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## anonymous one year ago Two pipe running together can fill a cistern in 3.5mins and one pipe takes 3 mins more than the other find the time in which each pipe would fill cistern

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1. anonymous

@Michele_Laino @pooja195 @perl

2. perl

Let x be the rate of the first pipe and y is the rate of the second pipe. And t be the time it takes for first pipe to fill cistern. Then we have the following equations, use d = r* t 1 cistern = (x+y) * 3.5 min 1 cistern = x * t 1 cistern = y * (t + 3)

3. anonymous

I can't understand the above.

4. perl

Have you used the formula distance = rate * time

5. anonymous

Yes..

6. perl

I am using a similar equation. 1 cistern = rate of pipe * time

7. anonymous

What next?

8. perl

Let x be the rate of the first pipe and y is the rate of the second pipe. And t be the time it takes for first pipe to fill cistern. 1 cistern = (x+y) * 3.5 min 1 cistern = x * t 1 cistern = y * (t + 3)

9. anonymous

3 cistern?

10. perl

they are three different equations

11. perl

"Two pipe running together can fill a cistern in 3.5mins" That gives us 1 cistern = (x+y) * 3.5 min , where x and y are the rates of the pipes

12. anonymous

Help PLEASE?

13. anonymous

@Michele_Laino

14. Michele_Laino

I think that @perl gave you the right explanation

15. Michele_Laino

nevertheless I can give you another way to solve your problem. The working rates of the two pipes are: $\Large\frac{W}{x},\quad \frac{W}{{x + 3}}$

16. Michele_Laino

where W is the work to be done

17. Michele_Laino

now, following the text of your problem, we can write: $\Large \frac{W}{x} + \frac{W}{{x + 3}} = \frac{W}{{3.5}}$ and, simplifying that expression for W, we get: $\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{1}{{3.5}}$ Please solve that equation for x

18. anonymous

x+3+x/x^2+3x = 1/3.5?

19. Michele_Laino

first step: we have: $\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{2}{7}$

20. Michele_Laino

yes! since 3.5= 7/2

21. Michele_Laino

now, the least common multiple, of x, x+3 and 7, is: x*(x+3)*7 am I right?

22. anonymous

Yes

23. Michele_Laino

is that equation is equivalent to this one? $\Large 1 \times 7 \times \left( {x + 3} \right) + 7x = 2x\left( {x + 3} \right)$

24. Michele_Laino

oops..is that equation equivalent to this one?

25. anonymous

I guess..

26. Michele_Laino

more steps: $\Large \frac{{1 \times 7\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}} + \frac{{1 \times 7x}}{{7x\left( {x + 3} \right)}} = \frac{{2x\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}}$

27. Michele_Laino

now?

28. anonymous

I'm lost...

29. Michele_Laino

as denominator, of each fraction, we find the least common multiple

30. anonymous

Getting it.. Now?

31. Michele_Laino

now, if we consider the first fraction, for example, we have: |dw:1436624598950:dw|

32. Michele_Laino

similarly for the other 2 fractions

33. Michele_Laino

after that, we have to simplify the last equation, so we get: $\Large 7x + 21 + 7x = 2{x^2} + 6x$ I have applied the distributive property of multiplication over addition Please rewrite that equation in this form: $\Large bA{x^2} + Bx + C = 0$

34. Michele_Laino

oops.. in this form: $\Large A{x^2} + Bx + C = 0$

35. Michele_Laino

what do you get?

36. anonymous

It should be 7x^2, right?

37. Michele_Laino

hint: here are more steps: $\Large \begin{gathered} 7x + 21 + 7x = 2{x^2} + 6x \hfill \\ 14x + 21 = 2{x^2} + 6x \hfill \\ \end{gathered}$

38. Michele_Laino

now I subtract 14 x from both sides so I can write: $\Large 14x + 21 - 14x = 2{x^2} + 6x - 14x$ please simplify

39. Michele_Laino

hint: $\Large \begin{gathered} 14x - 14x = 0 \hfill \\ 6x - 14x = - 8x \hfill \\ \end{gathered}$

40. anonymous

2x^2-8x-21=0

41. Michele_Laino

perfect!

