Time and Work
- One man, 3 women and 4 boys can do a piece of work in 96 hours, 2 men and 8 boys can do it in 80 hours, 2 men and 3 women can do it in 120 hours. 5 men and 12 boys can do it in
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View Hint View Answer Discuss in Forum
Using Rule 1,
1 hour’s work of 1 man and 4 boys= 1 160
[∵ 2 men and 8 boys can do the work in 80 hrs.]
1 hour’s work of 1 man 3 women and 4 boys= 1 96
1 hour’s work of 3 women= 1 - 1 = 10 - 6 = 1 96 160 960 240
1 hour’s work of 2 men1 - 1 = 1 120 240 240
1 hour’s work of 4 boys= 1 - 1 160 480 = 3 - 1 = 1 480 240
∴ 2 men = 3 women = 4 boys
∴ 2 men + 8 boys = 12 boys
5 men + 12 boys = 22 boys
∴ By M1D1 = M2D2
⇒ 12 × 80 = 22 × D2⇒ D2 = 12 × 80 22 = 480 = 43 7 hours 11 11 Correct Option: C
Using Rule 1,
1 hour’s work of 1 man and 4 boys= 1 160
[∵ 2 men and 8 boys can do the work in 80 hrs.]
1 hour’s work of 1 man 3 women and 4 boys= 1 96
1 hour’s work of 3 women= 1 - 1 = 10 - 6 = 1 96 160 960 240
1 hour’s work of 2 men1 - 1 = 1 120 240 240
1 hour’s work of 4 boys= 1 - 1 160 480 = 3 - 1 = 1 480 240
∴ 2 men = 3 women = 4 boys
∴ 2 men + 8 boys = 12 boys
5 men + 12 boys = 22 boys
∴ By M1D1 = M2D2
⇒ 12 × 80 = 22 × D2⇒ D2 = 12 × 80 22 = 480 = 43 7 hours 11 11
- If x men can do a piece of work in x days, then the number of days in which y men can do the same work is
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View Hint View Answer Discuss in Forum
Using Rule 1,
M1D1 = M2D2
⇒ x . x = y . D2⇒ D2 = x2 days y Correct Option: C
Using Rule 1,
M1D1 = M2D2
⇒ x . x = y . D2⇒ D2 = x2 days y
- 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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View Hint View Answer Discuss in Forum
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
- 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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View Hint View Answer Discuss in Forum
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
- 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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View Hint View Answer Discuss in Forum
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
