7 × 6 = 42 hours 1 hour's work of A = $$\frac{1}{{42}}$$ B can complete the work in 7 × 8 = 56 hours 1 hour's work of B = $$\frac{1}{{56}}$$ (A + B)'s 1 hour's work $$\eqalign{ & = \frac{1}{{42}} + \frac{1}{{56}} \cr & = \frac{{4 + 3}}{{168}} \cr & = \frac{7}{{168}} \cr} $$ ∴ Time taken by (A + B) working 8 hours daily $$\frac{{168}}{{7 \times 8}} = 3\,{\text{days}}$$

P can fill cistern in 12 minutes P fills cistern in 1 minute = $$\frac{1}{{12}}$$ Q can fill cistern in 15 minutes Q fills cistern in 1 minute = $$\frac{1}{{15}}$$ part P and Q together can fill cistern in 1 minute, $$ = \frac{1}{{12}} + \frac{1}{{15}} \Rightarrow \frac{9}{{60}}$$ part So, they can together fill cistern in 3 minute, $$ = 3 \times \frac{9}{{60}} \Rightarrow \frac{9}{{20}}$$ part Rest Cistern = $$1 - \frac{9}{{20}}$$ = $$\frac{{11}}{{20}}$$ part $$\frac{{11}}{{20}}$$ part cistern must be filled by Q in $$\frac{{\frac{{11}}{{20}}}}{{\frac{1}{{15}}}} = 8\frac{1}{4}$$ minutes Alternatively, Pipe P can fill cistern in one minute = $$\frac{{100}}{{12}}$$ = 8.33% Pipe Q can fill cistern in one minute = $$\frac{{100}}{{15}}$$ = 6.66% (P + Q) together can fill the cistern in one minute = 15%; In 3 minutes cistern is filled = 45% P become off, then rest of cistern will be filled by Pipe Q in = $$\frac{{55}}{{6.66}}$$ = $$8\frac{1}{4}$$ minutes.

Let capacity of the tank = x gallons; Part of the tank filled in 1 minute $$\eqalign{ & = {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 \cr & {\text{Or}}, {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 = \frac{x}{{60}} \cr & {\text{Or}}, \frac{x}{{24}} + \frac{x}{{40}} - \frac{x}{{60}} = 30 \cr & {\text{Or}}, {\frac{{ {10x + 6x - 4x} }}{{240}}} = 30 \cr & {\text{Or}},\,12x = 30 \times 240 \cr & {\text{Or}},\,x = 600\,{\text{gallons}} \cr} $$

% Cistern is filled by 1 st pipe in one minute = $$\frac{{100}}{{20}}$$ = 5% % Cistern is filled by 2 nd pipe in one minute = $$\frac{{100}}{{30}}$$ = 3.33% % cistern filled by 1 st and 2 nd pipes in one minute = 8.33% According to question, Cistern is totally filled by 2 nd pipe in last 10 minute. That means 2 nd pipe filled 33.3% of the cistern in last 10 minute and 66.66% of cistern is filled by 1 st and 2 nd pipe together in = $$\frac{{66.66}}{{8.33}}$$ = 8 minutes Thus, after 8 minute, 1 st pipe must be turned off.

Let the pipes fill the tank in x minutes. Part of tank filled by pipes A and B in one minute: $$\frac{1}{{36}}$$ Part of the tank filled by pipe A in 1 minute = $$\frac{1}{{36}} - \frac{1}{x}$$ According to the question, $$\eqalign{ & 30 \times \frac{1}{{\text{x}}} + 40 \left( {\frac{1}{{36}} - \frac{1}{{\text{x}}}} \right) = 1 \cr & \Rightarrow \frac{{30}}{{\text{x}}} + \frac{{40}}{36} - \frac{{40}}{{\text{x}}} = 1 \cr & \Rightarrow \frac{{30}}{{\text{x}}} + \frac{{10}}{9} - \frac{{40}}{{\text{x}}} = 1 \cr & \Rightarrow \frac{{10}}{{\text{x}}} = \frac{1}{9} \cr} $$ Hence, x = 90 minutes. Alternatively, % of tank filled by (A+B) in 1 minute = $$\frac{{100}}{{36}}$$ = 2.78% % of tank filled in 30 minutes = 2.78 × 30 = 83% That means 17% of tank can be filled by A in last 10 minutes Then, % tank filled by A in 1 minute = $$\frac{{17}}{{10}}$$ = 1.7% % of tank filled by B in 1 minute = 2.78 - 1.7 = 1.1% So, B alone can fill 100% tank in, = $$\frac{{100}}{{1.1}}$$ = 90 minutes (approx.)

