Time and work questions

(Ganesh + Ram + Sohan)'s 1 day's work \(=\frac{1}{16}\)

(Ganesh + Ram)'s 1 day's work \(=\frac{1}{24}\)

\(\therefore\) Sohan's 1 day's work

\(=\frac{1}{16}-\frac{1}{24}=\frac{3-2}{48}=\frac{1}{48}\)

\(\therefore\) Required time \(=48\) days

A does \(30 \%\) of work in 30 days.

\(\therefore\) A will do the whole work in

\(=\frac{30}{30} \times 100=100\) days

Let \(B\) alone do the same work in \(x\) days.

Both together do \(70 \%=\frac{7}{10}\) work in 20 days.

\(\therefore \frac{20}{100}+\frac{20}{x}=\frac{7}{10}\)

\(\Rightarrow \frac{1}{5}+\frac{20}{x}=\frac{7}{10}\)

\(\Rightarrow \frac{20}{x}=\frac{7}{10}-\frac{1}{5}=\frac{7-2}{10}=\frac{5}{10}\)

\(\Rightarrow \frac{20}{x}=\frac{1}{2} \Rightarrow x=20 \times 2\)

\(=40\) days

According to question,

• (A + B)’s one day’s work= \(\frac{1}{72}\)
• (B + C)’s one day’s work=\( \frac{1}{120}\)
• (C + A)’s one day’s work\(= \frac{1}{90}\)

• On adding all three,

  2 (A + B + C)’s one day’s work \( = \frac{1}{72}+ \frac{1}{120}+ \frac{1}{90}\)

  \( \frac{5+3+4}{360} = \frac{1}{30}\)

• ∴ (A + B + C)’s one day’s work\( = \frac{1}{30\times 2}\) = \(\frac{1}{60}\)
• ∴ A, B and C together can finish the whole work in 60 days.
• Hence option C is the right answer.

Work done by A in one day = \(\frac{1}{15}\) of the work

Work done by B in one day = \(\frac{1}{20}\) of the work

Work done by A and B together in one day = \(\frac{1}{15}+\frac{1}{20}\) = \(\frac{7}{60}\) of the work

Work done by A and B together in four days = \(\frac{7}{60}\) x 4 = \(\frac{7}{15}\)

Therefore, work left =\( 1 - \frac{7}{15}\) = \(\frac{8}{15}\) of the work

Remaining work \(=1-\frac{2}{3}=\frac{1}{3}\) Remaining days \(=124-64=60\)

\(\therefore \frac{ M {1} D {1}}{ W {1}}=\frac{ M {2} D {2}}{ W {2}}\)

\(\Rightarrow \frac{120 \times 64}{\frac{2}{3}}=\frac{M_{2} \times 60}{\frac{1}{3}}\)

\(\Rightarrow M _{2}=\frac{120 \times 64}{2 \times 60}=64\)

\(\therefore\) No. of men can be discharged

\(=120-64=56\) men

\(M _{1} D _{1}= M _{2} D _{2}\)

\(\Rightarrow x \cdot x=y \cdot D _{2}\)

\(\Rightarrow D _{2}=\frac{x^{2}}{y}\) days

\( \frac{ M _{1} D _{1}}{ W _{1}}=\frac{ M _{2} D _{2}}{ W _{2}} \)

\( \therefore \quad \frac{ W _{1}}{ W _{2}}=\frac{ M _{1} D _{1}}{ M _{2} D _{2}} \)

\(\therefore \frac{9}{10}=\frac{(x-1)(x+1)}{(x+2)(x-1)}=\frac{x+1}{x+2}\)

\(\Rightarrow 10 x+10=9 x+18\)

\(\Rightarrow x=18-10=8\)

\(\Rightarrow \frac{12}{18}=\frac{8}{x}\)

where \(x\) is hours/days \(\Rightarrow 12 x=18 \times 8\)

\(\Rightarrow x=\frac{18 \times 8}{12}=12\) hours

\(\Rightarrow\) no. of days required \(=\frac{32}{2}\) \(=16\) days

One day's work done by A and B =\(\frac{1}{A}+\frac{1}{B}\)

