ratio calculation problem, can someone help?

Question

the problem is like this : an online recipe calls for 2 liter of 3% bleach to 50 liter of water. so, how many liter of 11% bleach to 50 liter of water ?

anybody able to teach me ?

thanks andrew

Answer 1

Concentration and volume are inversely proportional to each other. So if the volume of the bleach increases by a multiple of $x$(say) , then the concentration should be decreased to the initial concentration divided by $x$, to maintain equivalence of the amount of bleach(net) and vice-versa.[Provided the volume of water is same.]

Now, $V_1 = 2L$ , $C_1 = 3 \%$

$V_2 = ? \ , \ C_2 = 11 \%$

So, $$\frac{C_2}{C_1} = \frac{11}{3}$$

Now for volume, the ratio should be inverted. $$\frac{V_2}{V_1} = \frac{3}{11}$$ $$V_2 = \frac{3}{11}\cdot2 = \frac{6}{11} \approx 0.545\cdots$$

Answer 2

There are two different ways to answer this question. The first is simply to consider that we need to add the same amount of pure bleach. In the first case, we add $2\cdot.03=.06$ liter of bleach so in the second case, if we add $k$ liters of $11\%$ bleach we want $$.11k=.06\implies k={6\over11}=.545454...$$

The other way to solve the problem is to say that when we add $2$ liters of $3\%$ bleach we add $60$ cc of bleach and $1940$ cc of water, so that the concentration of bleach is ${60\over51940}={6\over5194}$ Now if we add $k$ liters of $11\%$ bleach we will add $110$ cc of bleach and $890$ cc of water, so to get the same concentration, we must have$${110k\over50000+890k}={6\over5194}$$

Solving this gives $k={300\over566}\approx.5300$