**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.