= \(\frac{1}{11}+\frac{1}{20}\) = \(\frac{31}{220}\)
One day's work done by A and C =\(\frac{1}{A}+\frac{1}{C}\)
\( = \frac{1}{11}+\frac{1}{55}\) = \(\frac{5+1}{55}=\frac{6}{55}\)
Work done in first two days \( \frac{31}{220}+\frac{6}{55} \) = \( \frac{31+24}{220}=\frac{55}{220}=\frac{1}{4}\)
Therefore required time = \( 2 \times 4 =8\) days

Earning in the first additional-hour \(=\frac{2000}{50}=40\)

Total extra earning \(=2300-2000=₹ 300\)

Hours \(=\frac{\text { Earning }}{\text { Earning / hour }}\)

\(=\frac{300}{40 \times \frac{3}{2}}=5 hrs\)

Work done by B

\(=1-\frac{19}{23}=\frac{23-19}{23}=\frac{4}{23}\)

\(\therefore(A+C): B=\frac{19}{23}: \frac{4}{23}=19: 4\)

\(\therefore\) Sum of ratios \(=19+4=23\)

\(\therefore\) B's share

\(=\frac{4}{23} \times 575=₹ 100\)

Skilled : half skilled : unskilled

\(=\frac{1}{3}: \frac{1}{4}: \frac{1}{6}\)

\(=\) \(\left(\frac{1}{3} \times 12\right):\left(\frac{1}{4} \times 12\right):\left(\frac{1}{6} \times 12\right)\)

\(=4: 3: 2\)

Share of the trained labourer

\(=\frac{28}{(7 \times 4+8 \times 3+2 \times 10)} \times 369\)

\(=\frac{28}{(28+24+20)} \times 369\)

\(=\frac{28}{72} \times 369=₹ 143.50\)

Work done by \((A+B)\) in 5 days

\(=5\left(\frac{1}{12}+\frac{1}{20}\right)=5\left(\frac{5+3}{60}\right)\)

\(=\frac{40}{60}=\frac{2}{3}\)

Remaining work \(=1-\frac{2}{3}=\frac{1}{3}\)

\(\because\) Time taken by \(\mathrm{C}\) in \(\operatorname{doing} \frac{1}{3}\) work \(=3\) days

\(\therefore\) Required time \(=3 \times 3\) \(=9\) days

Time taken by P alone in doing work \(=x\) days

\(\therefore\) Time taken by \(Q\) alone \(=4 x\) days According to the question \(\therefore 4 x-x=45\)

\(\Rightarrow 3 x=45\)

\(\Rightarrow x=\frac{45}{3}=15\) days

\(\therefore\) Time taken by \(Q =60\) days

\(\therefore( P + Q )^{\prime}\) s 1 day's work

\(=\frac{1}{15}+\frac{1}{60}=\frac{4+1}{60}\)

\(=\frac{5}{60}=\frac{1}{12}\)

\(\therefore\) Required time \(=12\) days

Time taken by Rahul \(=42\) hours

Raj is \(100 \%\) more efficient.

Time taken by Raj \(=21\) hours

\(\therefore\) Work done by Rahul and Raj in 1 hour

\(=\frac{1}{42}+\frac{1}{21}=\frac{1+2}{42}=\frac{3}{42}=\frac{1}{14}\)

\(\therefore\) Required time \(=14\) hours.

Time taken by Shashi in doing 1 work \(=20\) days Tanya is \(25 \%\) more efficient than Shashi.

Time taken by Tanya

\(=\frac{100}{125} \times 20=16\) days

In second case, the efficiency of a man is twice to that in the first case.

