# Fast Application Help!!!

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• Jan 24th 2006, 06:46 PM
dvnswl
Fast Application Help!!!
hi im in a hurry and i dont get a few problems please help...

-The diffrence between two numbers is 16. Three times the larger number is nine times smaller. What are the numbers? Show some work if you can please....

-A gardener has two kinds of solutions containing weed killer and water. One solution is 5% weed killer and the other solutiion is 15% weed killer. The gardener needs 100 liters of a 12% solution of weed killer and wants to make it by mixing it himself. How much of each solution should he use?

THX FAST PLX!
• Jan 24th 2006, 06:53 PM
ThePerfectHacker
Problem 1
Let $\displaystyle x>y$ be the 2 numbers thus,
$\displaystyle x-y=16$ (1)
$\displaystyle 3x=9y$ (2)
Divide equation (2) by 3 thus,
$\displaystyle x=3y$
Substitute that into equation (1) thus,
$\displaystyle 3y-y=16$
Thus,
$\displaystyle 2y=16$
Thus,
$\displaystyle y=8$
Thus,
$\displaystyle x=24$
Q.E.D.
• Jan 24th 2006, 07:00 PM
ThePerfectHacker
Problem 2:
Let $\displaystyle x,y$ be the amout of weedkiller solution for 5% and 15% respectively.
Thus, he want to make 100 thus,
$\displaystyle x+y=100$
Now for the second part since the first solution has 5% the amount of actual weedkiller is $\displaystyle .05x=\frac{1}{20}x$ similarily the amount of actual weekiller for the second one is $\displaystyle .15y=\frac{3}{20}$.
Now he mixes them in such a way that he has 12% of weedkiller thus, there are
12%(100) of actual weedkiller thus, 12. Thus, the two equations are:
$\displaystyle x+y=100$
$\displaystyle \frac{1}{20}x+\frac{3}{20}y=12$
Multiply the second by 20 thus,
$\displaystyle x+y=100$
$\displaystyle x+3y=240$
From the second subtract the first,
$\displaystyle 2y=140$
Thus, $\displaystyle y=70$ of 15% solution.
Thus, $\displaystyle x=30$ of 5% solution.
• Jan 24th 2006, 07:06 PM
dvnswl
I dont get how you got 1/20 and 3/20 for the equations....
Quote:

Originally Posted by ThePerfectHacker
Problem 2:
Let $\displaystyle x,y$ be the amout of weedkiller solution for 5% and 15% respectively.
Thus, he want to make 100 thus,
$\displaystyle x+y=100$
Now for the second part since the first solution has 5% the amount of actual weedkiller is $\displaystyle .05x=\frac{1}{20}x$ similarily the amount of actual weekiller for the second one is $\displaystyle .15y=\frac{3}{20}$.
Now he mixes them in such a way that he has 12% of weedkiller thus, there are
12%(100) of actual weedkiller thus, 12. Thus, the two equations are:
$\displaystyle x+y=100$
$\displaystyle \frac{1}{20}x+\frac{3}{20}y=12$
Multiply the second by 20 thus,
$\displaystyle x+y=100$
$\displaystyle x+3y=240$
From the second subtract the first,
$\displaystyle 2y=140$
Thus, $\displaystyle y=70$ of 15% solution.
Thus, $\displaystyle x=30$ of 5% solution.

• Jan 24th 2006, 08:32 PM
CaptainBlack
Quote:

Originally Posted by dvnswl
I dont get how you got 1/20 and 3/20 for the equations....

5%=5/100=1/20,

so 1/20 th of the first solution is weedkiller.

15%=15/100=3/20,

so 3/20 ths of the second solution is weedkiller.

RonL