A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. The distance between two cities is :
A. 540 km
B. 420 km
C. 600 km
D. 620 km
Answer: Option B
Solution(By Examveda Team)
Let distance = x km and usual rate = y kmph. Then,$$\eqalign{ & \frac{x}{y} - \frac{x}{{y + 10}} = 1 \cr & {\text{or,}}\,y\left( {y + 10} \right) = 10x\,.\,.\,.\,.\,.\,.\,.\,.\,\,\left( 1 \right) \cr} $$
Now, in the 2nd scenario with a further increase in speed the driver could have saved another 45 min = $$\frac{3}{4}$$ hrs.
Therefore, total time saved
$$ = 1 + \frac{3}{4} = \frac{7}{4}\,{\text{hrs}}{\text{.}}$$
Putting it in equation, we get
$$\eqalign{ & \frac{x}{y} - \frac{x}{{y + 20}} = \frac{7}{4} \cr & {\text{or,}}\,y\left( {y + 20} \right) = \frac{{80x}}{7}\,.\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
On dividing (1) by (2), we get y = 60
Substituting y = 60 in (1), we get :
x = 420 km.
Related Questions on Speed Time and Distance
A. 48 min.
B. 60 min.
C. 42 min.
D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
E. None of these
A. 4 hours
B. 4 hours 30 min.
C. 4 hours 45 min.
D. 5 hours
Join The Discussion