A person has to cover a distance of 90 km. If he increases his speed by 6 km/hr, he reaches 4 hours earlier. Find his initial speed (in km/hr).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9
The time at the usual speed minus the time at the higher speed must equal the 4 hours saved.
Let the initial speed be \(v\) km/hr
\(\displaystyle \frac{90}{v}-\frac{90}{v+6}=4\)
\(\displaystyle \frac{90\times6}{v(v+6)}=4\), so \(v(v+6)=135\)
\(v^{2}+6v-135=0\), that is \((v+15)(v-9)=0\)
Speed is positive, so \(v=9\) km/hr
Check: \(\displaystyle \frac{90}{9}-\frac{90}{15}=10-6=4\) hours