Good day! Can someone help me with a math problem?
Boxes X, Y, and Z contain 14.7 liters of water in total.
2/7 of the water in Box X is poured into Box Y.
Then, 1/3 of the water in Box Y is poured into Box Z.
After that, the volume of water in each box was the same.
a) Find the volume of water in each box in the end.
b) Find the volume of water in Box Y at first.
What I can worked out is only:
a) Find the volume of water in each box in the end.
It's actually 14.7 / 3 = 4.9
b) Find the volume of water in Box Y at first.
X + Y + Z = 14.7
X after poured: 1 - 2/7 X = 5/7 X
Y after poured: (Y + 2/7 X) - 1/3 Y
5/7 X = (Y + 2/7 X) - 1/3 Y = 4.9
5/7 X = 14.7 / 3 = 4.9
X = 4.9 * 7 /5 = 6.86
5/7 * 6.86 = (Y + 2/7 * 6.86) - 1/3 Y = 4.9
(Y + 2/7 * 6.86) - 1/3 Y = 4.9
2/3 Y + 1.96 = 4.9
Y = 4.41
Am I getting it right? Thank you.
I think it's not correct.You can solve this using equations.
1.How much water is in y before the second pouring:
2/3 of y before=4.9
y before=4.9*3/2
ybefore=7.35
2.How much water is in y at the start:
Because x now contains 5/7 of x at start,2/7 of x at start are 2/5 of x now.X now is 4.9 so:
y before=y at start +2/5 of x now
y before=y at start +2/5 of4.9
y before=y at start +1.96
y at start=y before -1.96
y at start=7.35-1.96
y at start=5.39
If you sum all the boxes at the start you should get 14.7:
(x+y+z) at start=6.86+2.45+5.39=14.7=(x+y+z) in the end.
I think this is right,but I'm not completely sure.Try it yourself!