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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
esaktasynov wrote:
T - total volume of water
x - amount removed

1) 5 liters is equal to (T/2)*(2/3) hence from there we could obtain T, which is equal to 15 liters

2) Now for the x equation would look like (T/2 + 5) - x = 2T/5, plug the T into equation and you will find x, and it equals to 6.5


Please explain how you reached "1)"
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
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That is a translation into words from math.

it is one of the critical skills you should have for this exam.

The stem says

A barrel is all full so if we put the capacity as C

The barrel is C/2. Half the capacity

you add 5 liters and the resulòt is that the barrel is full to 75% or 2/3 of its capacity

C/2+5=2/3 C or 75%

Hope now is more clear

C or T it does not matter
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
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Doesn't the word increase by 2/3 mean? (1 + 2/3) = 5/3

If it were increases to 2/3 then I would understand that the equation would then be 5 + 1/2 C = 2/3C

Hence it should be 5 + 1/2 C = (1/2) C * (5/3)

5 = (5/6)C - 3/6 C
5 = 2/6 C
5 = 1/3 C
C = 15

Then since this is all hypothetical and it says decreases to 2/5, which means the final capacity is 2/5 it becomes (5 + 1/2 C) - x = 2/5 * C

5 + 1/2 * 15 - x = 2/5 * 15
5 + 7.5 - x = 6
12.5 - x = 6
x = 6.5
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
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Salsanousi wrote:
Doesn't the word increase by 2/3 mean? (1 + 2/3) = 5/3

If it were increases to 2/3 then I would understand that the equation would then be 5 + 1/2 C = 2/3C

Hence it should be 5 + 1/2 C = (1/2) C * (5/3)

5 = (5/6)C - 3/6 C
5 = 2/6 C
5 = 1/3 C
C = 15

Then since this is all hypothetical and it says decreases to 2/5, which means the final capacity is 2/5 it becomes (5 + 1/2 C) - x = 2/5 * C

5 + 1/2 * 15 - x = 2/5 * 15
5 + 7.5 - x = 6
12.5 - x = 6
x = 6.5


I understood it would increase by 2/3 of the amount already existing

that is: (T/2) + 5 = (2/3)(T/2) + (T/2)
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
katerjigeorge wrote:
Salsanousi wrote:
Doesn't the word increase by 2/3 mean? (1 + 2/3) = 5/3

If it were increases to 2/3 then I would understand that the equation would then be 5 + 1/2 C = 2/3C

Hence it should be 5 + 1/2 C = (1/2) C * (5/3)

5 = (5/6)C - 3/6 C
5 = 2/6 C
5 = 1/3 C
C = 15

Then since this is all hypothetical and it says decreases to 2/5, which means the final capacity is 2/5 it becomes (5 + 1/2 C) - x = 2/5 * C

5 + 1/2 * 15 - x = 2/5 * 15
5 + 7.5 - x = 6
12.5 - x = 6
x = 6.5


I understood it would increase by 2/3 of the amount already existing

that is: (T/2) + 5 = (2/3)(T/2) + (T/2)


Hi, yeah that is my understanding that it increases by (2/3) of the amount existing which is another way of saying (1 + 2/3)

In your equation if (T/2) is a common factor then it becomes (T/2) * (2/3 + 1).

Thanks!
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
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Carcass wrote:
If 5 liters of water are added to a barrel when it is half full of water, the amount of water in the barrel will increase by \(\frac{2}{3}\). If x liters of water are then removed from the barrel, the amount of water in the barrel will decrease to \(\frac{2}{5}\) of the capacity of the barrel. What is the value of X?


Show: :: OA_x liters
6.5


Let v = the ORIGINAL volume of water in the tank (i.e., when the tank was HALF full)
If 5 liters of water are added to a barrel when it is half full of water, the amount of water in the barrel will increase by \(\frac{2}{3}\)
We can write: v + 5 = v + (2/3 of v)
In other words: v + 5 = v + 2v/3
Subtract v from both sides of the equation: 5 = 2v/3
Multiply both sides by 3 to get: 15 = 2v
Solve: v = 15/2 = 7.5

Important: Since v = the volume of water when the tank was HALF full, we can conclude that of the total CAPACITY of the barrel = 15 liters
Also important: Since we added 5 liters of water to the tank, the tank NOW contains 12.5 liters of water

If x liters of water are then removed from the barrel, the amount of water in the barrel will decrease to \(\frac{2}{5}\) of the capacity of the barrel.
We can write: 12.5 - x = 2/5 of 15
In other words: 12.5 - x = 6
Solve: x = 6.5

Answer: 6.5

Cheers,
Brent
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
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Re: If 5 liters of water are added to a barrel when it is half f [#permalink]
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