Asolution of salt and water is 20% salt. if 25% of the water is removed, what is the percentage of salt in the new solution?

So lets say that this salt and water solution consists of 100 liters.

Then that means 20 liters is salt, and 80 liters out of the 100 is water.

Now, we remove 25% of the water. 25% of 80 liters is 20 liters. So, we are left with 80-20, or 60 liters.

Now in all we have 20 liters of salt, and 60 liters of water.

The percentage of salt in the new mixture will be 20/(20+60) = 20/80 = 25%.

Your answer is 25%

+1,000)=g7e=mc2-g=7 because if you know algebra you should know g=7 to get your answer

Answer: b 2 = 1.

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

therefore, b 2= 1.