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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

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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. $$\frac{V_2}{V_1} = \frac{3}{11}$$ $$V_2 = \frac{3}{11}\cdot2 = \frac{6}{11} \approx 0.545\cdots$$

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  • $\begingroup$ Hii AK19, Your method is hard for me to understand. whats your calc method called ? how come $$\frac{V_2}{V_1} = \frac{3}{11}$$ aren't they are nothing the same ? if you are using ratio to calc, isn't that you suppose to cross multiply ? but instead, you linked $$\frac{V_2}{V_1} to \frac{11}{3}$$ which 11 and 3 suppose to be concentration of the volume1 and volume2. i understand they are directly inverse proportionate.. but linking concentration to volume... is weird. i just can't understand. maybe you can further explain a bit ? appreciate your help. thank you. $\endgroup$ – andrew_ysk May 18 at 20:41
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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 $k={300\over566}\approx.5300$

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  • $\begingroup$ thankyou for your help. i do understand the 1 approach of solving, but the 2nd approach made me scratch my head. how do u know we will have to add 110cc of 11% bleach to 890cc of water to have the same concentration ? $\endgroup$ – andrew_ysk May 18 at 20:21
  • $\begingroup$ Every liter of $11\%$ bleach contains $110$ cc of pure bleach and $890$ cc of pure water. $\endgroup$ – saulspatz May 18 at 20:36

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