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anonymous

  • one year ago

I'll give you a medal and be a fan if you HELP PLEASE 1. The pump used to fill the water tank on a fire truck takes 40 minutes to fill the tank. If the pump sending water to the hose is on, it takes 60 minutes to fill the tank. If the pump filling the water tank is off, how long does the pump sending water to the hose take to empty the tank? A. 120 minutes B. 60 minutes C. 30 minutes D. 24 minutes 2. Annmarie can plow a field in 240 minutes. Gladys can plow a field 80 minutes faster. If they work together, how many minutes does it take them to plow the field? A. 96 minutes B. 160 minutes C. 400 minutes D. 480 minutes 3.Brody can fill a bowl with candy in 3 minutes. While Brody fills the bowl, Hudson takes the candy out of the bowl. With Hudson taking candy out of the bowl, it takes 5 minutes for Brody to fill the bowl. Which of the following can be used to determine the amount of time it takes for Hudson to empty the bowl if Perry does not add candy? A. 1 over 3 minus 1 over x equals 1 over 5 B. 1 over 3 minus 1 over 5 equals 1 over x C. 1 over 5 minus 1 over x equals 1 over 3 D. 1 over 3 minus 1 over x equals x over 5 4. Leanne can clean a fish tank in 30 minutes. Karl can clean a fish tank in 20 minutes. If they work together, how many minutes does it take them to clean a fish tank? A. 6 minutes B.10 minutes C. 12 minutes D. 50 minutes 5.Genna can build a cabinet in 4 hours. Claude can build a cabinet in only 3 hours. Which of the following can be used to determine the amount of time it would take for Genna and Claude to build a cabinet together? A. 1 over 4 plus 1 over x equals 1 over 3 B. 1 over 4 plus 1 over 3 equals 1 over x C. 1 over x plus 1 over 3 equals 1 over 4 D. 1 over 3 plus 1 over 4 equals x over 7

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  1. anonymous
    • one year ago
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    @ganeshie8

  2. anonymous
    • one year ago
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    @iGreen

  3. anonymous
    • one year ago
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    @Michele_Laino

  4. anonymous
    • one year ago
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    @TheSmartOne

  5. Michele_Laino
    • one year ago
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    Question 5) the working rate of Genna is W/4, the working rate of Claude is W/3, where W is the work to do. When Gwenna and Claude work together, then the working rate is: W/4 + W/3 so we can write: \[\Large \left( {\frac{W}{4} + \frac{W}{3}} \right)t = W\] where t is time needed to Genna and Claude to do the work W Please solve that equation for T, what do you get?

  6. Michele_Laino
    • one year ago
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    oops.. for t, what do you get?

  7. anonymous
    • one year ago
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    3 hrs?

  8. Michele_Laino
    • one year ago
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    hint: what is: \[\frac{1}{4} + \frac{1}{3} = ...?\]

  9. anonymous
    • one year ago
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    \[\frac{ 7 }{ 12 }\]

  10. Michele_Laino
    • one year ago
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    ok! so, from preceding equation, we can write: \[\frac{{7W}}{{12}}t = W\] or: \[\frac{7}{{12}}t = 1\]

  11. anonymous
    • one year ago
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    ok

  12. Michele_Laino
    • one year ago
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    namely: \[t = \frac{{12}}{7}\]

  13. anonymous
    • one year ago
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    so the answer is D then?

  14. Michele_Laino
    • one year ago
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    yes!

  15. anonymous
    • one year ago
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    Thanks ! can you help with the others ?

  16. Michele_Laino
    • one year ago
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    ok! Question 4)

  17. Michele_Laino
    • one year ago
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    the working rate of Leanne is W/30, where the working rate of Karl is W/20, where as usual, W is the work to do. So when Leanne and Karl work together the working rate is: W/30+ W/20, and we can write: \[\Large \left( {\frac{W}{{30}} + \frac{W}{{20}}} \right)t = W\] where t is time neeeded to Leanne and Karl to finish the work W. Please solve for t

  18. Michele_Laino
    • one year ago
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    oops...whereas the working rate of Karl

  19. Michele_Laino
    • one year ago
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    as before, we can write: \[\Large \left( {\frac{1}{{30}} + \frac{1}{{20}}} \right)t = 1\]

  20. Michele_Laino
    • one year ago
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    what is t?

  21. anonymous
    • one year ago
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    \[\frac{ 5 }{ 15 } ?\]

  22. Michele_Laino
    • one year ago
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    we have: \[\frac{1}{{30}} + \frac{1}{{20}} = \frac{{2 + 3}}{{60}} = ...?\]

  23. anonymous
    • one year ago
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    i entered the wrong thing my bad!

  24. anonymous
    • one year ago
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    \[\frac{ 1 }{ 12 }\]

  25. Michele_Laino
    • one year ago
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    ok! So we have: \[\frac{t}{{12}} = 1\]

  26. Michele_Laino
    • one year ago
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    and therefore: \[t = 12 \times 1 = ...?\]

  27. anonymous
    • one year ago
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    12?

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

  29. Michele_Laino
    • one year ago
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    it is 12 minutes

  30. anonymous
    • one year ago
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    Ok Thanks!

  31. anonymous
    • one year ago
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    Have time for another one?

  32. Michele_Laino
    • one year ago
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    yes!

  33. Michele_Laino
    • one year ago
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    question 3)

  34. anonymous
    • one year ago
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    ok

  35. Michele_Laino
    • one year ago
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    I'm pondering....

  36. Michele_Laino
    • one year ago
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    the working rate of Brody is: W/3, when he works alone whereas the working rate of Brody is W/5, when Brody and Hudson work together W is the work to be done Now the working rate of Hudson is: W/x, where x is the unknown quantity. So, we can write: \[\Large \frac{W}{3} - \frac{W}{x} = \frac{W}{5}\] or: \[\Large \frac{1}{3} - \frac{1}{x} = \frac{1}{5}\] so, what is the right option?

