CAT 2024 Slot 3 · Quant · Question 15

Let the actual speed be s km/h and the actual time t hours, so the distance is D=st.

Faster by 6 and quicker by 4 hours: \left( s+6 \right)\left( t-4 \right)=st.

Expanding, st-4s+6t-24=st, so 6t-4s=24, that is 3t-2s=12.

Slower by 6 and longer by 6 hours: \left( s-6 \right)\left( t+6 \right)=st.

Expanding, st+6s-6t-36=st, so 6s-6t=36, that is s-t=6.

From the second relation, s=t+6.

Substituting into the first, 3t-2\left( t+6 \right)=12, so t-12=12 and t=24.

Then s=24+6=30 km/h.

The distance is D=30\times 24=720 km, the second option.

Shortcut. Each condition keeps the distance fixed, so expanding always cancels the st term and leaves a clean linear equation. Two linear equations in s and t is all this question ever is — no quadratic appears at any stage.