**Ans . **

(3) 10 days

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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~~
~~

**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

**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

**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

**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

**Explanation :**(1) (A + B)’s 1 day’s work =\( \frac{1}{8} \)

br> C alone will complete the work in 60 days

(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 =\( \frac{17}{120} \)-\( \frac{1}{8} \)=\( \frac{17-15}{120} \)=\( \frac{1}{60} \)

**Ans . **

(1) 24 days

**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

**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

**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

**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

**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

**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

**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~~
~~

**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

**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

**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

**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

**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

**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

**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/ 2x^{2}+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

**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

**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

**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

**Explanation :**(2) Using Rule 1, M

_{1}D_{1}T_{1}= M_{2}D_{2}T_{2}

=>15 × 20 × 8 = 20 × 12 × T_{2}

=> T_{2}= \( \frac{15*20*8}{20*12} \) =10 hours.

**Ans . **

(4) 60 days

**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

**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

**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

=> M_{1}D= M _{2}D_{2}

=>15 × 21 = M_{2}× 5

=>M_{2}= \( \frac{15*21}{5} \) = 63 days.s

**Ans . **

(2) 160 minutes

**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

**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

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

**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

**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

**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

**Explanation :**

**Ans . **

(2) 8 days

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

**Explanation :**(3)Using Rule 1,

(M^{1}D^{1}T^{1}) / (W^{1}) = (M^{2}D^{2}T^{2}) / (W^{2})

= \( \frac{90*16*12}{1} \) = (70*24*8 ) / (W^{2})

W^{2}= \( \frac{90*16*12}{70*24*8 }\) = \( \frac{7}{9} \) parts

**Ans . **

(3)12 days

**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

**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

**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

**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

**Explanation :**

**Ans . **

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

**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

**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

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

**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

**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

**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

**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

**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

**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

**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

**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

**Explanation :**

**Ans . **

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

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

**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

**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

**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

**Explanation :**M

_{1}D_{1}= M_{2}D_{2}

=>24 × 12 = 36 × D_{2}

= D_{2}=\( \frac{24*12}{36} \) = 8 days.

**Ans . **

(1)18 days

**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

**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

**Explanation :**

**Ans . **

(2)2

**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

**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

**Explanation :**

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

**Ans . **

(2)20 days

**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

**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

**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

**Explanation :**

**Ans . **

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

**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

**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

**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

**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

**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

**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

**Explanation :**

**Ans . **

(3) 10 days

**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

**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

**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

**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

**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

**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

**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

**Explanation :**

**Ans . **

(3)16 days

**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

**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

**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

**Explanation :**

**Ans . **

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

**Explanation :**

**Ans . **

(4)9 days

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**Explanation :**

**Ans . **

(3) 120 days

**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~~
~~

**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

**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

**Explanation :**

**Ans . **

(4) 4

**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

**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

**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

**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

**Explanation :**

**Ans . **

(1) 44

**Explanation :**12 men =24 boys

=>1 man = 2 boys

=> 15 men + 6 boys

= 30 boys + 6 boys = 36 boys

=>M_{1}D_{1}=M_{2}D_{2}

=>24 × 66 = 36 × D_{2}

D_{2}=\( \frac{24*66}{36} \) = 44 days.

**Ans . **

(3) 100 days

**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

**Explanation :**(2) Number of men originally = x (let)

=>M_{1}D_{1}=M_{2}D_{2}

=> × 60 = (x + 8) × 50

=>6x = 5x + 40

=>6x – 5x = 40

=>x = 40 men

**Ans . **

(4) 60

**Explanation :**(4) Using Rule 1,,

Number of men originally = x (let)

=>M_{1}D_{1}=M_{2}D_{2}

=> x × 18 = (x – 6) × 20

=>x × 9 = (x – 6) × 10

= 10x – 60

=>10x – 9x = 60

=> x = 60 men

**Ans . **

(4)75

**Explanation :**(4) Original number of men= x (let)

=>M_{1}D_{1}=M_{2}D_{2}

=>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

**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

**Explanation :**

**Ans . **

(4) 11

**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

**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

**Explanation :**(2) Number of men initially = x (let)

=>M_{1}D_{1}=M_{2}D_{2}

=> x × 40 = (x + 8) × 30

=>4x = 3x + 24

=>4x – 3x = 24

=>x = 24 men

**Ans . **

(3) 36 days

**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

**Explanation :**

**Ans . **

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

**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

**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

**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

**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

**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

**Explanation :**(2) Let the number of working men be x.

=> M_{1}D_{1}=M_{2}D_{2}

=> x × 60 = (x + 6) × 40

=> 3x = 2x + 12

=> 3x – 2x = 12

=> x = 12

**Ans . **

(4) 12 days

**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

**Explanation :**. (3) Let 5 men leave the work after x days.

=>M_{1}D_{1}=M_{2}D_{2}+M_{3}D_{3}

=>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

**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

**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

**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

**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

**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

**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

**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

**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, M_{1}D_{1}= M_{2}D_{2}=> 4 * 43 = \( \frac{43}{3} \)*D_{2}, where D_{2}= number of days

=> D_{2}= \( \frac{4*3*43}{43} \) = 12 days.

**Ans . **

(3) 15 days

**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

**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

**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

**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

**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

**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

**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

**Explanation :**(3) 1 man = 2 women = 3 boys

1 man + 1 woman + 1 boy = [3+\( \frac{3}{2} \)+1] boys = \( \frac{11}{2} \) boys

M_{1}D_{1}= M_{2}D_{2}

=> 3*88 = \( \frac{11}{2} \)*D_{2}

=> D_{2}= \( \frac{2*3*88}{11} \) = 48 days

**Ans . **

(2) 20 days

**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

**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

M_{1}D_{1}= M_{2}D_{2}

=> 21*10 = 12*D_{2}

=> D_{2}= \( \frac{21*10}{12} \) = \( \frac{35}{2} \) = 17\( \frac{1}{2} \) days.

**Ans . **

(4) 8 months

**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

M_{1}D_{1}= M_{2}D_{2}

40*7 = 35*D_{2}

D_{2}= \( \frac{40*7}{35} \) = 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

**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

**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

**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

M_{1}D_{1}= M_{2}D_{2}

=> 10*10 = 8*D_{2}

=> D_{2}= \( \frac{25}{2} \) = 12\( \frac{1}{2} \) days

**Ans . **

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

**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

M_{1}D_{1}= M_{2}D_{2}

=> 21*10 = 12*D_{2}

=>D_{2}= \( \frac{21*10}{12} \) = \( \frac{35}{2} \) = 17\( \frac{1}{2} \) days

**Ans . **

(2) 7 days

**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

**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

**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

\( \frac{M1D1}{W1} \) = \( \frac{M2D2}{W2} \)

D_{2}= \( \frac{80*6*13}{130} \) = \( \frac{144}{13} \) months

Half of the work can do

= \( \frac{144}{13} \)*\( \frac{1}{2} \) = \( \frac{72}{13} \) = 5\( \frac{7}{13} \) months

**Ans . **

(1) 49

**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

M_{1}D_{1}= M_{2}D_{2}

=> 7*7 = 1*D_{2}

=> D_{2}= 49 days

**Ans . **

(3) 48 days

**Explanation :**(3) 1 man = 2 women = 3 boys

1 man + 1 woman + 1 boy

= [3+\( \frac{3}{2} \)+1] boys

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

= \( \frac{11}{2} \)

M_{1}D_{1}= M_{2}D_{2}

3*88 = \( \frac{11}{2} \)*D_{2}

D_{2}= \( \frac{3*2*88}{11} \) = 48 days

**Ans . **

(2) 21 days

**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

**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 M_{1}D_{1}= M_{2}D_{2}

3*44 = \( \frac{11}{2} \)*D_{2}

=> D_{2}= \( \frac{2*3*44}{11} \) = 24 days

**Ans . **

(3) 12 days

**Explanation :**(3) Using Rule 1,

2 children = 1 man

8 children + 12 men = 16 men

M_{1}D_{1}= M_{2}D_{2},

16*9 = 12*D_{2}

D_{2}= \( \frac{16*9}{12} \) = 12 days.

**Ans . **

(1) 2:1

**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

**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

**Explanation :**(4) 18 men = 36 boys

=> 1 man = 2 boys

24 men + 24 boys

= (24 + 12) men

= 36 men

M_{1}D_{1}T_{1}= M_{2}D_{2}T_{2}

=> 18*24*6 = 36*D_{2}*9

=> D_{2}= \( \frac{1}{72} \) = 8 days

**Ans . **

(3) 10 days

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

**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

**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

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

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

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

**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

**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

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

**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

**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.

**Ans . **

(2) Q

**Explanation :**(2) Time taken by P in completing

1 work = 10*4 = 40 days

Time taken by Q in completing 1

work = \( \frac{15*5}{2} \) = \( \frac{75}{2} \) days

Time taken by R in completing 1

work = 13*3 = 39 days

Time taken by S in completing 1

work = 7*6 = 42 days

Clearly, Q took the least time i.e. \( \frac{75}{2} \) or 35\( \frac{1}{2} \) days.

