CAT Exam Model Paper 5 with solutions for free online practice

Let the %age of honey and milk in the two solutions be (a%, b%) and (c%, d%) respectively. According to the question:
$\frac{\left(\mathrm{a}-10\right)}{\left(10-\mathrm{c}\right)}=\frac{\left(\mathrm{b}-16\right)}{\left(16-\mathrm{d}\right)}$
(It does not make a difference whether a is greater than c or whether b is greater than d, because taking -1 common from both the numerator and the denominator will nullify the effect.)
Solving, we get,
$16\mathit{a}-\mathit{a}\mathit{d}-160+10\mathit{d}=10\mathit{b}-160-\mathit{b}\mathit{c}+16\mathit{c}$
So, $\mathit{a}\mathit{d}-\mathit{b}\mathit{c}=16\mathit{a}-10\mathit{b}-16\mathit{c}+10\mathit{d}$ ...(1)
Similarly from the second and third proportions we can say that $\mathit{a}\mathit{d}-\mathit{b}\mathit{c}=12\mathit{a}-12\mathit{b}-12\mathit{c}+12\mathit{d}$. ...(2)
and $\mathit{a}\mathit{d}-\mathit{b}\mathit{c}=\mathit{x}\mathit{a}-16\mathit{b}-\mathit{x}\mathit{c}+16\mathit{d}$ ...(3)
From (1) and (2), we get
$\frac{\left(\mathit{a}-\mathit{c}\right)}{\left(\mathit{d}-\mathit{b}\right)}=\frac{1}{2}$
Also from (2) and (3), we get
$\frac{\left(\mathit{a}-\mathit{c}\right)}{\left(\mathit{d}-\mathit{b}\right)}=\frac{4}{\left(12-\mathit{x}\right)}$
Hence,$\frac{4}{\left(12-\mathit{x}\right)}=\frac{1}{2}$
$8=12-\mathit{x}\mathrm{and}\mathit{x}=4$