# Strength of salt in a mixture is p%. Various concentration of slats are added

The strength of a salt solution is $$p$$% if $$100$$ ml of the solution contains $$p$$ grams of salt. If three salt solutions A, B, C are mixed in the proportion $$1 : 2: 3$$, then the resulting solution has strength $$20$$%. If instead, the proportion is $$3 : 2: 1$$, then the resulting solution has strength $$30$$%. A fourth solution, D, is produced by mixing B and C in the ratio $$2: 7$$. The ratio of the strength of D to that of A is

1) $$2:5$$

2) $$3:10$$

3) $$1:4$$

4) $$1:3$$

My attempt: Let's say A has $$a$$%, B has $$b$$% and C has $$c$$%. And we take $$1$$, $$2$$ and $$3$$ kgs of them respectively. This will give us $$20$$% solution. Therefore, $$\frac{a}{100}$$+$$\frac{2b}{100}+\frac{3c}{100}$$=$$1.2$$ kgs This means $$a+2b+3c=120$$…(1)

Similarly $$\frac{3a}{100}+\frac{2b}{100}+\frac{c}{100}$$=$$1.8$$ kgs So, $$3a+2b+c=180$$…(2)

Adding (1) and (2) we get, $$a+b+c=75$$ and subtracting them we get $$a-c=30$$. How to proceed further?

$$a+b+c=75$$ & $$a−c=30$$ give $$c=a-30$$ and $$b=105-2a$$.
Required ratio: $$\frac{2b+7c}{9a}=\frac{210-4a+7a-210}{9a}=1:3$$