CAT 2023 Slot 2 · Quant · Question 21

Three expressions are all equal, so pair them off starting with the two that share the term pq.

From 2pq-20=52-2pq, collect: 4pq=72.

So pq=18.

Substituting back, the common value is 2\times 18-20=16.

The first expression must also equal 16, so {p}^{2}+{q}^{2}-29=16.

Therefore {p}^{2}+{q}^{2}=45.

Now {\left( p-q \right)}^{2}={p}^{2}+{q}^{2}-2pq=45-36=9.

So p-q=3 or p-q=-3, both of which are achievable with real p and q.

For the cube difference, use {p}^{3}-{q}^{3}=\left( p-q \right)\left({p}^{2}+pq+{q}^{2}\right).

The second bracket is 45+18=63, the same in both cases.

So {p}^{3}-{q}^{3}=63\left( p-q \right), giving 189 when p-q=3 and -189 when p-q=-3.

The difference between the maximum and the minimum is 189-\left( -189 \right)=378 — the third option.

Shortcut. Since p and q are interchangeable, the two cube differences are negatives of each other, so the answer is simply twice the positive value. That is 2\times 63\times 3=378, and noticing the symmetry saves computing the negative case at all.