Time and Work
- A can complete a work in 24 days, B in 32 days and C in 64 days. They start together. A works for 6 days and leaves and B leaves 6 days before the work is finished. In how many days was the work finished?
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Let the work was completed in x days. Hence, A worked for 6 days, B worked for (x – 6) days and C worked for x days.
Now, A’s 1 day’s work = 1 24
∴ A’s 6 days’ work= 1 × 6 = 1 24 4 B’s 1 day’s work = 1 32
∴ B’s (x – 6) days’ work= 1 × (x - 6) = x - 6 32 32 C’s 1 day’s work = 1 64
∴ C’s x days’ work= 1 × x = x 64 64 ∴ 1 + x - 6 + x = 1 4 32 64 ⇒ x - 6 + x 32 64 = 1 - 1 = 3 4 4
⇒ 3x – 12 = 48
⇒ 3x = 48 + 12 = 60⇒ x = 60 = 20 3
Hence, the work was completed in 20 daysCorrect Option: A
Let the work was completed in x days. Hence, A worked for 6 days, B worked for (x – 6) days and C worked for x days.
Now, A’s 1 day’s work = 1 24
∴ A’s 6 days’ work= 1 × 6 = 1 24 4 B’s 1 day’s work = 1 32
∴ B’s (x – 6) days’ work= 1 × (x - 6) = x - 6 32 32 C’s 1 day’s work = 1 64
∴ C’s x days’ work= 1 × x = x 64 64 ∴ 1 + x - 6 + x = 1 4 32 64 ⇒ x - 6 + x 32 64 = 1 - 1 = 3 4 4
⇒ 3x – 12 = 48
⇒ 3x = 48 + 12 = 60⇒ x = 60 = 20 3
Hence, the work was completed in 20 days
- A, B and C can complete a work separately in 24, 36 and 48 days respectively. They started together but C left after 4 days of start and A left 3 days before the completion of work. In how many days will the work be completed?
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View Hint View Answer Discuss in Forum
Let the work be completed in x days. Therefore, A worked for x – 3 days, B for x days and C for 4 days.
A’s 1 day’s work = 1 24 B’s 1 day’s work = 1 36 and, C’s 1 day’s work = 1 48 ∴ (x - 3) × 1 + x × 1 + 4 × 1 = 1 24 36 48 ⇒ x - 3 + x + 1 = 1 24 36 12 ⇒ 3x - 9 + 2x + 6 = 1 72
⇒ 5x – 3 = 72
⇒ 5x = 75⇒ x = 75 = 15 5
Hence, the work was completed in 15 days.Correct Option: D
Let the work be completed in x days. Therefore, A worked for x – 3 days, B for x days and C for 4 days.
A’s 1 day’s work = 1 24 B’s 1 day’s work = 1 36 and, C’s 1 day’s work = 1 48 ∴ (x - 3) × 1 + x × 1 + 4 × 1 = 1 24 36 48 ⇒ x - 3 + x + 1 = 1 24 36 12 ⇒ 3x - 9 + 2x + 6 = 1 72
⇒ 5x – 3 = 72
⇒ 5x = 75⇒ x = 75 = 15 5
Hence, the work was completed in 15 days.
- A and B together can complete a piece of work in 72 days, B and C together can complete it in 120 days, and A and C together in 90 days. In what time can A alone complete the work ?
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(A + B)’s 1 day’s work = 1 72
(B + C)’s 1 day’s work = 1 120 (C + A)’s 1 day’s work = 1 90
Adding all three,2(A + B + C)’s 1 day’s work = 1 + 1 + 1 72 120 90 2(A + B + C)’s 1 day’s work = 5 + 3 + 4 = 12 = 1 360 360 30 ∴ (A + B + C)’s 1 day’s work = 1 60
Now, A’s 1 day’s work = (A + B + C)’s 1 day’s work - (B + C)’s 1 day’s workA’s 1 day’s work = 1 - 1 60 120 A’s 1 day’s work = 2 - 1 = 1 120 120
∴ A alone can complete the work in 120 days.
Second method to solve this question ,
Here , x = 72 , y = 120 , z = 90A alone can do in = 2xyz xy + yz - zx
Correct Option: C
(A + B)’s 1 day’s work = 1 72
(B + C)’s 1 day’s work = 1 120 (C + A)’s 1 day’s work = 1 90
Adding all three,2(A + B + C)’s 1 day’s work = 1 + 1 + 1 72 120 90 2(A + B + C)’s 1 day’s work = 5 + 3 + 4 = 12 = 1 360 360 30 ∴ (A + B + C)’s 1 day’s work = 1 60
Now, A’s 1 day’s work = (A + B + C)’s 1 day’s work - (B + C)’s 1 day’s workA’s 1 day’s work = 1 - 1 60 120 A’s 1 day’s work = 2 - 1 = 1 120 120
∴ A alone can complete the work in 120 days.
