Ans .

(3) 10 days

1. Explanation :

   (3) According to question, A and B can do a work in 12 days .
   (A + B)s one days work = \( \frac{1}{12} \) Similarly,
   (B + C)s one days work = \( \frac{1}{15} \) and (C + A)s one days work = \( \frac{1}{20} \) .
   On adding all three,
   2 (A + B + C)s one dayss work =\( \frac{1}{12} \) +\( \frac{1}{15} \) +\( \frac{1}{20} \) = \( \frac{10+8+6}{120} \) = \( \frac{1}{5} \)
   => (A + B + C)s one days`s work = \( \frac{1}{10} \)
   A, B and C together can finish the whole work in 10 days.

Ans .

(3) 60 days

1. Explanation :

   (3) (A+B)`s 1 days work = \( \frac{1}{72} \) .
   (B+C)`s 1 days work =\( \frac{1}{120} \) and (C+A)`s 1 days work = \( \frac{1}{90} \) .
   On adding all three,
   2 (A + B + C)`s 1 days work =\( \frac{1}{72} \) +\( \frac{1}{120} \) +\( \frac{1}{90} \) = \( \frac{5+3+4}{360} \) = \( \frac{1}{30} \)
   =>(A+B+C)`s 1 day`s work = \( \frac{1}{60} \)
   (A+B+C) will do the work in 60 days.

Ans .

(1) 4 days

1. Explanation :

   (1) According to question, 10 mens one days work =\( \frac{1}{12} \)
   1 man one days work =\( \frac{1}{12*10} \) = \( \frac{1}{120} \)
   Similarly, 1 woman one days work =\( \frac{1}{6*10} \) = \( \frac{1}{60} \)
   (1 man + 1 woman)s one days work =\( \frac{1}{120} \) +\( \frac{1}{60} \) =\( \frac{1+2}{120} \) = \( \frac{3}{120} \) = \( \frac{1}{40} \)
   (10 men + 10 women)s one days work = \( \frac{10}{40} \)= \( \frac{1}{4} \)
   Therefore, both the teams can finish the whole work in 4 days.

Ans .

(3) 3.6 days

1. Explanation :

   (3) According to question, A can finish the whole work in 6 days. A`s one days work=\( \frac{1}{6} \)
   Similarly, B`s one days work = \( \frac{1}{9} \)
   (A + B)`s one days work=\( \frac{1}{6} \) +\( \frac{1}{9} \) = \( \frac{3+2}{18} \) +\( \frac{5}{18} \)
   Therefore, (A + B)s can finish thewhole work in \( \frac{18}{5} \) days i.e., 3.6 days. .

Ans .

(2)120 days

1. Explanation :

   (2) According to the question Work done by A and B together in one day = \( \frac{1}{10} \) part Work done by B and C together in one day = \( \frac{1}{15} \) part Work done by C and A together in one day = \( \frac{1}{20} \) part.
   So, A + B = \( \frac{1}{10} \) ....(I)
   B + C = \( \frac{1}{15} \) ...(II)
   C + A = \( \frac{1}{20} \) ....(III)
   Adding I, II, III, we get 2 (A + B + C) = \( \frac{1}{10} \) +\( \frac{1}{15} \) +\( \frac{1}{20} \)
   2 (A + B + C) =\( \frac{6+4+3}{60} \) =\( \frac{13}{60} \)
   A + B + C = 13 120 ....(IV)
   Putting the value of eqn. (I) in eqn. (IV) \( \frac{1}{10} \)+c =\( \frac{13}{120} \) Work done in 1 day by C is \( \frac{1}{120} \) part.
   Hence, C will finish the whole work in 120 days

Ans .

(2)12 hours

1. Explanation :

   (2) A`s 1 hours work = \( \frac{1}{4} \)
   (B + C)s 1 hours work = \( \frac{1}{3} \) and (A + C)s 1 hours work = \( \frac{1}{2} \)
   C`s 1 hours work = \( \frac{1}{2} \) - \( \frac{1}{4} \) = \( \frac{2-1}{4} \) =\( \frac{1}{4} \)
   and B`s 1 hours work = \( \frac{1}{3} \) - \( \frac{1}{4} \) = \( \frac{4-3}{12} \) =\( \frac{1}{12} \)
   Hence, B alone can do the work in 12 hours.

Ans .

(3)3\( \frac{3}{7} \)days

1. Explanation :

   (3) A`s 1 days work =\( \frac{1}{24} \)
   B`s 1 days work = \( \frac{1}{6} \) and C`s 1 day+s work = \( \frac{1}{12} \)
   (A + B + C)s 1 days work =\( \frac{1}{24} \) + \( \frac{1}{6} \) +\( \frac{1}{12} \) =\( \frac{1+4+2}{24} \) =\( \frac{7}{24} \)
   The work will be completed by them in \( \frac{24}{7} \) i.e.. 3\( \frac{3}{7} \)days.

Ans .

(3) 15 days

1. Explanation :

   (3) (A + B)s 1 days work =\( \frac{1}{10} \)
   A`s 1 days work =\( \frac{1}{30} \)
   B`s 1 days work =\( \frac{1}{10} \)-\( \frac{1}{30} \) = \( \frac{3-1}{30} \)=\( \frac{2}{30} \)=\( \frac{1}{15} \)
   Hence, B, alone can complete the work in 15 days.

Ans .

(3) 120 days

1. Explanation :

   (3) (A + B)s 1 days work=\( \frac{1}{72} \)
   (B + C)s 1 days work =\( \frac{1}{120} \) and (C + A)s 1 days work =\( \frac{1}{90} \)
   Adding all three,
   2(A + B + C)s 1 days work =\( \frac{1}{72} \) + \( \frac{1}{120} \)+ \( \frac{3-1}{90} \) = \( \frac{5+3+4}{360} \)= \( \frac{12}{360} \) =\( \frac{1}{30} \)
   (A + B + C)s 1 days work =\( \frac{1}{60} \)
   Now, A`s 1 day`s work = (A + B + C)`s 1 day`s work - (B + C)s 1 days work=\( \frac{1}{60} \)-\( \frac{1}{120} \)=\( \frac{2-1}{120} \)=\( \frac{1}{120} \) A alone can complete the work in 120 days. .

Ans .

(3) 5\( \frac{5}{47} \)days

1. Explanation :

   (3) (A + B)s 1 days work \( \frac{1}{8} \)
   (B + C)s 1 days work =\( \frac{1}{6} \) and (C + A)s 1 days work =\( \frac{1}{10} \)
   On adding, 2(A + B + C)s 1 days work= \( \frac{1}{8} \)+\( \frac{1}{6} \)+\( \frac{1}{10} \)
   =\( \frac{15+20+12}{120} \)=\( \frac{47}{120} \)
   => (A + B + C)'s 1 days work =\( \frac{47}{240} \).
   (A + B + C) together will complete the work in \( \frac{240}{47} \)=5\( \frac{5}{47} \) = days.

Ans .

(4)48 days

1. Explanation :

   (4) (A + B)s 1 days work =\( \frac{1}{12} \)(i)
   (B + C)s 1 days work=\( \frac{1}{8} \) (ii) and (C + A)s 1 days work=\( \frac{1}{6} \)(iii)
   On adding, 2(A + B + C)s 1 days work =\( \frac{1}{12} \) +\( \frac{1}{8} \) +\( \frac{1}{6} \) =\( \frac{2+3+4}{24} \) =\( \frac{9}{24} \)
   (A+ B + C)'s 1 days work =\( \frac{9}{24*2} \)=\( \frac{9}{48} \) ...(iv)
   On, subtracting (iii) from (iv),
   Bs 1 days work =\( \frac{9}{48} \) -\( \frac{1}{6} \) =\( \frac{9-8}{48} \) =\( \frac{1}{48} \)
   => B can complete the work in 48 days.

Ans .

(4)13\( \frac{1}{3} \)days

1. Explanation :

   (4) Work done by (A + B) in 1 day =\( \frac{1}{30} \)
   Work done by (B + C) in 1 day = \( \frac{1}{20} \) and Work done by (C + A) in 1 day = \( \frac{1}{15} \)
   On adding, Work done by 2 (A +B + C) in 1 day =\( \frac{1}{30} \)+\( \frac{1}{20} \)+\( \frac{1}{15} \)=\( \frac{2+3+4}{60} \)=\( \frac{9}{60} \)=\( \frac{3}{20} \)
   Work done by (A + B + C) in 1 day =\( \frac{3}{40} \)
   (A + B + C) will do the work in \( \frac{4}{30} \) = 13\( \frac{1}{3} \)days

Ans .

(1)8 days

1. Explanation :

   (1) Let A and C complete the work in x days
   (A + B)s 1 days work =\( \frac{1}{8} \)
   (B + C)s 1 days work =\( \frac{1}{12} \)and (C + A)s 1 days work =\( \frac{1}{x} \)
   Then (A + B + B + C + C + A)s 1 day s work =\( \frac{1}{8} \)+\( \frac{1}{12} \)+\( \frac{1}{x} \)
   2(A + B + C)s 1 days work =\( \frac{5x+24}{24x*2} \)
   According to the question,
   (A + B + C)s 1 days work =\( \frac{1}{6} \)=\( \frac{5x+24}{48x} \)
   30x + 144 = 48x
   x =\( \frac{144}{18} \)= 8 days.

Ans .

(1)24 days

1. Explanation :

   A`s 1 days work = \( \frac{1}{12} \)
   (A+B)s 1 days work = \( \frac{1}{8} \)
   B`s 1 days work=\( \frac{1}{8} \) -\( \frac{1}{12} \) =\( \frac{3-2}{24} \) = \( \frac{1}{24} \)
   B alone can do the work in 24 days.

Ans .

(2)24 days

1. Explanation :

   (2) (A + B)s 1 days work =\( \frac{1}{18} \)
   (B + C)s 1 days work =\( \frac{1}{9} \) and (A + C)s 1 days work =\( \frac{1}{12} \)
   Adding all the above three.
   2 (A + B + C)s 1 days work =\( \frac{1}{18} \) +\( \frac{1}{9} \) +\( \frac{1}{12} \) =\( \frac{2+4+3}{36} \) =\( \frac{9}{36} \) =\( \frac{1}{4} \)
   (A + B + C)s 1 days work =\( \frac{1}{8} \)
   B`1s 1 days work = (A + B + C)s 1 days work - (A + C)s 1 days work =\( \frac{1}{8} \) -\( \frac{1}{12} \)=\( \frac{3-2}{24} \)=\( \frac{1}{24} \)
   Hence, B alone can do the work in 24 days.

Ans .

(1)3 days

1. Explanation :

   (1) A alone can complete the work in 42 days working 1 hour daily. Similarly, B will take 56 days working 1 hour daily.
   A`s 1 days work = \( \frac{1}{42} \)
   Bs 1 day’s work = \( \frac{1}{56} \)
   (A + B)s 1 days work =\( \frac{1}{42} \) +\( \frac{1}{56} \) =\( \frac{4+3}{168} \) =\( \frac{7}{168} \)
   = Time taken by (A + B) working 8 hours daily = \( \frac{168} {7}\)= 3 days.

Ans .

(3)40 days

1. Explanation :

   (3) (A + B)s 1 days work =\( \frac{1}{10} \).............. (i)
   (B + C)s 1 days work=\( \frac{1}{12} \)............. (ii) and (C + A)s 1 days work = \( \frac{1}{15} \)............... (iii)
   On adding all these, 2(A + B + C)s 1 days work=\( \frac{1}{10} \)+\( \frac{1}{12} \)+\( \frac{1}{15} \) =\( \frac{6+5+4}{60} \) =\( \frac{1}{4} \)
   (A + B + C)s 1 day work=\( \frac{1}{8} \)................ (iv)
   C`s 1 days work =\( \frac{1}{8} \) -\( \frac{1}{10} \)=\( \frac{5-4}{40} \)=\( \frac{1}{40} \)
   C will finish the work in 40 days.

Ans .

(1)60 days

1. Explanation :

   (1) (A + B)s 1 days work =\( \frac{1}{15} \)
   B`s 1 days work =\( \frac{1}{20} \)
   A`s 1 days work =\( \frac{1}{15} \) - \( \frac{1}{20} \) =\( \frac{4-3}{60} \) =\( \frac{1}{60} \)
   A alone will do the work in 60 days.

Ans .