**Ans . **

(3) \( \frac{1}{20} \)

**Explanation :**(3) Using basics of Rule 5,

(A + B)s 1 days work = \( \frac{1}{72} \)

(B + C)s 1 days work = \( \frac{1}{120} \)

(C + A)s 1 days work = \( \frac{1}{60} \)

On adding all three, 2(A + B + C)s 1 days work

= \( \frac{1}{72} \)+\( \frac{1}{120} \)+\( \frac{1}{60} \) = \( \frac{5+3+4}{360} \) = \( \frac{1}{30} \)

(A + B + C)s 1 days work = \( \frac{1}{60} \)

(A + B + C)s 3 days work = \( \frac{3}{60} \) = \( \frac{1}{20} \)

**Ans . **

(1) 12 days

**Explanation :**(1) Using basics of Rule 2,

Time taken by A to finish the work

= 5*6 = 30 days

Time taken by B to complete the

work = \( \frac{8*5}{2} \) = 20 days

(A + B)s 1 days work

= \( \frac{1}{30} \)+\( \frac{1}{20} \) = \( \frac{2+3}{60} \) = \( \frac{1}{12} \)

Required time = 12 days

**Ans . **

(1) \( \frac{5}{8} \)

**Explanation :**(1) Using basics of Rule 2,

(A + B)s 5 days work

= 5*[\( \frac{1}{20} \)+\( \frac{1}{40} \)]

= 5*\( \frac{2+1}{40} \) = \( \frac{15}{40} \) = \( \frac{3}{8} \)

Remaining work = 1-\( \frac{3}{8} \) = \( \frac{5}{8} \)

**Ans . **

(4) 6 days

**Explanation :**(4) (A+B)s 1 days work

= \( \frac{1}{20} \)+\( \frac{1}{30} \) = \( \frac{3+2}{60} \) = \( \frac{1}{12} \)

Work done in 6 days

= \( \frac{6}{12} \) = \( \frac{1}{2} \)

**Ans . **

(1) 30 days

**Explanation :**(1) Using basics of Rule 2,

Let B completes the work in x days.

Work done by A in \( \frac{3x}{4} \) days = \( \frac{1}{2} \)

=> Time taken by A in completing

the work = 2*\( \frac{3x}{4} \) = \( \frac{3x}{2} \) days

(A + B)s 1 days work

= \( \frac{1}{x} \)+\( \frac{2}{3x} \) = \( \frac{3+2}{3x} \) = \( \frac{5}{3x} \)

\( \frac{5}{3x} \) = \( \frac{1}{18} \) => 3x = 90

=> x = 30

Hence, time taken by B in completing

the work = 30 days

**Ans . **

(2) 25 days

**Explanation :**(2) Using basics of Rule 2,

If B completes a work in x days,

A will complete the same in

\( \frac{2x}{3} \) days.

\( \frac{1}{x} \)+\( \frac{3}{2x} \) = \( \frac{1}{10} \)

=> \( \frac{2+3}{2x} \) = \( \frac{1}{10} \) => 2x = 50

=> x = 25 days

**Ans . **

(4) 7\( \frac{1}{5} \) days

**Explanation :**(4) Using basics of Rule 2,

Ratio of efficiency of A and B

= 3:2

Ratio of time taken = 2:3

Time taken by A

= \( \frac{2}{3} \)*18 = 12 days

(A + B)s 1 days work

= \( \frac{1}{12} \)+\( \frac{1}{18} \) = \( \frac{3+2}{36} \) = \( \frac{5}{36} \)

Required time

= \( \frac{36}{5} \) = 7*\( \frac{1}{5} \) days

**Ans . **

(4) 13\( \frac{5}{7} \) days

**Explanation :**(4) Using basics of Rule 2,

A does \( \frac{7}{8} \) work in 28 days.

A will complete the work in

28*\( \frac{8}{7} \) = 32 days

B does \( \frac{5}{6} \) work in 20 days.

B will complete the work in

\( \frac{20*6}{5} \) = 24 days

(A + B)s 1 days work

= \( \frac{1}{32} \)+\( \frac{1}{24} \) = \( \frac{3+4}{96} \) = \( \frac{7}{96} \)

Required time

= \( \frac{96}{7} \) = 13\( \frac{5}{7} \) days

**Ans . **

(4) 13\( \frac{5}{7} \) days

**Explanation :**(4) Time taken by A and B = x

hours (let).

According to the question,

Time taken by A alone

= (x + 8) hours.

Time taken by B alone

= [x+\( \frac{9}{2} \)] days

\( \frac{1}{x+8} \)+\( \frac{2}{x+9} \) = \( \frac{1}{x} \)

=> \( \frac{1}{x+8} \)+\( \frac{2}{2x+9} \) = \( \frac{1}{x} \)

=> \( \frac{2x+9+2x+16}{(x+8)*(2x+9)} \) = \( \frac{1}{x} \)

=> 4x+25 / 2x^{2}+16x+9x+72 = \( \frac{1}{x} \)

=> 4x^{2}+25x = 2x^{2}+25x+72

=> 2x^{2}= 72 => 2x^{2}= \( \frac{72}{2} \) = 36

=> x = 6 hours

**Ans . **

(1) 100

**Explanation :**

**Ans . **

(2) 9\( \frac{3}{5} \) days

**Explanation :**(2) Using basics of Rule

x does \( \frac{1}{4} \) work in 6 days

x does 1 work in 24 days

Similarly,

y does \( \frac{3}{4} \) work in 12 days

y does 1 work in \( \frac{12*4}{3} \)

= 16 days

(x + y)s 1 days work

= \( \frac{1}{24} \)+\( \frac{1}{16} \) = \( \frac{2+3}{48} \) = \( \frac{5}{48} \)

Required time = \( \frac{48}{5} \)

= 9*\( \frac{3}{5} \) days

**Ans . **

(4) 30 days

**Explanation :**(4) Let the time taken by B in

doing the work alone = x days

According to the question,

Time taken by A

= 2*\( \frac{3x}{4} \) = \( \frac{3x}{2} \) days

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

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

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

=> 3x = 18*5

=> x = \( \frac{18*5}{3} \) = 30 days

**Ans . **

(2) Rs. 3,450

**Explanation :**(2) Part of work done by A and

B = \( \frac{19}{23} \)

Part of work done by C

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

Part of work done by B and C

= \( \frac{8}{23} \)

Part of work done by B

= \( \frac{8}{23} \)-\( \frac{4}{23} \) = \( \frac{4}{23} \)

Part of work done by A

= \( \frac{19}{23} \)-\( \frac{4}{23} \) = \( \frac{15}{23} \)

Ratio of the shares of wages of

A, B and C

= \( \frac{15}{23} \) : \( \frac{4}{23} \) : \( \frac{4}{23} \) = 15:4:4

As share

= \( \frac{15}{23} \)*5290 = Rs. 3450

**Ans . **

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

**Explanation :**(4) Using basics of Rule 2

Work done by A and B in 1 day

= \( \frac{1}{10} \)+\( \frac{1}{20} \) = \( \frac{2+1}{20} \) =\( \frac{3}{20} \)

(A + B)s 5 days work

= \( \frac{5*3}{20} \) = \( \frac{3}{4} \)

Remaining work

= 1-\( \frac{3}{4} \) = \( \frac{1}{4} \)

**Ans . **

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

**Explanation :**(2) According to the question,

(4*8) men + (6*8) women =

(2*8) men + (9*8) women

=> 4 men + 6 women = 2 men +

9 women

=> (4 - 2) men = (9 - 6) women

=> 2 men = 3 women

4 men + 6 women = 12 women

M_{1}D_{1}= M_{2}D_{2}

=> 12*8 = 18*D_{2}

=> D_{2}= \( \frac{12*8}{18} \) = \( \frac{16}{3} \) = 5\( \frac{1}{3} \) days

**Ans . **

(1) 90 days

**Explanation :**(1) Let time taken by A alone

in doing work be x days.

Time taken by B alone

= 3x days

A and B together finish \( \frac{2}{5} \)

work in 9 days.

TIme taken by A and B in

doing whole work

= \( \frac{9*5}{2} \) = \( \frac{45}{2} \) days

\( \frac{1}{x} \)+\( \frac{1}{3x} \) = \( \frac{2}{45} \)

=> \( \frac{3+1}{3x} \) = \( \frac{2}{45} \)

=> \( \frac{4}{3x} \) = \( \frac{2}{45} \) => 2*3x = 4*45

=> x = \( \frac{4*45}{2*3} \) = 30 days

Time taken by B = 3x days

= 3*30 = 90 days

**Ans . **

(4) 8

**Explanation :**(4) Using Rule 1,

=> 16*\( \frac{15}{2} \)*x = 12*8*10

=> 8*15*x = 12*8*10

=> x = \( \frac{12*8*10}{8*15} \) = 8 days

**Ans . **

(4) 16

**Explanation :**

**Ans . **

(3) 9 days

**Explanation :**

**Ans . **

(4) 2\( \frac{8}{11} \)

**Explanation :**(4) According to the question,

John does \( \frac{1}{2} \) work in 3 hours.

Time taken by John in doing

whole work = 6 hours

Joe does \( \frac{1}{8} \) work in 1 hour.

