Time and Work
- A and B can do a piece of work in 15 days. B and C can do a similar work in 12 days and C and A in 10 days. How many days will A take to do the work by himself ?
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(A + B)’s 1 day’s work
1 ..........(i) 15
(B + C)’s 1 day’s work1 ..........(ii) 12
(C + A)’s 1 day’s work1 ..........(iii) 10
On adding all three equations,
2 (A + B + C)’s 1 day’s work= 1 + 1 + 1 15 12 10 = 4 + 5 + 6 = 15 = 1 60 60 4
∴ (A + B + C)’s 1 day’s work1 ..........(iv) 8
By equation (iv) – (ii),
A’s 1 day’s work1 - 1 8 12 3 - 2 = 1 24 24
∴ Required time = 24 days
Aliter : Using Rule 19,A alone can do in = 2xyz xy + yz - zx = 2× 15 × 12 × 10 15 × 12 + 12 × 10 - 10 × 15 = 2 × 3 × 12 × 10 36 + 24 - 30 = 720 24 days 60 - 30 Correct Option: B
(A + B)’s 1 day’s work
1 ..........(i) 15
(B + C)’s 1 day’s work1 ..........(ii) 12
(C + A)’s 1 day’s work1 ..........(iii) 10
On adding all three equations,
2 (A + B + C)’s 1 day’s work= 1 + 1 + 1 15 12 10 = 4 + 5 + 6 = 15 = 1 60 60 4
∴ (A + B + C)’s 1 day’s work1 ..........(iv) 8
By equation (iv) – (ii),
A’s 1 day’s work1 - 1 8 12 3 - 2 = 1 24 24
∴ Required time = 24 days
Aliter : Using Rule 19,A alone can do in = 2xyz xy + yz - zx = 2× 15 × 12 × 10 15 × 12 + 12 × 10 - 10 × 15 = 2 × 3 × 12 × 10 36 + 24 - 30 = 720 24 days 60 - 30
- If 35 men can finish a piece of work in 8 days, then the number of men who can do the same work in 10 days is :
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M1D1 = M2D2
⇒ 35 × 8 = M2 × 10⇒ M2 = 35 × 8 = 28 men 10 Correct Option: B
M1D1 = M2D2
⇒ 35 × 8 = M2 × 10⇒ M2 = 35 × 8 = 28 men 10
- A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days, B had to leave and A alone completed the remaining work. The whole work was completed in :
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Work done by (A + B) in 1 day
= 1 + 1 = 2 + 3 = 1 15 10 30 6 ∴ (A + B)’s 2 days’ work = 2 = 1 6 3 Remaining work = 1 - 1 = 2 3 3
This part is done by A alone.
∵ one work is done by A in 15 days.∴ 2 work is done in 15 × 2 = 10 days. 3 3
∴ Total number of days
= 10 + 2 = 12 days
Aliter : Using Rule 20,
Here, m = 15, n = 10, p =2
A alone completed the work in= mn - p(m + n) days n = 15 × 10 - 2(15 + 10) 10 = 150 - 50 = 10 days 10
Total time taken= 10 + 2 = 12 daysCorrect Option: C
Work done by (A + B) in 1 day
= 1 + 1 = 2 + 3 = 1 15 10 30 6 ∴ (A + B)’s 2 days’ work = 2 = 1 6 3 Remaining work = 1 - 1 = 2 3 3
This part is done by A alone.
∵ one work is done by A in 15 days.∴ 2 work is done in 15 × 2 = 10 days. 3 3
∴ Total number of days
= 10 + 2 = 12 days
Aliter : Using Rule 20,
Here, m = 15, n = 10, p =2
A alone completed the work in= mn - p(m + n) days n = 15 × 10 - 2(15 + 10) 10 = 150 - 50 = 10 days 10
Total time taken= 10 + 2 = 12 days
- A and B can do a piece of work in 28 and 35 days respectively. They began to work together but A leaves after sometime and B completed remaining work in 17 days. After how many days did A leave ?
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Let A worked for x days.
According to questionx + ( x + 17 ) = 1 28 35 ⇒ 5x + 4(x + 17) = 1 140
⇒ 5x + 4x + 68 = 140
⇒ 9x = 140 – 68 = 72
⇒ x = 8
∴ A worked for 8 daysCorrect Option: C
Let A worked for x days.
According to questionx + ( x + 17 ) = 1 28 35 ⇒ 5x + 4(x + 17) = 1 140
⇒ 5x + 4x + 68 = 140
⇒ 9x = 140 – 68 = 72
⇒ x = 8
∴ A worked for 8 days
- A and B working separately can do a piece of work in 10 days and 15 days respectively. If they work on alternate days beginning with A, in how many days will the work be completed ?
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Work done by 2 (A + B) in one day
= 1 + 1 = 3 + 2 = 1 10 15 30 6
∴ Work done by (A + B) in one day= 1 12
∴ (A + B) can complete the work in 12 daysCorrect Option: C
Work done by 2 (A + B) in one day
= 1 + 1 = 3 + 2 = 1 10 15 30 6
∴ Work done by (A + B) in one day= 1 12
∴ (A + B) can complete the work in 12 days
