SOLUTION: 1. Mr. Gardener has two kinds of solutions containing fertilizer and water. One of the solutions is 10% fertilizer and the other is 30% fertilizer. Mr. Gardener needs 200 liters of

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: 1. Mr. Gardener has two kinds of solutions containing fertilizer and water. One of the solutions is 10% fertilizer and the other is 30% fertilizer. Mr. Gardener needs 200 liters of      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1117445: 1. Mr. Gardener has two kinds of solutions containing fertilizer and water. One of the solutions is 10% fertilizer and the other is 30% fertilizer. Mr. Gardener needs 200 liters of a 24% fertilizer solution. How much of the 10% solution should Mr. Gardener use in order to have the 24% fertilizer solution?

2. The top floor of the school building is a rectangle whose perimeter is 320 feet. The width is 30 feet less than the length. What is the width of the rectangle?

3. Jane can clean the living room in 3 hours, Kai in 6 hours, and Dana in 8 hours. If they work together, in how many minutes can they clean the entire room?

4. Rachelle invested part of $185 000 at 7% and the rest 9%. If the interest from 9% investment is $1 450 more than the interest earned at the 7% investment. How much did she invest at the 9% rate?

5. Joy's house, the supermarket, and Marty's office are on a straight road. Joy rode her car and traveled with a speed of 60 kph for 10 minutes. Marty left his office to meet Joy at the supermarket. He used his car and traveled at the speed of 90 kph for 5 mins. How far is Marty's office from Joy's house?

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
1. Mr. Gardener has two kinds of solutions containing fertilizer and water. One of the solutions is 10% fertilizer and the other is 30% fertilizer. Mr. Gardener needs 200 liters of a 24% fertilizer solution. How much of the 10% solution should Mr. Gardener use in order to have the 24% fertilizer solution?
x = 10% solution
y = 30% solution
x + y = 200
.1x + .3y = .24 * (x + y) = .24 * 200 = 48
equations become:
x + y = 200
.1x + .3y = 48
solve for x from first equation to get x = 200 - y
replace x with 200 - y in second equation to get:
.1(200 - y) + .3y = 48
simplify to get 20 - .1y + .3y = 48
combine like terms to get 20 + .2y = 48
solve for y to get y = (48 - 20) / .2 = 140
that makes x = 60
you have x = 60 and y = 140
60 + 140 = 200
.1 * 60 + .3 * 140 = 48
48 / 200 = .24
that's your solution.
you need 60 liters of the 10% solution mixed with 140 liters of the 30% solution.

2. The top floor of the school building is a rectangle whose perimeter is 320 feet. The width is 30 feet less than the length. What is the width of the rectangle?

perimeter= 2L +2W, therefore 2L + 2W = 320
W = L - 30
replace W with L - 30 to get:
2L + 2(L - 30) = 320
simplify to get 4L - 60 = 320
add 60 to both sides of the equaiton to get 4L = 380
solve for L to get L = 95
W = L - 30 makes W = 65
2L = 2 * 95 = 190
2W = 2 * 65 = 130
2L + 2W = 320
solution is that the width of the rectangle is 65 feet.

3. Jane can clean the living room in 3 hours, Kai in 6 hours, and Dana in 8 hours. If they work together, in how many minutes can they clean the entire room?
rate * time = quantity
quantity = 1 living room.
time = hours.
rate = fraction of the job done in 1 hour.
jane takes 3 hours.
her rate is therefore 1/3 of the job in 1 hour.
kai takes 6 hours
here rate is therefore 1/6 of the job in 1 hour.
dana takes 8 hours.
her rate is therefore 1/8 of the job in 1 hour.
when they work together, their rates are added together.
you get (1/3 + 1/6 + 1/8) * time = 1
least common multiple is 24.
formula becomes (8/24 + 4/24 + 3/24) * time = 1
combine like terms to get 15/24 * time = 1
solve for time to get time = 24/15 hours.
there are 60 minutes in an hour, therefore 24/15 * 60 = 96 minutes
working together, it will take them 96 minutes to clean the living room.

4. Rachelle invested part of $185000 at 7% and the rest at 9%. If the interest from 9% investment is $1450 more than the interest earned at the 7% investment. How much did she invest at the 9% rate?

x = amount invested at 7%
y = amount imvested at 9%
x + y = 185000
interest earned from 9% investment is 1450 more than interest earned from 7% investment.
this means .09y = .07x + 1450
solve for y to get y = (.07x + 1450)/.09
in x + y = 185000, replace y with (.07x + 1450) / .09 to get:
x + (.07x + 1450)/.09 = 185000
multiply both sides of this equation by .09 to get:
.09x + .07x + 1450 = 16650
combine like terms and subtract 1450 from both sides to get:
.16x = 15200.
solve for x to get x = 15200 / .16 = 95000
that makes y = 185000 - 95000 = 90000
x + y becomes 95000 + 90000 = 185000
interest earned on 7% investment = .07 * 95000 = 6650
interest earned on 9% investment = .09 * 90000 = 8100
8100 - 6650 = 1450
interested earned on 9% investment is 1450 more than interest earned on 9% investment.
everything looks good.
your solution is that she invested 90000 at 9% rate of return.


5. Joy's house, the supermarket, and Marty's office are on a straight road. Joy rode her car and traveled with a speed of 60 kph for 10 minutes. Marty left his office to meet Joy at the supermarket. He used his car and traveled at the speed of 90 kph for 5 mins. How far is Marty's office from Joy's house?

rate * time = distance
minutes / 60 = hours
joy traveled 60 kph * 10/60 hours = 10 kilometers.
marty traveled 90 * 5/60 = 7.5 kilometers.
they met at the supermarket.
marty's office is 10 + 7.5 = 17.5 kilometers from joy's house.

any questions, send email to dtheohilis@gmail.com