Question

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 1

Hint: First find the distance Robert travelled to reach point A. After that find the time taken by him to reach point A at the speed of 10 kmph. Then, subtract 1 hour from that time to reach the destination at 1 P.M. Now, calculate the speed of the cycle by dividing the total distance to travel by the time calculated to reach at 1 P.M.

Complete step by step answer:
Given: - 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.
Let the distance travelled by Robert be $x$ km.
Time taken at the speed of 10 kmph,
$ \Rightarrow \dfrac{x}{{10}}$
Time taken at the speed of 15 kmph,
$ \Rightarrow \dfrac{x}{{15}}$
Since the time taken by Robert on travelling at the speed of 15kmph is 2 hours less than the time taken on travelling at the speed of 10kmph. So,
$ \Rightarrow \dfrac{x}{{10}} - \dfrac{x}{{15}} = 2$
Take L.C.M. on the left side,
$ \Rightarrow \dfrac{{3x - 2x}}{{30}} = 2$
Subtract the values in the numerator,
$ \Rightarrow \dfrac{x}{{30}} = 2$
Multiply both sides by 30,
\[ \Rightarrow x = 2 \times 30\]
Multiply the values on the right side,
$ \Rightarrow x = 60$ kmph
Now, the time taken by Robert to travel at 10 kmph,
$ \Rightarrow \dfrac{{60}}{{10}} = 6$ hr
To reach point A at 1 P.M., subtract 1 hour from the time taken at the speed of 10 kmph,
$ \Rightarrow 6 - 1 = 5$ hr
Now, calculate the speed to reach point A in 5 hours,
$ \Rightarrow \dfrac{{60}}{5} = 12$ kmph
Thus, the speed must be 12 kmph to reach point A at 1 P.M.

Hence, option (C) is the correct answer.

Note: The formula for speed distance time is mathematically given as:
Speed = Distance/Time
where, x = Speed,
d = Distance travelled,
t= time taken.
Distance travelled formula is $d=xt$.