Two pipes running together can fill the cistern in 3 hrs. If one
pipe takes 3 minutes more than the other to fill it then find the time
in which each pipe would fill the cistern.
Answers
Let faster pipe takes x min to fill the cistern.
To fill the cistern slower pipe take (x+3) min.
In one minute the faster pipe filled the cistern= 1/x
In 3 1/13= 40/13 min the faster pipe filled the cistern= 40/13 × (1/x) = 40/13x
In 3 1/13= 40/13 min the slower pipe filled the cistern= 40/13 × (1/x+3) = 40/13(x+3).
ATQ
40/13x + 40/13(x+3) = 1
40/13 [ 1/x + 1/(x+3)] = 1
40 [ (x +3+x) / x(x+3)] =13
40(2x +3) =13 x(x+3)]
80x + 120 = 13x² +39x
13x² +39x -80x -120= 0
13x² - 41x -120= 0
13x² - 65x +24x -120= 0
13x(x -5) + 24(x -5)= 0
(13x +24)(x -5)= 0
(13x +24)= 0 or (x -5)= 0
x =- 24/13 or x = 5
Time cannot be negative, so x = 5
Hence, Faster pipe takes 5 min to fill the cistern while slower pipe takes (x+3) = 5+3= 8 min to fill the cistern.
Answer:
Step-by-step explanation:
Let faster pipe takes x min to fill the cistern.
To fill the cistern slower pipe take (x+3) min.
In one minute the faster pipe filled the cistern= 1/x
In 3 1/13= 40/13 min the faster pipe filled the cistern= 40/13 × (1/x) = 40/13x
In 3 1/13= 40/13 min the slower pipe filled the cistern= 40/13 × (1/x+3) = 40/13(x+3).
ATQ
40/13x + 40/13(x+3) = 1
40/13 [ 1/x + 1/(x+3)] = 1
40 [ (x +3+x) / x(x+3)] =13
40(2x +3) =13 x(x+3)]
80x + 120 = 13x² +39x
13x² +39x -80x -120= 0
13x² - 41x -120= 0
13x² - 65x +24x -120= 0
13x(x -5) + 24(x -5)= 0
(13x +24)(x -5)= 0
(13x +24)= 0 or (x -5)= 0
x =- 24/13 or x = 5
Time cannot be negative, so x = 5
Hence, Faster pipe takes 5 min to fill the cistern while slower pipe takes (x+3) = 5+3= 8 min to fill the cistern.