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Suppose that the faster pipe takes x hours to fill the tank
slower pipe will take x+10 hours to fill the tank
since faster tap (larger diameter tap) takes x hours to fill the tank,
portion of tank filled in 1 hour = 1/x
portion of tank filled in 75/8 hours = 75/8x --------------------(i)
similarly portion of tank filled by slower pipe in x+10 hours to fill the tank is 1/(x+10)
portion of tank filled in 75/8 hours = 75/[ 8* (x+10) ] -----------------------(ii)
it is given that the tank is filled in 75/8 hours
so
75/8x + 75/[ 8* (x+10) ] = 1
taking 75/8 common outside and moving it to the RHS we get
1/x + 1/(x+10) = 8/75
( x + 10 + x )/ x(x+10) = 8/75
75 * (2x + 10) = 8 * (x2+10x)
150x + 750 = 8x2 + 80x
8x2 - 70x - 750 = 0
dividing by 2
4x2 - 35x - 375 = 0
now solve the equation and ignore the negative answer obtained !
you will get the time taken by the tap of larger diameter, add 10 to it and you will get the time taken by the smaller diametered tap
PLEASE APPROVE !
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