Question

Have two tanks of 100 liters each, with salt solution. One is at 9% and 4% other. How many liters at least must be removed from each container, so that upon mixing which is removed to obtain a 7% solution of salt concentration?

Answer 1

Mathmate showed this clever solution some time ago. If I can do it right it's like this.
......4%........7%........9%
......|.7-4=3...||..9-7=2..|
So you take 2 part of 4% and mix with 3 parts of 9% to give you 5 parts of 7%.

Answer 2

.09*H+.04L=.07(H+L)

Where H is the number liters high concentration, and L is number of liters of low concentraton.

Now, H<100, and L<100,

because the problem does not say if they have to be mixed in one of the 100 L contaners....I think it meant that.

If it meant that, then 100=H+L
H=100-L
.09(100-L)+.04L=.07*100
-.05L=7-9
L= 40 H=60
so you remove 60 liters of Low concentration, and pour from the high concentration 40 liters to fill the orignally low concentration, and you have 7 percent solution.

Now if you mix the two you didn't use in the other barrel, you have
.04*60+.09*40=X*100
X=(2.4+3.6)/100=six percent solution in that barrel.