\(\therefore \quad M _{1} D _{1}=2 M _{2} D _{2}\)

\(\Rightarrow 10 \times 20=2 \times 20 \times D_{2}\)

\(\Rightarrow D _{2}=\frac{10 \times 20}{2 \times 20}=5\) days

Let time taken by \(A =x\) days

\(\therefore\) Time taken by \(B =2 x\) days Time taken by \(C =3 x\) days According to the question,

\(\frac{1}{x}+\frac{1}{2 x}+\frac{1}{3 x}=\frac{1}{6}\)

\(\Rightarrow \frac{6+3+2}{6 x}=\frac{1}{6}\)

\(\Rightarrow \frac{11}{6 x}=\frac{1}{6}\)

\(\Rightarrow 6 x=6 \times 11\)

\(\Rightarrow x=\frac{6 \times 11}{6}=11\)

\(\therefore\) Time taken by \(C\) alone \(=3 x\) \(=3 \times 11=33\) days

Time taken by Sonia \(=3 x\) days (let)

\(\therefore\) Time taken by Pratibha \(=x\) days

\(\therefore 3 x-x=60 \Rightarrow 2 x=60\)

\(\Rightarrow x=30\) days

\(\therefore\) Time taken by Sonia

\(=3 x\) days \(=3 \times 30=90\) days.

If Anuj alone does the work in \(x\) days and Vijay alone does the work in \(y\) days, then

\(\frac{1}{x}+\frac{1}{y}=\frac{1}{5}\)

Again, \(\frac{2}{x}+\frac{1}{3 y}=\frac{1}{3}\)

By equation (ii) \(\times 3-(i)\)

\(\frac{6}{x}+\frac{1}{y}-\frac{1}{x}-\frac{1}{y}=1-\frac{1}{5}\)

\(\Rightarrow \frac{6}{x}-\frac{1}{x}=\frac{4}{5}\)

\(\Rightarrow \frac{6-1}{x}=\frac{4}{5}\)

\(\Rightarrow x=\frac{25}{4}=6 \frac{1}{4}\) days

Let time taken by \(P =x\) days Then, time taken by \(Q =3 x\) days \(\therefore 3 x-x=48 \Rightarrow x=24\)

\(\therefore( P + Q )^{\prime}\) s 1 day's work

\(=\frac{1}{24}+\frac{1}{72}=\frac{3+1}{72}=\frac{1}{18}\)

\(\therefore\) Required time \(=18\) days

Work done by A and B in 5 days

\(=5\left(\frac{1}{12}+\frac{1}{15}\right)=5\left(\frac{5+4}{60}\right)\)

\(=\frac{9}{12}=\frac{3}{4}\)

Time taken by \(C\) in \(\operatorname{doing} \frac{1}{4}\) work \(=5\) days

C will complete in 20 days.

\(\therefore\) Ratio of wages \(=\frac{1}{12}: \frac{1}{15}: \frac{1}{20}\)

\(=5: 4: 3\)

\(\therefore\) Amount received by A

\(=\frac{5}{12} \times 960=Rs. 400\)

Ratio of wages of \(A, B\) and \(C\) respectively \(=5 \times 6: 6 \times 4: 4 \times 9\)

\(=30: 24: 36=5: 4: 6\)

\(\therefore\) Amount received by A

\(=\frac{5}{5+4+6} \times 1800\)

\(=\frac{5}{15} \times 1800=₹ 600\)

56 × 24 = 42 × x (x = no. of days required)
\( \Rightarrow \,\,\,\,\,x\, = \,\frac{{56 \times 24}}{{42}}\)
x = 32 day
Concept-
man × day = man × days (work) = (work)

Man : boy \(=3: 1\)

\(\therefore\) Boy's share \(=\frac{1}{4} \times 800=\) 200

\(\therefore\) The daily wages of boy

\(=₹\left(\frac{200}{5}\right)=₹ 40\)

C's 1 day's work

\(=\frac{1}{4}-\left(\frac{1}{8}+\frac{1}{12}\right)=\frac{1}{4}-\left(\frac{3+2}{24}\right)\)

\(=\frac{1}{4}-\frac{5}{24}=\frac{6-5}{24}=\frac{1}{24}\)

\(A: B: C=\frac{1}{8}: \frac{1}{12}: \frac{1}{24}=3: 2: 1\)

C's share \(=\left(\frac{1}{6} \times 4500\right)=\) Rs. 750

• Ratio of Suman's and Sumati's 1 day's work \(=\)

  \(\frac{1}{3}: \frac{1}{2}=2: 3\)

• Sum of the ratios \(=2+3=5\)

• Suman's share \(=\frac{2}{5} \times 150=Rs. 60\)

\(\because\) Pramod does \(\frac{1}{4}\) work in 9 days.