Time taken by Joe in doing

whole work = 8 hours

Remaining work = \( \frac{1}{2} \)-\( \frac{1}{8} \)

= \( \frac{4-1}{8} \) = \( \frac{3}{8} \) parts

Time taken by George

= \( \frac{8*5}{3} \) = \( \frac{40}{3} \) hours

Work done by all three in 1 hour

= \( \frac{1}{6} \)+\( \frac{1}{8} \)+\( \frac{3}{40} \)

= \( \frac{20+15+9}{120} \) = \( \frac{44}{120} \)

= \( \frac{11}{30} \)

Required time = \( \frac{30}{11} \)

= 2\( \frac{8}{11} \) hours

**Ans . **

(4) 18 days

**Explanation :**(4) Remaining work = 1-\( \frac{2}{5} \)

= \( \frac{3}{5} \) parts

(A + B) together do \( \frac{3}{5} \)th part

of work in 6 days.

Time taken by A and B in doing

whole work = \( \frac{6*5}{3} \)

= 10 days

A does \( \frac{2}{5} \)th part of work in 9 days.

Time taken by A in doing whole

work = \( \frac{9*5}{2} \) = 45 days

Bs 1 days work = \( \frac{1}{10} \)-\( \frac{2}{45} \)

= \( \frac{9-4}{90} \) = \( \frac{5}{90} \) = \( \frac{1}{18} \)

Required time = 18 days

**Ans . **

(1) 20

**Explanation :**(1) Remaining work

= 1-\( \frac{37}{100} \)

= \( \frac{100-37}{100} \) = \( \frac{63}{100} \)

Time taken by (A + B) in doing

\( \frac{63}{100} \) part of work

= 7 days

Time taken by them in doing

whole work = \( \frac{100}{63} \)*7

= \( \frac{100}{9} \) days

Respective ratio of time taken by

A and B in doing the work

= 5:4

\( \frac{1}{4x} \)+\( \frac{1}{5x} \) = \( \frac{9}{100} \)

=> \( \frac{5+4}{20x} \) = \( \frac{9}{100} \)

=> 20x = 100 => x = 5

Required time

= 4*5 = 20 days

**Ans . **

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

**Explanation :**(4) Dhiru digs \( \frac{1}{a} \) part of field

in 20 hours.

Dhiru digs 1 part of field in 20a

hours.

= \( \frac{1}{60} \)-\( \frac{1}{20a} \) = \( \frac{a-3}{60a} \)

Part of field dug by Kaku in 1

hour

= \( \frac{20(a-3)}{60a} \) = \( \frac{a-3}{3a} \)

**Ans . **

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

**Explanation :**(2) A can do a work in 12 days.

B is 60% more efficient than A.

Time taken by B

= [\( \frac{100}{160} \)*12] days

= \( \frac{15}{2} \) = 7\( \frac{1}{2} \) days

**Ans . **

(3) 48 days

**Explanation :**(3) B completes \( \frac{1}{3} \) work in 12

days.

B will complete 1 work in 12*3 = 36 days.

Bs 1 days work = \( \frac{1}{36} \)

(A + B)s 1 days work = \( \frac{1}{24} \)

A’`s 1 days work = \( \frac{1}{24} \)-\( \frac{1}{36} \)

= \( \frac{3-2}{72} \) = \( \frac{1}{72} \)

Time taken by A in doing 1

work = 72 days

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

Time taken by A in doing \( \frac{2}{3} \)

work = \( \frac{2}{3} \)*72 = 48 days

**Ans . **

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

**Explanation :**(3) A does \( \frac{1}{3} \)rd part of work

in 5 days.

A will do 1 work in 5*3

= 15 days.

B does \( \frac{2}{5} \)th of work in 10

days.

B will do 1 work in \( \frac{10*5}{2} \) = 25 days

(A + B)s 1 days work

= \( \frac{1}{15} \)+\( \frac{1}{25} \)

= \( \frac{5+3}{75} \) = \( \frac{8}{75} \)

Required time = \( \frac{75}{8} \)

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

**Ans . **

(4) Rs. 80

**Explanation :**(4) Work done by A and B together

= \( \frac{9}{11} \) parts

Work done by C

= 1-\( \frac{9}{11} \) = \( \frac{2}{11} \) parts

Total amount = Rs. 440

Cs share = Rs. [\( \frac{2}{11} \)*440]

= Rs. 80

**Ans . **

(3) R

**Explanation :**(3) P does \( \frac{1}{4} \)th work in 10

days.

P will do 1 work in 10*4

= 40 days

Q, does 40% part of work in

40 days

Q will do 100% work in

\( \frac{40*100}{40} \) = 100 days

R, does \( \frac{1}{3} \)rd work in 13 days.

R will do 1 work in 13*3

= 39 days

**Ans . **

(3) 40 days

**Explanation :**(3) Let B does the whole work in x days

Work done by B in 1 day = \( \frac{1}{x} \)

According to question A does the \( \frac{1}{2} \) work in \( \frac{x}{6} \) days

A does the whole work in \( \frac{2x}{6} \) or \( \frac{x}{13} \) days Work done by A in one day = \( \frac{3}{x} \) days.

Work done by A and B together in one day = \( \frac{1}{x} \) +\( \frac{3}{x} \) +\( \frac{4}{x} \)

Time taken to complete the whole work by A and B together = \( \frac{0.25}{x} \) days

Again, given that , \( \frac{x}{4} \) =10 => x = 40 days.

**Ans . **

(3) 11 days

**Explanation :**(3) Ratio of efficiency of Babu and Asha = 1 : \( \frac{7}{4} \) = 4 : 7.

As the time taken is inversely proportional to efficiency, therefore, if Babu takes 7x days to complete work, Asha will take 4x days.

\( \frac{1}{7x} \) +\( \frac{1}{4x} \) =\( \frac{1}{7} \) =>\( \frac{4+7}{28x} \) =\( \frac{1}{7} \)

=>28x = 11 * 7

=> x= \( \frac{11*7}{28} \)=\( \frac{11}{4} \)

Asha will complete the work in 4x = 4 *\( \frac{11}{4} \) = 11 days

**Ans . **

(2) 8 days

**Explanation :**(2) Using Rule 1,

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

Jyothi can do 1 job in \( \frac{12*4}{3} \) =16 days.

As Mala is twice as efficient as Jyothi,

Mala will finish the job in 8 days.

**Ans . **

(3) 50 days

**Explanation :**(3) A : B = D

_{2}: D_{1}

=>100 : 140 =D_{2}: 70

=>100 * 70 = 140 * D_{2}

D_{2}= \( \frac{100*70}{140} \) =50 days.

**Ans . **

(4) 8 days

**Explanation :**

**Ans . **

(1) 30 days

**Explanation :**(1) Using basics of Rule 2,

Let B alone can do the work in x days.

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

According to the question, \( \frac{1}{x} \)+ \( \frac{2}{3x} \)= \( \frac{1}{18} \)=> \( \frac{3+2}{3x} \)= \( \frac{1}{18} \)

= >\( \frac{5}{3x} \) = \( \frac{1}{18} \) => 3x= 18* 5

=> x =\( \frac{18*5}{3} \) =30 days.

**Ans . **

(3) 24 days

**Explanation :**(3) Using basics of Rule 2,

According to the question,

If A takes x days to complete the work, B will take 2x days and C will take 4x days, Now, (A + B)'s 1 days work = \( \frac{1}{4} \)

=>\( \frac{1}{x} \)+\( \frac{1}{2x} \)=\( \frac{1}{4} \)=>\( \frac{2+1}{2x} \)=\( \frac{1}{4} \)

=> 2x = 12 => x = 6

C will complete the work in 4x i.e. 24 days.

**Ans . **

(3) 25 days

**Explanation :**(*) Ratio of the work of A and B done in 1 day = 3 : 2

[ B`s work doen = x (let), then A`s work done = \( \frac{x+50}{100} \)x =\( \frac{3}{2} \)x So, (A : B)s work done =\( \frac{3}{2} \)x:x or 3 : 2 ]

Work done by A and B together in 1 day =\( \frac{1}{15} \)

A`s 1 days work =\( \frac{1}{15} \)*\( \frac{3}{5} \) =\( \frac{1}{25} \)

Hence, A alone will finish the work in 25 days.

**Ans . **

(2)18 days

**Explanation :**(2) Using Rule 2,

If Tapas alone takes x days to complete the work, then \( \frac{1}{x} \) +\( \frac{1}{2x} \) =\( \frac{1}{12} \)

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

=>2x = 36 => x = 18 days.

**Ans . **

(3) 20 days

**Explanation :**(3) (A + B)s 1 days work = \( \frac{1}{12} \)....... (i)

(B + C)s 1 days work =\( \frac{1}{15} \).......... (ii)

Difference between A and Cs 1 days work =\( \frac{1}{12} \)-\( \frac{1}{15} \)=\( \frac{5-4}{60} \)=\( \frac{1}{60} \)

If A alone completes the work in x days, C will do the same in 2x days.

