Question

Two taps together can fill a tank in 9 3/8 hrs. D tap of larger diameter takes 10hrs less the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank

Answer 1

Let the time taken by the tap of larger diameter be x hrs.
,,     ,,     ,,      ,,,,    ,,      ,,    ,, ,,  smaller diameter be y hrs.
ATQ....
1.)     x+y=9 3/8
2.)     x=y-10
          x-y=-10
Sub. 1 & 2 , we get,
y=9.6875 hrs

Answer 2

let the time taken by the two taps separately be x hrs and (x+10) hrs .
total time = 9 3/8 = 75/8 hrs .
according to the problem ,
                     1/x + 1/(x+10) = 8/75
                      x+10+x/x(x+10) = 8/75
                      2x+10/ x2 +10x = 8/75
                      8(x2+10) = 75(2x+10)
                      8x2+80x = 150x + 750
                      8x2 -70x -750 =0
                      4x2 -35x -375 =0
                      (x-15)(4x+25) =0
                      if (4x+25) =0
                                    x =-25/4 is rejected because time will not be in negative .
                      if (x-15) =0
                                  x =15 
            x =15 & x+10 = 15+10 =25 
         therefore , time taken by the two taps separately are 15hrs & 25hrs .
HOPE it helps !