Question

How many liters of 60% alcohol solution should be added to 40 liters of a 20% alcohol solution to make a 50% solution?

Answer 1

Answer:

120 liters of 60% alcohol solution is needed

Step-by-step explanation:

Let's assume

x liters of 60% alcohol solution  is added

so, we  get solution as

$=\frac{60}{100}\times x$

$=0.6x$

we are given

40 liters of a 20% alcohol solution is

$=40\times\frac{20}{100}$

$=40\times\frac{20}{100}$

$=8$

so, total alcohol solution is

$=8+0.6x$

now, we can find total solutions

total solutions is

$=x+40$

now, it is making 50% solution

so, we get

$\frac{8+0.6x}{x+40} =\frac{50}{100}$

now, we can solve for x

$\left(8+0.6x\right)\cdot \:100=\left(x+40\right)\cdot \:50$

$800+60x=50x+2000$

$10x=1200$

$x=120$

So,

120 liters of 60% alcohol solution is needed