\( \frac{1}{x} \)-\( \frac{1}{2x} \)=\( \frac{1}{60} \)

=> \( \frac{2-1}{x} \) =\( \frac{1}{60} \)=>\( \frac{1}{2x} \)=\( \frac{1}{60} \)

=> x = 30

B`s 1 days work = \( \frac{1}{12} \)-\( \frac{1}{30} \) [From equation (i)]

\( \frac{5-2}{60} \) =\( \frac{3}{60} \) =\( \frac{1}{20} \)

Hence, B alone will complete the work in 20 days

**Ans . **

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

**Explanation :**(2) If B alone completes the work in x days, A will do the same in 2x days.

(A + B)s 1 days work

\( \frac{1}{x} \)+\( \frac{1}{2x} \)=\( \frac{2+1}{2x} \)=\( \frac{3}{2x} \)and C`s 1 days work =\( \frac{3}{4x} \)

\( \frac{3}{4x} \)=\( \frac{1}{20} \)

=>4x = 3 * 20

=> x =\( \frac{3*20}{4} \) =15

(A + B + C)s 1days work =\( \frac{1}{2x} \)+\( \frac{1}{x} \)+\( \frac{3}{4x} \)=\( \frac{1}{30} \)+\( \frac{1}{15} \)+\( \frac{1}{20} \)

. \( \frac{2+4+3}{60} \)=\( \frac{9}{60} \)=\( \frac{3}{20} \)

Hence, all three together will complete the work in\( \frac{20}{3} \) or 6\( \frac{2}{3} \)days.

**Ans . **

(2)22\( \frac{1}{2} \) days

**Explanation :**(2) Using Rule 2,

If A completes the work in x days, B will do the same in 3x days.

=>3x - x = 60

=> 2x = 60

=> x = 30 and 3x = 90

(A + B)s 1 days work =\( \frac{1}{30} \)+\( \frac{1}{90} \)=\( \frac{3+1}{90} \)=\( \frac{4}{90} \)=\( \frac{2}{45} \)

A and B together will do the work in\( \frac{45} {2}\) or 22\( \frac{1}{2} \) days

**Ans . **

(2) 6 hours

**Explanation :**(2) A does 20% less work than B.

Ratio of time taken = 5 : 4

A completes a work in \( \frac{15}{2} \) hours

Time taken by B to do the same work =\( \frac{15}{2} \)*\( \frac{4}{5} \)= 6 hours.

**Ans . **

(3) 10 days

**Explanation :**(3) Using Rule 15,

Efficiency and time taken are inversely proportional Bimal : Kamal = 150 : 100 (work)

=> 100 : 150 (Time) = 2 : 3

=> 3 units => 15 days

=> 2 units => \( \frac{15}{3} \)*2=10 days.

**Ans . **

(3) 6 days

**Explanation :**(3) Let time taken by C to complete the work = x days

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

Time taken by A to complete the work = 3x days and time taken by B to complete the work = \( \frac{3x}{2} \)days

According to the question,

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

=>\( \frac{6}{3x} \) =1

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

=>x = 2

Time taken by A= 3x = 3 * 2 = 6 days

**Ans . **

(4) 15 days

**Explanation :**(4) Using Rule 2,

Time taken by A to complete the work = x days

Time taken by B to complete the work = 3x days

So, 3x - x = 2x = 40

=> x = 20 and 3x = 60

=> (A + B)s 1 days work= \( \frac{1}{20} \)+\( \frac{1}{60} \)=\( \frac{3+1}{60} \)=\( \frac{4}{60} \)=\( \frac{1}{15} \)

A and B together will completethe work in 15 days.

**Ans . **

(2) 21 days

**Explanation :**(2) Using Rule 2,

If A completes the work in x days, B will take 2x days.

\( \frac{1}{x} \)+\( \frac{1}{2x} \)=\( \frac{1}{14} \)=>\( \frac{2+1}{2x} \)=\( \frac{1}{14} \)

=> 2x = 42 => x = 21 day

**Ans . **

(3) 15 days

**Explanation :**(3) Time taken by B

\( \frac{21*100}{140} \)=15 days.

**Ans . **

(3) 25

**Explanation :**(3) Using Rule 2,

If the time taken by B to complete the work be x days, then time taken by A = (x-5) days.

\( \frac{1}{x} \)+\( \frac{1}{x-5} \)=\( \frac{9}{100} \)

=>\( \frac{x-5+x}{x(x-5)} \) =\( \frac{9}{100} \)

=>9x^{2}- 45x =200x -500

=>9x^{2}-245x + 500 = 0

=>9x^{2}-225x-20x + 500 = 0

=>9x(x-25)-20 (x-25) = 0

=>(x-25) (9x-20) = 0

=>x = 25 because x !=\( \frac{20}{9} \)

**Ans . **

(1) 30 days

**Explanation :**(1) Using Rule 2,

Let time taken by B in completing the work = x days

Time taken by A = (x-10) days

\( \frac{1}{x} \)+\( \frac{1}{x-10} \)=\( \frac{1}{12} \)

\( \frac{x-10+x}{x(x-10)} \)=\( \frac{1}{12} \)

=>24x-120 =x^{2}- 10x

=>x^{2}-34x + 120 = 0

=>x^{2}-30x-4x+120 = 0

=>x(x-30)-4(x-30) = 0 =>(x-4) (x-30) = 0 =>x = 30 because x is not equal to 4.

**Ans . **

(3) 6 days

**Explanation :**(3) Time taken by B=9*\( \frac{100}{150} \)= 6 days.

**Ans . **

(2)13 days

**Explanation :**(2) Using Rule 2,

Time taken by B=\( \frac{130}{100} \)*23=\( \frac{299}{10} \)days

(A + B)s 1 days work=\( \frac{1}{23} \)+\( \frac{10}{299} \)=\( \frac{13+10}{299} \)=\( \frac{23}{299} \)=\( \frac{1}{13} \)

Time taken by (A + B) = 13 days.

**Ans . **

(2) 2 : 1

**Explanation :**5m + 2w = 4m + 4w

=>m = 2w

=> Required ratio = 2 : 1

**Ans . **

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

**Explanation :**(1) Time taken by B =12*\( \frac{100}{160} \)

=\( \frac{15}{2} \)=7\( \frac{1}{2} \) days

**Ans . **

(4)\( \frac{60}{13} \) days

**Explanation :**(4) Time taken by B in completing the work

=12*\( \frac{100}{160} \)=\( \frac{15}{2} \) days.

(A+B)s 1 days work =\( \frac{1}{12} \)+\( \frac{2}{15} \) =\( \frac{5+8}{60} \) =\( \frac{13}{60} \)

Hence, the work will be completed in \( \frac{60}{13} \) days.

**Ans . **

(3)36 days

**Explanation :**(3) Using Rule 2,

If A alone completes the work in x days, B will complete the same in 2x days.

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

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

=>2x = 36

=>B alone will complete the work in 36 days (i.e. 2x ).

**Ans . **

(1)18 days

**Explanation :**(1) Using Rule 2,

Let time taken by P = x days Then, time taken by Q = 3x days

=> 3x - x = 48 => x = 24 => (P + Q)s 1 days work =\( \frac{1}{24} \)+\( \frac{1}{72} \)=\( \frac{3+1}{72} \)=\( \frac{1}{18} \)

Required time = 18 days.

**Ans . **

(2)24 days

**Explanation :**(2) Using Rule 2 and 3,

If B does the work in 3x days, (A + C) will do the same work in x days.

If C does that work in 2y days.(A + B) will do it in y days.

\( \frac{1}{x} \)+\( \frac{1}{3x} \)=\( \frac{1}{10} \)

=>\( \frac{4}{3x} \)=\( \frac{1}{10} \)

=>3x = 40

=>x=\( \frac{4}{5} \)

Again,\( \frac{1}{y} \)+ \( \frac{1}{2y} \)=\( \frac{1}{10} \)

=> \( \frac{3}{2y} \)=\( \frac{1}{10} \) => y=15

(A + B + C)s 1 days work =\( \frac{1}{10} \)

=>\( \frac{1}{A} \)+\( \frac{1}{40} \)+\( \frac{1}{30} \)=\( \frac{1}{10} \)

=>\( \frac{1}{A} \)= \( \frac{1}{10} \)-\( \frac{1}{40} \)-\( \frac{1}{30} \)

=\( \frac{12-3-4}{120} \)=\( \frac{1}{24} \)

A alone will complete the work in 24 days

**Ans . **

(1)4\( \frac{4}{5} \)days

**Explanation :**(1) Ratio of A's and B's efficiency = 4 : 5

Ratio of time taken = 5 : 4

Time taken by B =\( \frac{6*4}{5} \) =\( \frac{24}{5} \)=4\( \frac{4}{5} \)days.

**Ans . **

(2)6\( \frac{1}{4} \) days

**Explanation :**(2) Using Rule 2,

If A alone does the work in x days and B alone does the work in y days, then

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

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

By equation (ii) * 3 - (i)

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

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

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

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

**Ans . **

(4)1\( \frac{5}{19} \) days

**Explanation :**(4) Using Rule 3,

Time taken by Ramesh

= 4*\( \frac{2}{3} \)=\( \frac{8}{3} \)days.

