Time and Work
- 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?
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According to the question, 20 men + 30 boys = 24 men + 16 boys
∴ 4 men = 14 boys
⇒ 2 men = 7 boys
⇒ 2 men + 1 boy = 8 boys
⇒ 2 men + 3 boys = 10 boys
By M1 D1 = M2 D2
⇒ 10 × 10 = 8 × D2⇒ D2 = 10 × 10 = 25 = 12 1 days 8 2 2
Aliter :
Using Rule 11,
Here, A1 = 2, B1 = 3, D1 = 10
A2 = 3, B2 = 2, D2 = 8
A3 = 2, B3 = 1Required time = D1D2(A1B2 - A2B1) days D1(A1B3 - A3B1) - D2(A2B3 - A3B2) = 10(2 × 2 - 3 × 3) days 10(2 × 1 - 2 × 3) - 8(3 × 1 - 2 × 2) = 80 × -5 = 12 1 days -40 + 8 2 Correct Option: C
According to the question, 20 men + 30 boys = 24 men + 16 boys
∴ 4 men = 14 boys
⇒ 2 men = 7 boys
⇒ 2 men + 1 boy = 8 boys
⇒ 2 men + 3 boys = 10 boys
By M1 D1 = M2 D2
⇒ 10 × 10 = 8 × D2⇒ D2 = 10 × 10 = 25 = 12 1 days 8 2 2
Aliter :
Using Rule 11,
Here, A1 = 2, B1 = 3, D1 = 10
A2 = 3, B2 = 2, D2 = 8
A3 = 2, B3 = 1Required time = D1D2(A1B2 - A2B1) days D1(A1B3 - A3B1) - D2(A2B3 - A3B2) = 10(2 × 2 - 3 × 3) days 10(2 × 1 - 2 × 3) - 8(3 × 1 - 2 × 2) = 80 × -5 = 12 1 days -40 + 8 2
- If 8 men or 12 boys can do a piece of work in 16 days, the number of days required to complete the work by 20 men and 6 boys is
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∵ 8 men ≡ 12 boys
∴ 4 men ≡ 6 boys
⇒ 20 men ≡ 30 boys
⇒ 20 men + 6 boys = 36 boys
∴ M1D1 = M2D2
⇒ 12 × 16 = 36 × D2⇒ D2 = 12 × 16 = 16 = 5 1 days 36 3 3
Aliter :
Using Rule 12,
Here, A = 8, B = 12, a = 16
A1 = 20, B1 = 6,Required number of days = a = 16 A1 + B1 20 + 6 A B 8 12 = 16 = 16 × 2 = 5 1 days 5 + 1 12 3 2 2 Correct Option: A
∵ 8 men ≡ 12 boys
∴ 4 men ≡ 6 boys
⇒ 20 men ≡ 30 boys
⇒ 20 men + 6 boys = 36 boys
∴ M1D1 = M2D2
⇒ 12 × 16 = 36 × D2⇒ D2 = 12 × 16 = 16 = 5 1 days 36 3 3
Aliter :
Using Rule 12,
Here, A = 8, B = 12, a = 16
A1 = 20, B1 = 6,Required number of days = a = 16 A1 + B1 20 + 6 A B 8 12 = 16 = 16 × 2 = 5 1 days 5 + 1 12 3 2 2
- If 10 men or 20 women or 40 children can do a piece of work in 7 months, then 5 men, 5 women and 5 children together can do half of the work in :
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10 men ≡ 20 women
1 man = 2 women = 5 children
1 woman = 2 children
∴ 5 men + 5 women + 5 children = 20 + 10 + 5 = 35 children
∴ M1D1 = M2D2
⇒ 40 × 7 = 35 × D2⇒ D2 = 40 × 7 = 8 months 2
∴ 5 men, 5 women and 5 children can do half of the work in 8 months Required time = 4 months.
Aliter :
Using Rule 13, Here, A = 10, B= 20, C = 40, a = 7
A1 = 5, B1 = 5, C1 = 5Time taken to do same work = a A1 + B1 + C1 A B C = 7 5 + 5 + 5 10 20 30 = 7 1 + 1 + 1 2 4 8 = 7 4 + 2 + 1 8
= 8 months Half of the work they do in 4 months.Correct Option: D
10 men ≡ 20 women
1 man = 2 women = 5 children
1 woman = 2 children
∴ 5 men + 5 women + 5 children = 20 + 10 + 5 = 35 children
∴ M1D1 = M2D2
⇒ 40 × 7 = 35 × D2⇒ D2 = 40 × 7 = 8 months 2
∴ 5 men, 5 women and 5 children can do half of the work in 8 months Required time = 4 months.