Water filled by (boy + girl) in one minute, $$\eqalign{ & \frac{4}{3} + \frac{3}{4} = \frac{{16 + 9}}{{12}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{25}}{{12}}\,{\text{liter}} \cr} $$ Hence, time taken to fill 100 liter, $${\text{ = 100}} \times \frac{{12}}{{25}} = 48\,{\text{minutes}}$$

Let the capacity of the tank = q litres According to the question, Quantity of water emptied by the leak in 1 hour = $$\frac{\text{q}}{10}$$ litres Qunatity of water filled by the tap in 1 hour = 240 litres According to the question, $$\eqalign{ & \Rightarrow \frac{{\text{q}}}{{10}} - \frac{{\text{q}}}{{15}} = 240 \cr & \Rightarrow \frac{{3{\text{q}} - 2{\text{q}}}}{{30}} = 240 \cr & \Rightarrow \frac{9}{{30}} = 240 \cr & \Rightarrow {\text{q}} = 240 \times 30 \cr & \Rightarrow {\text{q}} = 7200\,{\text{litres}} \cr} $$

Ratio of their daily wages = 5 : 6 : 4 Total wage = Rs. 1800 6 × 5x + 4 × 6x + 4 × 9x = 1800 [As A, B, C works for 6, 4, 9 days respectively] Or, 90x = 1800 Or, x = $$\frac{{1800}}{{90}}$$ Total amount of A = $$\frac{{1800 \times 30}}{{90}}$$ = Rs. 600

Man = 3 boy Daily wages for them = $$\frac{{800}}{5}$$ = 160 4 boy (1 man + 1 boy) = 160 4 boy = 160 Or, boy = Rs. 40

Number of page copied by (Subhash + Prakash) in 1 hour = $$\frac{{300}}{{40}}$$ = 7.5 pages; Subhash copied pages in one hour = $$\frac{{50}}{{10}}$$ = 5 pages Hence, Prakash copied pages in one hour = 7.5 - 5 = 2.5 Thus, Prakash can copied 30 pages in = $$\frac{{30}}{{2.5}}$$ = 12 hour

Variation Method: In 100 days only 2.5 km road i.e. 16.66 % of work has been completed Men Days Road (km) 45 100↓ 2.5 x↑ 200 12.5↑ Arrows show the directions of variation of quantity with respect to each other $$\frac{{\text{x}}}{{45}} = \frac{{100 \times 12.5}}{{200 \times 2.5}}$$ x = 113 men; Required men to be increased, = 113 - 45 = 68

Work equivalence method: 1200 × 45 = 1000 × x Hence, x = 54 days Variation Method: After 15 days 200 soldiers leaved. Soldiers Food for days 1200 ↓ 45 1000 ↑ x (let) Arrows show the opposite variation to each other $$\frac{{1200}}{{1000}} = \frac{{\text{x}}}{{45}}$$ Or, x = $$\frac{{1200 \times 45}}{{1000}}$$ = 54 days.