(4)20 days

1. Explanation :

   (4) (A + B)s 1 days work =\( \frac{1}{12} \) ,/br> (B + C)s 1 days work =\( \frac{1}{15} \) and (C + A)s 1 days work =\( \frac{1}{20} \)
   On adding, 2 (A + B + C)s 1 days work =\( \frac{1}{12} \) +\( \frac{1}{15} \)+\( \frac{1}{20} \) =\( \frac{5+4+3}{60} \) =\( \frac{1}{5} \)
   (A+B+C)s 1 days work =\( \frac{1}{10} \)
   B`s 1 days work =\( \frac{1}{10} \) - \( \frac{1}{20} \) \( \frac{2-1}{20} \) \( \frac{1}{20} \)
   B alone can do the work in 20 days.

Ans .

(3)30 days

1. Explanation :

   (3) (P + Q)s 1 days work=\( \frac{1}{12} \)...(i)
   (Q + R)s 1 days work =\( \frac{1}{15} \)..(ii) and (R + P)s 1 days work =\( \frac{1}{20} \) ...(iii)
   Adding all three equations, 2 (P + Q + R)s 1 days work= \( \frac{1}{12} \) +\( \frac{1}{15} \)+\( \frac{1}{20} \) =\( \frac{5+4+3}{60} \)=\( \frac{12}{60} \)=\( \frac{1}{5} \)
   (P + Q + R)s 1 days work=\( \frac{1}{10} \)
   P`s 1 days work=\( \frac{1}{10} \)-\( \frac{1}{15} \)=\( \frac{3-2}{30} \)\( \frac{1}{30} \)
   P alone will complete the work in 30 days.

Ans .

(3)6 days

1. Explanation :

   (A + B)s 1 days work =\( \frac{1}{8} \)
   (B + C)s 1 days work =\( \frac{1}{12} \) and (C + A)s 1 days work =\( \frac{1}{8} \)
   On adding, 2 (A + B + C)s 1 days work =\( \frac{1}{8} \)+\( \frac{1}{12} \)+\( \frac{1}{8} \)=\( \frac{3+2+3}{24} \)=\( \frac{8}{24} \)=\( \frac{1}{3} \)
   (A + B + C)s 1 days work =\( \frac{1}{6} \) Hence, the work will be completed in 6 days.

Ans .

(3)5\( \frac{5}{7} \) days

1. Explanation :

   (3) (A + B)s 1 days work =\( \frac{1}{10} \)
   (B + C)s 1 days work = \( \frac{1}{6} \) and (C + A)s 1 days work = \( \frac{1}{12} \)
   Adding all three 2 (A + B + C)s 1 days work = \( \frac{1}{10} \)+\( \frac{1}{6} \)+\( \frac{1}{12} \)=\( \frac{6+10+5}{60} \)=\( \frac{21}{60} \)=\( \frac{7}{20} \)
   (A + B + C)s 1 days work = \( \frac{7}{40} \)
   All three together will complete the work in= \( \frac{40}{7} \) = 5\( \frac{5}{7} \)days.

Ans .

(2) 2 hours

1. Explanation :

   (2) (A + B)s 1 hours work =\( \frac{2}{9} \) .....(i)
   (B + C)s 1 hours work =\( \frac{1}{3} \) .....(ii) and (C + A)s 1 hours work =\( \frac{4}{9} \)...(iii)
   Adding all three equations, 2 (A + B + C)s 1 hours work= \( \frac{2}{9} \)+\( \frac{1}{3} \)+\( \frac{4}{9} \)=\( \frac{2+3+4}{9} \)=1
   A, B and C together will complete the work in 2 hours.

Ans .

(1)16 days

1. Explanation :

   (1) (A + B)s 1 days work =\( \frac{1}{18} \)
   (B + C)s 1 days work = \( \frac{1}{24} \) and (A + C)s 1 days work =\( \frac{1}{36} \)
   Adding all three, 2 (A + B + C)s 1 days work= \( \frac{1}{18} \)+\( \frac{1}{24} \)+\( \frac{1}{36} \)=\( \frac{4+3+2}{71} \)=\( \frac{1}{8} \)
   (A + B + C) 1 days work =\( \frac{1}{16} \)
   A, B and C together will complete the work in 16 days.

Ans .

(3)13\( \frac{1}{3} \)days

1. Explanation :

   (3) (A + B)`s 1 day`s work =\( \frac{1}{5} \) and A`s 1 day`s work = \( \frac{1}{8} \)
   B`s 1 days work =\( \frac{1}{5} \)-\( \frac{1}{8} \)=\( \frac{8-5}{40} \)
   B alone will complete the work in \( \frac{40}{3} \)=13\( \frac{1}{3} \) days.

Ans .

(2)75 minutes

1. Explanation :

   (2) Work done by (A + B + C) in 1 minute =\( \frac{1}{30} \)
   Work done by (A + B) in 1 minute =\( \frac{1}{50} \)
   Work done by C alone in 1 minute =\( \frac{1}{30} \)-\( \frac{1}{50} \)=\( \frac{5-3}{150} \)=\( \frac{2}{150} \)=\( \frac{1}{75} \).
   C alone will complete the work in 75 minutes.

Ans .

(1) 60 day

1. Explanation :

   (1) (A + B)’s 1 day’s work =\( \frac{1}{8} \)
   (B + C)s 1 days work =\( \frac{1}{24} \) and (C + A)s 1 days work =\( \frac{7}{60} \)
   On adding all three, 2 (A + B + C)’s 1 day’s work=\( \frac{1}{8} \)+\( \frac{1}{24} \)+\( \frac{7}{60} \)=\( \frac{15+5+14}{120} \)=\( \frac{34}{120} \),/br> (A + B + C)s 1 day’s work =\( \frac{17}{120} \)
   C’s 1 day’s work =\( \frac{17}{120} \)-\( \frac{1}{8} \)=\( \frac{17-15}{120} \)=\( \frac{1}{60} \)

   br> C alone will complete the work in 60 days

Ans .

(1) 24 days

1. Explanation :

   (1) (A+B)s 1 days work =\( \frac{1}{10} \) and (B + C)s 1 days work =\( \frac{1}{12} \)
   (C + A)s 1 days work = \( \frac{1}{15} \)
   On adding all three, 2(A + B + C)s 1 days work=\( \frac{1}{10} \)+\( \frac{1}{12} \)+\( \frac{1}{15} \)=\( \frac{6+5+4}{60} \)=\( \frac{15}{60} \)=\( \frac{1}{4} \)
   (A + B + C)s 1 days work = \( \frac{1}{8} \)
   A`s 1 days work = \( \frac{1}{8} \)-\( \frac{1}{12} \)=\( \frac{3-2}{24} \)=\( \frac{1}{24} \)
   A will complete the work in 24 days.

Ans .

(3)8\( \frac{4}{7} \) days

1. Explanation :

   (3) (A + B)s 1 days work =\( \frac{1}{20} \)
   (B + C)s 1 days work =\( \frac{1}{10} \) and (C + A)s 1 days work =\( \frac{1}{12} \)
   On adding all three, 2 (A + B + C)’s 1 days work=\( \frac{1}{20} \)+\( \frac{1}{10} \)+\( \frac{1}{12} \)=\( \frac{3+6+5}{60} \)=\( \frac{14}{60} \)=\( \frac{7} {30} \)
   (A + B + C)s 1 days work =\( \frac{7}{60} \)
   Hence, the work will be completed in \( \frac{60}{7} \)= 8\( \frac{4}{7} \)days.

Ans .

(3)4 days

1. Explanation :

   (3) Work done by A, B and C in 1 day=\( \frac{1}{10} \) +\( \frac{1}{12} \) +\( \frac{1}{15} \) =\( \frac {6+5+4}{60} \) =\( \frac{15}{60} \) =\( \frac{1}{4} \)
   Required time = 4 days

Ans .

(1) 20 days

1. Explanation :

   (1) A`s 1 days work =\( \frac{1}{12} \) -\( \frac{1}{30} \) =\( \frac{5-2}{60} \) =\( \frac{3}{60} \) = \( \frac{1}{20} \)
   Hence, A alone will complete the work in 20 days.

Ans .

(3)6\( \frac{6}{11} \)days

1. Explanation :

   (3) (A + B + C)s 1 days work =\( \frac{1}{12} \)+\( \frac{1}{24} \)+\( \frac{1}{36} \)=\( \frac{6+3+2}{72} \)=\( \frac{11}{72} \)
   (A + B + C) together will complete the work in\( \frac{72}{11} \)days=6\( \frac{6}{11} \)days

Ans .

(4)4 days

1. Explanation :

   (4) (A + B)s 1 days work =\( \frac{1}{6} \)+\( \frac{1}{12} \)=\( \frac{2+1}{12} \)=\( \frac{1}{4} \)
   A and B together will complete the work in 4 days.

Ans .

(3) 120 days

1. Explanation :

   (2) (A + B)s 1 days work =\( \frac{1}{36} \)
   (B + C)s 1 days work =\( \frac{1}{60} \) and (C + A)s 1 days work =\( \frac{1}{45} \)
   1 45 Adding all three, 2(A + B + C)s 1 days work=\( \frac{1}{36} \)+\( \frac{1}{60} \)+\( \frac{1}{45} \)=\( \frac{5+3+4}{180} \)=\( \frac{1}{15} \)
   (A + B + C)s 1 days work =\( \frac{1}{30} \)
   C`s 1 days work =\( \frac{1}{30} \)-\( \frac{1}{36} \)=\( \frac{6-5}{180} \)=\( \frac{1}{180} \)
   Hence, C alone will finish the work in 180 days

Ans .

(3) 8 hrs. 15 min

1. Explanation :

   (3) Ronald’s 1 hour’s work=\( \frac{32}{6} \)=\( \frac{16}{3} \)pages
   [Pages typed in 6 hrs. = 32 pages typed in 1 hr =\( \frac{32}{6} \)]
   Elans 1 hours work = 8 pages 1 hours work of the both=\( \frac{16}{3} \)+8=\( \frac{40}{3} \)pages.
   Required time=\( \frac{110*3}{40} \)=\( \frac{33}{4} \)hours
   =8 hours 15 minutes

Ans .

(4)12 days

1. Explanation :

   (4) A`s 1days work =\( \frac{1}{20} \)
   B`s 1days work =\( \frac{1}{30} \)
   (A + B)s 1 days work =\( \frac{1}{20} \)+\( \frac{1}{30} \)=\( \frac{3+2}{60} \)=\( \frac{1}{12} \)
   Hence, the work will be completed in 12 days. When worked together.

Ans .

(2) 24 hrs

1. Explanation :

   (2) 9 hours 36 minutes=9+\( \frac{36}{60} \) =9\( \frac{3}{5} \)hours=
   =\( \frac{48}{5} \)
   (A + B)s 1 hours work =\( \frac{5}{48} \) and C`s 1 hours work =\( \frac{1}{48} \)
   (A + B + C)s 1 hours work = \( \frac{5}{48} \)+\( \frac{1}{48} \)=\( \frac{1}{8} \)....(i)
   A`s 1 hours work = (B + C)’s 1 hours work .....(ii)
   From equations (i) and (ii), 2 × (A`s 1 hours work) = \( \frac{1}{8} \)
   A`s 1 hours work = \( \frac{1}{16} \)
   B`s 1 hours work =\( \frac{5}{48} \)-\( \frac{1}{16} \)=\( \frac{5-3}{48} \)=\( \frac{1}{24} \)
   B alone will finish the work in 24 hours .

Ans .

(1)3

1. Explanation :

   (1) Work done by A and B in 5 days =5(\( \frac{1}{12} \)+\( \frac{1}{15} \))=5(\( \frac{5+4}{60} \))
   =5*\( \frac{9}{60} \)=\( \frac{9}{12} \)=\( \frac{3}{4} \)
   Remaining work = 1-\( \frac{3}{4} \)=\( \frac{1}{4} \)
   Time taken by A = \( \frac{1}{4} \)*12 = 3 days.

Ans .

(4)21 days

1. Explanation :

   (2) B`s 1 days work = (A + B)s 1 days work - A`s 1 day`s work=\( \frac{1}{12} \)-\( \frac{1}{28} \)= \( \frac{7-3}{84} \)=\( \frac{4}{84} \)=\( \frac{1}{21} \). Required time = 21 days.

Ans .

(4) (\( \frac{mn}{m+n} \)) days

1. Explanation :

   4) A`s 1 days work =\( \frac{1}{m} \)and B`s 1 days work =\( \frac{1}{n} \)
   (A + B)s 1 days work =\( \frac{1}{m} \)+\( \frac{1}{n} \)=\( \frac{n+m}{m} \)=\( \frac{m+n}{mn} \)
   Required time = (\( \frac{mn}{m+n} \)) days

Ans .