Work done by all three in 1 day

\( \frac{1}{4} \)+\( \frac{1}{6} \)+\( \frac{3}{8} \)=\( \frac{6+4+9}{24} \)=\( \frac{19}{24} \)

Required time =\( \frac{24}{19} \)=1\( \frac{5}{19} \) days.

**Ans . **

(3)30 days, 90 days

**Explanation :**(3) Time taken by Sonia = 3x days (let)

=>Time taken by Pratibha = x days

=>3x - x = 60 => 2x = 60

=>x = 30 days

=>Time taken by Sonia = 3x days = 3 * 30 = 90 days.

**Ans . **

(1)33 days

**Explanation :**(1) Using Rule 3,

Let time taken by A = x days

Time taken by B = 2x days

Time taken by C = 3x days

According to the question,

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

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

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

=>6x = 6 * 11

=> x=\( \frac{6*11}{6} \)=11

=>Time taken by C alone = 3x= 3 * 11 = 33 days.

**Ans . **

(4)24 days

**Explanation :**(4) A is twice as good as B.

Time taken by A = x days

Time taken by B = 2x days

According to the question,

\( \frac{1}{x} \)+\( \frac{1}{2x} \)=\( \frac{1}{16} \)

=>\( \frac{2+1}{2x} \) =\( \frac{1}{16} \)

=> \( \frac{3}{x} \)=\( \frac{1}{8} \)

=> x = 3 * 8 = 24 days.

**Ans . **

(2) 16

**Explanation :**(2) According to the question,

1 man = 2 boys

3 men + 4 boys

=(3 + 2) men = 5 men

= M_{1}D_{1}= M_{2}D_{2}

=> 5 * D_{1}= 10 * 8

=>D1_{1}= \( \frac{10*8}{5} \) = 16 days.

**Ans . **

(1)4

**Explanation :**(1) A is twice efficient than B.

Time taken by B = 12 days

=>Time taken by A = 6 days

(A + B)s 1 days work =\( \frac{1}{6} \)+\( \frac{1}{12} \)=\( \frac{2+1}{12} \)=\( \frac{1}{4} \)

Required time = 4 days.

**Ans . **

(1)5 days

**Explanation :**(1) In second case, the efficiency of a man is twice to that in the first case

=>M_{1}D_{1}= 2M_{2}D_{2}

=>10 * 20 = 2 * 20 * D_{2}

D_{2}=\( \frac{10*20}{2*20} \) =5 days.

**Ans . **

(2)16

**Explanation :**(2) Time taken by Shashi in doing 1 work = 20 days

Tanya is 25% more efficient than Shashi.

Time taken by Tanya =\( \frac{100}{125} \) * 20= 16 days.

**Ans . **

(2)13 days

**Explanation :**

**Ans . **

(1)30

**Explanation :**

**Ans . **

(1)15

**Explanation :**(1) Using Rule 1,

7 men = 10 women

or 1 man =\( \frac{10}{7} \)women

14 men + 20 women

(\( \frac{10*14}{7} \)+20) women = 40 women

Now, more work, more days More women, less days

Work 1:6 women 40:10 } ::10:x

Where x = number of days

=> 1 * 40 * x = 6 * 10 * 10 or x = \( \frac{600}{40} \)=15

**Ans . **

(1)10 hours

**Explanation :**

**Ans . **

(4)16

**Explanation :**

x=\( \frac{8*8*4}{4*4} \)=16.

**Ans . **

(1) 6 days

**Explanation :**

**Ans . **

(1)3

**Explanation :**

**Ans . **

(4) n^{3}/p^{2}

**Explanation :**4)P men working P hours/ day for P days produce P units of work.

1 man working 1 hour/day for 1 day produce

= p/p^{3}= 1/p^{2}units of work

n men working n hours a da for n days produce n^{3}/p^{2}units of work

**Ans . **

(2)10 days

**Explanation :**

**Ans . **

(3)14

**Explanation :**

**Ans . **

(3)12

**Explanation :**

**Ans . **

(2)20

**Explanation :**(2) Using Rule 1,

Let the original number of carpenters be x.

=>M_{1}D_{1}=M_{2}D_{2}

=>x * 9 = (x-5) * 12

=>9x = 12x-60

=>3x = 60 => x = 20

**Ans . **

(2)16 days

**Explanation :**(2) Using Rule 1,

=>2 men + 3 women = 4 men => 2 men = 3 women \ 3 men + 3 women = 5 men =>M_{1}D_{1}=M_{2}D_{2}

=> 4 * 20 = 5 *D_{2}

=>D_{2}=\( \frac{4*20}{5} \) =16 days.

**Ans . **

(1)12 hours

**Explanation :**

**Ans . **

(4)3 days

**Explanation :**

**Ans . **

(4)8

**Explanation :**

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

=>10x + 10 = 9x + 18

=> x = 18-10 = 8

**Ans . **

(3)8 hrs.

**Explanation :**(3) Using Rule 1,

=>M_{1}D_{1}T_{1}=M_{2}D_{2}T_{2}

=>80*16*6 = 64*15*T_{2}

=>T_{2}=\( \frac{80*16*6}{64*15} \)=8 hours.

**Ans . **

(1)16 days

**Explanation :**(1) Using Rule 1,

=>M_{1}D_{1}=M_{2}D_{2}

=>18 * 24 = 27 * D_{2}

D_{2}=\( \frac{18*24}{27} \)= 16 days.

**Ans . **

(3)43\( \frac{7}{11} \) hours

**Explanation :**

**Ans . **

(3)x^{2}/y days

**Explanation :**(3) Using Rule 1,

=>M_{1}D_{1}= M_{2}D_{2}

=> x.x = y.D_{2}

=>D_{2}= x^{2}/y days.

**Ans . **

(2)15 days

**Explanation :**Using Rule 1,

=>M_{1}D_{1}= M_{2}D_{2}

=> 30 * 18 = 36 * D_{2}

=>D_{2}= \( \frac{30*18}{36} \) = 15 days.

**Ans . **

(1)10 days

**Explanation :**1) 20 men = 24 women

=> 5 men = 6 women

=> 30 men + 12 women

= 40 men

=>M_{1}D_{1}= M_{2}D_{2}

=> 20 * 20 = 40 * D_{2}

D_{2}=\( \frac{20*20}{40} \)= 10 days.

**Ans . **

(3)34 days

**Explanation :**

**Ans . **

(4)20 days

**Explanation :**(4) 3 * 5 men + 7 * 5 women

= 4 * 4 men + 6 * 4 women

=> 16 men - 15 men = 35 women - 24 women

=>1 man = 11 women

=> 3 men + 7 women = 40 women

=>M_{1}D_{1}= M_{2}D_{2}

=> 40 * 5 = 10 * D_{2}

=>D_{2}= 20 days.

**Ans . **

(4)30

**Explanation :**

**Ans . **

(2)100

**Explanation :**(2) Using Rule 1,

200 men do \( \frac{1}{4} \) work in 50 days.

**Ans . **

(2)56

**Explanation :**

**Ans . **

(2)\( \frac{125}{49} \)

**Explanation :**

**Ans . **

(1)300

**Explanation :**(1) Using Rule 1,

=>M_{1}D_{1}=M_{2}D_{2}

=>75 * 90 = M_{2}*18

=>M_{2}=\( \frac{75*90}{18} \)=375

=> Number of additional men = 375-75 = 300.

**Ans . **

(3)5 days

**Explanation :**(*) 4 men = 8 women

=> 1 man = 2 women

=> 6 men + 12 women

=> 12 women + 12 women = 24 women

=>M_{1}D_{1}=M_{2}D_{2}

=> 8 * 15 = 24 * D_{2}

=>D_{2}=\( \frac{8*15}{24} \)= 5days.

**Ans . **

(1)8

**Explanation :**=>M

_{1}D_{1}=M_{2}D_{2}

=>24 * 17 = M_{2}*51

=>M_{2}=\( \frac{24*17}{51} \) =8 men.

**Ans . **

(2)₹60

**Explanation :**2) Ratio of Suman`s and Sumati`s 1 days work = \( \frac{1}{3} \):\( \frac{1}{2} \) =2:3

Sum of the ratios = 2 + 3 = 5

Suman`s share =\( \frac{2}{5} \)*150=₹60.

**Ans . **

(2)₹200.20

**Explanation :**(2) Total wages of 500 workers = 500 × 200 = 100000

Now, according to question,

Correct Average =\( \frac{100000-180-20+80+220}{500} \)

=> \( \frac{100100}{500} \) =₹200.20.

**Ans . **

(3)₹750

**Explanation :**(3) Using Rule 25,

C`s 1 days work =\( \frac{1}{4} \)-(\( \frac{1}{8} \)+\( \frac{1}{12} \))=\( \frac{1}{4} \)-(\( \frac{3+2}{24} \))

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

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

=> C's share =₹(\( \frac{1}{6} \)*4500)=₹750.

