CAT 2022 Slot 3 · Quant · Question 11

Start by writing both sides of the first equation as powers of a single ratio.

Factorising, 875={5}^{3}\times 7 and 2401={7}^{4}.

So \frac{875}{2401}=\frac{{5}^{3}\times 7}{{7}^{4}}=\frac{{5}^{3}}{{7}^{3}}={\left( \frac{5}{7}\right)}^{3}.

That is the same as {\left( \frac{7}{5}\right)}^{-3}.

The left side is {\left( \frac{7}{5}\right)}^{\frac{3x-y}{2}}, since the square root halves the exponent.

Equating exponents, \frac{3x-y}{2}=-3, so 3x-y=-6.

Now the second equation. Writing t=6x-y, it reads {\left( \frac{4a}{b}\right)}^{t}={\left( \frac{2a}{b}\right)}^{-t}.

Multiplying both sides by {\left( \frac{2a}{b}\right)}^{t} gives {\left( \frac{8{a}^{2}}{{b}^{2}}\right)}^{t}=1.

This has to hold for every non-zero a and b, and the base \frac{8{a}^{2}}{{b}^{2}} changes as they change.

The only exponent that gives 1 for every base is t=0, so 6x-y=0 and y=6x.

Substituting into 3x-y=-6 gives 3x-6x=-6, so -3x=-6 and x=2.

Then y=12, and x+y=14.

Shortcut. The phrase "for all non-zero real values" is the whole clue in the second equation: a varying base forces the exponent to vanish. That gives y=6x in one step, with no algebra on a or b.