Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
A. 8 kmph
B. 11 kmph
C. 12 kmph
D. 14 kmph
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{by}}\,x\,{\text{km}} \cr & {\text{Then}},\,\frac{x}{{10}} - \frac{x}{{15}} = 2 \cr & \Rightarrow 3x - 2x = 60 \cr & \Rightarrow x = 60\,km \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{travel}}\,60\,km\,{\text{at}}\,10\,{\text{km/hr}} \cr & = {\frac{{60}}{{10}}} \,hrs = 6\,hrs \cr & {\text{So,}}\,{\text{Robert}}\,{\text{started}}\,{\text{6}}\,{\text{hours}}\,{\text{before}}\,2\,P.M.\,i.e.,\,at\,A.M. \cr & \therefore {\text{Required}}\,{\text{speed}} \cr & = {\frac{{60}}{5}} \,kmph = 12\,kmph \cr} $$Join The Discussion
Comments ( 1 )
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
This is how i do.
Let time taken to reach home at 2= y
Case 1: Distance covered= 10y
case 2: Distance covered= 15(y-2)
Since distance is same,
10y=15(y-2)
y=6hrs
and, Distance= speed X time= 10 X 6= 60 kms
To reach @1, (y-1)= (6-1)= 5 hrs
required speed= 60÷5= 12 kms/hr