\(\therefore\) Pramod will do 1 work in \(=9 \times 4=36\) days

\(\because\) Saroj does \(\frac{3}{4}\) work in 27 days.

\(\therefore\) Saroj will do 1 work in

\(=\frac{27 \times 4}{3}=36\) days

\(\therefore\) Work done by both in 1 day

\(=\frac{1}{36}+\frac{1}{36}=\frac{2}{36}=\frac{1}{18}\)

\(\therefore\) Required time \(=18\) days

Let the work be completed in \(x\) days. According to the question,

\(\frac{x}{16}+\frac{x-8}{32}+\frac{x-6}{48}=1\)

\(\Rightarrow \frac{6 x+3 x-24+2 x-12}{96}=1\)

\(\Rightarrow 11 x-36=96\)

\(\Rightarrow 11 x=96+36=132\)

\(\Rightarrow x=\frac{132}{11}=12\) days

Nuts cut by Ram and Hari in 1 day \(=\frac{12}{2} kg .=6 kg\)

Nuts cut by them in 5 days \(=30 kg\)

Amount of nuts cut by Ram alone \(=58-30=28 kg\)

Time \(=8\) days

\(\therefore\) Nuts cut by Ram in 1 day

\(=\frac{28}{8}=3.5 kg\)

\(\therefore\) From equation (i), Nuts cut by Hari in 1 day \(=(6-3.5) kg .=2.5 kg\)

\(\therefore\) Time taken by Hari in cutting

\(10 kg .\) of nuts \(=\frac{10}{2.5}=4\) days

1 man can do the work in \((16 \times 12 \times 14)\) hours Time taken by 28 men \(=\frac{16 \times 12 \times 14}{28 \times 12}\) days

\(=8\) days

\((A+B)\) 's 1 day's work

\(=\frac{1}{25}+\frac{1}{30}=\frac{6+5}{150}=\frac{11}{150}\)

\(\therefore(A+B)^{\prime} s 5\) days' work

\(=\frac{5 \times 11}{150}=\frac{11}{30}\)

\(\therefore\) Remaining work \(=1-\frac{11}{30}=\frac{30-11}{30}=\frac{19}{30}\)

Number of pages copied by \(x\) in hour \(=\frac{80}{20}=4\)

Number of pages copied by \(x\) and \(y\) in 1 hour \(=\frac{135}{27}=5\)

\(\therefore\) Number of pages copied by

\(y\) in 1 hour \(=5-4=1\)

\(\therefore\) Required time \(=20\) hours.

Let B alone can do the work in \(x\) days.

\(\therefore\) A can do the work in \(\frac{3 x}{2}\) days.

According to the question,

\(\frac{1}{x}+\frac{2}{3 x}=\frac{1}{18} \Rightarrow \frac{3+2}{3 x}=\frac{1}{18}\)

\(\Rightarrow \frac{5}{3 x}=\frac{1}{18} \Rightarrow 3 x=18 \times 5\)

\(\Rightarrow x=\frac{18 \times 5}{3}=30\) days

Calculation :

• Time taken by A to finish the work \(=40\) days

  Time taken by B to finish the work \(=60\) days

• Total time taken \(=\frac{x \times y}{x+y}\) days

  \(=\frac{40 \times 60}{40+60}\)

  \(=\frac{40 \times 60}{100}=24\) days

Hence the answer is option C

\(\because\) Jyothi can do \(\frac{3}{4}\) th of a job in 12 days.

\(\therefore\) Jyothi can do 1 job in

\(\frac{12 \times 4}{3}=16\) days

As Mala is twice as efficient as Jyothi,

\(\therefore\) Mala will finish the job in 8 days.

