Subtracting ratios from each other to find which solution is more concentrated.

Question

3 litres of orange concentrate were mixed with 5 litres of water to make a drink. Later, 2 litres of orange were mixed with 3 litres of water. Which mix is more concentrated? Consider the following strategy. To compare 3 Orange and 5 Water with 2 Orange and 3 Water

remove the second from the first and compare

1 Orange and 2 Water with 2 Orange and 3 Water.

Remove the first from the second and compare

1 Orange and 2 Water with 1 Orange and 1 Water.

Now you can see that the second was the more concentrated.

Will this strategy always work?

My attempt:

First I used examples to test this strategy by comparing 4 litres of orange concentrate with 5L of water with 2L of orange concentrate with 3L of water.

By subtracting the second from the first I have, 2L of orange and 2 L of water with 2 L of organ and 3 litres of water.

Now I subtract the the first from the second which results to 2L of orange with 2L of water compared to 0 litres of orange with 1 litre of water.

This shows that the first is more concentrated because either the orange or the water from the remaining ratio is not zero.

I tried to prove if this strategy works more generally by setting up ratios with variables but I can't seem to prove why this strategy works.

Can anyone help?

Answer 1

The way to solve this is to first compute the ratios of the amount of orange juice to the amount of the solution. In the first case you mix $3$ liters of orange juice with $5$ liters of water so the ratio of orange to the whole is: $3/(3+5) = 3/8$. In the second case you have $2/(2+3) = 2/5$. Then you compare the two: $3/8 ? 2/5 \rightarrow 15/40 <16/40$, which means that the second solution is more concentrated.

Answer 2

Let $a:b$ be the first ratio and $c:d$ be the second ratio. $a:b < c:d$ if and only if $bc > ad$ (you can see this by writing the ratios as fractions and simplifying the inequality).

Now write the rule using variables:

compare a:b to c:d

remove the second from the first and compare

$a-c:b-d$ with $c:d$

Remove the first from the second and compare

$a-c:b-d$ and $c-(a-c):d-(b-d) = 2c-a:2d-b$.

Consider the case where, after applying the rule, the second ratio is greater than the first. This statement can be written as $a-c:b-d < 2c-a:2d-b$. Start from that inequality and simplify to $bc > ad$, proving that if the second ratio is greater after applying the rule, then the original second ratio is greater.