A car covers a distance of 750 km at a constant speed. Had the car increased the speed by 10 kmph, it would have taken 2.5 hours less to cover the same distance. Find the original speed of the car.
50 kmph
Let the original speed be \(v\) kmph. Time at original speed \(=\frac{750}{v}\) h, and at the higher speed \(=\frac{750}{v+10}\) h.
Given the difference is 2.5 h: \(\frac{750}{v}-\frac{750}{v+10}=2.5 \Rightarrow \frac{7500}{v(v+10)}=2.5\).
So \(v(v+10)=3000 \Rightarrow v^{2}+10v-3000=0 \Rightarrow (v+60)(v-50)=0\).
Speed cannot be negative, so \(v = 50\) kmph. Check: 750/50 = 15 h and 750/60 = 12.5 h, a saving of 2.5 h.
Hence, the answer is 50 kmph.
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