Results 1 to 5 of 5

Math Help - fluid concentration, workhour problem

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    5

    Wink fluid concentration, workhour problem

    I need help for these problems :

    1. 1000 kgs of chemical is stored in a container. The chemical is made up of 99% water and 1% oil. Some water is evaporated from the chemical until the water content is reduced to 96%. How much does the chemical weigh now ?

    2. A piece of pasture grow at a constant rate everyday.
    200 sheep will eat up the grass in 100 days.
    150 sheep will eat up the grass in 150 days.
    How many days does it take for 100 sheep to eat up the grass?

    Waiting for expert advice. Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    600
    Hello, agus hendro!

    Here's the first one . . .


    1. 1000 kgs of chemical is stored in a container.
    The chemical is made up of 99% water and 1% oil.
    Some water is evaporated until the water content is reduced to 96%.
    How much does the chemical weigh now?
    Consider the number of kgs of water at each stage.

    We start with 1000 kgs of solution which is 99% water.
    . . It contains: . 99\% \times 1000 \:=\:990 kgs of water.

    We remove x kgs of water.
    . . It now contains: . \boxed{990 - x} kgs of water.


    Start again . . . We have 1000 kgs of solution.
    . . We remove x kgs of water.
    So we have: 1000 - x kgs of stuff.

    But this is supposed to be 96% water.
    . . So it contains: . \boxed{0.96(1000-x)} kgs of water.


    We just described the final amount of water in two ways.

    There is our equation! . . . . {\color{blue}990 - x \:=\:0.96(1000-x)}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    5

    Wink

    I found another problem like the second problem.

    Teams X and Y work separately on two different projects.
    On sunny days, team X can complete the work in 12 days, while team Y needs 15 days.
    On rainy days, team X's efficiency decreases by 50%, while team Y's efficiency decreases by 25%.
    Given that the two teams started and ended the projects at the same time, how many rainy days are there ?

    I think if I got the clue for the second problem, this new problem will be easy for me. Please help......
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    600
    Hello, agus hendro!

    I think I've solved it . . .


    Teams X and Y work separately on two different projects.
    On sunny days, team X can complete the work in 12 days,
    . . while team Y needs 15 days.
    On rainy days, team X's efficiency decreases by 50%,
    . . while team Y's efficiency decreases by 25%.

    Given that the two teams started and ended the projects
    at the same time, how many rainy days are there ?

    In sunny weather, team X can do the job in 12 days.
    In one sunny day, team X can do \frac{1}{12} of the job.
    . . In S sunny days, X can do \frac{S}{12} of the job.

    In rainy weather, team X can do the job in 24 days.
    In one rainy day, can do \frac{1}{24} of the job.
    . . In R rainy days, can do \frac{R}{24} of the job.

    X's equation is: . \frac{S}{12} + \frac{R}{24} \:=\:1\;\;{\color{blue}[1]}



    In sunny weather, team Y can do the job in 15 days.
    In one sunny day, Y can do \frac{1}{15} of the job.
    . . In S sunny days, Y can do \frac{S}{15} of the job.

    In rainy weather, team Y can do the job in: . 125\% \times 15 \:=\:\frac{75}{4} days.
    In one rainy day, Y can do \frac{4}{75} of the job.
    . . In R rainy days, Y can do \frac{4R}{75} of the job.

    Y's equation is: . \frac{S}{15} + \frac{4R}{75} \:=\:1\;\;{\color{blue}[2]}



    \begin{array}{ccccc}\text{Multiply {\color{blue}[1]} by 96:} & 8S + 4R &=& 96 & {\color{blue}[3]} \\<br />
\text{Multiply {\color{blue}[2]} by 75:} & 5S + 4R &=& 75 & {\color{blue}[4]} \end{array}

    Subtract [4] from [3]: . 3S \:=\:21\quad\Rightarrow\quad S\,=\,7

    Substitute into [4]: . 5(7) + 4R \:=\:75\quad\Rightarrow\quad R \:=\:10


    Therefore, there were \boxed{10\text{ rainy days}}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2008
    Posts
    5

    Wink still don't understand

    Thank you for the answer. But please explain why do you add s/12 and r/24 and how come the result of the addition equal to 1?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Adding Concentration Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 27th 2011, 12:09 AM
  2. Replies: 0
    Last Post: February 7th 2010, 10:34 AM
  3. Concentration Problem Revise Please
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 11th 2009, 04:27 PM
  4. Mixed Concentration Problem, please revise.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 11th 2009, 12:35 PM
  5. concentration problem
    Posted in the Algebra Forum
    Replies: 13
    Last Post: October 3rd 2008, 10:13 AM

Search Tags


/mathhelpforum @mathhelpforum