CAT 2018 Question Paper Shift 1 Solved Paper

The selling price of the mixture is $\mathit{R}\mathit{s}.40∕$k. g
Let a be the quantity of tea $\mathit{A}$ in the mixture and $\mathit{b}$ be the quantity of tea $\mathit{B}$ in the mixture.
It has been given that the profit is $10\%$ if the 2 varieties are mixed in the ratio 3: 2
Let the cost price of the mixture be $\mathit{x}$.
It has been given that $1.1\mathit{x}=40$
$\mathit{x}=40/1.1$
$\frac{3\mathit{a}+2\mathit{b}}{5}=\frac{40}{1.1}$
$3.3\mathit{a}+2.2\mathit{b}=200$.....(1)
The profit is $5\%$ if the 2 varieties are mixed in the ratio 2: 3 .
$\frac{2\mathit{a}+3\mathit{b}}{5}=\frac{40}{1.05}$
$2.1\mathit{a}+3.15\mathit{b}=200$.....(2)
Equating (1) and (2), we get,
$3.3\mathit{a}+2.2\mathit{b}=2.1\mathit{a}+3.15\mathit{b}$
$1.2\mathit{a}=0.95\mathit{b}$
$\frac{\mathit{a}}{\mathit{b}}=\frac{0.95}{1.2}$
$\frac{\mathit{a}}{\mathit{b}}=\frac{19}{24}$