B is more efficient than A and they together can complete a work in 18 days. Had A done 60% of the w

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21 Apr 2022 Time and Work

B is more efficient than A and they together can complete a work in 18 days. Had A done 60% of the work and then B, the remaining work, then the work would have been completed in 39 days. B alone will complete 60% of the same work in

OPtion

1) 21 days
2) 24 days
3) 30 days
4) 27 days
5) 36 days
6) 45 days
7) 12 days
8) 18 days
9) 15 days
10) None of these

Solution

Let A and B alone takes 'a' and 'b' days respectively, then 1 day work of A & B = 1/a + 1/b = 1/18 ---(i)
When A does 60% or 3/5 work and B remaining 2/5 work, total days required = 39
⇒ (3/5)/(1/a) + (2/5)/(1/b) = 39 ---(ii)
Substituting 1/a=(1/18 - 1/b) from (i) into Eqn (ii), we get (3/5)/[(1/18 - 1/b)] + (2/5)/(1/b) = 39
54b/[5(b-18)] + 2b/5 = 39
54b + 2b(b-18) = 39*5(b-18)
2b² - 177b + 3510 = 0
2b² - 117b - 60b + 3510 = 0
b(2b - 117) - 30(2b - 117) = 0
(b - 30) (2b - 117) = 0
Considering +ve integer value of the days, b=30
⇒ B alone can complete the work in 30 days.
.'. For 60% of the work B require = (60/100)*30 = 18 days.

Correct Option: 8 Best Solution (0)