  37. Michele_Laino
    • one year ago
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    hint: \[\Large \frac{1}{x} = \frac{1}{3} - \frac{1}{5}\]

  38. Michele_Laino
    • one year ago
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    @amielli

  39. anonymous
    • one year ago
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    2/15

  40. Michele_Laino
    • one year ago
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    yes! What is the right option?

  41. anonymous
    • one year ago
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    D?

  42. Michele_Laino
    • one year ago
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    I think that option D is wrong

  43. Michele_Laino
    • one year ago
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    please look at this equation: \[\Large \frac{1}{3} - \frac{1}{x} = \frac{1}{5}\] do you recognize to which option corresponds to?

  44. anonymous
    • one year ago
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    yes that would be A

  45. Michele_Laino
    • one year ago
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    yes! correct!

  46. anonymous
    • one year ago
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    I dont get it though

  47. Michele_Laino
    • one year ago
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    since Brody is filling the bowl, he is adding candies to that bowl. Hudson is subtracting candies to that bowl. Here is why I used the minus sign in the subsequent equation: \[\large \frac{W}{3} - \frac{W}{x} = \frac{W}{5}\]

  48. Michele_Laino
    • one year ago
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    now, is it clear?

  49. anonymous
    • one year ago
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    Oh yes I do

  50. anonymous
    • one year ago
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    Thanks

  51. Michele_Laino
    • one year ago
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    ok!

  52. Michele_Laino
    • one year ago
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    question 2)

  53. Michele_Laino
    • one year ago
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    here we can write: working rate of Annmarie is W/240, whereas the working rate of Gladies is W/80. Now when Annmarie and Gladies work together, the working rate is: W/240 + W/80. As usual W is the work to be done. So we can write: \[\Large \left( {\frac{W}{{240}} + \frac{W}{{80}}} \right)t = W\] or: \[\Large \left( {\frac{1}{{240}} + \frac{1}{{80}}} \right)t = 1\] Please solve for t

  54. Michele_Laino
    • one year ago
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    hint: \[\Large \frac{1}{{240}} + \frac{1}{{80}} = \frac{{1 + 3}}{{240}} = ...?\]

  55. Michele_Laino
    • one year ago
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    please wait a moment, I think to have made an error

  56. anonymous
    • one year ago
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    ok

  57. Michele_Laino
    • one year ago
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    the working rate of Gladies is: \[\Large \frac{W}{{240 - 80}} = \frac{W}{{160}}\]

  58. Michele_Laino
    • one year ago
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    so the right equation is: \[\Large \left( {\frac{W}{{240}} + \frac{W}{{240 - 80}}} \right)t = W\]

  59. Michele_Laino
    • one year ago
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    or: \[\Large \left( {\frac{1}{{240}} + \frac{1}{{160}}} \right)t = 1\]

  60. Michele_Laino
    • one year ago
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    please solve for t, keep in mind that: \[\Large \frac{1}{{240}} + \frac{1}{{160}} = \frac{{2 + 3}}{{480}} = \frac{5}{{480}}\]

  61. Michele_Laino
    • one year ago
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    after a substitution, we have: \[\Large \frac{5}{{480}}t = 1\] what is t?

  62. anonymous
    • one year ago
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    just simplify it?

  63. Michele_Laino
    • one year ago
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    hint: \[\Large t = \frac{{480}}{5} = ...?\]

  64. anonymous
    • one year ago
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    96

  65. anonymous
    • one year ago
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    I get that one

  66. anonymous
    • one year ago
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    So the last one

  67. Michele_Laino
    • one year ago
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    ok! question 1)

  68. Michele_Laino
    • one year ago
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    this question is similar to the question of candies, namely question 3). So the rate of working for the pump is W/40, when the pump works alone the working rate of the firehose is W/x, where x is the unknown quantity and W/60 is the working rate of the pump, when the pump and the firehose work together. So we can write: \[\Large \frac{W}{{40}} - \frac{W}{x} = \frac{W}{{60}}\] or: \[\Large \frac{1}{{40}} - \frac{1}{x} = \frac{1}{{60}}\]

  69. anonymous
    • one year ago
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    so the answer is 60?

  70. Michele_Laino
    • one year ago
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    no, since we have to compute the value of x

  71. Michele_Laino
    • one year ago
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    hint: \[\Large \frac{1}{x} = \frac{1}{{40}} - \frac{1}{{60}} = \frac{{3 - 2}}{{120}} = ...\]

  72. Michele_Laino
    • one year ago
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    please continue my computation

  73. anonymous
    • one year ago
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    1/120

  74. Michele_Laino
    • one year ago
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    ok! so we have: \[\Large x = \frac{{120}}{1} = ...{\text{minutes}}\]

  75. anonymous
    • one year ago
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    thanks very much I got 80%

  76. anonymous
    • one year ago
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    number 5 was wrong

  77. Michele_Laino
    • one year ago
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    are you sure?

  78. Michele_Laino
    • one year ago
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    I'm very sorry, the correct answer is option B. Sorry again

  79. anonymous
    • one year ago
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    That is totally fine

  80. anonymous
    • one year ago
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    Thanks!

  81. Michele_Laino
    • one year ago
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    since we can write this: \[\Large \left( {\frac{W}{4} + \frac{W}{3}} \right)t = W\] and then: \[\Large \frac{1}{4} + \frac{1}{3} = \frac{1}{t}\] Sorry again

  82. Michele_Laino
    • one year ago
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    thanks for your comprehension!

  83. anonymous
    • one year ago
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    Thanks for helping me

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