Question

How many liters of water must be added to 20 liters of 10 percent solution of salt and water to produce 6 percent solution

Answer 1

Answer:

Step-by-step explanation:

we have 10% of 20 L

SO salt concentration = 10÷100×20

=2L

So left water =20L solution - 2L salt concentration

Amount of water added to change the concentration to 6%=

suppose water added is x.

2 6

_____ = ____

20 +x 100

2 3

____= ___

20 +x 50

cross multiply

2×50=3(20+x)

60 + x = 100

x = 100 - 60

x = 40

hence 40L water is added to change 10% concentration of salt solution to 6%

Answer 2

Answer:

Let x be amount of 6% solution

20 x 0.10 = x X 0.06

X =. 20 X 0.10 / 0.06

=. 2 / 0.06

= 200 / 6

= 33.33

So the amount of water added

= 33.33 - 20

= 13.33

Thanks