SOLUTION: A dairy farmer had 54 liters of milk in 3 churns. He wanted to make sure each churn contained 18 liters. First, he poured 1/4 of the firs churn into the second churn. Then he poure

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: A dairy farmer had 54 liters of milk in 3 churns. He wanted to make sure each churn contained 18 liters. First, he poured 1/4 of the firs churn into the second churn. Then he poure      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 251219: A dairy farmer had 54 liters of milk in 3 churns. He wanted to make sure each churn contained 18 liters. First, he poured 1/4 of the firs churn into the second churn. Then he poured 1/2 of the second churn into the third churn. Finally, he poured 1/3 of the third churn into the first churn. How many liters did each churn contain before he started pouring?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A dairy farmer had 54 liters of milk in 3 churns.
He wanted to make sure each churn contained 18 liters.
First, he poured 1/4 of the firs churn into the second churn.
Then he poured 1/2 of the second churn into the third churn.
Finally, he poured 1/3 of the third churn into the first churn.
How many liters did each churn contain before he started pouring?
:
Let x, y, z = the amts in each churn originally
:
"A dairy farmer had 54 liters of milk in 3 churns."
x + y + z = 54
:
"he poured 1/4 of the first churn into the second churn.
then
1st churn = .75x
2nd churn = (y+.25x)
:
"Then he poured 1/2 of the second churn into the third churn."
then
2nd churn = .5(y+.25x) = (.5y+.125x)
3rd churn = z+.5y+.125x
:
" he poured 1/3 of the third churn into the first churn."
then
3rd churn = 2%2F3(z+.5y+ .125x)
1st churn = .75x + 1%2F3(z+.5y+.125x)
:
At this point we assume all churns = 18 gal
:
1st churn
.75x + 1%2F3(z+.5y+.125x) = 18
Multiply by 3:
2.25x + z + .5y + .125x = 54
2.375x + .5y + z = 54
:
2nd churn
.5y + .125x = 18
.125x + .5y = 18
.5y = 18 - .125x
multiply by 2, can use this for substitution
y = (36 - .25x)
:
3rd churn
2%2F3(z+.5y+ .125x) = 18
multiply by 3
2(z+.5y + .125x) = 54
2z + y + .25x = 54
.25x + y + 2z = 54
Replace y with (36-.25x)
.25x + (36-.25x) + 2z = 54
.25x - .25x + 2z = 54 - 36; (x is conveniently eliminated)
2z = 18
z = 9 gal originally
:
Using the total equation
x + y + 9 = 54
x + y = 54 - 9
x + y = 45
we also know, (use these two equation for elimination)
.125x + .5y = 18
Multiply by 2, subtract x + y = 45
.25x + y = 36
x + y = 45
-----------------subtraction eliminates y
-.75x = -9
x = %28-9%29%2F%28-.75%29
x = 12 gal originally
then
12 + y + 9 = 54
y = 54 - 21
y = 33 gal originally
;
:
See if that works
"he poured 1/4 of the first churn into the second churn.
then
1st churn has 9 gal
2nd churn has 36 gal
:
"poured 1/2 of the second churn into the third churn."
then
2nd churn has 18 gal
3rd churn has 27 gal
:
" he poured 1/3 of the third churn into the first churn."
then
3rd churn = 18
1st churn = 18
:
How many liters did each churn contain before he started pouring?
x = 12; y = 33, z = 9
:
Merry Christmas, hope this made sense!!