Time and Work
- A can do a job in 10 days and B can do the same job in 15 days. They start working together, but B leaves after 5 days. How many more days A want to finish the work ?
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Work done by A and B in 5 days
⇒ 5 
1 + 1 
= 5 3 + 2 10 15 30 = 5 × 5 = 5 30 6 Remaining work = 1 – 5 = 1 6 6
∴ Time taken by A= 1 × 10 = 5 days = 1 2 days 6 3 3 Correct Option: B
Work done by A and B in 5 days
⇒ 5 1 + 1 = 5 3 + 2 10 15 30 = 5 × 5 = 5 30 6 Remaining work = 1 – 5 = 1 6 6
∴ Time taken by A= 1 × 10 = 5 days = 1 2 days 6 3 3
- 12 men can complete a work in 90 days. 30 days after they started work, 2 men left and 8 men joined. How many days will it take to complete the remaining work ?
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Work done in 30 days = W2
∴ M1D1 = M2D2 W1 W2 ⇒ 12 × 90 = 12 × 30 1 W2 ⇒ W2 = 12 × 30 = 1 12 × 90 3 Remaining work = 1 – 1 = 2 3 3
New number of men = 18∴ M1D1 = M2D2 1 W ⇒ 12 × 90 = 18 × D2 1 2/3 ⇒ 18 × D2 = 12 × 90 × 2 = 12 × 60 3 ⇒ D2 = 12 × 60 = 40 days 18 Correct Option: C
Work done in 30 days = W2
∴ M1D1 = M2D2 W1 W2 ⇒ 12 × 90 = 12 × 30 1 W2 ⇒ W2 = 12 × 30 = 1 12 × 90 3 Remaining work = 1 – 1 = 2 3 3
New number of men = 18∴ M1D1 = M2D2 1 W ⇒ 12 × 90 = 18 × D2 1 2/3 ⇒ 18 × D2 = 12 × 90 × 2 = 12 × 60 3 ⇒ D2 = 12 × 60 = 40 days 18
- X can do a piece of work in 24 days. When he had worked for 4 days, Y joined him. If complete work was finished in 16 days, Y can alone finish that work in:
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Let Y alone complete the work in x days.
According to the question,
X’s 16 days’ work + Y’s 12 days’ work = 1⇒ 16 + 12 = 1 24 x ⇒ 2 + 12 = 1 3 x ⇒ 12 = 1 - 2 = 1 x 3 3
⇒ x = 12 × 3 = 36 daysCorrect Option: C
Let Y alone complete the work in x days.
According to the question,
X’s 16 days’ work + Y’s 12 days’ work = 1⇒ 16 + 12 = 1 24 x ⇒ 2 + 12 = 1 3 x ⇒ 12 = 1 - 2 = 1 x 3 3
⇒ x = 12 × 3 = 36 days
- A certain number of men can do a work in 40 days. If there were 8 men more, it could be finished in 10 days less. How many men were there initially?
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Number of men initially = x (let)
∴ M1 D1 = M2 D2
⇒ x × 40 = (x + 8) × 30
⇒ 4x = 3x + 24
⇒ 4x – 3x = 24
⇒ x = 24 men
Aliter : Using Rule 23,
Here, D = 40, a = 8, d = 10
Required number= a(D - d) men d = 8(40 - 10) = 24 men 10 Correct Option: B
Number of men initially = x (let)
∴ M1 D1 = M2 D2
⇒ x × 40 = (x + 8) × 30
⇒ 4x = 3x + 24
⇒ 4x – 3x = 24
⇒ x = 24 men
Aliter : Using Rule 23,
Here, D = 40, a = 8, d = 10
Required number= a(D - d) men d = 8(40 - 10) = 24 men 10
- Raja can do a piece of work in 20 days while Ramesh can finish it in 25 days. Ramesh started working and Raja joined him after 10 days. The whole work is completed in
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Work done by Raja and Ramesh in 1 day
= 1 + 1 = 5 + 4 = 9 20 25 100 100
Work done by Ramesh in 10 days= 10 = 2 25 5 Remaining work = 1 – 2 = 3 5 5
∴ This part is done by Raja and Ramesh.
∴ Time taken= 3 × 100 = 20 = 6 2 days. 5 9 3 3
∴ Required time= 10 + 6 2 = 16 2 days 3 3 Correct Option: B
Work done by Raja and Ramesh in 1 day
= 1 + 1 = 5 + 4 = 9 20 25 100 100
Work done by Ramesh in 10 days= 10 = 2 25 5 Remaining work = 1 – 2 = 3 5 5
∴ This part is done by Raja and Ramesh.
∴ Time taken= 3 × 100 = 20 = 6 2 days. 5 9 3 3
∴ Required time= 10 + 6 2 = 16 2 days 3 3