(4) \( \frac{4}{3} \)hour

1. Explanation :

   (4) Let A, B and C together do the work in x hours.
   Time taken by A = (x + 6) hours Time taken by B= (x + 1) hours Time taken by C = 2x hours
   =>\( \frac{1}{x+6} \)+\( \frac{1}{x+1} \)+\( \frac{1}{2x} \)=\( \frac{1}{x} \)
   =>\( \frac{1}{x+6} \)+\( \frac{1}{x+1} \)=\( \frac{1}{x} \)-\( \frac{1}{2x} \)=\( \frac{1}{2x} \)
   =>\( \frac{1}{x+6} \)=\( \frac{1}{2x} \)-\( \frac{1}{x+1} \)=\( \frac{x+1-2x}{2x(x+1)} \)
   =>\( \frac{1}{x+6} \)=1-x/ 2x²+2x
   =>2x ²+ 2x = x + 6 – x2 – 6x
   =>3x ² + 7x – 6 = 0
   => 3x ² + 9x – 2x – 6 = 0
   => 3x (x + 3) – 2 (x + 3) = 0
   =>(3x – 2) (x +3) = 0
   =>3x – 2 = 0 as x + 3 != 0
   => x= \( \frac{2}{3} \)
   Time taken by A =6+\( \frac{2}{3} \)=\( \frac{18+2}{3} \)=\( \frac{20}{3} \)hours
   Time taken by B =1+\( \frac{2}{3} \)=\( \frac{5}{3} \)hours
   (A +B)’s 1 hour’s work=\( \frac{3}{20} \)+\( \frac{5}{3} \)=\( \frac{3+12}{20} \)=\( \frac{15}{20} \)=\( \frac{3}{4} \)
   Required time =\( \frac{4}{3} \)hour.

Ans .

(2) 96

1. Explanation :

   (2) Time taken by B and C = x days (let)
   Time taken by A = 3x days
   Part of work done by A, B and C in 1 day= \( \frac{1}{x} \) +\( \frac{1}{3x} \)=\( \frac{3+1}{3x} \)=\( \frac{4}{3x} \)
   => \( \frac{4}{3x} \)=\( \frac{1}{24} \) => 3x = 4 × 24
   => x= \( \frac{4*24}{3} \)=32days
   Time taken by A = 32 × 3 = 96 days

Ans .

(4) 3 days

1. Explanation :

   (3) (4) A`s 1 day`s work =\( \frac{1}{4} \)
   B`s 1 days work = \( \frac{1}{12} \)
   (A + B)s 1 days work =\( \frac{1}{4} \) +\( \frac{1}{12} \) =\( \frac{3+1}{12} \) =\( \frac{4}{12} \) =\( \frac{1}{3} \)
   Required time = 3 days.

Ans .

(3) 24 days

1. Explanation :

   (3) A does \( \frac{1}{4} \) work in 10 days
   A will do 1 work in 10 × 4 = 40 days
   Similarly, B will do the same work in 20 × 3 = 60 days
   (A + B)s 1 days work =\( \frac{1}{40} \)+\( \frac{1}{60} \) = \( \frac{3+2}{120} \)=\( \frac{5}{120} \) = \( \frac{1}{24} \)
   Required time = 24 days .

Ans .

(2)10 hours

1. Explanation :

   (2) Using Rule 1, M₁ D₁ T₁ = M₂D₂T₂
   =>15 × 20 × 8 = 20 × 12 × T₂
   => T₂ = \( \frac{15*20*8}{20*12} \) =10 hours.

Ans .

(4) 60 days

1. Explanation :

   4) (Raj + Ram)s 1 days work =\( \frac{1}{10} \)
   Raj`s 1 days work = \( \frac{1}{12} \)
   Ram`s 1 day’s work = \( \frac{1}{10} \)-\( \frac{1}{12} \) = \( \frac{6-5}{60} \) = \( \frac{1}{60} \)
   Required time = 60 days

Ans .

(2) 11 days

1. Explanation :

   (2) A`s 1 days work =\( \frac{1}{9} \)
   B`s 1 days work = \( \frac{1}{15} \)
   Work done in first 2 days = A`s 1 days work + B`s 1 days work = \( \frac{1}{9} \)+\( \frac{1}{15} \)= \( \frac{5+3}{45} \) = \( \frac{8}{45} \)
   Work done in first 10 days = \( \frac{8*5}{45} \) =\( \frac{8}{9} \)
   Remaining work = 1-\( \frac{8}{9} \) =\( \frac{1}{9} \)
   Now, it is turn of 'A' for the eleventh day.
   Time taken by 'A' in doing \( \frac{1}{9} \) work = \( \frac{1}{9} \) *9 = 1 day
   Required time = 10 + 1 = 11 days.

Ans .

(4) 63

1. Explanation :

   (4) Using Rule 1,
   15 men complete \( \frac{1}{3} \)work in 7 days.
   Time taken in doing 1 work = 3 × 7 = 21 days
   => M₁D = M₂D₂
   =>15 × 21 = M₂ × 5
   =>M₂= \( \frac{15*21}{5} \) = 63 days.s

Ans .

(2) 160 minutes

1. Explanation :

   (2) (x and y)s 1 hour work = \( \frac{1}{4} \) +\( \frac{1}{8} \) = \( \frac{2+1}{8} \) = \( \frac{3}{8} \)
   Required time =\( \frac{8}{3} \) hours
   = (\( \frac{8}{3} \)*60) minutes. => 160 minutes.

Ans .

(1) 20 hours

1. Explanation :

   (1) 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
   Number of pages copied by y in 1 hour = 5 – 4 = 1
   Required time = 20 hours.

Ans .

(2)24

1. Explanation :

   (2) (A + B)s 1 days work = \( \frac{1}{15} \)....(i)
   (B + C)s 1 days work = \( \frac{1}{12} \) .... (ii) and (C + A)s 1 days work = \( \frac{1}{10} \).... (iii)
   On adding all three equations, 2 (A + B + C)’s 1 days work = \( \frac{1}{15} \)+\( \frac{1}{12} \)+\( \frac{1}{10} \)
   =\( \frac{4+6+5}{60} \) =\( \frac{15}{60} \) = \( \frac{1}{4} \)
   (A + B + C)s 1 days work = \( \frac{1}{8} \)....(iv)
   By equation (iv) – (ii), A`s 1 days work = \( \frac{1}{8} \)- \( \frac{1}{12} \) = \( \frac{3-2}{24} \) = \( \frac{1}{24} \)
   Required time = 24 days.

Ans .

(3)\( \frac{19}{30}\)

1. Explanation :

   (3)(A + B)`s 1 days work =\( \frac{1}{25} \)+\( \frac{1}{30} \) = \( \frac{6+5}{150} \)=\( \frac{11}{150} \)
   (A + B)s 5 days work =\( \frac{5*11}{150} \)=\( \frac{11}{30} \)
   Remaining work =1-\( \frac{11}{30} \)=\( \frac{30-11}{30} \) =\( \frac{19}{30} \)

Ans .

(3) 9

1. Explanation :

   (3)(A + B)s 1 days work = \( \frac{1}{6} \) A`s 1 days work =\( \frac{1}{18} \)
   B`s 1 days work =\( \frac{1}{6} \)-\( \frac{1}{18} \)= \( \frac{3-1}{18} \) = \( \frac{2}{18} \) = \( \frac{1}{9} \)
   Required time = 9 days

Ans .

(2) 12 days

1. Explanation :

   (2) A`s 2 days work = B`s 3 days work
   Time taken by A = 8 days
   Time taken by B =\( \frac{8}{2} \) *3 =>12 days.

Ans .

(3) 8 days

1. Explanation :

   

Ans .

(2) 8 days

1. Explanation :

   (2)(A + B)s 1 days work =\( \frac{1}{8} \)
   (B + C)s 1 days work =\( \frac{1}{12} \) and (A + B + C)s 1 days work =\( \frac{1}{6} \) C`s 1 days work = \( \frac{1}{6} \)-\( \frac{1}{8} \)=\( \frac{4-3}{24} \)=\( \frac{1}{24} \)
   A`s 1 days work =\( \frac{1}{6} \)-\( \frac{1}{12} \)= \( \frac{2-1}{12} \)= \( \frac{1}{12} \)
   (A + C)s 1 days work =\( \frac{1}{12} \)+\( \frac{1}{24} \)=\( \frac{2+1}{24} \)=\( \frac{1}{8} \)
   Required time = 8 days

Ans .

(3)\( \frac{7}{9} \)

1. Explanation :

   (3)Using Rule 1,
   (M¹D¹T¹) / (W¹) = (M²D²T²) / (W²)
   = \( \frac{90*16*12}{1} \) = (70*24*8 ) / (W²)
   W² = \( \frac{90*16*12}{70*24*8 }\) = \( \frac{7}{9} \) parts

Ans .

(3)12 days

1. Explanation :

   (3) Let the work be completed in x days.
   According to the question,
   \( \frac{x}{16} \)+\( \frac{x-8}{32} \)+\( \frac{x-6}{48} \) =1
   \( \frac{6x+3x-24+2x-12 }{96} \) =1
   11x – 36 = 96
   11x = 96 + 36 = 132
   x =\( \frac{132}{11} \) =12 days.

Ans .

(3) 8 days

1. Explanation :

   (3) (A+B)s 1 days work =\( \frac{1}{15} \)
   (B+C)s 1 days work =\( \frac{1}{10} \) and (A+C)s 1 days work =\( \frac{1}{12} \) On adding all three, 2(A+B+C)s 1 days work =\( \frac{1}{15} \)+\( \frac{1}{10} \)+\( \frac{1}{12} \) =\( \frac{4+6+5}{60} \) =\( \frac{15}{60} \)=\( \frac{1}{4} \)
   (A + B + C)s 1 days work =\( \frac{1}{8} \)
   Required time = 8 days.

Ans .

(3 7

1. Explanation :

   (3) Let the whole work be completed in x days
   A`s 1 days work =\( \frac{1}{10} \)
   B`s 1 days work =\( \frac{1}{12} \) and C`s 1 days work =\( \frac{1}{15} \)
   According to the question, A`s (x – 5) days work + B`s (x – 3) days work + Cs x days work = 1
   =>\( \frac{x-5}{10} \) +\( \frac{x-3}{12} \) +\( \frac{x}{15} \) = 1
   => \( \frac{x-5}{10} \)+\( \frac{x-3}{12} \)+\( \frac{x}{15} \) = 1
   => \( \frac{6(x-5)+5(x-3)+4x}{60} \) =1
   => 6x – 30 + 5x – 15 + 4x = 60
   =>15x – 45 = 60
   => 15x = 60 + 45 = 105
   => x=\( \frac{105}{15} \) = 7 days.

Ans .

(2)3\( \frac{1}{13} \)days

1. Explanation :

   A`s 1 days work =\( \frac{1}{24} \)
   B`s 1 days work =\( \frac{1}{5} \) and C 1 days work =\( \frac{1}{12} \)
   (A + B + C)s 1 days work = \( \frac{1}{24} \) +\( \frac{1}{5} \) +\( \frac{1}{12} \) =\( \frac{5+24+10}{120} \) =\( \frac{39}{120} \) =\( \frac{13}{40} \)
   Required Time=\( \frac{40}{13} \) =3\( \frac{1}{13} \) days.

Ans .

(4)10 days

1. Explanation :

   

Ans .

(1)5\( \frac{1}{7} \)days

1. Explanation :

   (1) A`s 1 days work =\( \frac{1}{9} \)
   B`s 1 days work = 5\( \frac{1}{12} \)
   (A + B)s 1 day’s work = \( \frac{1}{9} \)+\( \frac{1}{12} \)= \( \frac{4+3}{36} \) = \( \frac{7}{36} \)
   Required time = \( \frac{36}{7} \) =5\( \frac{1}{7} \)days.

Ans .

(1) 60 hours

1. Explanation :

   (1) Let time taken by son be x hours.
   Father’s and son`s 1 days work = \( \frac{1}{30} \)+\( \frac{1}{x} \)
   \( \frac{1}{30} \)+\( \frac{1}{x} \) =\( \frac{1}{20} \)
   =>\( \frac{1}{x} \)= \( \frac{1}{20} \)-\( \frac{1}{30} \)
   =\( \frac{3-2}{60} \)=\( \frac{1}{60} \)
   => x = 60 hour.

