Question #25389

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

Expert's answer

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.

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.

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