One Day's work of A and B together, $$\frac{1}{{\text{A}}} + \frac{1}{{\text{B}}} = \frac{1}{5}\,........({\text{i}})$$ When A works with twice efficiency,Then, $$\frac{2}{{\text{A}}} + \frac{1}{{3{\text{B}}}} = \frac{1}{3}\,........({\text{ii}})$$ on solving equations (i) and (ii), we get $$A = \frac{{25}}{4} = 6\frac{1}{4}$$

Let the obstruction remain for x min. Hence, Part of cistern filled in X min + part of cistern filled in 3 min = full cistern $$\left[ {\frac{{7{\text{x}}}}{{8 \times 12}} + \frac{{5{\text{x}}}}{{6 \times 16}}} \right]$$ $$ + $$ $$\left[ {\frac{3}{{12}} + \frac{3}{{16}}} \right]$$ = 1 $$\frac{{12{\text{x}}}}{{96}} + \frac{7}{{16}} = 1$$ Thus, X = 4.5 min.

Let pipe C alone can empty the cistern in x min. A fills cistern in 1 min = $$\frac{1}{{10}}$$ B fills cistern in 1 min = $$\frac{1}{{15}}$$ A and B together fill in 1 min $$ = \frac{{10 \times 15}}{{10 + 15}} = \frac{{150}}{{25}} = 6\,{\text{min}}$$ Since, waste pipe was left open for 6 min then, 6 min, $$\frac{6}{x}$$ part of cistern will be emptied by C Now, $$\frac{6}{x}$$ part of the cistern would be filled by A and B in 2 min. Hence, cistern will be filled in $$\frac{3}{x}$$ min. And $$\frac{x}{3}$$ = 6 x = 18 min.

Let 1 boy's 1 day's work = x And 1 girl's 1 day's work = y Now, (5 boys + 2 girls)'s work = 5x + 2y Given , 5x + 2y is equal to 4 times work done by a boy and a girl Thus, 5x + 2y = 4(x + y) 5x + 2y = 4x + 4y x = 2y $$\frac{{\text{x}}}{{\text{y}}} = \frac{2}{1}$$ Hence, the required ratio is 2 : 1

Let x be the required number of days Given,12 men and 12 women can complete a work separately in 14 days and 21 days respectively Then, 12 men's 1 day work = $$\frac{1}{{14}}$$ And, 12 women's 1 day work = $$\frac{1}{{21}}$$ Then ,12 women's 3 days work = $$\frac{3}{{21}}$$ = $$\frac{1}{7}$$ The remaining work = $$1 - \frac{1}{7}$$ = $$\frac{6}{7}$$ Man's group leaves 3 days before the completion of work That is, they were working together for x - 3 days Thus, we have $$\frac{1}{7}$$ work left to be done in last 3 days by the women's group. This also means $$\frac{6}{7}$$ th of work has been done by both the groups (before men left) Now, (12 men + 12 women)'s 1 day work = $$\frac{1}{{14}} + \frac{1}{{21}}$$ = $$\frac{5}{{42}}$$ i.e., $$\frac{5}{{42}}$$ work is done by 2 groups in 1 day. So, $$\frac{6}{7}$$ of work is done by 2 groups together in $$\frac{{42}}{5} \times \frac{6}{7}$$ = $$\frac{{36}}{5}$$ days Total time take to complete the work will be = $$\frac{{36}}{5}$$ + 3 = $$\frac{{51}}{5}$$

Vimal's 1 day work = $$\frac{1}{{20}}$$ Since, Vimal and Kamal can together complete in 12 days i.e. (Vimal + Kamal)'s 1 day work = $$\frac{1}{{12}}$$ Then, Kamal's 1 day work, $$ = \frac{1}{{12}} - \frac{1}{{20}} \Rightarrow \frac{2}{{60}} \Rightarrow \frac{1}{{30}}$$ If Kamal Works only for half a day daily, then his 1 day work becomes $$\frac{1}{2} \times \frac{1}{{30}}$$ = $$\frac{1}{{60}}$$ Therefore, 1 day work of both Vimal and Kamal, $$ = \frac{1}{{20}} + \frac{1}{{60}} \Rightarrow \frac{4}{{60}} \Rightarrow \frac{1}{{15}}$$ Hence, the work will be completed in 15 days.