Ans .

(2)9days

1. Explanation :

   (2) Work done by (A + B) in 5 days = 5(\( \frac{1}{12} \)+\( \frac{1}{20} \) )
   =5(\( \frac{5+3}{60} \) ) = \( \frac{40}{60} \) =\( \frac{2}{3} \)
   Remaining work = 1-\( \frac{2}{3} \) =\( \frac{1}{3} \)
   Time taken by C in doing \( \frac{1}{3} \) work = 3 days
   Required time = 3 × 3 = 9 days.

Ans .

(3)3\( \frac{15}{16} \)

1. Explanation :

   (3) A`s 1 days work =\( \frac{1}{7} \)
   B`s 1 days work =\( \frac{1}{9} \)
   (A + B)s 1 days work =\( \frac{1}{7} \)+\( \frac{1}{9} \) =\( \frac{9+7}{63} \) =\( \frac{16}{63} \)
   \ Required time =\( \frac{63}{16} \) days =3\( \frac{15}{16} \)days.

Ans .

(1) 12 days

1. Explanation :

   (1) (A + B)s 1 days work =\( \frac{1}{24} \)
   (A + B + C)s 1 days work =\( \frac{1}{8} \)
   C`s 1 days work = \( \frac{1}{24} \) -\( \frac{1}{8} \) =\( \frac{3-1}{24} \) =\( \frac{2}{24} \) =\( \frac{1}{12} \)
   Required time = 12 days.

Ans .

(4) 8 days

1. Explanation :

   (4) (A + B)’s 1 day’s work=\( \frac{1}{12} \)+\( \frac{1}{24} \)=\( \frac{2+1}{24} \)=\( \frac{3}{24} \)=\( \frac{1}{8} \)
   Required time = 8 days.

Ans .

(4)8

1. Explanation :

   (4) (A + B)s 1 days work =\( \frac{1}{11} \)+\( \frac{1}{20} \) =\( \frac{20+11}{220} \)=\( \frac{31}{220} \)
   (A + C)’s 1 days work =\( \frac{1}{11} \)+\( \frac{1}{55} \)=\( \frac{5+1}{55} \)=\( \frac{6}{55} \)
   Work done in first two days = \( \frac{31}{220} \)+\( \frac{6}{55} \) =\( \frac{31+24}{220} \)=\( \frac{55}{220} \)=\( \frac{1}{4} \)
   Required time = 2 × 4 = 8 days.

Ans .

(1)18 days

1. Explanation :

   (1) (A + B)s 1 days work =\( \frac{1}{6} \)
   A`s 1 days work =\( \frac{1}{9} \)
   B’s 1 day’s work =\( \frac{1}{6} \)-\( \frac{1}{9} \)=\( \frac{3-2}{18} \)=\( \frac{1}{18} \)
   Required time = 18 days.

Ans .

(2) 24 days

1. Explanation :

   A`s 1 days work =\( \frac{1}{18} \)
   A`s 12 days work =\( \frac{12}{18} \)=\( \frac{2}{3} \)
   =>Remaining work =1-\( \frac{2}{3} \) =\( \frac{1}{3} \)
   Time taken by B in doing \( \frac{1}{3} \) work = 8 days
   Time taken by B in doing whole work = 3 × 8 = 24 days .

Ans .

(3)8 days

1. Explanation :

   (3) (A + B)s 1 days work=\( \frac{1}{8} \) ... (i)
   (B + C)s 1 days work= \( \frac{1}{12} \).... (ii) and (A + B + C)s 1 days work =\( \frac{1}{6} \)... (iii)
   By equations (i) + (ii) – (iii), Bs 1 days work =\( \frac{1}{8} \)+\( \frac{1}{12} \)-\( \frac{1}{6} \) = \( \frac{3+2-4}{24} \) = \( \frac{1}{24} \) .... (iv)
   By equations (iii) – (iv), (A + C)’s 1 days work =\( \frac{1}{6} \)-\( \frac{1}{24} \) = \( \frac{4-1}{24} \)= \( \frac{3}{24} \) =\( \frac{1}{8} \)
   Required time = 8 days

Ans .

(4)12 days

1. Explanation :

   (4) Let time taken by A be x days.
   Time taken by B = 3x days According to the question,
   \( \frac{1}{x} \)+\( \frac{1}{3x} \)=\( \frac{1}{9} \)
   =>\( \frac{3+1}{3x} \) =\( \frac{1}{9} \)
   =>3x = 4 × 9
   => x = \( \frac{4*9}{3} \) =12 days.

Ans .

(4) 100 days

1. Explanation :

   

Ans .

(3)(\( \frac{pq}{p+q} \)

1. Explanation :

   (3) X`s 1 days work =\( \frac{1}{p} \)
   Y`s 1 day’s work =\( \frac{1}{q} \)
   (X + Y)s 1 days work= \( \frac{1}{p} \)+\( \frac{1}{q} \)=\( \frac{q+p}{pq} \)
   Required time =(\( \frac{pq}{p+q} \).

Ans .

(2)\( \frac{40}{9} \)

1. Explanation :

   (2) A`s 1 days work =\( \frac{1}{8} \) B`s 1 days work =\( \frac{1}{10} \)
   (A + B)s 1 days work = \( \frac{1}{8} \)+\( \frac{1}{10} \) =\( \frac{5+4}{40} \)=\( \frac{9}{40} \)
   Required time = \( \frac{40}{9} \) days

Ans .

(2) 16 days

1. Explanation :

   (2) (A + B)s 1 days work =\( \frac{1}{36} \)
   (B + C)s 1 days work = \( \frac{1}{24} \) and (A + C)s 1 days work =\( \frac{1}{18} \)
   On adding all three, 2 (A + B + C)s 1 days work =\( \frac{1}{36} \)+\( \frac{1}{24} \)+\( \frac{1}{18} \) = \( \frac{2+3+4}{72} \) =\( \frac{9}{72} \) =\( \frac{1} {8} \)
   (A + B + C)’s 1 day’s work =\( \frac{1}{36} \)
   Required time = 16 days .

Ans .

\( \frac{xy}{x+y} \) days

1. Explanation :

   (3) Koushik`s 1 days work =\( \frac{1}{x} \)
   Krishnu`s 1 days work =\( \frac{1}{y} \)
   One days work of both =\( \frac{1}{x} \)+\( \frac{1}{y} \) =\(\frac{x+y}{xy} \)
   Required time =\( \frac{xy}{x+y} \) days.

Ans .

(2)8

1. Explanation :

   M₁D₁ = M₂D₂
   =>24 × 12 = 36 × D₂
   = D₂ =\( \frac{24*12}{36} \) = 8 days.

Ans .

(1)18 days

1. Explanation :

   (3) (A + B)s 1 days work =\( \frac{1}{18} \)
   (B + C)s 1 days work =\( \frac{1}{24} \) and (C + A)s 1 days work =\( \frac{1}{36} \)
   On adding all three, 2 (A + B + C)s 1 day’s work =\( \frac{1}{18} \)+\( \frac{1}{24} \)+\( \frac{1}{36} \)=\( \frac{4+3+2}{72} \)=\( \frac{9}{72} \)=\( \frac{1}{8} \)
   (A + B + C)’s 1 days work =\( \frac{1}{16} \)
   Required time = 16 days .

Ans .

(1) 10\( \frac{5}{24} \) days

1. Explanation :

   (1) A`s 4 days work = Bs 5 days work
   => A : B = 4 : 5
   Again, B : C = 6 : 7
   => A : B : C = 4 × 6 : 5 × 6 : 5 × 7 = 24 : 30 : 35
   Q Time taken by A = 7 days
   Time taken by C =\( \frac{35}{24} \) * 7 =\( \frac{245}{24} \) =10\( \frac{5}{24} \) days.

Ans .

(2)7 days

1. Explanation :

   

Ans .

(2)2

1. Explanation :

   (2) Work done by two sons in an hour =\( \frac{1}{3} \)+\( \frac{1}{6} \)=\( \frac{2+1}{6} \)=\( \frac{1}{2} \)
   Work done by father in an hour =\( \frac{1}{2} \)
   Required time = 2 hours

Ans .

(4) 4 days

1. Explanation :

   A`s 1 days work =\( \frac{1}{10} \)
   B`s 1 days work =\( \frac{1}{12} \) and C`s 1 day’s work =\( \frac{1}{15} \)
   (A + B + C)’s 1 days work = \( \frac{1}{10} \) +\( \frac{1}{12} \) +\( \frac{1}{15} \) =\( \frac{6+5+4}{60} \) =\( \frac{15}{60} \) =\( \frac{1}{4} \)
   Required time = 4 days.

Ans .

(1) 20 mats

1. Explanation :

   
   =>5 × 5 × x = 10 × 10 × 5 => x =\( \frac{10*10*5}{5*5} \) = 20 mats.

Ans .

(2)20 days

1. Explanation :

   (2) (A + B)s 1 days work =\( \frac{1}{12} \)
   As 1 days work =\( \frac{1}{30} \)
   B`s 1 day’s work =\( \frac{1}{12} \)-\( \frac{1}{30} \) =\( \frac{5-2}{60} \) =\( \frac{1}{20} \) Required time = 20 days

Ans .

(2)48

1. Explanation :

   (2) (Ganesh + Ram + Sohan)s 1 days work =\( \frac{1}{16} \)
   (Ganesh + Ram)s 1 days work = \( \frac{1}{24} \)
   Sohan`s 1 days work = \( \frac{1}{16} \) -\( \frac{1}{24} \) =\( \frac{3-2}{48} \) =\( \frac{1}{48} \)
   Required time = 48 days.

Ans .

(1) 120 days

1. Explanation :

   (1) (A + B)s 1 days work= \( \frac{1}{72} \) ..... (i)
   (B + C)s 1 days work = \( \frac{1}{120} \).... (ii) and (C + A)’s 1 day’s work = \( \frac{1}{90} \)..... (iii)
   On adding all three, 2 (A + B + C)s 1 days work =\( \frac{1}{72} \)+\( \frac{1}{120} \)+\( \frac{1}{90} \)
   =\( \frac{5+3+4}{360} \) =\( \frac{12}{360} \)=\( \frac{1}{30} \)
   (A + B + C)s 1 days work= \( \frac{1}{60} \)..... (iv)
   A`s 1 days work = Equation (iv) – (ii),
   \( \frac{1}{60} \)-\( \frac{1}{120} \)=\( \frac{2-1}{120} \)=\( \frac{1}{120} \)
   Required time = 120 days .

Ans .

(2)28

1. Explanation :

   

Ans .

(3)17\( \frac{1}{7} \) days

1. Explanation :

   (3) A`s 1 days work =\( \frac{1}{30} \)
   B`s 1 days work =\( \frac{1}{40} \)
   (A + B)’s 1 day’s work =\( \frac{1}{30} \)+\( \frac{1}{40} \)= \( \frac{4+3}{120} \)=\( \frac{7}{120} \)
   Required time =\( \frac{120}{7} \)=17\( \frac{1}{7} \).

Ans .

(2) 5\( \frac{1}{3} \)days

1. Explanation :

   (2) A can finish the work in 18 days.
   A`s one days work = \( \frac{1}{18} \)
   Similarly, B`s one days work = \( \frac{1}{24} \)
   (A + B)s 8 days work =( \( \frac{1}{18} \)+\( \frac{1}{24} \)) *8 =\( \frac{7}{72} \) *8=\( \frac{7}{9} \)
   Remaining work = 1-\( \frac{7}{9} \)=\( \frac{2}{9} \)
   Time taken to finish the remaining work by B is \( \frac{2}{9} \) *24 = \( \frac{16}{3} \) =5\( \frac{1}{3} \)days.

Ans .

(4) 13 days

1. Explanation :

   (4) (A+B)s 2 days work =2(\( \frac{1}{12} \) +\( \frac{1}{18} \) ) =\( \frac{10}{36} \)
   Remaining work = 1 - \( \frac{10}{36} \) =\( \frac{26} {36} \)
   Time taken by B to complete \( \frac{26}{36} \) part of work
   =>\( \frac{26}{36} \) *18= 13 days.

Ans .

(3) 6 days

1. Explanation :

   (3) A1s one days work = \( \frac{1}{6} \)
   B`s one days work =\( \frac{1}{12} \)
   (A + B)s one days work =\( \frac{1}{6} \) +\( \frac{1}{12} \) =\( \frac{2+1}{12} \) =\( \frac{1}{4} \)
   (A + B)s three days work =\( \frac{3}{4} \)
   Remaining work = 1-\( \frac{3}{4} \)= \( \frac{1}{4} \)
   Total required number of days = \( \frac{1}{4} \)*\( \frac{12}{1} \) +3= 3 + 3 = 6 days.

