Time and Work
- A contractor undertook to finish a work in 92 days and employed 110 men. After 48 days, he found that he had already done 3/5 part of the work, the number of men he can withdraw so that the work may still be finished in time is :
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Using Rule 1,
M1D1 = M2D2 W1 W2 ⇒ 110 × 48 = M2 × 44 3 2 5 5
⇒ M2 × 44 × 3 = 110 × 48 × 2⇒ M2 = 110 × 48 × 2 = 80 44 × 3
∴ Number of men can be withdrawn
= 110 – 80 = 30Correct Option: D
Using Rule 1,
M1D1 = M2D2 W1 W2 ⇒ 110 × 48 = M2 × 44 3 2 5 5
⇒ M2 × 44 × 3 = 110 × 48 × 2⇒ M2 = 110 × 48 × 2 = 80 44 × 3
∴ Number of men can be withdrawn
= 110 – 80 = 30
- 3 men and 7 women can do a job in 5 days, while 4 men and 6 women can do it in 4 days. The number of days required for a group of 10 women working together, at the same rate as before, to finish the same job is :
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3 × 5 men + 7 × 5 women
= 4 × 4 men + 6 × 4 women
⇒ 16 men – 15 men = 35 women – 24 women
∴ 1 man = 11 women
⇒ 3 men + 7 women = 40 women
∴ M1D1 = M2D2
⇒ 40 × 5 = 10 × D2
⇒ D2 = 20 days
Aliter : Using Rule 11,
Here, A1 = 3, B1 = 7, D1 = 5
A2 = 4, B2 = 6, D2 = 4
A3 = 0, B3 = 10
Required days= D1D2(A1B2 - A2B1) days D1(A1B3 - A3B1) - D2(A2B3 - A3B2) = 5 × 4(3 × 6 - 4 × 7) 5(3 × 10 - 0) - 4(4 × 10 - 0) = 20 × (-10) = 20 days 150 - 160 Correct Option: D
3 × 5 men + 7 × 5 women
= 4 × 4 men + 6 × 4 women
⇒ 16 men – 15 men = 35 women – 24 women
∴ 1 man = 11 women
⇒ 3 men + 7 women = 40 women
∴ M1D1 = M2D2
⇒ 40 × 5 = 10 × D2
⇒ D2 = 20 days
Aliter : Using Rule 11,
Here, A1 = 3, B1 = 7, D1 = 5
A2 = 4, B2 = 6, D2 = 4
A3 = 0, B3 = 10
Required days= D1D2(A1B2 - A2B1) days D1(A1B3 - A3B1) - D2(A2B3 - A3B2) = 5 × 4(3 × 6 - 4 × 7) 5(3 × 10 - 0) - 4(4 × 10 - 0) = 20 × (-10) = 20 days 150 - 160
- Either 8 men or 17 women can paint a house in 33 days. The number of days required to paint three such houses by 12 men and 24 women working at the same rate is :
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8 men ≡ 17 women
⇒ 12 men ≡ 17 × 12 8 = 51 women 2
∴ 12 men + 24 women= 51 + 24 = 99 women 2 2 By M1D1 = M2D2 W1 W2 = 17 × 33 = 99 × D2 1 2 × 3 ⇒ D2 = 17 × 33 × 6 = 34 days 99
Aliter : Using Rule 12,
Here, A = 8, B = 17, a = 33
A1 = 12, B1 = 24Number of days= a(A × B) A1B + B1A = 33(8 × 17) 12 × 17 + 24 × 8 = 4488 396
No. of days to paint 3 houses= 4488 × 3 = 34 days 396 Correct Option: C
8 men ≡ 17 women
⇒ 12 men ≡ 17 × 12 8 = 51 women 2
∴ 12 men + 24 women= 51 + 24 = 99 women 2 2 By M1D1 = M2D2 W1 W2 = 17 × 33 = 99 × D2 1 2 × 3 ⇒ D2 = 17 × 33 × 6 = 34 days 99
Aliter : Using Rule 12,
Here, A = 8, B = 17, a = 33
A1 = 12, B1 = 24Number of days= a(A × B) A1B + B1A = 33(8 × 17) 12 × 17 + 24 × 8 = 4488 396
No. of days to paint 3 houses= 4488 × 3 = 34 days 396
- 20 men or 24 women can complete a piece of work in 20 days. If 30 men and 12 women undertake to complete the work, the work will be completed in
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20 men ≡ 24 women
⇒ 5 men ≡ 6 women
∴ 30 men + 12 women = 40 men
∴ M1D1 = M2D2
⇒ 20 × 20 = 40 × D2⇒ D2 = 20 × 20 = 10 days 40
Aliter : Using Rule 12,
Here, A = 20, B = 24, a = 20
A1 = 30, B1 = 12Required time = a(A × B) A1B + B1A = 20(20 × 24) 30 × 24 + 12 × 20 = 9600 720 + 240 = 9600 = 10 days 960 Correct Option: A
20 men ≡ 24 women
⇒ 5 men ≡ 6 women
∴ 30 men + 12 women = 40 men
∴ M1D1 = M2D2
⇒ 20 × 20 = 40 × D2⇒ D2 = 20 × 20 = 10 days 40
Aliter : Using Rule 12,
Here, A = 20, B = 24, a = 20
A1 = 30, B1 = 12Required time = a(A × B) A1B + B1A = 20(20 × 24) 30 × 24 + 12 × 20 = 9600 720 + 240 = 9600 = 10 days 960
- 30 men can repair a road in 18 days. They are joined by 6 more workers. Now the road can be repaired in
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Using Rule 1,
M1D1 = M2D2
⇒ 30 × 18 = 36 × D2⇒ D2 = 30 × 18 = 15 days 36 Correct Option: B
Using Rule 1,
M1D1 = M2D2
⇒ 30 × 18 = 36 × D2⇒ D2 = 30 × 18 = 15 days 36