**Ans . **

(3)₹9450

**Explanation :**

**Ans . **

(2)₹400

**Explanation :**(2) Using Rule 25,

A`s 1 days work =\( \frac{1}{6} \)

B`s 1 days work =\( \frac{1}{8} \) and (A + B + C)s 1 days work =\( \frac{1}{3} \)

C`s 1days work =\( \frac{1}{3} \)-\( \frac{1}{6} \)-\( \frac{1}{8} \)=\( \frac{8-4-3}{24} \)=\( \frac{1}{24} \)

Ratio of their one day’s work respectively=\( \frac{1}{6} \):\( \frac{1}{8} \):\( \frac{1}{24} \)=4:3:1

Sum of the ratios = 4 + 3 + 1 = 8

=> C`s share =\( \frac{1}{8} \)*3200=₹400.

**Ans . **

(4)₹12,000

**Explanation :**(4) A`s 1 days work =\( \frac{1}{15} \)

B`s 1 days work =\( \frac{1}{10} \)

Ratio =\( \frac{1}{15} \):\( \frac{1}{10} \)=2:3

Sum of the ratios = 2 + 3 = 5

A`s share =₹(\( \frac{2}{5} \)*300000) =₹12,000.

**Ans . **

(4)₹40

**Explanation :**(4) Man : boy = 3 : 1

Boys share = \( \frac{1}{4} \)*800 = 200.

The daily wages of boy=₹(\( \frac{200}{5} \))=₹40.

**Ans . **

(2)₹250

**Explanation :**

**Ans . **

(2)₹600

**Explanation :**(2) Using Rule 25,

Ratio of wages of A, B and C respectively

= 5 * 6 : 6 * 4 : 4 * 9

= 30 : 24 : 36 = 5 : 4 : 6

Amount received by A =\( \frac{5}{5+4+6} \)*1800

=\( \frac{5}{15} \)*1800=₹600.

**Ans . **

(3)4

**Explanation :**(3) Total salary for 20 days= (75 × 20) = 1500

Actual salary received = 1140

Difference = (1500 – 1140) = 360

Money deducted for 1 day’s absencefrom work= (15 + 75) = 90

Number of days he was absent =\( \frac{360}{90} \)=4 days.

**Ans . **

(3)₹275

**Explanation :**(3) Using Rule 25,

First mans 1 days work =\( \frac{1}{7} \)

Second mans 1 days work =\( \frac{1}{8} \)

Let, Boys 1 days work =\( \frac{1}{x} \)

=>\( \frac{1}{7} \)+\( \frac{1}{8} \)+\( \frac{1}{x} \)=\( \frac{1}{3} \)

=>\( \frac{1}{x} \) = \( \frac{1}{3} \)-\( \frac{1}{7} \)-\( \frac{1}{8} \)

=>\( \frac{56-24-21}{168} \)=\( \frac{11}{168} \)

=> Ratio of their one days work=\( \frac{1}{7} \):\( \frac{1}{8} \):\( \frac{11}{168} \)=24:21:11

=>Sum of the ratios = 24 + 21 + 11= 56

=> Boy`s share in wages =\( \frac{11}{56} \)*1400 =₹275.

**Ans . **

(4)₹17,100

**Explanation :**(4) 5 men = 7 women

[Both earn same amount in 1 day]

7 men =\( \frac{7}{5} \)*7 =\( \frac{49}{5} \) women

7 men + 13 women

=>\( \frac{49}{5} \)+13 =\( \frac{114}{5} \) women

Now,7 women = 5250 => \( \frac{114}{5} \) women

=>\( \frac{5250}{7} \)*\( \frac{114}{5} \)=₹17,100.

**Ans . **

(1)₹400

**Explanation :**(1) According to the question,

(2 × 14) men + 14 women

= 16 men + 32 women

=>(28 – 16) men =(32–14) women

=> 12 men = 18 women

=> 2 men = 3 women

1 woman =\( \frac{2}{3} \)man

Amount received by 1 woman per day= \( \frac{2}{3} \)*600=₹400.

**Ans . **

(3)₹160

**Explanation :**(3) Using Rule 25,

Work done by the third person in 1 day=\( \frac{1}{8} \)-\( \frac{1}{16} \)-\( \frac{1}{24} \)=\( \frac{6-3-2}{48} \)=\( \frac{1}{48} \)

Ratio of their 1 day’s work=\( \frac{1}{16} \):\( \frac{1}{24} \):\( \frac{1}{48} \)=3:2:1

Share of the third person =\( \frac{1}{3+2+1} \)*960=\( \frac{960}{6} \)

=>₹160.

**Ans . **

(1)6:5

**Explanation :**(1) Using Rule 25,

Required ratio = 15 * 22 : 11 * 25 = 6 : 5

**Ans . **

(4)₹12000

**Explanation :**(4) Expert`s 1 days work=\( \frac{1}{12} \)-\( \frac{1}{36} \)-\( \frac{1}{48} \)

=>\( \frac{12-4-3}{144} \)=\( \frac{5}{144} \)

Ratio of their respective work for 1 day =\( \frac{1}{36} \):\( \frac{1}{48} \):\( \frac{5}{144} \)=4:3:5

Experts share =\( \frac{5}{12} \)*28800=₹12000.

**Ans . **

(1)₹300

**Explanation :**(1) Using Rule 25,

According to the question

\( \frac{1}{15} \) +\( \frac{1}{12} \) +\( \frac{1}{C} \) =\( \frac{1}{5} \)

Let C1s work in day be \( \frac{1}{C} \)

\( \frac{1}{C} \)=\( \frac{1}{5} \)-\( \frac{1}{15} \)-\( \frac{1}{12} \)=\( \frac{12-4-5}{60} \)=\( \frac{1}{20} \)

=>A:B:C=\( \frac{1}{15} \):\( \frac{1}{12} \):\( \frac{1}{20} \)=4:5:3

=> C`s share =\( \frac{3}{12} \)*1200 = ₹300.

**Ans . **

(4)12 days

**Explanation :**(4) A`s 1 days work = \( \frac{1}{21} \)

B`s 1 days work = \( \frac{1}{28} \)

Total work done by both=\( \frac{1}{21} \)+\( \frac{1}{28} \)=\( \frac{4+3}{84} \)=\( \frac{1}{12} \)

Amount is sufficient to pay 12 days wages of both.

**Ans . **

(4)₹400

**Explanation :**(4) Rule 2 and Rule 25,

Work done by A and B in 5 days = 5(\( \frac{1}{12} \)+\( \frac{1}{15} \))

=5(\( \frac{5+4}{60} \))=\( \frac{9}{12} \)=\( \frac{3}{4} \)

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

C will complete in 20 days.

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

=>5:4:3

Amount received by A= \( \frac{5}{12} \)*960 = ₹400

**Ans . **

(2)₹20

**Explanation :**(2) The daily earning of 'C' = Daily earning of (A + C) and (B + C) - Daily earning of (A + B + C) = 94 + 76 - 150 = 20

**Ans . **

(3)₹225

**Explanation :**(3) Rule 3 and Rule 25,

If the fourth person completes the work in x days, then

\( \frac{3}{8} \)+\( \frac{3}{12} \)+\( \frac{3}{16} \)+\( \frac{3}{x} \)=1

=>\( \frac{1}{x} \) = \( \frac{1}{3} \)-\( \frac{1}{8} \)-\( \frac{1}{12} \)-\( \frac{1}{16} \)

=>\( \frac{16-6-4-3}{48} \) =>\( \frac{1}{16} \)

x = 16

Ratio of wages =\( \frac{1}{8} \):\( \frac{1}{12} \):\( \frac{1}{16} \):\( \frac{1}{16} \)=6:4:3:3

Sum of ratios = 6 + 4 + 3+3 = 16

Fourth person`s share =\( \frac{3}{16} \)*1200=₹225.

**Ans . **

(1) A : 150, B : 100, C : 150

**Explanation :**(1) Rule 3 and Rule 25,

If C alone completes the work in x days, then

\( \frac{1}{16} \)+\( \frac{1}{24} \)+\( \frac{1}{x} \)=\( \frac{1}{6} \)

=> \( \frac{1}{x} \) = \( \frac{1}{6} \)-\( \frac{1}{16} \)-\( \frac{1}{24} \)

=>\( \frac{8-3-2}{48} \)=\( \frac{1}{16} \)

=>x = 16 days

Ratio of their remuneration =\( \frac{1}{16} \):\( \frac{1}{24} \):\( \frac{1}{16} \)= 3:2:2

A`s remuneration =\( \frac{3}{8} \)*400 = ₹150

B`s remuneration =\( \frac{2}{8} \)*400 =₹100

C`s remuneration =\( \frac{3}{8} \)*400 =₹150

=> A : 150, B : 100, C : 150

**Ans . **

(4)143.50~~
~~

**Explanation :**(4) Using Rule 25,

Skilled : half skilled : unskilled =\( \frac{1}{3} \):\( \frac{1}{4} \):\( \frac{1}{6} \)

=>(\( \frac{1}{3} \)*12) :(\( \frac{1}{4} \)*12):(\( \frac{1}{6} \)*12)

= 4 : 3 : 2 Share of the trained labourer =\( \frac{28}{(7*4+8*3+2*10)} \) *369 =\( \frac{28}{28+24+20} \)*369

=>\( \frac{28}{72} \)*369 =143.50.