Ans .

(1) 18 days

1. Explanation :

   (1) (A + B)s days work =\( \frac{1}{30} \)
   (B + C)s 1 days work =\( \frac{1}{24} \) and (C + A)s 1 days work =\( \frac{1}{20} \)
   2 (A + B + C)s 1 days work =\( \frac{1}{30} \)+\( \frac{1}{24} \)+\( \frac{1}{20} \) =\( \frac{4+5+6}{120} \)=\( \frac{15}{120} \)=\( \frac{1}{8} \)
   (A + B + C)s 1 days work =\( \frac{1}{16} \)
   (A + B + C)s 10 days’ work =\( \frac{10}{16} \)=\( \frac{5}{8} \)
   Remaining work = 1 -\( \frac{5}{8} \) =\( \frac{3}{8} \)
   This part of work is done by A alone.
   Now A`s 1 day’s work =\( \frac{1}{16} \) -\( \frac{1}{24} \) =\( \frac{3-2}{48} \) =\( \frac{1}{48} \)
   The required no. of days =\( \frac{3}{8} \)* 3= 18 days.

Ans .

(2)60 days

1. Explanation :

   2) (A+B)s 1 days work =\( \frac{1}{30} \)
   (A + B)s 20 days work =\( \frac{20}{30} \) =\( \frac{2}{3} \)
   Remaining work =1-\( \frac{2}{3} \)=\( \frac{1}{3} \)
   Now,\( \frac{1}{3} \)part of work is done by A in 20 days.
   Whole work will be done by A alone in 20 × 3 = 60 days

Ans .

(3) 4

1. Explanation :

   

Ans .

(3) 10 days

1. Explanation :

   (3) Work done by (B + C) in 3 days = 3* (\( \frac{1}{9} \) +\( \frac{1}{12} \))
   = \( \frac{1}{3} \)+\( \frac{1}{4} \)=\( \frac{4+3}{12} \)=\( \frac{7}{12} \)
   Remaining work = 1-\( \frac{7}{12} \) =\( \frac{5}{12} \)
   This part of work is done by A alone.
   Now, \( \frac{1}{24} \) part of work is done by A in 1 day.
   => \( \frac{5}{12} \) part of work will be done by A in = 24 ´*\( \frac{5}{12} \) = 10 days.

Ans .

(2) 24 days

1. Explanation :

   (2) Originally, let there be x men Now, more men, less days (x + 6) : x : : 55 : 44
   so,\( \frac{x+6}{x} \)=\( \frac{55}{44} \) =\( \frac{5}{4} \)
   or 5x = 4x + 24 or x = 24.

Ans .

(3) 12 days

1. Explanation :

   (3) Work done by 2 (A + B) in one day =( \frac{1}{10} \)+( \frac{1}{15} \) =( \frac{3+2}{30} \)=( \frac{5}{30} \)=( \frac{1}{6} \)
   Work done by (A + B) in oneday =( \frac{1}{12} \)
   (A + B) can complete the work in 12 days

Ans .

(3)8 days

1. Explanation :

   (3) Let A worked for x days.
   According to question \( \frac{x}{28} \)+\( \frac{(x+17)}{35} \) =1
   =>\( \frac{5x+4(x+17)}{140} \) =1
   =>5x + 4x + 68 = 140
   =>9x = 140 – 68 = 72
   => x = 8
   A worked for 8 days

Ans .

(3)12 days

1. Explanation :

   (3) Work done by (A + B) in 1 day =\( \frac{1}{15} \)+\( \frac{1}{10} \)=\( \frac{2+3}{30} \)=\( \frac{1}{6} \)
   (A + B)s 2 days work =\( \frac{2}{6} \) =\( \frac{1}{3} \)
   Remaining work =1 -\( \frac{1}{3} \) =\( \frac{2}{3} \)
   This part is done by A alone.
   one work is done by A in 15 days.
   \( \frac{2}{3} \) work is done in 15*\( \frac{2}{3} \)
   = 10 days.
   Total number of days = 10 + 2 = 12 days

Ans .

(4)10 days

1. Explanation :

   (1) A`s 1 day`s work =\( \frac{1}{20} \).
   A`s 4 days work =\( \frac{4}{20} \) =\( \frac{1}{5} \)
   This part is completed by A and B together.
   Now, (A + B)s 1 days work = \( \frac{1}{20} \)+\( \frac{1}{12} \)=\( \frac{3+5}{60} \)=\( \frac{8}{60} \)=\( \frac{2}{15} \)
   Now,\( \frac{2}{15} \)work is done by (A +B) in 1 day.
   \( \frac{4}{5} \) work is done in .
   =\( \frac{15}{2} \)*\( \frac{4}{5} \) = 6 days
   Hence, the work lasted for 4 + 6 = 10 days.

Ans .

(2)9 days

1. Explanation :

   (2) (A + B)s 1 days work =(\( \frac{1}{45} \) +\( \frac{1}{40} \) ) =\( \frac{8+9}{360} \) =\( \frac{17}{360} \)
   Work done by B in 23 days = \( \frac{1}{4} \) * 23 =\( \frac{23}{40} \)
   Remaining work = 1- \( \frac{23}{40} \)= \( \frac{17}{40} \)
   Now,\( \frac{17}{40} \)work was done by (A + B) in 1 day
   \( \frac{17}{40} \) work was done by (A + B) in 1 * \( \frac{360}{17} \) *\( \frac{17}{40} \) = 9 days.
   Hence, A left after 9 days.

Ans .

(1)72 days

1. Explanation :

   

Ans .

(3)16 days

1. Explanation :

   (3) Time taken by A = \( \frac{8*12}{4} \) = 24days.
   Work done of by B = \( \frac{4}{12} \)=\( \frac{1}{3} \)
   Remaining work =1-\( \frac{1}{3} \) = \( \frac{2}{3} \)
   A can complete a work in 24 days
   A can complete \( \frac{2}{3} \) part of work in 24* \( \frac{2}{3} \) = 16 days.

Ans .

(1)14\( \frac{1}{3} \) days

1. Explanation :

   (1) A`s 1 days work = \( \frac{1}{12} \)
   B`s 1 days work = \( \frac{1}{18} \)
   Part of work done by A and B in first two days = \( \frac{1}{12} \)+ \( \frac{1}{18} \)= \( \frac{3+2}{36} \)= \( \frac{5}{36} \)
   Part of work done by A and B in 14 days =\( \frac{35}{36} \)
   [14 days to be taken randomly] Remaining work = 1-\( \frac{35}{36} \) =\( \frac{1}{36} \)
   Now A will work for 15th day. A will do the \( \frac{1}{36} \) work in \( \frac{1}{36} \)*12 =\( \frac{1}{3} \) day.
   Total Work will be done in 14\( \frac{1}{3} \) days.

Ans .

(3)7 days

1. Explanation :

   (3) Let the work be completed in x days.
   According to the question ,\( \frac{x-5}{10} \)+\( \frac{x-3}{12} \)+\( \frac{x}{15} \)=1
   =>\( \frac{6x-30+5x-15+4x}{60} \) =1
   =>15x – 45 = 60 ,/br> => 15x = 105
   => x =\( \frac{105}{15} \) = 7 days.

Ans .

(4)8 days

1. Explanation :

   

Ans .

(1)56\( \frac{2}{3} \) days

1. Explanation :

   
   

Ans .

(4)9 days

1. Explanation :

   (4) Let the work be finished in x days.
   According to the question,
   A worked for x days while B worked for (x – 3) days
   \( \frac{x}{18} \)+\( \frac{x-3}{12} \)=1
   =>\( \frac{2x+3x-9}{36} \) = 1
   => 5x – 9 = 36
   => 5x = 45
   => x = \( \frac{45}{5} \)= 9
   Hence, the work was completed in 9 days.

Ans .

(1)6 days

1. Explanation :

   (1) Let A and B worked together for x days
   According to the question,
   Part of work done by A for (x + 10) days + part of work done by B for x days = 1
   => \( \frac{x+10}{20} \) +\( \frac{x}{30} \) = 1
   => \( \frac{3x+30+2x}{60} \) = 1
   =>5x + 30 = 60
   => 5x = 30
   => x= \( \frac{30}{5} \) = 6 days.

Ans .

(2)8 days

1. Explanation :

   (2) Let the work be completed in x days.
   According to the question,
   A worked for (x –3) days, while B worked for x days.
   \( \frac{x-3}{9} \) +\( \frac{x}{18} \) = 1
   =>\( \frac{2x-6+x}{18} \) = 1 => 3x–6 = 18
   =>3x = 18 + 6 = 24
   => x =\( \frac{24}{3} \) = 8 days.

Ans .

(3)15 days

1. Explanation :

   (3) (B + C)s 2 days work= 2(\( \frac{1}{30} \) + (\( \frac{1}{20} \))
   = 2(\( \frac{2+3}{60} \)) = \( \frac{1}{6} \) part
   Remaining work = 1- \( \frac{1}{6} \)= \( \frac{5}{6} \) part.
   Time taken by A to complete this part of work \( \frac{5}{6} \) *18 = 15 days.

Ans .

(3)6 days

1. Explanation :

   (3) Part of work done by B in 10 days = 10*\( \frac{1}{15} \) = \( \frac{2}{3} \)
   Remaining work =1 - \( \frac{1}{2} \) = \( \frac{1}{3} \)
   Time taken by A = \( \frac{1}{3} \)*18 = 6 days.

Ans .

(3)10\( \frac{1}{4} \)days

1. Explanation :

   (3) Part of work done by A and B in first two days \( \frac{1}{9} \) +\( \frac{1}{12} \) = \( \frac{4+3}{36} \) = \( \frac{7}{36} \)
   Part of work done in first 10 days = \( \frac{35}{36} \)
   Remaining work= 1-\( \frac{35}{36} \) = \( \frac{1}{36} \)
   Now it is the turn of A.
   Time taken by A =\( \frac{1}{36} \)*9 = \( \frac{1}{4} \)day
   Total time = 10 +\( \frac{1}{4} \) =10\( \frac{1}{4} \)days.

Ans .

(4)15 days

1. Explanation :

   (4) B`s 1 days work =\( \frac{1}{12} \)-\( \frac{1}{20} \)= \( \frac{5-3}{60} \) =\( \frac{1}{30} \)
   B`s \( \frac{1}{2} \)days work = \( \frac{1}{60} \)
   (A + B)s 1 days work = \( \frac{1}{20} \) + \( \frac{1}{60} \) = \( \frac{3+1}{60} \)= \( \frac{1}{15} \)
   [ B works for half day daily]
   Hence, the work will be completed in 15 days

Ans .

(3)6 days

1. Explanation :

   . (3) Part of the work done by A and B in 4 days = 2*(\( \frac{1}{12} \)+\( \frac{1}{15} \)) = 4\( \frac{5+4}{60} \)
   = 4*\( \frac{9}{60} \) =\( \frac{3}{5} \)
   Remaining work = 1-\( \frac{3}{5} \) =\( \frac{2}{5} \)
   Time taken by B to complete the remaining work =\( \frac{2}{5} \) *15 =6 days

Ans .

(1)13\( \frac{1}{3} \) days

1. Explanation :

   (1) Part of the work done by X in 8 days=\( \frac{8}{40} \)=\( \frac{1}{5} \)
   Remaining work = 1-\( \frac{1}{5} \)=\( \frac{4}{5} \)
   This part of work is done by Y in 16 days.
   Time taken by Y in doing 1 work =\( \frac{16*5}{4} \) = 20 days.
   Work done by X and Y in 1 day = \( \frac{1}{40} \)+\( \frac{1}{20} \)=\( \frac{1+2}{40} \)=\( \frac{3}{40} \)
   Hence, both together will complete the work in\( \frac{40}{3} \) i.e13\( \frac{1}{3} \) days

Ans .

