Math, asked by BrainlyHelper, 1 year ago

Two pipes running together can fill a cistern in 3 1/13 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern?

Answers

Answered by nikitasingh79
243
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.  

HOPE THIS WILL HELP YOU...
Answered by Anonymous
11

Answer:

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.  

HOPE THIS WILL HELP YOU...

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