Math, asked by DiyaDENNY958, 11 months ago

A, b and C can do a piece of work in 30,x and x+ 28 days respectively. They started working together but after 4 days A met with an accident and left the job for few days. 8 days before completion of work C also left the job. 5 days before completion of the work A recovered and joined the work again With his double efficiency. The whole work completed in 17 days. Find in how many days the work is completed if both A and C work together.

Answers

Answered by nausheen54
0

I don't know what is the answer I m sorry

Answered by sushant2586
2

Answer:

21 days

Step-by-step explanation:

A completes work in 30 days alone.

Speed of A of working Wa = 1/30

Similarly,

Speed of B of working = 1/x  and

Speed of C of working = 1/(x+28)

With given condition, lets find out each worked how many days and completed how much of work.

For 1st 4 days, everyone worked togater. Work completed in 1st 4 days

W1 = 4 × [1/30 + 1/x + 1/(x+28) ]

W1 = 4× {[x(x+28) + 30(x+28) + 30x] / 30x(x+28)}

W1 = [4x² + 352x + 3360] / [30x²+840x]  .....(1)

Last 8 days, C was not working and last 5 days, A was working with twice efficiency.

As total work was completed in 17 days,

∴ 17 - 4 - 8 = 5

for 5 days, B and C were working togater.

after that for next 3 days, only B is woriking.

and for last 5 days, A and B working with A working with twice efficiency.

∴  17 days work can be split in 4 parts.

W1 = 4 days + W2 = 5 days + W3 = 3 days + W4 = 5 days

For W2, B and C working togater

W2 = 5× [1/x + 1/(x+28)]

∴ W2 = 5× (2x + 28) / (x(x+28) ...(2)

For W3, when B working alone

W3 = 3 × (1/x)

∴ W3 = 3/x   ....(3)

For W4, A and B working togater,

W4 = 5 × (1/x + 2/30)     .....(A works with twice efficiency)

∴ W4 = 5× (15+x)/15x

∴ W4 = (15+x)/3x  .....(4)

From eqn 1,2,3 and 4, we can write,

W1 + W2 + W3 + W4 = 1

[4x² + 352x + 3360] / [30x(x+28)] + 5× (2x + 28) / (x(x+28) + 3/x + (15+x)/3x = 1

∴{(4x² + 352x + 3360) + [30×5×(2x+28)] + 30×(x+28)×3 + 10×(x+28)×(15+x)} ÷ 30x(x+28) = 1

∴ 4x² + 352x + 3360 + 300x + 4200 + 90x + 2520 + 10x² + 430x + 4200 = 30x² + 840x

∴ 16x² - 332x -14280 = 0

∴ 4x² - 83x - 3570 = 0

∴ 4x² - 168x + 85x - 3570 = 0

∴ 4x (x - 42) + 85(x-42) = 0

∴ (x-42) × (4x + 85) = 0

∴ x = 42    or    x = -85/4

But number of days cannot be negative.

∴ x = 42

B takes 42 days to complete work alone

C takes 42 + 28 = 70 days to complete the work alone.

If both A and C worked togater, work completed in one day

Wac = 1/30 + 1/70

∴ Wac = (7 + 3) / 210

∴ Wac = 10/210

∴ Wac = 1/21

∴ Number of days to complete work = 1 / work done in one day

∴ Number of days = 1/ (1/21) = 21 days

A and C combined will take 21 days to complete work

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