Aliter :
Using Rule 13, Here, A = 10, B= 20, C = 40, a = 7
A1 = 5, B1 = 5, C1 = 5Time taken to do same work = a A1 + B1 + C1 A B C = 7 5 + 5 + 5 10 20 30 = 7 1 + 1 + 1 2 4 8 = 7 4 + 2 + 1 8
= 8 months Half of the work they do in 4 months.
- 3 men and 4 boys can complete a piece of work in 12 days. 4 men and 3 boys can do the same work in 10 days. Then 2 men and 3 boys can finish the work in
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12 (3 men + 4 boys)
≡ 10 (4 men + 3 boys)
⇒ 36 men + 48 boys = 40 men + 30 boys
⇒ 4 men = 18 boys
⇒ 2 men = 9 boys
∴ 4 men + 3 boys
= 21 boys, who do the work in 10 days and 2 men + 3 boys = 12 boys
∴ M1D1 = M2D2
⇒ 21 × 10 = 12 × D2⇒ D2 = 21 × 10 = 35 = 17 1 days 12 2 2
Aliter :
Using Rule 11, Here, A1 = 3, B1 = 4, D1 = 12
A2 = 4, B2 = 3, D2 = 10
A3 = 2, B3 = 3Required time = D1D2(A1B2 - A2B1) days D1(A1B3 - A3B1) - D2(A2B3 - A3B2) = 12 × 10(3 × 3 - 4 × 4) days 12(3 × 3 - 2 × 4) - 10(4 × 3 - 2 × 3) = 120 × - 7 days 12(9 - 8) - 10 × 6 = - 840 = 17 1 days - 48 2 Correct Option: A
12 (3 men + 4 boys)
≡ 10 (4 men + 3 boys)
⇒ 36 men + 48 boys = 40 men + 30 boys
⇒ 4 men = 18 boys
⇒ 2 men = 9 boys
∴ 4 men + 3 boys
= 21 boys, who do the work in 10 days and 2 men + 3 boys = 12 boys
∴ M1D1 = M2D2
⇒ 21 × 10 = 12 × D2⇒ D2 = 21 × 10 = 35 = 17 1 days 12 2 2
Aliter :
Using Rule 11, Here, A1 = 3, B1 = 4, D1 = 12
A2 = 4, B2 = 3, D2 = 10
A3 = 2, B3 = 3Required time = D1D2(A1B2 - A2B1) days D1(A1B3 - A3B1) - D2(A2B3 - A3B2) = 12 × 10(3 × 3 - 4 × 4) days 12(3 × 3 - 2 × 4) - 10(4 × 3 - 2 × 3) = 120 × - 7 days 12(9 - 8) - 10 × 6 = - 840 = 17 1 days - 48 2
- 6 men and 8 women can do a work in 10 days, Then 3 men and 4 women can do the same work in
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6m + 8w ≡ 10 days
⇒ 2 (3m + 4w) ≡ 10 days
⇒ 3m + 4w ≡ 20 days
[Since the workforce has become half of the original force, so number of days must be double].
Aliter :
Using Rule 14,
Let us assume efficiency of 6 men
= efficiency of 8 men.
A = 6, a = 20
B = 8, b = 20
A1 = 3, B1 = 4∴ Required time = 1 A1 + B1 A × a B × b = 1 1 + B1 40 40 = 40 = 20days 2 Correct Option: B
6m + 8w ≡ 10 days
⇒ 2 (3m + 4w) ≡ 10 days
⇒ 3m + 4w ≡ 20 days
[Since the workforce has become half of the original force, so number of days must be double].
Aliter :
Using Rule 14,
Let us assume efficiency of 6 men
= efficiency of 8 men.
A = 6, a = 20
B = 8, b = 20
A1 = 3, B1 = 4∴ Required time = 1 A1 + B1 A × a B × b = 1 1 + B1 40 40 = 40 = 20days 2
