A and B can complete a work in 2 hours and 55 minutes.A alone can do the same work two hours faster than B alone can.how long will B take to do the work alone ?
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siddhartharao77
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Note: 1 hour = 60 minutes.
According to the given condition,
Given that A and B can do work in 72 minutes.
⇒ 1/A + 1/B = 1/72
Let the time taken by A to finish the work is 'x' hours.
⇒ 1/A = 1/x.
Given that B takes 1 hour more than A takes to do the same work
⇒ 1/B = 1/x + 60.
Now,
Frame the equations and solve, we get
⇒ (1/x) + (1/x + 60) = (1/72)
LCM = 72x(x + 60)
⇒ 72(x + 60) + 72x = x(x + 60)
⇒ 72x + 4320 + 72x = x^2 + 60
⇒ 144x + 4320 = x^2 + 60
⇒ x^2 - 84x - 4320 = 0
⇒ x^2 - 120x + 36x - 4320 = 0
⇒ x(x - 12) + 36(x - 12) = 0
⇒ (x - 12)(x + 36) = 0
⇒ x = 120,-36{Neglect -ve values}
⇒ x = 120 minutes
⇒ x = 2 hours.
Then:
⇒ 1/B = 1/x + 60
= 1/120 + 60
= 1/180
= 180 minutes.
= 3 hours.
Therefore:
⇒ Time taken by A to do the same work = 2 hours.
⇒ Time taken by B to do the same work = 3 hours.
Hope this helps!