two workers undertake to do a job. 2 hours after the first worker started to work the second worker joined in to work. 5 hours after the second worker has begun working there is still 9/20 of the work to be done. when the job is completed , it turns out that the first worker has done 60% of the work.
Answers
Answer:
total job completed in 12 hrs
Step-by-step explanation:
Let say job done by 1st in 1 hr = a
& job done by 2nd in 1 hr = b
job done by 1st in 2 hr = 2a
Job done in next 5 hr by a & b = 5a + 5b
Total work done = 2a + 5a + 5b = 7a + 5b
Work left = 9/20
Work done = 1 - 9/20 = 11/20
7a + 5b = 11/20
140a + 100b = 11 - eq 1
Let say x+2 hrs taken to complete job
(x + 2)a + xb = 1
(x + 2)a = 6/10 ( 60 % of job)
xb = 4/10 ( remaining job)
10xa + 20a = 6 - eq 2
10xb = 4
10x((11-140a)/100) = 4 ( putting value of b from eq 1)
=> 11x - 140ax = 40 - eq 3
14 Eq2 + Eq 3
280a + 11x = 84 + 40
=> 11x = 124 - 280a
=> x = (124 - 280a)/11
10a(124 - 280a)/11 + 20a = 6
=> 1240a - 2800a² + 220a = 66
=> 2800a² - 1460a + 66 = 0
=> a = (1460 ± √(2131600 - 739200) )/(2 * 2800)
a = (1460 ± √1392400) )/5600
a = (1460 ± 1180 )/5600
a = 2640/5600 , 280/5600
a = 33/70 , 1/20
a = 33/70
140a + 100b = 11
=> 66 + 100b = 11
=> b = -55/100 ( -ve is not possible so this is eliminated)
a = 1/20
7 + 100b = 11
b = 4/100
b = 1/25
x = (124 - 280a)/11 = (124 - 14)/11 = 110/11 = 10
x +2 = 12
a's 1 hr job = 1/20
b's 1 hr job = 1/25
total job completed in 12 hrs
work done by a = 12/20 = 3/5
work done by b = 10/25 = 2/5