# fluid concentration, workhour problem

• April 17th 2008, 05:17 AM
agus hendro
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.
(Doh)
• April 17th 2008, 09:19 AM
Soroban
Hello, agus hendro!

Here's the first one . . .

Quote:

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.

. . 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! . . . . ${\color{blue}990 - x \:=\:0.96(1000-x)}$

• April 18th 2008, 07:34 AM
agus hendro
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......
• April 20th 2008, 11:19 AM
Soroban
Hello, agus hendro!

I think I've solved it . . .

Quote:

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: . $\frac{S}{15} + \frac{4R}{75} \:=\:1\;\;{\color{blue}[2]}$

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

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 $\boxed{10\text{ rainy days}}$

• April 22nd 2008, 05:54 AM
agus hendro
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?
(Wondering)