**Ans . **

(2)₹100

**Explanation :**(2) Work done by B= 1-\( \frac{19}{23} \) = \( \frac{23-19}{23} \) = \( \frac{4}{23} \)

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

Sum of ratios = 19 + 4 = 23

B`s share =\( \frac{4}{23} \)*575 =₹100

**Ans . **

(4)5 hours

**Explanation :**Rate of earning of the man = 2000/50 = Rs. 40 per hour

Rate of earning for additional hours = 40 × 3/2 = Rs. 60 per hour

Let the man has to work for n additional hours.

Then, 2000 + n × 60 = 2300

⇒ n × 60 = 300

⇒ n = 5h.

**Ans . **

(3)Rs.120

**Explanation :**(3) (2 men + 1 woman)s 14 days work = (4 women + 2 men)s 8 days work

=> 28 men + 14 women

=> 32 women + 16 men

=> (28 - 16) = 12 men

=> (32 - 14) = 18 women

=> 2 men = 3 women

1 woman = \( \frac{2}{3} \) man

=> Wages per day of 1 man = Rs. 180

=> Wages per day of 1 woman\( \frac{2}{3} \)*180 =Rs.120.

**Ans . **

(1)Rs. 67.50

**Explanation :**(1) Time taken by A =\( \frac{63}{3.50} \)= 18 days

Time taken by B =\( \frac{75}{2.5} \) =30 days.

(A + B)s 1 days work =\( \frac{1}{18} \)+\( \frac{1}{30} \) = \( \frac{5+3}{90} \) =\( \frac{8}{90} \) = \( \frac{4}{45} \)

Required time =\( \frac{45}{4}\) days

Total wages =\( \frac{45}{4}\) * (3.50 + 2.50)

=>Rs(\( \frac{45}{4}\)*6) = )Rs. 67.5.

**Ans . **

(3) Rs.250

**Explanation :**(3) Ratio of A`s and B`s 1 days work =\( \frac{1}{12} \):\( \frac{1}{15} \) = 15:12 => 5:4

Sum of the terms of ratio = 5 + 4 = 9

A`s share = Rs.(\( \frac{5}{9} \)*450) = Rs.250.

**Ans . **

(1)Rs. 200

**Explanation :**(1) Part of work done by C

=> 1- \( \frac{7}{11} \)-\( \frac{4}{11} \)

=> Total amount received = Rs. 550

=> C`s share = Rs(\( \frac{4}{11} \)*550) =Rs. 200.

**Ans . **

(1)Rs.50

**Explanation :**(1) Let C alone complete the work in x days.

According to the question,

\( \frac{1}{5} \)+\( \frac{1}{15} \)+\( \frac{1}{x} \)=\( \frac{1}{3} \)

=>\( \frac{1}{x} \) =\( \frac{1}{3} \)-\( \frac{1}{5} \)-\( \frac{1}{15} \)

=>\( \frac{5-3-1}{15} \) =>\( \frac{1}{15} \)

\ => x = 15 days = Time taken by C alone.

Ratio of the 1 day’s work of A, B and C =\( \frac{1}{5} \):\( \frac{1}{15} \):\( \frac{1}{15} \) => 3:1:1

Sum of the terms of ratio = 3 + 1 + 1 = 5

C’s share = Rs(\( \frac{1}{5} \)*250) => Rs.50

**Ans . **

(1) Man ₹2.75, Woman ₹2.25

**Explanation :**(1) Let daily wages of a man be Rs. x

Daily wages of a woman = Rs.(x-\( \frac{1}{2} \))

According to the question, 600x + 400(x-\( \frac{1}{2} \))

= 1000 × 2.55

=> 600x + 400x - 200 = 2550

=> 1000x = 2550 + 200 = 2750

=> x=\( \frac{2750}{1000} \) =Rs. 2.75

=> Daily wages of a woman

=>Rs. (2.75 – 0.5)

=>Rs. 2.25

**Ans . **

(1) 10

**Explanation :**(1) Let initially the number of men

be x.

=> According to question,

M_{1}D_{1}W_{2}= M_{2}D_{2}W_{1}

x*30 = (x + 5)*(30 - 10)

x*30 = 20x + 100

30x - 20x = 100

10x = 100

x = 10

**Ans . **

(4) 8

**Explanation :**(4) Using Rule 1,

=> 450*10*5*x

= 625*8*6*6

=> x = \( \frac{625*8*6*6}{450*10*5} \) = 8

**Ans . **

(1) 24 days

**Explanation :**(1) Work done by A in 15 days

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

Remaining work = 1-\( \frac{1}{4} \) = \( \frac{3}{4} \)

Now, \( \frac{3}{4} \) work is done by B in 30

days

Whole work will be done by B in

\( \frac{30*4}{3} \) = 40 days

As 1 days work = \( \frac{1}{60} \) and Bs 1

days work = \( \frac{1}{40} \)

(A + B)s 1 days work

= \( \frac{1}{60} \)+\( \frac{1}{40} \) = \( \frac{2+3}{120} \) = \( \frac{5}{120} \) = \( \frac{1}{24} \)

Hence, both will finish the work in 24 days.

**Ans . **

(2) 25 days

**Explanation :**(2) As 1 days work

= (B +C)s 1 days work ...(i)

(A + B)s 1 days work = \( \frac{1}{10} \)

Cs 1 days work = \( \frac{1}{50} \)

(A + B + C)s 1 days work

= \( \frac{1}{10} \)+\( \frac{1}{50} \) = \( \frac{5+1}{50} \) = \( \frac{6}{50} \) = \( \frac{3}{25} \) ...(iii)

(A + A)s 1 days work = \( \frac{3}{25} \)

(By (i) & (iii)

As 1 days work = \( \frac{3}{50} \)

Bs 1 days work = \( \frac{1}{10} \)-\( \frac{3}{50} \)

= \( \frac{5-3}{50} \) = \( \frac{2}{50} \) = \( \frac{1}{25} \)

Hence, B alone will complete the

work in 25 days

**Ans . **

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

**Explanation :**(3) Using Rule 2,

Let the son take x days to do the

work.

\( \frac{1}{5} \)+\( \frac{1}{x} \) = \( \frac{1}{3} \)

=> \( \frac{x+5}{5x} \) = \( \frac{1}{2} \)

=> 3x + 15 = 5x

=> 2x = 15 => x = \( \frac{15}{2} \) = 7\( \frac{1}{2} \) days

**Ans . **

(4) 40

**Explanation :**(4) Let the number of men in the

beginning = x

Then, \( \frac{x+8}{x} \) = \( \frac{60}{50} \)

=> \( \frac{x+8}{x} \) = \( \frac{6}{5} \)

=> 6x = 5x + 40 => x = 40

**Ans . **

(1) 192

**Explanation :**(1) 12 persons can complete a

work in 4 days.

=> 24 persons can complete the work in 2 days.

=> 24 persons can complete the 8 times work in 16 days

=> 24*8 persons = 192 persons can complete the 8 times work in 2 days.

**Ans . **

(2) 110

**Explanation :**(2) Let the original number of

workers = x. Then,

x*100 = (x -10)*110

=> 10x = 11x - 110

=> x = 110

**Ans . **

(3) 12 days

**Explanation :**(3) Work done by 12 men in 6

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

Remaining work

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

6 men leave the work.

Time taken = \( \frac{12*12}{6*2} \) = 12 days

**Ans . **

(2) 15

**Explanation :**(2) Using Rule 1,

60 men can complete a work in

250 days.

Work done by 60 men in 1 day

= \( \frac{1}{250} \)

=> Work done by 60 men in 200

days = \( \frac{200}{250} \) = \( \frac{4}{5} \)

Remaining work = 1-\( \frac{4}{5} \) = \( \frac{1}{5} \)

Work is stopped for 10 days.

**Ans . **

(1) 3 days

**Explanation :**(1) Using Rule 2,

Working 5 hours a day, A can complete a work in 8 days.

i.e. A can complete the work in 40 hours.

Similarly, B will complete the same work in 60 hours.

(A + B)s 1 hours work

= \( \frac{1}{40} \)+\( \frac{1}{60} \) = \( \frac{3+2}{120} \)

= \( \frac{5}{120} \) = \( \frac{1}{24} \)

Hence, A and B together will complete the work in 24 hours.

They can complete the work in 3 days working 8 hours a day.

**Ans . **

(4) 2 days

**Explanation :**(4) According to the question,

2 persons with equal abilities can do 1 job in 1 day

Time taken by 1 man to complete 1 job = 2 days

=> Time taken by 100 persons in completing 100 jobs = 2 days

**Ans . **

(2) 6.30 p.m.

**Explanation :**(2) Part of the field mowed by

Ganga and Saraswati in first 2

hours

= \( \frac{1}{8} \)+\( \frac{1}{12} \) = \( \frac{3+2}{24} \) = \( \frac{5}{24} \)

Part of the field mowed in first

8 hours = \( \frac{5*4}{24} \) = \( \frac{20}{24} \) = \( \frac{5}{6} \)

Remaining work = 1 -\( \frac{5}{6} \) = \( \frac{1}{6} \)

Now, it is the turn of Ganga, part

of work done by Ganga in 1 hour = \( \frac{1}{8} \)

Remaining work = \( \frac{1}{6} \)-\( \frac{1}{8} \) = \( \frac{1}{24} \)

Now, time taken by Saraswati in

completing this part of work

= \( \frac{1}{24} \)*12 = \( \frac{1}{2} \) hour

Total time = 9*\( \frac{1}{2} \) hour

The mowing starts at 9 am.