(1)9\( \frac{3}{8} \) days

1. Explanation :

   (1) Work done in first two days =\( \frac{2}{30} \)+\( \frac{1}{10} \)+\( \frac{1}{20} \)=\( \frac{1}{15} \)+\( \frac{1}{20} \)+\( \frac{1}{10} \)
   =\( \frac{4+3+6}{60} \) = \( \frac{13}{60} \)
   Work done in first 8 days =\( \frac{52}{60} \) Remaining work = 1-\( \frac{52}{60} \) = \( \frac{8}{60} \) = \( \frac{2}{15} \)
   (A + B)s 1 days work = \( \frac{1}{30} \) +\( \frac{1}{20} \)=\( \frac{2+3}{60} \)=\( \frac{1}{12} \)
   Remaining work =\( \frac{2}{15} \) -\( \frac{1}{12} \) =\( \frac{8-5}{60} \)=\( \frac{3}{60} \)=\( \frac{1}{20} \)
   (A + C)s 1 days work =\( \frac{1}{30} \)+\( \frac{1}{10} \)=\( \frac{1+3}{30} \)=\( \frac{2}{15} \)
   Time taken =\( \frac{1}{20} \)*\( \frac{15}{2} \)
   =\( \frac{3}{8} \) day.
   Total time = 9 +\( \frac{3}{8} \) =9\( \frac{3}{8} \) days

Ans .

(1)5 days

1. Explanation :

   (1) Work done by B in 9 days = \( \frac{9}{12} \) =\( \frac{3}{4} \)part
   Remaining work = 1- \( \frac{3}{4} \)= \( \frac{1}{4} \)which is done by A
   Time taken by A =\( \frac{1}{4} \)*20 = 5days.

Ans .

(2) 7\( \frac{1}{3} \) days

1. Explanation :

   (2) Work done by A in 6 days =\( \frac{6}{8} \)=\( \frac{3}{4} \)part
   Work destroyed by B in 2 days =\( \frac{2}{3} \)part
   Remaining work after destruction = \( \frac{3}{4} \)- \( \frac{2}{3} \)= \( \frac{9-8}{12} \)=\( \frac{1}{12} \)
   Now, time taken by A in doing \( \frac{11}{12} \) part
   \( \frac{11}{12} \) * 8 = \( \frac{22}{3} \) =7\( \frac{1}{3} \)days.

Ans .

(1)6 days

1. Explanation :

   (1) Work done by B in 10 days =\( \frac{10}{15} \)=\( \frac{2}{3} \)
   Remaining work = 1-\( \frac{2}{3} \)=\( \frac{1}{3} \)
   Time taken by A to complete the work = \( \frac{1}{3} \) *18 = 6 days.

Ans .

(1)48 days

1. Explanation :

   
   

Ans .

(3) 120 days

1. Explanation :

   (3) (A + B + C)s 1 days work = \( \frac{1}{20} \)+\( \frac{1}{30} \)+\( \frac{1}{60} \) = \( \frac{3+2+1}{60} \) = \( \frac{1}{10} \)
   A`s 2 days work =\( \frac{2}{20} \)=\( \frac{1}{10} \)
   Work done in first three days= \( \frac{1}{10} \)+\( \frac{1}{10} \)=\( \frac{2}{10} \)=\( \frac{1}{5} \)
   [A`s work for 2 days + (A + B + C) work on 3rd day]
   Hence, the work will be finished in 15 days.

Ans .

(3)6 days

1. Explanation :

   (3) (A + B)`s 2 days work =\( \frac{2}{3} \)
   Remaining Work =1 - \( \frac{2}{3} \) = \( \frac{1}{3} \)
   Time taken by A in destroying \( \frac{1}{3} \)work = 2 days
   Time taken by A in completing the work = 6 days
   B`s 1 days work = \( \frac{1}{3} \)-\( \frac{1}{6} \) =\( \frac{2-1}{6} \) =\( \frac{1}{6} \)
   B alone will complete the work in 6 days.

Ans .

(3)24 days

1. Explanation :

   (3) Work done by A and B in 7 days=\( \frac{7}{20} \)+\( \frac{7}{30} \)=\( \frac{21+14}{60} \)=\( \frac{35} {60} \)=\( \frac{7}{12} \)
   So, Remaining work = 1-\( \frac{7}{12} \) =\( \frac{5}{12} \)
   Time taken by C= \( \frac{12}{5} \)*10 = 24 days.

Ans .

(3)6\( \frac{2}{3} \)days

1. Explanation :

   

Ans .

(4) 4

1. Explanation :

   (4) Work done by A and B in first 6 days
   = (A + B)s 4 days work + B`s 2 days work =4*\( \frac{1}{8} \)+\( \frac{1}{12} \)
   =\( \frac{1}{2} \)+\( \frac{1}{6} \)= \( \frac{3+1}{6} \) =\( \frac{2}{3} \)
   Remaining work = 1-\( \frac{2}{3} \) =\( \frac{1}{3} \)
   Time taken by C =\( \frac{1}{3} \)*12 = 4days.

Ans .

(1)60 days

1. Explanation :

   (1) (A + B) together do the work in 30 days
   (A + B)s 1 days work =\( \frac{1}{30} \)
   (A + B)s 20 days work =\( \frac{20}{30} \) =\( \frac{2}{3} \)
   Remaining work =1-\( \frac{2}{3} \) = \( \frac{1}{3} \)
   Time taken by A in doing \( \frac{1}{3} \) work = 20 days
   Time taken in doing 1 work= 20 × 3 = 60 days.

Ans .

(3) 3\( \frac{1}{2} \)days

1. Explanation :

   (3) Remaining work =1-\( \frac{1}{8} \) =\( \frac{7}{8} \)
   (A + B)s 1 days work =\( \frac{1}{6} \)+\( \frac{1}{12} \)=\( \frac{2+1}{12} \)=\( \frac{3}{12} \)=\( \frac{1}{4} \)
   Time taken in doing \( \frac{7}{8} \)part of work =\( \frac{7}{8} \)*4= \( \frac{7}{2} \)= 3\( \frac{1}{2} \)days

Ans .

(2)6

1. Explanation :

   (2) Work done by 12 men in 8 days = Work done by 16 women in 12 days.
   => 12 × 8 men
   => 16 × 12 women
   =>1 man => 2 women
   Now, work done by 12 men in 1 day =\( \frac{1}{8} \)
   1 mans 1 days work = \( \frac{1}{12*8} \) =\( \frac{1}{96} \)
   16 mens 3 days work
   

Ans .

(2)8

1. Explanation :

   

Ans .

(1) 44

1. Explanation :

   12 men =24 boys
   =>1 man = 2 boys
   => 15 men + 6 boys
   = 30 boys + 6 boys = 36 boys
   =>M₁D₁ =M₂D₂
   =>24 × 66 = 36 × D₂
   D₂ =\( \frac{24*66}{36} \) = 44 days.

Ans .

(3) 100 days

1. Explanation :

   (3) A, B and C together complete the work in 40 days.
   (A + B + C)s 1 days work = \( \frac{1}{40} \)
   (A + B + C)s 16 days work=\(\frac{16}{40} \)=\(\frac{2}{5} \)
   Remaining Work = 1-\(\frac{2}{5} \)=\(\frac{3}{5} \)
   This part of work is done by B and C in 40 days.
   =>Time taken in doing \(\frac{3}{5} \) work = 40 days. =>Time taken in doing in 1 work = \(\frac{40 *5}{3} \)=\(\frac{200}{3} \)days.
   A`s days work = (A + B + C)s 1 days work - (B + C)s 1 days work = \(\frac{1}{40} \)+ \(\frac{3}{200} \)=\(\frac{5-3}{200} \)= \(\frac{2}{200} \)= \(\frac{1}{100} \)
   Required time = 100 days.

Ans .

(2)40

1. Explanation :

   (2) Number of men originally = x (let)
   =>M₁D₁=M₂D₂
   => × 60 = (x + 8) × 50
   =>6x = 5x + 40
   =>6x – 5x = 40
   =>x = 40 men
   

Ans .

(4) 60

1. Explanation :

   (4) Using Rule 1,,
   Number of men originally = x (let)
   =>M₁D₁=M₂D₂
   => x × 18 = (x – 6) × 20
   =>x × 9 = (x – 6) × 10
   = 10x – 60
   =>10x – 9x = 60
   => x = 60 men

Ans .

(4)75

1. Explanation :

   (4) Original number of men= x (let)
   =>M₁D₁=M₂D₂
   =>x × 40 = (x + 45) × 25
   =>8x = (x + 45) × 5
   =>8x = 5x + 225
   => 8x – 5x = 225
   => 3x = 225
   => x= \( \frac{225}{3} \)= 7 men.

Ans .

(2) 9 days

1. Explanation :

   (2) Let A left the work after x days.
   According to the question,
   Work done by A in x days + work done by B in (23 + x ) days = 1
   => \( \frac{x}{45} \) +\( \frac{23+x}{40} \)=1
   =>\( \frac{8x+207+9x}{360} \) = 1
   =>17x + 207 = 360
   => 17x = 360 – 207 = 153
   => x = \( \frac{153}{17} \) = 9days.

Ans .

(2) 12

1. Explanation :

   

Ans .

(4) 11

1. Explanation :

   (4) Let the work be completed in x days.
   According to the question,
   C worked for (x – 4) days.
   = \( \frac{x}{24} \) +\( \frac{x}{30} \)=\( \frac{x-4}{40} \)=1
   =>\( \frac{5x+4x+3(x-4)}{120} \) =1
   =>\( \frac{12x-12}{120} \) = 1
   =>\( \frac{12(x-1)}{120} \)= 1
   =>\( \frac{x-1}{10} \) =>x – 1 = 10
   => x = 10 + 1 = 11 days.

Ans .

(2)24

1. Explanation :

   (2) (A + B)s 1 days work = \( \frac{1}{15} \)....(i)
   (B + C)s 1 days work = \( \frac{1}{12} \) .... (ii) and (C + A)s 1 days work = \( \frac{1}{10} \).... (iii)
   On adding all three equations, 2 (A + B + C)’s 1 days work = \( \frac{1}{15} \)+\( \frac{1}{12} \)+\( \frac{1}{10} \)
   =\( \frac{4+6+5}{60} \) =\( \frac{15}{60} \) = \( \frac{1}{4} \)
   (A + B + C)s 1 days work = \( \frac{1}{8} \)....(iv)
   By equation (iv) – (ii), A`s 1 days work = \( \frac{1}{8} \)- \( \frac{1}{12} \) = \( \frac{3-2}{24} \) = \( \frac{1}{24} \)
   Required time = 24 days.

Ans .

(2)24

1. Explanation :

   (2) Number of men initially = x (let)
   =>M₁D₁=M₂D₂
   => x × 40 = (x + 8) × 30
   =>4x = 3x + 24
   =>4x – 3x = 24
   =>x = 24 men

Ans .

(3) 36 days

1. Explanation :

   (3) Let Y alone complete the work in x days.
   According to the question,
   X`s 16 days work + Y`s 12 days work = 1
   => \( \frac{16}{24} \) +\( \frac{12}{x} \) =1
   =>\( \frac{2}{3} \) +\( \frac{12}{x} \) =1
   =>\( \frac{12}{x} \) = 1 -\( \frac{2}{3} \) =\( \frac{1}{3} \)
   =>x = 12 × 3 = 36 days

Ans .

(3)40 days

1. Explanation :

   

Ans .

(2)1\( \frac{2}{3} \)days

1. Explanation :

   (2) Work done by A and B in 5 =5(\( \frac{1}{10} \)+\( \frac{1}{15} \)) =5(\( \frac{3+2}{30} \))
   = 5*\( \frac{5}{30} \) =\( \frac{5}{6} \)
   Remaining Work = 1- \( \frac{5}{6} \) =\( \frac{1}{6} \)
   Time taken by A =\( \frac{1}{6} \)*10 = \( \frac{5}{3} \) days.=1\( \frac{2}{3} \) days.

Ans .

(4)60 days

1. Explanation :

   (4) (A + B)’s 1 day’s work = \( \frac{1}{30} \)
   (A + B)s 20 days’ work =\( \frac{20}{30} \) =\( \frac{2}{3} \)
   Remaining Work =1- \( \frac{2}{3} \) =\( \frac{1}{3} \)
   Time taken by A in doing \( \frac{1}{3} \) of work = 20 days
   Time taken by A in doing whole work = 3 × 20 = 60 days

Ans .

(4)4 days

1. Explanation :

   (4) Nuts cut by Ram and Hari in 1 day =\( \frac{12}{2} \)kg. = 6 kg. ....(i)
   Nuts cut by them in 5 days = 30 kg.
   Amount of nuts cut by Ram alone = 58 – 30 = 28 kg.
   Time = 8 days
   Nuts cut by Ram in 1 day =\( \frac{28}{8} \) = 3.5 kg.
   From equation (i),
   Nuts cut by Hari in 1 day = (6 – 3.5) kg. = 2.5 kg.
   Time taken by Hari in cutting 10 kg. of nuts = \( \frac{10}{2.5} \)= 4 days

Ans .

