Simultaneous equation problem (metric system)?

Question

There were 60 litres of wine in two barrels. When 1/4 of wine was transferred from the first barrel to the second one, the amount of wine in the second barrel became three times as great a sthe new amount of wine in the first barrel. How much wine was there in each barrel originally?

And... can u explain how u did it?

Answer 1

barrelA started with 20litres..barrel B had 40litres originally. A+B=60
3/4A=(B+1/4A)/3
9/4A=B+1/4A
8/4A=B
2A=B

I did it in my head the hard part was the equation.

Answer 2

Barrel A has x liters of wine.
Barrel B has 60-x liters of wine.

Take out 1/4 of the wine from A by subtracting 1/4 x.
Add 1/4 x to Barrel B to show that the wine went into B.
Now the wine in Barrel A .. times 3 .... equals the wine in Barrel B.

3(x - 1/4 x) = (60 - x) + 1/4 x Simplify.
3x - 3/4 x = 60 - x + 1/4 x
3x - 3/4x + x - 1/4x = 60
3x = 60
x = 20

Start with 20 liters in A; 40 liters in B.
Take out a fourth of 20 from A, leaving 15 liters in A.
Take the 5 liters you removed from A and pour it into B, so B has 40 + 5 = 45 liters.

Now B has 3 times as much as A because 45 liters is 3 times 15 liters.

Answer 3

let x and y be amounts in each barrel
x + y = 60
(3/4)x becomes new amount in first barrel and
(9/4)x becomes that in second barrel.
(3/4)x +(9/4)x = 60
(12/4)x = 60
3x = 60
x = 20 so y = 40
20 litres in one barrel and 40 litres in other barrel originally.

Answer 4

suppose, one barrel contains x litre and other one y litre
so, x+y=60
now, when we remove 1/4 th from first barrel then it contain 3/4x litres
and other one now contains y+1/4x
so, form 2nd condition 3/4x=3(y+1/4x)
1/4x=y+1/4x
so, y=0
and x=60

so, inituially in first barrel there is 60 litres and nothing is in 2nd barrel