Answer to Question #25389 in Algebra for herin

Question

ONE TANK CAN BE FILLED UP BY TWO TAPS IN 6 HOURS. THE SMALLER TAP ALONE TAKES 5 HRS MORE THAN THE BIGGER TAP ALONE. FIND THE TIME REQUIRED BY EACH TAP TO FILL THE TANK SEPARATELY

Answer 1

Suppose, the smaller tap alone fill tank in X hours and the bigger - in Y hours:
X-Y=5
If they work together we need to add speeds of filling:
V/X + V/Y = V/6,
V - volume of tank
Finally, we have a system of equations:
X-Y=5
1/X+1/Y=1/6
And X>0 and Y>0 ( because it's time)
From the first: X=Y+5
From the second:
1/(Y+5)+1/Y=1/6 *6*Y*(Y+5)
6Y+6*(Y+5)=Y(Y+5)
Y^2+5Y-12Y-30=0
Y^2-7Y-30=0
Y1 = -3
Y2=10
X = 10+5 = 15
Answer: 10 and 15 hours.