(4)30 days

1. Explanation :

   (4) Ramesh`s 1 days work =\( \frac{1}{20} \)
   Rahman`s 1 days work = \( \frac{1}{25} \)
   (Ramesh + Rahman)s 1 days work = \( \frac{1}{20} \)+\( \frac{1}{25} \)= \( \frac{5+4}{100} \)=\( \frac{9}{100} \)
   Their 10 days work =\( \frac{90}{100} \)=\( \frac{9}{10} \)
   Remaininf Work = 1- \( \frac{9}{10} \) =\( \frac{1}{10} \)
   Suresh does\( \frac{1}{10} \) work in 3 days.
   Time taken by Suresh in doing 1 work = 3 × 10 = 30 days

Ans .

(3) 40 days

1. Explanation :

   (3) Let C alone complete the work in x days.
   According to the question,
   A`s 7 days work + B`s 3 days work + C`s 2 days work = 1
   => \( \frac{7}{10} \)+\( \frac{3}{12} \)+\( \frac{2}{x} \) =1
   =>\( \frac{2}{x} \)= 1 -\( \frac{7}{10} \)-\( \frac{1}{4} \)
   =>\( \frac{20-14-5}{20} \) =\( \frac{1}{20} \)
   => x = 2 × 20 = 40 days.

Ans .

(2) 12

1. Explanation :

   (2) Let the number of working men be x.
   => M₁D₁=M₂D₂
   => x × 60 = (x + 6) × 40
   
   => 3x = 2x + 12
   => 3x – 2x = 12
   => x = 12

Ans .

(4) 12 days

1. Explanation :

   . (4) A`s 1 days work =\( \frac{1}{20} \)
   B`s 1 days work =\( \frac{1}{15} \)
   (A + B + C)s 1 days work =\( \frac{1}{5} \)
   C`s 1 day`s work =\( \frac{1}{5} \)-\( \frac{1}{20} \)-\( \frac{1}{15} \)
   =\( \frac{12-3-4}{60} \)=\( \frac{5}{60} \)=\( \frac{1}{12} \)
   Required time = 12 days.

Ans .

(3)30

1. Explanation :

   . (3) Let 5 men leave the work after x days.
   =>M₁D₁=M₂D₂+M₃D₃
   =>15 × 40 = 15 × x + 10 × (45 – x)
   => 600 = 15x + 450 – 10x
   => 600 – 450 = 5x
   => 5x = 150
   => x =\( \frac{150}{5} \)= 30 days.

Ans .

(1)24days

1. Explanation :

   (1)(A + B)s 1 days work =\( \frac{1}{12} \)
   (A + B)’s 5 days’ work = \( \frac{5}{12} \)
   Remaining work = 1- \( \frac{5}{12} \) = \( \frac{7}{12} \)
   A does\( \frac{7}{12} \) work in 14 days.
   A will do 1 work in =\( \frac{14*12}{7} \)= 24 days.

Ans .

(2) 4 days

1. Explanation :

   (3) According to question, (6M + 8B)10 = (26M + 48B)2
   60M + 80B = 52M + 96B or, 1M = 2B ; 5M + 20B = (30 + 20)B = 50 boys and 6M + 8B => (12 + 8) boys = 20 boys
   20 boys can finish the work in 10 days, 50 boys can finish the work in \( \frac{20*10}{50} \) days = 4 days

Ans .

(2) 5 days

1. Explanation :

   (2) 5*6 men = 10*5 women => 3 men = 5 women;
   5 women + 3 men = 6 men; 5 men complete the work in 6 days
   6 men will complete the work in \( \frac{5*6}{6} \) = 5 days.

Ans .

(3) 3 days

1. Explanation :

   . (3) 3m = 6w => 1m = 2w; 12m + 8w = (12*2w) + 8w = 32w;
   6 women can do the work in 16 days
   32 women can do the work in \( \frac{16*6}{32} \) = 3 days

Ans .

(4) 41 days

1. Explanation :

   . (4) 1 mans 1 days work = \( \frac{1}{3} \) 1 womans 1 days work = \( \frac{1}{4} \)
   1 boys 1 days work = \( \frac{1}{4} \)
   (1 man + 1 woman)s days work = \( \frac{1}{4} \)*[\( \frac{1}{3} \) + \( \frac{1}{4} \)] =\( \frac{7}{48} \)
   Remaining work = 1-\( \frac{7}{48} \)=\( \frac{41}{48} \). Now;
   1 boys \( \frac{1}{4} \) days work = \( \frac{1}{4} \)*\( \frac{1}{12} \)= \( \frac{1}{48} \)
   \( \frac{41}{48} \) work will be done by \( \frac{41}{48} \)*48 = 41 boys

Ans .

(4) 10 days

1. Explanation :

   4) 16 men = 20 women => 4 men = 5 women. Now, according to question, 16 men complete the work in 25 days.
   1 man one days work = \( \frac{1}{25*16} \) => 4 men one days work = \( \frac{4}{25*16} \) = \( \frac{1}{100} \). Similarly,
   1 woman one days work = \( \frac{1}{25*20} \) => 5 women one days work = \( \frac{5}{25*20} \) = \( \frac{1}{100} \) => 28 men
   = \( \frac{28}{4} \)*5 = 35 women => [28 men + 15 women] => 50 women one days work = \( \frac{50}{25*20} \) = \( \frac{1}{10} \)
   Therefore, 28 men and 15 women can complete the whole work in 10 days.

Ans .

(3) 13\( \frac{1}{3} \) days

1. Explanation :

   (3) According to the question, 5 men = 8 women
   2 men = \( \frac{8}{5} \)*2 = \( \frac{16}{5} \)
   Total women = \( \frac{16}{5} \)+4 = \( \frac{36}{5} \)
   No. of days to do the same work = \( \frac{8*12*5}{36} \) = \( \frac{40}{3} \) = 13\( \frac{1}{3} \)

Ans .

(4) 12 days

1. Explanation :

   (4) 3 men = 4 women
   1 man = \( \frac{4}{3} \) women
   7 men = \( \frac{7*4}{3} \) = \( \frac{28}{3} \)
   7 men + 5 women = \( \frac{28}{3} \)+5 = \( \frac{28+15}{3} \) = \( \frac{43}{3} \) Women
   Now, M₁D₁ = M₂D₂ => 4 * 43 = \( \frac{43}{3} \)*D₂, where D₂ = number of days
   => D₂ = \( \frac{4*3*43}{43} \) = 12 days.

Ans .

(3) 15 days

1. Explanation :

   (3) 6 men = 12 women
   1 man = 2 women
   Now, 8 men + 16 women = (8*2*16) women = 32 women
   12 women can do a work in 20 days. 32 women can do the twice work in \( \frac{20*12*2}{32} \) = 15 days.

Ans .

(1) 9 days

1. Explanation :

   (1) Work done by 1 woman in 1
   day = \( \frac{1}{3} \)-\( \frac{1}{6} \)-\( \frac{1}{18} \) = \( \frac{6-3-1}{18} \) = \( \frac{1}{9} \)
   Woman will do the work in 9 days.

Ans .

(3) 3 mens work = 5 womens work
1 mans work = \( \frac{5}{3} \) womens work
6 mens work = \( \frac{5}{3} \)*6 = 10 womens work
6 men + 5 women = 15 women
5 women can do work in 12 days.
Hence, 15 women can do it in \( \frac{5*12}{15} \) = 4 days.

Ans .

(2) 204 days

1. Explanation :

   (1) 10 men = 20 boys
   1 man = 2 boys
   8 men + 4 boys = (16 + 4) boys = 20 boys
   Hence, 8 men and 4 boys will make 260 mats in 20 days.

Ans .

(2) 2 days

1. Explanation :

   (2) Work done in two days = \( \frac{1}{6} \)*2
   = \( \frac{1}{3} \) , remaining work = \( \frac{2}{3} \)
   => \( \frac{M1D1}{W1} \) = \( \frac{M2D2}{W2} \)
   => \( \frac{3*2*3}{1} \) = \( \frac{6*D2*3}{2} \)
   => D2 = \( \frac{3*2*2}{6} \) = 2 days

Ans .

(3) 40 days

1. Explanation :

   (3) Work done by 1 woman in 1 day
   = \( \frac{1}{8} \)-\( \frac{1}{10} \) = \( \frac{5-4}{40} \)-\( \frac{1}{40} \)
   One woman will complete the work in 40 days.

Ans .

(3) 40 days

1. Explanation :

   (3) Let 1 mans 1 days work = x and 1 womans 1 days work = y
   Then, 4x + 6y = \( \frac{1}{8} \) and
   3x+7y = \( \frac{1}{10} \) br/> From both equations, we get y = \( \frac{1}{400} \)
   10 womens 1 days work = \( \frac{10}{400} \) = \( \frac{1}{40} \)
   10 women will finish the work in 40 days.

Ans .

(1) 8 days

1. Explanation :

   (1) Part of work done by 2 men and 2 women in 2 days.
   = 2[\( \frac{2}{20} \)+\( \frac{8}{30} \)]
   = 2[\( \frac{1}{10} \)+\( \frac{8}{30} \)] = 2\( \frac{3+8}{30} \)
   = \( \frac{22}{30} \) = \( \frac{11}{15} \)
   Remaining work = 1-\( \frac{11}{15} \) = \( \frac{4}{15} \)
   Work done by 1 boy in 2 days = \( \frac{2}{60} \) = \( \frac{1}{30} \)
   Number of boys required to assist = \( \frac{4}{15} \)*30 = 8

Ans .

(3) 48 days

1. Explanation :

   (3) 1 man = 2 women = 3 boys
   1 man + 1 woman + 1 boy = [3+\( \frac{3}{2} \)+1] boys = \( \frac{11}{2} \) boys
   M₁D₁ = M₂D₂
   => 3*88 = \( \frac{11}{2} \)*D₂
   => D₂= \( \frac{2*3*88}{11} \) = 48 days
   

Ans .

(2) 20 days

1. Explanation :

   (2) 6m + 8w = 10 days
   => 2*(3m + 4w) = 10 days
   => 3m + 4w = 20 days
   [Since the workforce has become half of the original force, so number of days must be double].

Ans .

(1) 17\( \frac{1}{2} \) days

1. Explanation :

   (1) 12*(3 men + 4 boys) = 10*(4 men + 3 boys)
   => 36 men + 48 boys = 40 men + 30 boys => 4 men = 18 boys => 2 men = 9 boys
   4 men + 3 boys = 21 boys, who do the work in 10 days and 2 men + 3 boys = 12 boys
   M₁D₁ = M₂D₂
   => 21*10 = 12*D₂
   => D₂ = \( \frac{21*10}{12} \) = \( \frac{35}{2} \) = 17\( \frac{1}{2} \) days.

Ans .

(4) 8 months

1. Explanation :

   (4) 10 men = 20 women
   1 man = 2 women = 5 children
   1 woman = 2 children
   5 men + 5 women + 5 children = 20 + 10 + 5 = 35 children
   M₁D₁ = M₂D₂
   40*7 = 35*D₂
   D₂ = \( \frac{40*7}{35} \) = 8 months
   5 men, 5 women and 5 children can do half of the work in 8 months
   Required time = 4 months.

Ans .

(1) 5\( \frac{1}{3} \) days

1. Explanation :

   (1) 8 men = 12 boys
   4 men = 6 boys
   => 20 men = 30 boys
   => 20 men + 6 boys = 36 boys
   M1D1 = M2D2
   => 12*16 = 36*D2
   D2 = \( \frac{12*16}{36} \)= \( \frac{16}{3} \) = 5\( \frac{1}{3} \) days.

Ans .

(3) 12\( \frac{1}{2} \) days

1. Explanation :

   (3) According to the question, 20 men + 30 boys = 24 men + 16 boys
   4 men = 14 boys
   => 2 men = 7 boys
   => 2 men + 1 boy = 8 boys
   =>2 men + 3 boys = 10 boys
   By M1D1 = M2D2
   => 10*10 = 8*D2
   => D2 = \( \frac{10*10}{8} \) = \( \frac{25}{2} \)
   = 12\( \frac{1}{2} \) days

Ans .

