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