Two taps together can fill a tank completely in three into one upon thirteen min .The smaller tap take 3 min more than bigger tap to fill tank. How much time does each tap take to fill the tank completely?
Answers
Step-by-step explanation:
Let x be the number of minutes taken by bigger pipe,
So, the work done by bigger pipe in one minute = \frac{1}{x}
x
1
Then according to the question,
Time taken by smaller pipe = (x + 3) minutes,
So, the work done by smaller pipe in one minute = \frac{1}{x+3}
x+3
1
Thus, their combined work in one minute = \frac{1}{x}+\frac{1}{x+3}
x
1
+
x+3
1
∵ When they work together they took 3\frac{1}{13}3
13
1
minutes.
Their combined work in one minute = \frac{1}{3\frac{1}{13}}=\frac{1}{\frac{40}{13}}=\frac{13}{40}
3
13
1
1
=
13
40
1
=
40
13
\implies \frac{1}{x}+\frac{1}{x+3} = \frac{13}{40}⟹
x
1
+
x+3
1
=
40
13
\frac{x+3+x}{x(x+3)}=\frac{13}{40}
x(x+3)
x+3+x
=
40
13
\frac{2x+3}{x^2+3x}=\frac{13}{40}
x
2
+3x
2x+3
=
40
13
80x+120 = 13x^2 + 39x80x+120=13x
2
+39x
13x^2 -41x - 120=013x
2
−41x−120=0
13x^2 - 65x + 24x - 120=013x
2
−65x+24x−120=0
13x(x-5)+24(x-5)=013x(x−5)+24(x−5)=0
(13x+24)(x-5)=0(13x+24)(x−5)=0
13x + 24 =0\text{ or }x-5=013x+24=0 or x−5=0
\implies x = -\frac{24}{13}\text{ or }x=5⟹x=−
13
24
or x=5
∵ Number of minutes can not be negative,
Hence, the time taken by bigger pipe = 5 minutes,
And, the time taken by smaller pipe = 5 + 3 = 8 minutes,
First pipe take 5 min to fill and the second pipe take (5+3) = 8 min to fill the tank.
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