Hence, the mowing will be completed

at 6.30 pm.

**Ans . **

(3) 200

**Explanation :**(3) Using Rule 1,

Remaining work

= 5-\( \frac{7}{2} \) = \( \frac{3}{2} \)

M_{1}D_{1}W_{2}= M_{2}D_{2}W_{1}

=> 280*80*\( \frac{3}{2} \) = M_{2}*20*\( \frac{7}{2} \)

=> M_{2}= \( \frac{280*80*30}{20*7} \) = 480

Required number of additionalmen = 480 - 280 = 200

**Ans . **

(1) 6 days

**Explanation :**(1) Let B alone do the work in x

days.

6*\( \frac{1}{12} \) + 3*\( \frac{1}{x} \) = 1

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

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

**Ans . **

(4) 4:3

**Explanation :**(4) Using Rule 15,

Efficiency and time taken are inversely

proportional.

Required ratio = 4:3

**Ans . **

(3) 75

**Explanation :**(3) Scheduled time to complete the

work = 40 days

25 men in 24 days do \( \frac{1}{3} \) work

1 man in 1 day does \( \frac{1}{3*25*24} \) = \( \frac{1}{1800} \) work

Work remaining = 1-\( \frac{1}{3} \) = \( \frac{2}{3} \)

The work is to be completed 4

days before schedule i.e.,

in (40 - 4) = 36 days

No. of days left for \( \frac{2}{3} \)rd work

= 36 - 24 = 12 days

\( \frac{1}{1800} \) work is done in 1 day by

1 man.

\( \frac{2}{3} \)rd work will be done in 12 days

by

1800*\( \frac{2}{3} \)*\( \frac{1}{12} \) = 100 men

25 men are already working

Extra men to be employed

= 100 - 25 = 75

**Ans . **

(2) 4:3

**Explanation :**(2) 20*16 women

= 16*15 men

=> 4 women = 3 men

=> \( \frac{men}{women} \) = \( \frac{4}{3} \)

Hence, working capacity of man :

woman = 4:3

**Ans . **

(1) 45 days

**Explanation :**(1) Man : Woman (efficiency)

= 3:2

i.e., Woman completes \( \frac{2}{5} \)

th work in 18 days.

Time taken by the woman to

complete the whole work = \( \frac{18*5}{2} \) = 45 days

**Ans . **

(1) 3y:2x

**Explanation :**(1) 1 mans 1 days work = \( \frac{1}{2x} \)

1 womans 1 days work = \( \frac{1}{3y} \)

Required ratio = \( \frac{1}{2x} \) \( \frac{1}{3y} \)

= 3y:2x

**Ans . **

(2) 9 hrs

**Explanation :**(2) Using Rule 1,

D_{1}T_{1}= D_{}T_{2}

=> 18*6 = 12*T_{2}

=> T_{2}= \( \frac{18*6}{12} \) = 9 hrs

**Ans . **

(3) 138

**Explanation :**(3) Using Rule 1,

=> 12*6*240*x = 18*8*36*460

=> x = \( \frac{18*8*36*460}{12*6*240} \) = 138

**Ans . **

(3) 30

**Explanation :**(3) Using Rule 1,

=> 18*2*12*6*8x = 32*3*9*9*10*8

=> x = \( \frac{32*3*9*9*10*8}{18*2*12*6*8} \) = 30 days

**Ans . **

(2) 10

**Explanation :**(2) (P + Q)s 1 days work = \( \frac{1}{6} \)

(Q + R)s 1 days work = \( \frac{7}{60} \)

Let P alone do the work in x

days.

According to the question,

\( \frac{3}{x} \)+\( \frac{6*7}{60} \) = 1

=> \( \frac{3}{x} \) = 1-\( \frac{7}{10} \) = \( \frac{3}{10} \)

=> x = 10 days

Qs 1 days work = \( \frac{1}{6} \)-\( \frac{1}{10} \) = \( \frac{1}{15} \)

Rs 1 days work = \( \frac{7}{60} \)-\( \frac{1}{15} \) = \( \frac{1}{20} \)

Time taken by R = 20 days

Required answer = 20 - 10

= 10 days

**Ans . **

(3) 25

**Explanation :**(3) Let 150 workers complete the

work in x days.

150*x = 150 + 146 + .... to (x + 8) terms

On putting x = 17

LHS = 150*17 = 2550

RHS = 150 + 146 + .... to 25 terms

a = 150, d = - 4, n = 25

S = \( \frac{n}{2} \)*[2a+(n-1)d]

= \( \frac{25}{2} \)*[2*150 + 24*(-4)]

= \( \frac{25}{2} \)*(300-96) = 2550

Note : It is better to solve by options.

**Ans . **

(1) 20

**Explanation :**(1) Using Rule 1,

According to the question,

M_{1}D_{1}= M_{2}D_{2}

=>(x + 4)*(x + 5)

= (x - 5)*(x + 20)

=> x^{2}+ 5x + 4x + 20

= x^{2}- 5x + 20x - 100

=> 9x + 20 = 15x - 100

=> 15x - 9x = 100 + 20

=> 6x = 120 => x = 20

**Ans . **

(3) 10 days

**Explanation :**(3) Let the work be finished in x

days.

\( \frac{x}{50} \)+\( \frac{x-1}{50} \)+\( \frac{x-2}{50} \)+ ... + \( \frac{1}{50} \) = 1

=> x + x - 1 + x - 2 + .... + 1 = 50

i.e., 10 + 9 + 8 + .... + 1

= 55

9 + 8 + .... + 1 = 45

Required time = 10 days

**Ans . **

(3) 10

**Explanation :**(3)

=> 48*7*x = 20*21*8

=> x = \( \frac{20*21*8}{48*7} \) => x = 10

**Ans . **

(2) 6\( \frac{10}{33} \) days

**Explanation :**(2) Area of the four walls and ceiling

of the room

= 2h (l + b) + lb

= 2*10 (25 + 12) + 25*12

= (20*37 + 300) sq. metre

= (740 + 300) sq. metre

= 1040 sq. metre

Area painted by A in 1 day

= \( \frac{250}{2} \) = 125 sq. metre

Area painted by both in1 day

= (125 + 40) sq. metre

= 165 sq. metre

Required time = \( \frac{1040}{165} \) = \( \frac{208}{33} \) = 6\( \frac{10}{33} \) days

**Ans . **

(1) 54

**Explanation :**(1) Here, the length of wall is

same in both cases.

M_{1}D_{1}= M_{2}D_{2}

=> 36*21 = M_{2}*14

=> M_{2}= \( \frac{36*21}{14} \) = 54 days

**Ans . **

(2) 488 kg.

**Explanation :**(2) Number of days in April and

May = 30 + 31 = 61

Q Requirement of rice for 7 days = 56 kg.

Requirement of rice for 61 days

= \( \frac{56}{7} \)*61 = 488 kg.

**Ans . **

(1) 40 minutes

**Explanation :**(1) Total working time of school

= (45*8) minutes

= 360 minutes

If 9 periods are held per day,

Working time of each period

\( \frac{360}{9} \) = 40 minutes

**Ans . **

(3) 7

**Explanation :**

**Ans . **

(2) 45.

**Explanation :**

**Ans . **

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

**Explanation :**

**Ans . **

(4) 8 days

**Explanation :**

**Ans . **

(2)120 days

**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

**Ans . **

(1)42

**Explanation :**(1) Using Rule 1,

**Ans . **

(2)RS.163.04

**Explanation :**(2) Using Rule 1,

Amount received by Meeta =\( \frac{6}{23} \)*625 = Rs. 163.04

**Ans . **

(1)35

**Explanation :**

**Ans . **

(1)4 \( \frac{4}{5} \)day

**Explanation :**(1) A`s one days work =\( \frac{1}{12} \)

B`s one days work =\( \frac{1}{8} \)

(A + B)s one days work=\( \frac{1}{12} \)+\( \frac{1}{8} \) =\( \frac{2+3}{24} \) =\( \frac{5}{24} \)

Now,\( \frac{5}{24} \) work is done in 1day

1 work is done in = \( \frac{24}{5} \)days =4 \( \frac{4}{5} \) days.

**Ans . **

(2)9 days

**Explanation :**

**Ans . **

(3)12 days

**Explanation :**

**Ans . **

(4)30 days

**Explanation :**

**Ans . **

(1)36 days

**Explanation :**

**Ans . **

(2)15 days

**Explanation :**

**Ans . **

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

**Explanation :**

**Ans . **

(4)9 days

**Explanation :**

**Ans . **

(1)40 days

**Explanation :**

**Ans . **

(2)24 days

**Explanation :**

**Ans . **

(3)18 days

**Explanation :**

**Ans . **

(4)15 days

**Explanation :**

**Ans . **

(1)20 days

**Explanation :**

**Ans . **

(2)16.5 days

**Explanation :**

**Ans . **

(3)240 days

**Explanation :**