Ans .
(3) 10 days
(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
(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) 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
(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
(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
(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
(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
(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
(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
(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
(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
(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) 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
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
(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) 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
(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) (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
(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
(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
(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
(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
(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) (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
(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
(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) (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 daysAns .
(1) 24 days
(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
(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
(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) 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
(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
(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
(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
(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
(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
(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) 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.