Two water taps together can fill a tank in 11 1/9 hours. The tap of larger diameter takes 5 hours less thsn the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank
Answers
Answered by
40
hey buddy here is your answer
==> two taps together can fill a tank in = 100/9 hours
==> now let smaller tap can fill tank in = x hours
==> therefore,other tap will do it in = x-5 hours
==> X+(x-5) = 100/9 hours
==> so work done by both in 1 hour is =
1/x+ 1/x-5 = 1/100/9(9/100)hours
by take lcm and get value of X
==> two taps together can fill a tank in = 100/9 hours
==> now let smaller tap can fill tank in = x hours
==> therefore,other tap will do it in = x-5 hours
==> X+(x-5) = 100/9 hours
==> so work done by both in 1 hour is =
1/x+ 1/x-5 = 1/100/9(9/100)hours
by take lcm and get value of X
Answered by
32
take lcm and find value of x
l/x+1/x+5=9/100
2x-5/x^2-5x=9/100
200x-500=9x^2-45x
9x^2-45x-200x+500=0
9x^2-245x+500=0
l/x+1/x+5=9/100
2x-5/x^2-5x=9/100
200x-500=9x^2-45x
9x^2-45x-200x+500=0
9x^2-245x+500=0
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