Algebra word problem, help help!!!!?

Question

Two rainstorms occurred in one week in a certain area. The first storm lasted hours, and the second storm lasted hours, for a total of rain. What was the rate of rainfall for each of the two storms if the sum of the two rates was per hour?

Two rainstorms occurred in one week in a certain area. The first storm lasted 20 hours, and the second storm lasted 40 hours, for a total 1400 mL of rain. What was the rate of rainfall for each of the two storms if the sum of the two rates was 50 mL per hour?

Answer 1

X- RATE OF STORM1
Y- RATE OF STORM2

x+y=50
20x+40y=1400
x=50-y
1000-20y+40y=1400

20y=400
y=20 so x=30

So first storm had a rate of 30 mililiters per hour
second had 20 mililiters per hour

Answer 2

(1) x / 20 = rate of rainfall for the first storm
(2) x + y = 1400
(3) x = 1400 - y

sub (3) into (1)
(1400 - y) / 20 = rate of rainfall for the first storm

(1400 - y) / 20) + (y/40) = 50

now isolate y:
40 * (((1400 - y) / 20) + (y/40) = 40 * 50
2800 - 2y + y = 2000
-y = 2000 - 2800
(4) y = 800 mL

sub (4) into (2)
x + 800 = 1400
x = 600

CHECK:
x/20 + y/40 ?= 50
600/20 + 800/40 = 50
30 + 20 = 50 mL/h

Hope that helps.

Answer 3

Try setting eqn to zer0

20x+40(50-x)-1400=0
-------
20x=2000 - 40x-1400=0
-20x+600=0
x=30 mL
plug that into 20 x+40y= 1400

600+40y=800
y=20mL

Answer 4

I still think there's something missing, but you can think of it the physics way {use units}.

Total time it rained = 40h + 20h = 60h

Total rainfall = 1400mL

Rate of rainfall {mL/h} = ?

Answer 5

let x= rate of first storm
let y = rate of second storm
20(x)+40(y)=1400
x+y=50
.....
-20y+1000+40y=1400
20y=400
y=20
x=30

Answer 6

Insufficient data. Difficult to answer.

Answer 7

wel i hope some1 really helps u out of dis rain.............. ;-)

Answer 8

WTF I got as far as naming variables then i said F uc k this