Second method to solve this question ,
Here , x = 72 , y = 120 , z = 90A alone can do in = 2xyz xy + yz - zx A alone can do in = 2 × 72 × 120 × 90 72 × 120 + 120 × 90 - 72 × 90 A alone can do in = 2 × 72 × 120 × 90 8640 + 10800 – 6480 A alone can do in = 144 × 10800 = 120 days 12960
- A and B together can do a piece of work in 30 days, B and C together can do it in 20 days. A starts the work and works on it for 5 days, then B takes it up and works for 15 days. Finally C finishes the work in 18 days. In how many days can C do the work when doing it separately?
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Let us denote A’s 1 day’s work by A, B’s 1 day’s work by B and C’s work by C.
So, A + B = 1 30 and B + C = 1 20
Also, 5A + 15B + 18C = 1 work.
This can be written as,
5 (A + B) + 10 (B + C) + 8C= 1
Substituting the values of (A + B) and (B + C) we get,= 
5 × 1 
+ 10 × 1 + 8C = 1 30 20 or 1 + 1 + 8C = 1 6 2 or 8C = 1 - 1 - 1 6 2 or 8C = 6 - 1 - 3 6 or 8C = 2 6 or C = 2 = 1 6 × 8 24
Hence, C will complete the work in 24 days.Correct Option: B
Let us denote A’s 1 day’s work by A, B’s 1 day’s work by B and C’s work by C.
So, A + B = 1 30 and B + C = 1 20
Also, 5A + 15B + 18C = 1 work.
This can be written as,
5 (A + B) + 10 (B + C) + 8C= 1
Substituting the values of (A + B) and (B + C) we get,= 5 × 1 + 10 × 1 + 8C = 1 30 20 or 1 + 1 + 8C = 1 6 2 or 8C = 1 - 1 - 1 6 2 or 8C = 6 - 1 - 3 6 or 8C = 2 6 or C = 2 = 1 6 × 8 24
Hence, C will complete the work in 24 days.
- A and B together can finish a work in 15 days. A and C take 2 days more to complete the same work than that of B and C. A, B and C together complete the work in 8 days. In how many days will A finish it separately?
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(A + B)’s 1 day’s work = 1 15 (A + B + C)’s 1 day’s work = 1 8
∴ C’s 1 day’s work= 1 - 1 = 15 - 8 = 7 8 15 120 120
Let (B + C) can complete the work in x days.
∴ (A + C) can complete the work in (x + 2) days.∴ (B + C)’s 1 day’s work = 1 x (A + C)’s 1 day’s work = 1 x + 2
∴ B’s 1 day’s work= 1 - 7 = 120 - 7x x 120 120x
and, A’s 1 day’s work= 1 - 7 = 120 - 7(x + 2) x + 2 120 120(x + 2) = 106 - 7x 120(x + 2)
Now, A’s 1 day’s work + B’s 1 day’s work = (A + B)’s 1 day’s work⇒ 106 - 7x + 120 - 7x = 1 120(x + 2) 120x 15 ⇒ 106x - 7x2 + 120x + 240 - 7x2 - 14x = 1 120x(x + 2) 15
⇒ – 14x2 + 212x + 240 = 8x2 + 16x
⇒ 22x2 – 196x – 240 = 0
⇒ 11x2 – 98x – 120 = 0
⇒ 11x2 – 110x + 12x –120 = 0
⇒ 11x (x – 10) + 12 (x – 10)= 0
⇒ (x – 10) (11x + 12) = 0⇒ x =10, and - 12 11
But no. of days cannot be negative
∴ x = 10
∴ A’s 1 day’s work= 1 - 7 10 + 2 120 = 1 - 7 12 120 = 10 - 7 = 3 = 1 120 120 40
∴ A alone can complete the work in 40 days.Correct Option: A
(A + B)’s 1 day’s work = 1 15 (A + B + C)’s 1 day’s work = 1 8
∴ C’s 1 day’s work= 1 - 1 = 15 - 8 = 7 8 15 120 120
Let (B + C) can complete the work in x days.
∴ (A + C) can complete the work in (x + 2) days.∴ (B + C)’s 1 day’s work = 1 x (A + C)’s 1 day’s work = 1 x + 2
∴ B’s 1 day’s work= 1 - 7 = 120 - 7x x 120 120x
and, A’s 1 day’s work= 1 - 7 = 120 - 7(x + 2) x + 2 120 120(x + 2) = 106 - 7x 120(x + 2)
Now, A’s 1 day’s work + B’s 1 day’s work = (A + B)’s 1 day’s work⇒ 106 - 7x + 120 - 7x = 1 120(x + 2) 120x 15 ⇒ 106x - 7x2 + 120x + 240 - 7x2 - 14x = 1 120x(x + 2) 15
⇒ – 14x2 + 212x + 240 = 8x2 + 16x
⇒ 22x2 – 196x – 240 = 0
⇒ 11x2 – 98x – 120 = 0
⇒ 11x2 – 110x + 12x –120 = 0
⇒ 11x (x – 10) + 12 (x – 10)= 0
⇒ (x – 10) (11x + 12) = 0⇒ x =10, and - 12 11
But no. of days cannot be negative
∴ x = 10
∴ A’s 1 day’s work= 1 - 7 10 + 2 120 = 1 - 7 12 120 = 10 - 7 = 3 = 1 120 120 40
∴ A alone can complete the work in 40 days.
