Mixtures and solutions

Question

Suppose we have two solutions of sulphuric acid in water. The first is 40% strong and second is 60% strong. We mix the two solutions, add 5 kg of pure water and obtain a 20% solution. If instead of adding 5 kg of pure water, we were to add 5 kg of an 80% solution, we would get a 70% solution. How much of the 40% solution and 60% solution do we have?

a) 2 kg : 1 kg b) 1kg : 2 kg c) 2 kg : 3 kg d) 3 kg : 1kg

Answer 1

If this from a speed test, you can manage with only one equation,
as all the ratios in the choices are different

Let the ratio of $40\%:60\%$ solutions be $1:k$, then

$\dfrac{0.4 + 0.6k}{1+k+5} = 0.2,\;$ which yields $k=2,\; i.e. 1\; kg:\;2\; kg$

A double check just for your satisfaction, plugging into the second equation:

$\dfrac{0.4*1 + 0.6*2 + 0.8*5}{1+2+5} = 0.7$

Answer 2

HINT:

If the quantity of the solutions be $10x,10y$ KG respectively,

$$\dfrac{10x\cdot40\%+10y\cdot60\%}{10x+10y+5}=20\%$$

Similarly form another equation in $x,y$ from the second condition.

Now solve the two simultaneous linear equations in $x,y$