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: $$\frac{2b+7c}{9a}=\frac{210-4a+7a-210}{9a}=1:3$$