42. Michele_Laino

now you have to solve that equation, using the standard formula: $\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}$ where $\Large \begin{gathered} a = 2, \hfill \\ b = - 8, \hfill \\ c = - 21. \hfill \\ \end{gathered}$

43. anonymous

Um? I can't use this formula as it's npt stated in the question *not

44. Michele_Laino

have you studied that formula in your course of mathematics?

45. anonymous

Yep.. But my teacher would add "correct to 2 or 3 decimals" ONLY THEN can I use this formula.

46. Michele_Laino

then we can use that formula, here is the next step: $\Large \begin{gathered} x = \frac{{8 \pm \sqrt {{{\left( { - 8} \right)}^2} - 4 \times 2 \times \left( { - 21} \right)} }}{{2 \times 2}} = \hfill \\ \hfill \\ = \frac{{8 \pm \sqrt {64 + 168} }}{4} = ... \hfill \\ \end{gathered}$ please continue

47. anonymous

I would get this marked wrong if I use the formula... Please understand....

48. Michele_Laino

ok! I understand

49. anonymous

So I'll HAVE TO factor it out..

50. Michele_Laino

ok! I will write your next steps

51. Michele_Laino

If I divide both sides of your equation by 2, I get: $\Large {x^2} - 4x - \frac{{21}}{2} = 0$

52. Michele_Laino

Then I add and subtract 4 at left side, so I can write: $\Large {x^2} - 4x + 4 - 4 - \frac{{21}}{2} = 0$

53. Michele_Laino

is it ok, for you?

54. anonymous

Yes

55. Michele_Laino

now we note that: $\Large {x^2} - 4x + 4 = {\left( {x - 2} \right)^2}$

56. Michele_Laino

so we can rewrite my last expression as follows: $\Large {\left( {x - 2} \right)^2} - 4 - \frac{{21}}{2} = 0$

57. Michele_Laino

and, being: $\Large - 4 - \frac{{21}}{2} = \frac{{ - 8 - 21}}{2} = - \frac{{29}}{2}$

58. Michele_Laino

we have: $\Large {\left( {x - 2} \right)^2} - \frac{{29}}{2} = 0$

59. Michele_Laino

now I add -29/2 to both sides of that equation, so I get: $\Large \begin{gathered} {\left( {x - 2} \right)^2} - \frac{{29}}{2} + \frac{{29}}{2} = 0 + \frac{{29}}{2} \hfill \\ \hfill \\ {\left( {x - 2} \right)^2} = \frac{{29}}{2} \hfill \\ \end{gathered}$

60. Michele_Laino

am I right?

61. anonymous

Yes

62. Michele_Laino

ok! now I ask you, what is: $\Large \sqrt {\frac{{29}}{2}} = ...?$ nearest to tenth?

63. anonymous

3.8

64. anonymous

wow! @Michele_Laino perfect explanation :)

65. Michele_Laino

that's right! Now if I take the square rooth of both sides of my last equation, I get: $\Large x - 2 = \pm 3.8$ please remember we have 2 square roots

66. Michele_Laino

thanks!! @behappy

67. Michele_Laino

so we have 2 equations: $\Large \begin{gathered} x - 2 = 3.8 \hfill \\ \hfill \\ x - 2 = - 3.8 \hfill \\ \end{gathered}$

68. anonymous

5.8 and -1.8

69. Michele_Laino

yes! Nevertheless -1.8 can not be acceptable, since x represents a time. So only x=5.8 is the acceptable solution

70. Michele_Laino

finally the requested time are: $\Large \begin{gathered} {t_1} = 5.8\min \hfill \\ {t_2} = 5.8 + 3 = ...\min \hfill \\ \end{gathered}$

71. Michele_Laino

times*

72. anonymous

8.8

73. Michele_Laino

that's right!

74. anonymous

THANKS!!!!!! CAN YOU PLEASE HELP ME MORE?!

75. Michele_Laino

ok!

76. Michele_Laino

thanks again for your appreciation to my work @behappy :)

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