A mason’s helper requires 4 hours more to pave a concrete walk than it takes the mason. The two worked together for 3 hours when the mason was called away. The helper completed the job in 2 hours. How long would it take each to do the same job working alone?
Answers
Answer
The mason will finish the job in 6 hours while the helper will be finished in 10 hours.
Step-by-step explanation:
Mason does job in m hours. Helper does job in m+4 hours.
t=time in hrs
1/m + 1/(m+4)=(2m+4)/(m^2+4m) = 1/t
3 * (2m+4)/(m^2+4m) = (6m+12)/(m^2+4m) = 3/t
1 - (6m+12)/(m^2+4m) time remaining [1 - 3/t]
2 * 1/(m+4)= 1 - (6m+12)/(m^2+4m)
2/(m+4)=1 - (6m+12)/(m(m+4))
2m=m^2+4m-6m-12
m^2-4m-12=0
(m-6)(x+2)=0
m=6 hr
m+4=10 hr
Given : A mason’s helper requires 4 hours more to pave a concrete walk than it takes the mason.
The two worked together for 3 hours when the mason was called away. The helper completed the job in 2 hours.
To Find : How long would it take each to do the same job working alone?
Solution:
Mason complete the job alone in x hrs
Then Helper completed the job alone in x + 4 hrs
1 hr work of Mason = 1/x
1 hr work of helper = 1/(x + 4)
Mason worked for 3 hrs , and helper worker for (3 + 2) = 5 hrs
Total Work done
3/x + 5/(x + 4) = 1
=> 3(x + 4) + 5x = x(x + 4)
=> 3x + 12 + 5x = x² + 4x
=> x² - 4x - 12 = 0
=> (x - 6)(x + 2) = 0
=> x = 6 (x = -2 not possible )
x + 4 = 6 + 4 = 10
Mason complete the job alone in 6 hrs
Helper completes the job alone in 10 hrs
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