Time and Work
- 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
- 15 men can finish a piece of work in 40 days. The number of days after which 5 men should leave the work so that the work is finished in 45 days altogether is :
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Let 5 men leave the work after x days.
∵ M1D1 = M2D2 + M3D3
∴ 15 × 40 = 15 × x + 10 × (45 – x)
⇒ 600 = 15x + 450 – 10x
⇒ 600 – 450 = 5x Þ 5x = 150⇒ x = 150 30 days 5 Correct Option: C
Let 5 men leave the work after x days.
∵ M1D1 = M2D2 + M3D3
∴ 15 × 40 = 15 × x + 10 × (45 – x)
⇒ 600 = 15x + 450 – 10x
⇒ 600 – 450 = 5x Þ 5x = 150⇒ x = 150 30 days 5
- 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
- Some staff promised to do a job in 18 days, but 6 of them went on leave. So the remaining men took 20 days to complete the job. How many men were there originally ?
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Using Rule 1,
Number of men originally = x (let)
∴ M1 D1 = M2 D2
⇒ x × 18 = (x – 6) × 20
⇒ x × 9 = (x – 6) × 10
= 10x – 60
⇒ 10x – 9x = 60
⇒ x = 60 menCorrect Option: D
Using Rule 1,
Number of men originally = x (let)
∴ M1 D1 = M2 D2
⇒ x × 18 = (x – 6) × 20
⇒ x × 9 = (x – 6) × 10
= 10x – 60
⇒ 10x – 9x = 60
⇒ x = 60 men