Let \(A, B\) and \(C\) together do the work in \(x\) hours.

\(\therefore\) Time taken by \(\mathrm{A}\) \(=(x+6)\) hours

Time taken by \(\mathrm{B}=(x+1)\) hours

Time taken by \(\mathbf{C}=2 \times\) hours

\(\therefore \frac{1}{x+6}+\frac{1}{x+1}+\frac{1}{2 x}=\frac{1}{x}\)

\(\Rightarrow \frac{1}{x+6}+\frac{1}{x+1}=\frac{1}{x}-\frac{1}{2 x}\)

\(=\frac{1}{2 x}\)

\(\Rightarrow \frac{1}{x+6}=\frac{1}{2 x}-\frac{1}{x+1}\)

\(=\frac{x+1-2 x}{2 x(x+1)}\)

\(\Rightarrow \frac{1}{x+6}=\frac{1-x}{2 x^{2}+2 x}\)

\(\Rightarrow \frac{x+6}{x}=\frac{30}{20}=\frac{3}{2}\)

\(\Rightarrow 2 x+12=3 x\)

\(\Rightarrow 3 x-2 x=12\)

\(\Rightarrow x=12\)

Let \(Y\) alone complete the work in \(x\) days.

According to the question, X's 16 days' work \(+\) Y's 12 days' work \(=1\)

\(\Rightarrow \frac{16}{24}+\frac{12}{x}=1\)

\(\Rightarrow \frac{2}{3}+\frac{12}{x}=1\)

\(\Rightarrow \frac{12}{x}=1-\frac{2}{3}=\frac{1}{3}\)

\(\Rightarrow x=12 \times 3=36\) days

\(\because P\) men working \(P\) hours \(/\) day for \(P\) days produce \(P\) units of work.

\(\therefore 1\) man working 1 hour/day for 1 day produce

\(\frac{P}{P^{3}}=\frac{1}{P^{2}}\) units of work

\(\therefore n\) men working \(n\) hours a day for \(n\) day's produce \(\frac{n^{3}}{P^{2}}\) units of work

\(\quad(A+B)^{\prime}\) s 1 hour's work

\(=\frac{1}{36}\)

A's 1 hour's work \(=\frac{1}{63}\)

\(\therefore\) B's 1 hour's work \(=\frac{1}{36}-\frac{1}{63}\)

\(=\frac{7-4}{252}=\frac{3}{252}=\frac{1}{84}\)

\(\therefore\) Ratio of A's and B's 1 hour's

work \(=\frac{1}{63}: \frac{1}{84}\)

\(=\left(\frac{1}{63} \times 252\right):\left(\frac{1}{84} \times 252\right)\)

\(=4: 3\)

Sum of the terms of ratio \(=4+3=7\)

\(\therefore B ^{\prime}\) s share \(= Rs .\left(\frac{3}{7} \times 5950\right)\) \(=\) Rs. 2550

Ratio of the 1 day's work of \(X\) and \(Y\)

\(=\frac{1}{12}: \frac{1}{60}=\frac{1}{12} \times 60: \frac{1}{60} \times 60\) \(=5: 1\)

Sum of the terms of ratio \(=5+1=6\)

\(Xs\) share \(= Rs .\left(\frac{5}{6} \times 24000\right)\) \(=\) Rs. 20000

Ratio of 1 day's work of \(A , B\) and \(c\)

\(=\frac{1}{10}: \frac{1}{12}: \frac{1}{15}\)

\(=\frac{1}{10} \times 60: \frac{1}{12} \times 60: \frac{1}{15} \times 60\)

[LCM of 10,12 and \(15=601\) \(=6: 5: 4\)

Sum of the terms of ratio \(=6+5+4=15\)

\(\therefore\) C's share \(= Rs .\left(\frac{4}{15} \times 4500\right)\)

\(=\) Rs. 1200

8 unit (work done)
work remaining 8 unit
\(\therefore \) no. of days = 2 days
no. of women required \( = \,\frac{{8 \times 3}}{2}\) = 12 women