Time and Work
- 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
- A, B and C can do a piece of work in 24, 30 and 40 days respectively. They began the work together but C left 4 days before completion of the work. In how many days was the work done ?
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Let the work be completed in x days.
According to the question,
C worked for (x – 4) days∴ x + x + x - 4 = 1 24 30 40 ⇒ 5x + 4x + 3(x - 4) = 1 120 ⇒ 12x - 12 = 1 120 ⇒ 12 (x - 1) = 1 120 ⇒ x - 1 = 1 ⇒ x - 1 = 10 10
⇒ x = 10 + 1 = 11 daysCorrect Option: D
Let the work be completed in x days.
According to the question,
C worked for (x – 4) days∴ x + x + x - 4 = 1 24 30 40 ⇒ 5x + 4x + 3(x - 4) = 1 120 ⇒ 12x - 12 = 1 120 ⇒ 12 (x - 1) = 1 120 ⇒ x - 1 = 1 ⇒ x - 1 = 10 10
⇒ x = 10 + 1 = 11 days
- 20 men can do a piece of work in 18 days, They worked together for 3 days, then 5 men joined them. In how many more days is the work completed ?
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Work done by 20 men in 3 days
= 3 = 1 part 18 6
Remaining work= 1 – 1 = 5 part 6 6 ∴ M1D1 = M2D2 W1 W2 ∴ 20 × 18 = 25 × D2 1 5/6
⇒ 6 × 25 × D2 = 5 × 20 × 18⇒ D2 = 5 × 20 × 18 = 12 days 6 × 25
Aliter : Using Rule 10,
Here, A = 20, a = 18 b = 3. B = 5
Required number of days= A(a - b) = 20 (18 - 3) A + B 20 + 5 = 20 × 15 = 12 days 25 Correct Option: B
Work done by 20 men in 3 days
= 3 = 1 part 18 6
Remaining work= 1 – 1 = 5 part 6 6 ∴ M1D1 = M2D2 W1 W2 ∴ 20 × 18 = 25 × D2 1 5/6
⇒ 6 × 25 × D2 = 5 × 20 × 18⇒ D2 = 5 × 20 × 18 = 12 days 6 × 25
Aliter : Using Rule 10,
Here, A = 20, a = 18 b = 3. B = 5
Required number of days= A(a - b) = 20 (18 - 3) A + B 20 + 5 = 20 × 15 = 12 days 25
- A and B can do a piece of work in 45 and 40 days repectively. They began the work together but A left after some days and B finished the remaining work in 23 days. A left after
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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⇒ x + 23 + x = 1 45 40 ⇒ 8x + 207 + 9x = 1 360
⇒ 17x + 207 = 360
⇒ 17x = 360 – 207 = 153⇒ x = 153 = 9 days 17
Aliter : Using Rule 26,
Here, x = 45, y = 40, a = 23A left after = (y - a) × x x + y = (40 - 23) × 45 45 + 40 = 17 × 45 = 9 days 85 Correct Option: B
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⇒ x + 23 + x = 1 45 40 ⇒ 8x + 207 + 9x = 1 360
⇒ 17x + 207 = 360
⇒ 17x = 360 – 207 = 153⇒ x = 153 = 9 days 17
Aliter : Using Rule 26,
Here, x = 45, y = 40, a = 23A left after = (y - a) × x x + y = (40 - 23) × 45 45 + 40 = 17 × 45 = 9 days 85
- A certain number of men can do a piece of work in 40 days. If there were 45 men more the work could have been finished in 25 days. Find the original number of men employed in the work.
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Original number of men = x (let)
∴ M1 D1 = M2 D2
⇒ x × 40 = (x + 45) × 25
⇒ 8x = (x + 45) × 5
⇒ 8x = 5x + 225
⇒ 8x – 5x = 225
⇒ 3x = 225⇒ x = 225 = 75 men 3
Aliter : Using Rule 23,
Here, D = 40, a = 45, d = (40 – 25) = 15∴ Required number= a(D - d) d = 45(40 - 15) 15 = 45 × 25 = 15 × 5 = 75 15 Correct Option: D
Original number of men = x (let)
∴ M1 D1 = M2 D2
⇒ x × 40 = (x + 45) × 25
⇒ 8x = (x + 45) × 5
⇒ 8x = 5x + 225
⇒ 8x – 5x = 225
⇒ 3x = 225⇒ x = 225 = 75 men 3
Aliter : Using Rule 23,
Here, D = 40, a = 45, d = (40 – 25) = 15∴ Required number= a(D - d) d = 45(40 - 15) 15 = 45 × 25 = 15 × 5 = 75 15