Combined efficiency of all the three boats = 60 passengers /trip Now, consider option (A) 15 trips and 150 passengers means efficiency of A = 10 passengers per trip A's efficiency = 10 passengers per trip Then, (B + C) combined efficiency = 50 passengers per trip Since, combined efficiency is 60 so option (A) is correct

Efficiency of Raj = $$\frac{{100}}{{20}}$$ = 5% Work completed by Raj = 25% Rest work = 75% Efficiency of Abhijit = $$\frac{{75}}{{10}}$$ = 7.5% Combined efficiency = 5 + 7.5 = 12.5% They will complete the whole work by working together in,= $$\frac{{100}}{{12.5}}$$ = 8 days

By work equivalence method, man × days × work = MAN × DAYS × WORK 10 × 20 × 1 = 20 × 2 × x → x = 5 days

Let x kg of oil be used for the eating purpose, daily, then (x + 11) × 50 = (x +15) × 45 → x = 25 Thus, Total quantity of oil, = (25 +11) × 50 = 1800 Hence, required number of days, = $$\frac{{1800}}{{25}}$$ = 72 days

Work done by 2 men = 3 women = 4 boys 1 man = 2 boys 1 woman = $$\frac{4}{3}$$ boys Thus, Boys × days = 4 × 52 boys × days Again, 1 man + 1 women + 1 boys, $$ = 2 + \frac{4}{3} + 1$$ $$ = \frac{{13}}{3}$$ boys Using work equivalent method, boys × day = BOYS × DAYS $$4 \times 52 = \frac{{13}}{3} \times {\text{x}}\,{\text{(let)}}$$ x = 48 days

Efficiency of pipe A, $$\frac{{100}}{{20}}$$ = 5% 20% of efficiency of A = 1% Then, efficiency of 5 such pipes = 5%. Then,Time taken to fill the tank = $$\frac{{100}}{5}$$ = 20 min.

M 1 × D 1 = M 2 × D 2 mr = (m +n) × D 2 D 2 = $$\frac{{{\text{mr}}}}{{{\text{m}} + {\text{n}}}}$$

X completes $$\frac{1}{3}$$ rd in 4 days = 33.33% job in 4 days X One day work = 8.33% Y one day work = 5.55% [As he complete $$\frac{1}{6}$$ job = 16.66% job in 3 days] Z one day work = 10% Work done in 3 days by X, Y and Z = 25 + 16.66 + 30 = 71.66% Remaining work will be done by Y in, $$\frac{{28.33}}{{5.55}}$$ = 5.1 days

Working efficiency of both typist together, = $$\frac{{100}}{6}$$ = 16.66% per minute Now, let work efficiency of first typist be x and then second typist will be (16.66 - x) First typist typed alone for 4 minutes and second typed alone for 6 minutes and they left with $$\frac{1}{5}$$ (i.e 20%) of job, means they have completed 80% job Now, First Typist typed in 4 minute + Second typed in 6 minutes = 80% 4 × x + 6 × (16.66 - x) = 80% 4x + 100% - 6x = 80% x = 10% First Typist typed 10% per minutes. Then second typed (16.66 - 10) = 6.66% per minute Then, Second typist complete the whole job in $$\frac{{100}}{{6.66}}$$ = 15.01 = 15 minutes.

Assume work to be done 100%. First case, A + B = $$\frac{{100}}{5}$$ = 20% work done per day -------- (1) Second case, 2A + $$\frac{{\text{B}}}{2}$$ = $$\frac{{100}}{4}$$ = 25% work done per day ------ (2) On solving equation (1) and (2), we get A = 10 days

Let First division have x hectare of the land then second division will have (x - 12) hectare of land Now according to the question, 25 × (x - 12) = 21x + 300 or, x = 150 hectare Hence, the first division harvest 3150 tons tendu leaves.