(2) 12\( \frac{1}{2} \) days

1. Explanation :

   (2) 2*10 men + 3*10 women
   = 3*8 men + 2*8 women
   => 20 men + 30 women
   = 24 men + 16 women
   => 4 men = 14 women
   or 2 men = 7 women
   2 men + 3 women = 10 women
   2 men + 1 woman = 8 women
   M₁D₁ = M₂D₂
   => 10*10 = 8*D₂
   => D₂ = \( \frac{25}{2} \) = 12\( \frac{1}{2} \) days

Ans .

(1) 17\( \frac{1}{2} \) days

1. Explanation :

   (1) 12*(3 men + 4 boys) = 10*(4 men + 3 boys)
   => 36 men + 48 boys = 40 men + 30 boys => 4 men = 18 boys or 2 men = 9 boys
   4 men + 3 boys = 21 boys who do the work in 10 days and, 2 men + 3 boys = 12 boys
   M₁D₁ = M₂D₂
   => 21*10 = 12*D₂
   =>D₂ = \( \frac{21*10}{12} \) = \( \frac{35}{2} \) = 17\( \frac{1}{2} \) days

Ans .

(2) 7 days

1. Explanation :

   (2) Using Rule 1, 4 men = 6 women
   1 men = \( \frac{6}{4} \) = \( \frac{3}{2} \) women
   10 men + 3 women = 10\( \frac{3}{2} \)+3 = 18 women
   \( \frac{M1D1T1}{W1} \) = \( \frac{M2D2T2}{W2} \)
   => \( \frac{6*12*7}{1} \) = \( \frac{18*D2*8}{W2} \)
   => D2 = \( \frac{6*12*7*2}{18*8} \) = 7 days

Ans .

(3) 40 days

1. Explanation :

   (3) Time taken by boy = x days
   \( \frac{1}{10} \)+\( \frac{1}{24} \)+\( \frac{1}{x} \) = \( \frac{1}{6} \)
   => \( \frac{1}{x} \) = \( \frac{1}{6} \)+\( \frac{1}{10} \)+\( \frac{1}{24} \)
   = \( \frac{20-12-5}{120} \) = \( \frac{3}{120} \) = \( \frac{1}{40} \)
   => x = 40 days

Ans .

(3) 5\( \frac{7}{13} \) months

1. Explanation :

   (3) 40 men = 60 women º=80 children
   10 men = \( \frac{80}{40} \)*10
   = 20 children
   10 women = \( \frac{80}{60} \)*10
   = \( \frac{40}{3} \) children
   10 men + 10 women + 10 children
   = [20+\( \frac{40}{3} \)+10] children
   = \( \frac{60+40+30}{3} \) children
   = \( \frac{120}{3} \) children
   \( \frac{M1D1}{W1} \) = \( \frac{M2D2}{W2} \)
   D₂ = \( \frac{80*6*13}{130} \) = \( \frac{144}{13} \) months
   Half of the work can do
   = \( \frac{144}{13} \)*\( \frac{1}{2} \) = \( \frac{72}{13} \) = 5\( \frac{7}{13} \) months

Ans .

(1) 49

1. Explanation :

   (1) Using Rule 11,
   According to the question,
   1 man = 2 women = 4 boys
   1 man + 1 woman + 1 boy
   = (4+2+1) boys = 7 boys
   M₁D₁ = M₂D₂
   => 7*7 = 1*D₂
   => D₂ = 49 days

Ans .

(3) 48 days

1. Explanation :

   (3) 1 man = 2 women = 3 boys
   1 man + 1 woman + 1 boy
   = [3+\( \frac{3}{2} \)+1] boys
   = \( \frac{6+3+2}{2} \)
   = \( \frac{11}{2} \)
   M₁D₁ = M₂D₂
   
   3*88 = \( \frac{11}{2} \)*D₂
   D₂= \( \frac{3*2*88}{11} \) = 48 days

Ans .

(2) 21 days

1. Explanation :

   (2) 3 men = 7 women
   7 men = \( \frac{7*7}{3} \)
   = \( \frac{49}{3} \) women
   7 men + 5 women
   = [\( \frac{49}{3} \)+5] women
   = \( \frac{49+15}{3} \) women
   = \( \frac{64}{3} \) women
   \( \frac{M1D1}{W1} \) = \( \frac{M2D2}{W2} \)
   => \( \frac{7*32}{1} \) = \( \frac{64*D2}{3*2} \)
   => D2 = \( \frac{7*32*3*2}{64} \) = 21 days

Ans .

(2) 24 days

1. Explanation :

   (2) 1 man = 2 women = 3 boys
   1 man + 1 woman + 1 boy
   = 3 boys + \( \frac{3}{2} \) boys + 1 boy = \( \frac{11}{2} \) boys
   By M₁D₁ = M₂D₂
   3*44 = \( \frac{11}{2} \)*D₂
   => D₂ = \( \frac{2*3*44}{11} \) = 24 days
   

Ans .

(3) 12 days

1. Explanation :

   (3) Using Rule 1,
   2 children = 1 man
   8 children + 12 men = 16 men
   M₁D₁ = M₂D₂ ,
   16*9 = 12*D₂
   D₂ = \( \frac{16*9}{12} \) = 12 days.

Ans .

(1) 2:1

1. Explanation :

   (1) Work done by 12 men + 16 boys in 5 days
   = Work done 13 men + 24 boys in 4 days
   => (60 men + 80 boys)s 1 days work = (52 men + 96 boys)s 1 days work
   => (60 - 52) men = (96 -80) boys
   => 8 men = 16 boys
   => 1 man = 2 boys
   Required ratio = 2 : 1

Ans .

(2) 4:3

1. Explanation :

   (2) 20 women complete 1 work in 16 days
   16 men complete same work in 15 days
   16*15 men = 20*16 women
   => 3 men = 4 women
   Required ratio = 4:3

Ans .

(4) 8

1. Explanation :

   (4) 18 men = 36 boys
   => 1 man = 2 boys
   24 men + 24 boys
   = (24 + 12) men
   = 36 men
   M₁D₁T₁ = M₂D₂T₂
   => 18*24*6 = 36*D₂*9
   => D₂= \( \frac{1}{72} \) = 8 days

Ans .

(3) 10 days

1. Explanation :

   (3) 5 men can do 1 work in 14 days.
   3 men will do \( \frac{3}{5} \) work in 14 days.
   Remaining work = 1-\( \frac{3}{5} \) - \( \frac{2}{5} \)
   5 men will do \( \frac{2}{5} \) work in 14 days.
   Time taken by 5 women in doing 1 work
   = \( \frac{14*5}{2} \) = 35 days
   (5 men + 5 women)s 1 days work
   = \( \frac{1}{14} \)+\( \frac{1}{35} \) = \( \frac{5+2}{70} \) = \( \frac{7}{70} \) = \( \frac{1}{10} \)
   Required time = 10 days.

Ans .

(1) \( \frac{8}{15} \)

1. Explanation :

   (1) Using basics of Rule 2,
   As work per day = \( \frac{1}{15} \)
   Bs work per day = \( \frac{1}{20} \)
   (A+ B)s work per day
   = \( \frac{1}{15} \)+\( \frac{1}{20} \) = \( \frac{4+3}{60} \) = \( \frac{7}{60} \)
   (A + B)s work in 4 days
   = 4*\( \frac{7}{60} \) = \( \frac{7}{15} \)
   Left work = 1-\( \frac{7}{15} \) = \( \frac{15-7}{15} \) = \( \frac{8}{15} \)

Ans .

(3) 8 days

1. Explanation :

   (3) Using basics of Rule 2,
   The part of field cultivated by A in 1 day
   = \( \frac{2}{5*6} \) = \( \frac{1}{15} \)
   The part of field cultivated by B in 1 day
   = \( \frac{1}{3*10} \) = \( \frac{1}{30} \)
   The part of field cultivated by A and B together
   = \( \frac{1}{15} \)+\( \frac{1}{30} \) = \( \frac{3}{30} \) = \( \frac{1}{10} \)
   \( \frac{4}{5} \) part of field cultivated by A and B together in
   \( \frac{40}{5} \) days = \( \frac{4*10}{5} \) = 8 days

Ans .

(1) 37\( \frac{1}{2} \) days

1. Explanation :

   (1) Using basics of Rule 2,
   A can do the whole work in
   \( \frac{20*5}{4} \) = 25 days
   Remaining work = 1-\( \frac{4}{5} \) = \( \frac{1}{5} \)
   (A + B)s 1 days work = \( \frac{1}{15} \)
   As 1 days work = \( \frac{1}{25} \)
   Bs 1 days work
   = \( \frac{1}{15} \)-\( \frac{1}{25} \) = \( \frac{5-3}{75} \) = \( \frac{2}{75} \)
   B can finish the work in \( \frac{75}{2} \)
   days i.e., 37 \( \frac{1}{2} \) days

Ans .

(1) \( \frac{1}{6} \)

1. Explanation :

   (1) Using basics of Rule 2,
   As 1 days work = \( \frac{1}{18} \)
   Bs 1 days work = \( \frac{1}{9} \) /
   (A+B)s 1 days work
   = \( \frac{1}{18} \)+\( \frac{1}{69} \) = \( \frac{1+2}{18} \) = \( \frac{3}{18} \) = \( \frac{1}{6} \)

Ans .

(3) 13\( \frac{1}{3} \)

1. Explanation :

   (3) Using basics of Rule 2,
   Remaining work
   = 1-\( \frac{7}{10} \) = \( \frac{3}{10} \)
   (A + B) take 4 days to do \( \frac{3}{10} \) work
   (A + B) will do the work in 4\( \frac{10}{3} \) days
   = \( \frac{40}{3} \) = 13\( \frac{1}{3} \) days

Ans .

(2) \( \frac{2}{3} \)

1. Explanation :

   (2) Using basics of Rule 2,
   Time taken by A and B
   = \( \frac{6*12}{6+12} \) = \( \frac{6*12}{18} \) = 4
   Work done by A in 4 days
   = \( \frac{4}{6} \) = \( \frac{2}{3} \)

Ans .

(4) Using basics of Rule 3,
A can do \( \frac{1}{2} \) work in 5 days.
A can do 1 work in 10 days
Similarly,
B can do 1 work in \( \frac{5}{3} \)*9
= 15 days.
C can do 1 work in 8*\( \frac{3}{2} \) = 12 days.
Now,
As 1 days work = \( \frac{1}{10} \)
Bs 1 days work = \( \frac{1}{15} \)
Cs 1 days work = \( \frac{1}{12} \)
(A + B + C)s 1 days work
= \( \frac{1}{10} \)+\( \frac{1}{15} \)+\( \frac{1}{12} \)
= \( \frac{6+4+5}{60} \) = \( \frac{15}{60} \) = \( \frac{1}{4} \)
Hence, (A + B + C) together can complete the work in 4 days.

Ans .

(3) 4 days

1. Explanation :

   (3) Using basics of Rule 1,
   
   \( \frac{7}{8} \): \( \frac{7}{8} \) :: 28: x
   where x is no. of men
   => \( \frac{7}{8} \)*x = \( \frac{1}{8} \)*28
   => x = \( \frac{28*8}{7*8} \) = 4

Ans .

(2) 9\( \frac{3}{8} \) days

1. Explanation :

   (2) Using basics of Rule 2,
   Time taken by A alone in doing
   the work = 15 days
   Time taken by B alone in doing
   the work =
   \( \frac{10*5}{2} \) = 25 days
   (A + B)s 1 days work
   = \( \frac{1}{15} \)+\( \frac{1}{25} \) = \( \frac{5+3}{75} \) = \( \frac{8}{75} \)
   Hence, the work will be completed
   in \( \frac{75}{8} \) = 9*\( \frac{3}{8} \) days

Ans .

(3) 3\( \frac{3}{4} \)

1. Explanation :

   (3) Using basics of Rule 2,
   Time taken by A to complete the
   work = \( \frac{3}{4} \) = 6 days
   Time taken by B to complete the
   work = \( \frac{6*5}{3} \) = 10 days
   (A + B)s 1 days work
   = \( \frac{1}{6} \)+\( \frac{1}{10} \) = \( \frac{5+3}{30} \) = \( \frac{8}{30} \) = \( \frac{4}{15} \)
   A and B together will complete
   the work in \( \frac{15}{4} \) = 3*\( \frac{3}{4} \) days

Ans .

(2) 20

1. Explanation :

   (2) Using basics of Rule 1,
   
   => 30*\( \frac{3}{4} \)*x = 60*\( \frac{1}{4} \)*60
   => x = \( \frac{60*60}{30*3} \) = 40
   20 men should be discharged.