Time and Work
- If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do half of the work in
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40 men ≡ 60 women ≡ 80 children
∴ 10 men ≡ 80 × 10 = 20 children 40 ∴ 10 women ≡ 80 × 10 60 = 40 children 3
∴ 10 men + 10 women + 10 children= 
20 + 40 + 10 
children 3 = 60 + 40 + 30 children 3 = 130 children 3 ∴ M1D1 = M2D2 W1 W2 D2 = 80 × 6 × 13 = 144 months 130 13
∴ Half of the work can do= 144 × 1 13 2 = 72 = 5 7 months 13 13
Aliter : Using Rule 13,
Here, A = 40, B= 60, C = 80, a = 6
A1 = 10, B1 = 10, C1 = 10
Time taken= a A1 + B1 + C1 A B C = 6 10 + 10 + 10 40 60 80 = 6 1 + 1 + 1 4 6 8 = 6 6 + 4 + 3 24 = 144 13
Half of the work they do in= 1 × 144 months 2 13 = 72 = 5 7 months 13 13 Correct Option: C
40 men ≡ 60 women ≡ 80 children
∴ 10 men ≡ 80 × 10 = 20 children 40 ∴ 10 women ≡ 80 × 10 60 = 40 children 3
∴ 10 men + 10 women + 10 children= 20 + 40 + 10 children 3 = 60 + 40 + 30 children 3 = 130 children 3 ∴ M1D1 = M2D2 W1 W2 D2 = 80 × 6 × 13 = 144 months 130 13
∴ Half of the work can do= 144 × 1 13 2 = 72 = 5 7 months 13 13
Aliter : Using Rule 13,
Here, A = 40, B= 60, C = 80, a = 6
A1 = 10, B1 = 10, C1 = 10
Time taken= a A1 + B1 + C1 A B C = 6 10 + 10 + 10 40 60 80 = 6 1 + 1 + 1 4 6 8 = 6 6 + 4 + 3 24 = 144 13
Half of the work they do in= 1 × 144 months 2 13 = 72 = 5 7 months 13 13
- A man, a woman and a boy together finish a piece of work in 6 days. If a man and a woman can do the work in 10 and 24 days respectively. The days taken by a boy to finish the work is
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Time taken by boy = x days
∴ 1 + 1 + 1 = 1 10 24 x 6 ⇒ 1 = 1 - 1 - 1 x 6 10 24 = 20 - 12 - 5 = 3 = 1 ⇒ x = 40 days 120 120 40
Aliter : Using Rule 18,
Here , x = 6, y = 10, z = 24Number of days = xyz days yz - x(y + z) = 6 × 10 × 24 = 1440 = 1440 = 40 days 10 × 24 - 6(10 + 24) 240 - 204 36 Correct Option: C
Time taken by boy = x days
∴ 1 + 1 + 1 = 1 10 24 x 6 ⇒ 1 = 1 - 1 - 1 x 6 10 24 = 20 - 12 - 5 = 3 = 1 ⇒ x = 40 days 120 120 40
Aliter : Using Rule 18,
Here , x = 6, y = 10, z = 24Number of days = xyz days yz - x(y + z) = 6 × 10 × 24 = 1440 = 1440 = 40 days 10 × 24 - 6(10 + 24) 240 - 204 36
- P and Q together can do a job in 6 days. Q and R can finish the same job in 60/7 days. P started the work and worked for 3 days. Q and R continued for 6 days. Then the difference of days in which R and P can complete the job is
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(P + Q)’s 1 day’s work = 1 6 (Q + R)’s 1 day’s work = 7 60
Let P alone do the work in x days.
According to the question.= 3 + 6 × 7 = 1 x 60 ⇒ 3 = 1 - 7 = 3 x 10 10
⇒ x = 10 days
∴ Q’s 1 day’s work⇒ 1 - 1 = 5 - 3 = 1 6 10 30 15
R’s 1 day’s work⇒ 7 - 1 = 7 - 4 = 1 60 15 60 20
∴ Time taken by R = 20 days
∴ Required answer = 20 – 10 = 10 daysCorrect Option: B
(P + Q)’s 1 day’s work = 1 6 (Q + R)’s 1 day’s work = 7 60
Let P alone do the work in x days.
According to the question.= 3 + 6 × 7 = 1 x 60 ⇒ 3 = 1 - 7 = 3 x 10 10
⇒ x = 10 days
∴ Q’s 1 day’s work⇒ 1 - 1 = 5 - 3 = 1 6 10 30 15
R’s 1 day’s work⇒ 7 - 1 = 7 - 4 = 1 60 15 60 20
∴ Time taken by R = 20 days
∴ Required answer = 20 – 10 = 10 days
- A and B can do a piece of work in 10 days, B and C in 15 days and C and A in 20 days. C alone can do the work in :
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According to the question
Work done by A and B together in one day= 1 part 10
Work done by B and C together in one day= 1 part 15
Work done by C and A together in one day= 1 part 20
So,A + B = 1 .........(i) 10 B + C = 1 .........(ii) 15 C + A = 1 .........(iii) 20
Adding I, II, III, we get2 (A + B + C) = 1 + 1 + 1 10 15 20 2 (A + B + C) = 6 + 4 + 3 = 13 60 60 A + B + C = 13 .........(iv) 120
Putting the value of eqn. (I) in eqn. (IV)1 + C = 13 10 120 C = 13 - 1 = 13 - 12 = 1 120 10 120 120
∴ Work done in 1 day by C is1 part 120
Hence, C will finish the whole work in 120 days
Aliter : Using Rule 19,
C alone can do the work in= 2xyz xy - yz + zx = 2 × 10 × 15 × 20 10 × 15 - 15 × 20 + 20 × 10 = 6000 150 - 300 + 200 = 6000 = 120 days 50 Correct Option: B
According to the question
Work done by A and B together in one day= 1 part 10
Work done by B and C together in one day= 1 part 15
Work done by C and A together in one day= 1 part 20
So,A + B = 1 .........(i) 10 B + C = 1 .........(ii) 15 C + A = 1 .........(iii) 20
Adding I, II, III, we get2 (A + B + C) = 1 + 1 + 1 10 15 20 2 (A + B + C) = 6 + 4 + 3 = 13 60 60 A + B + C = 13 .........(iv) 120
Putting the value of eqn. (I) in eqn. (IV)1 + C = 13 10 120 C = 13 - 1 = 13 - 12 = 1 120 10 120 120
∴ Work done in 1 day by C is1 part 120
Hence, C will finish the whole work in 120 days
Aliter : Using Rule 19,
C alone can do the work in= 2xyz xy - yz + zx = 2 × 10 × 15 × 20 10 × 15 - 15 × 20 + 20 × 10 = 6000 150 - 300 + 200 = 6000 = 120 days 50
- A can do a work in 12 days and B can do it in 16 days. A and B started the work jointly and A left 2 days before the work is finished. Find the number of days they took to finish the work.
(1) (2)
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Let the required number of days be x.
∴ A’s work in (x – 2) days= x- 2 12 B’s work in x days = x 16 = x - 2 + x = 1 12 16 ⇒ 4x - 8 + 3x = 1 48
⇒ 7x = 48 + 8 = 56
⇒ x = 8 daysCorrect Option: D
Let the required number of days be x.
∴ A’s work in (x – 2) days= x- 2 12 B’s work in x days = x 16 = x - 2 + x = 1 12 16 ⇒ 4x - 8 + 3x = 1 48
⇒ 7x = 48 + 8 = 56
⇒ x = 8 days
