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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
    • one year ago
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    @Michele_Laino @pooja195 @perl

  2. perl
    • one year ago
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    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
    • one year ago
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    I can't understand the above.

  4. perl
    • one year ago
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    Have you used the formula distance = rate * time

  5. anonymous
    • one year ago
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    Yes..

  6. perl
    • one year ago
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    I am using a similar equation. 1 cistern = rate of pipe * time

  7. anonymous
    • one year ago
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    What next?

  8. perl
    • one year ago
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    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
    • one year ago
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    3 cistern?

  10. perl
    • one year ago
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    they are three different equations

  11. perl
    • one year ago
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    "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
    • one year ago
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    Help PLEASE?

  13. anonymous
    • one year ago
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    @Michele_Laino

  14. Michele_Laino
    • one year ago
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    I think that @perl gave you the right explanation

  15. Michele_Laino
    • one year ago
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    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
    • one year ago
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    where W is the work to be done

  17. Michele_Laino
    • one year ago
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    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
    • one year ago
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    x+3+x/x^2+3x = 1/3.5?

  19. Michele_Laino
    • one year ago
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    first step: we have: \[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{2}{7}\]

  20. Michele_Laino
    • one year ago
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    yes! since 3.5= 7/2

  21. Michele_Laino
    • one year ago
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    now, the least common multiple, of x, x+3 and 7, is: x*(x+3)*7 am I right?

  22. anonymous
    • one year ago
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    Yes

  23. Michele_Laino
    • one year ago
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    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
    • one year ago
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    oops..is that equation equivalent to this one?

  25. anonymous
    • one year ago
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    I guess..

  26. Michele_Laino
    • one year ago
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    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
    • one year ago
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    now?

  28. anonymous
    • one year ago
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    I'm lost...

  29. Michele_Laino
    • one year ago
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    as denominator, of each fraction, we find the least common multiple

  30. anonymous
    • one year ago
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    Getting it.. Now?

  31. Michele_Laino
    • one year ago
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    now, if we consider the first fraction, for example, we have: |dw:1436624598950:dw|

  32. Michele_Laino
    • one year ago
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    similarly for the other 2 fractions

  33. Michele_Laino
    • one year ago
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    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
    • one year ago
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    oops.. in this form: \[\Large A{x^2} + Bx + C = 0\]

  35. Michele_Laino
    • one year ago
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    what do you get?

  36. anonymous
    • one year ago
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    It should be 7x^2, right?

  37. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    hint: \[\Large \begin{gathered} 14x - 14x = 0 \hfill \\ 6x - 14x = - 8x \hfill \\ \end{gathered} \]

  40. anonymous
    • one year ago
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    2x^2-8x-21=0

  41. Michele_Laino
    • one year ago
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    perfect!

  42. Michele_Laino
    • one year ago
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    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
    • one year ago
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    Um? I can't use this formula as it's npt stated in the question *not

  44. Michele_Laino
    • one year ago
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    have you studied that formula in your course of mathematics?

  45. anonymous
    • one year ago
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    Yep.. But my teacher would add "correct to 2 or 3 decimals" ONLY THEN can I use this formula.

  46. Michele_Laino
    • one year ago
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    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
    • one year ago
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    I would get this marked wrong if I use the formula... Please understand....

  48. Michele_Laino
    • one year ago
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    ok! I understand

  49. anonymous
    • one year ago
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    So I'll HAVE TO factor it out..

  50. Michele_Laino
    • one year ago
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    ok! I will write your next steps

  51. Michele_Laino
    • one year ago
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    If I divide both sides of your equation by 2, I get: \[\Large {x^2} - 4x - \frac{{21}}{2} = 0\]

  52. Michele_Laino
    • one year ago
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    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
    • one year ago
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    is it ok, for you?

  54. anonymous
    • one year ago
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    Yes

  55. Michele_Laino
    • one year ago
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    now we note that: \[\Large {x^2} - 4x + 4 = {\left( {x - 2} \right)^2}\]

  56. Michele_Laino
    • one year ago
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    so we can rewrite my last expression as follows: \[\Large {\left( {x - 2} \right)^2} - 4 - \frac{{21}}{2} = 0\]

  57. Michele_Laino
    • one year ago
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    and, being: \[\Large - 4 - \frac{{21}}{2} = \frac{{ - 8 - 21}}{2} = - \frac{{29}}{2}\]

  58. Michele_Laino
    • one year ago
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    we have: \[\Large {\left( {x - 2} \right)^2} - \frac{{29}}{2} = 0\]

  59. Michele_Laino
    • one year ago
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    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
    • one year ago
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    am I right?

  61. anonymous
    • one year ago
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    Yes

  62. Michele_Laino
    • one year ago
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    ok! now I ask you, what is: \[\Large \sqrt {\frac{{29}}{2}} = ...?\] nearest to tenth?

  63. anonymous
    • one year ago
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    3.8

  64. anonymous
    • one year ago
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    wow! @Michele_Laino perfect explanation :)

  65. Michele_Laino
    • one year ago
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    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
    • one year ago
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    thanks!! @behappy

  67. Michele_Laino
    • one year ago
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    so we have 2 equations: \[\Large \begin{gathered} x - 2 = 3.8 \hfill \\ \hfill \\ x - 2 = - 3.8 \hfill \\ \end{gathered} \]

  68. anonymous
    • one year ago
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    5.8 and -1.8

  69. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    times*

  72. anonymous
    • one year ago
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    8.8

  73. Michele_Laino
    • one year ago
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    that's right!

  74. anonymous
    • one year ago
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    THANKS!!!!!! CAN YOU PLEASE HELP ME MORE?!

  75. Michele_Laino
    • one year ago
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    ok!

  76. Michele_Laino
    • one year ago
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    thanks again for your appreciation to my work @behappy :)

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