Two vessels contain milk and water in the ratio 5:2 and 4:1 Find the quantity of the mixture from the first vessel to be mixed ....

Question

Two vessels contain milk and water in the ratio 5:2 and 4:1 Find the quantity of the mixture from the first vessel to be mixed with 20 lts of the mixture from the second vessel. So that the mixture formed has milk and water in the ratio 3:1.

I am not getting any idea how to proceed on this question, request you to please guide on this.. will be of great help. Thanks.

Answer 1

Hint:

Let the required quantity be $(5+2)x$ lts

So, we have

$$\dfrac31=\dfrac{\dfrac5{5+2}\cdot7x+\dfrac4{4+1}\cdot20}{\dfrac2{5+2}\cdot7x+\dfrac1{4+1}\cdot20}$$

Answer 2

Let $x$ the quantity of the mixture from the first vassel to be mixed, I have: $$3=\frac{\frac{16\cdot 7l+5x}{7}}{\frac{28l+2x}{7}}$$ from this, I obtain: $16\cdot7l+5x=12\cdot7l+6x$. This has solution $x=28$.I hope it has helped you...

Answer 3

$$\begin{array}{c | c | c}{2\over 7}& &{1\over 20}\\\hline &{1\over 4}&\\\hline {1\over 5} &&{5\over 28}\end{array}$$ Using parts arithmetic, on total parts we get the above array. We then get 20L, being equivalent to five in 28. This makes 1 equivalent to 112L, and one in 20, equivalent to 5.6 L

Answer 4

Let the quantity sought be $q,$ in litres. Then we have that (assuming no chemical reaction occurs) $$\frac34(q+20)=\frac57(q)+\frac45(20)$$ and $$\frac14(q+20)=\frac27(q)+\frac15(20)$$ must be true.

Solving either of these gives $q=28,$ and one can confirm that it also satisfies the other.