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
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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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