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7.  What is the average speed if a person travels from A to B and then B to C at the speeds of 10 km/hr and 20 km/hr, respectively for equal intervals of time?
a. 15 kmph
b. 13.33 kmph
c. 12 kmph
d. 11.11 kmph

Answer: A

Explanation:
The average speed = $\displaystyle\frac{{\left( {10 + 20} \right)}}{2} = 15{\rm{ }}km/hr$

Note:
For equal intervals of time, the average speed is given as $\displaystyle\frac{{{{\rm{S}}_{\rm{1}}}{\rm{ + }}{{\rm{S}}_{\rm{2}}}{\rm{ + }}{{\rm{S}}_{\rm{3}}}{\rm{ + }}\,...{\rm{ + }}\,{{\rm{S}}_{\rm{n}}}}}{{\rm{n}}}$
, where $S_1, S_2, S_3, ... , S_n$ are the speeds and n is the number of observations.

8.  What is the average speed if a person travels from A to B and back at the speeds of 10 km/hr and 20 km/hr, respectively?
a. 15 kmph
b. 13.33 kmph
c. 12 kmph
d. 11.11 kmph

Answer: B

Explanation:
In this case, times are not given.
General formula for Average speed = $\dfrac{\text{Distance}}{\text{Total Time}}$
Let the total distance = 2D.
Total time taken = T = $\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{10}}}}} \right){\rm{ + }}\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{20}}}}} \right){\rm{}}$
So, Average speed = $\displaystyle\frac{{{\rm{2D}}}}{{\rm{T}}}{\rm{ = }}\displaystyle\frac{{{\rm{2D}}}}{{\left( {\displaystyle\frac{{\rm{D}}}{{{\rm{10}}}}{\rm{ + }}\frac{{\rm{D}}}{{{\rm{20}}}}} \right)}}{\rm{ = }}\displaystyle\frac{{{\rm{2D}}}}{{{\rm{D}}\left( {\displaystyle\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{ + }}\displaystyle\frac{{\rm{1}}}{{{\rm{20}}}}} \right)}}{\displaystyle\rm{ = 13}}\displaystyle\frac{{\rm{1}}}{{\rm{3}}}{\rm{ }}k{\rm{m/hr}}$

Alternative Method:
Use formula = $\dfrac{{2xy}}{{x + y}}$

9.  A car covers four successive three km stretches at speeds of 10 km/hr, 20 km/hr, 30 km/hr and 60 km/hr respectively. Its average speed over this distance :
a. 10 km/hr
b. 20 km/hr
c. 30 km/hr
d. 25 km/hr

Answer: B

Explanation:
Total time taken = $\left( {\displaystyle\frac{3}{{10}} + \displaystyle\frac{3}{{20}} + \displaystyle\frac{3}{{30}} + \displaystyle\frac{3}{{60}}} \right)$ hrs = $\displaystyle\frac{3}{5}$ hrs
Average speed = $\left\{ {\displaystyle\frac{{12}}{{3/5}}} \right\}$ km/hr = $\left( {\displaystyle\frac{{12 \times 5}}{3}} \right)$ km/hr = 20 km/hr.

10.  A train covers a distance in 50 minutes. If it runs at a speed of 48 km per hour on an average, the speed at which the train must run to reduce the time of journey to 40 minutes will be :
a. 50 km/hr
b. 55 km/hr
c. 60 km/hr
d. 70 km/hr

Answer: C

Explanation:
Distance = $\left( {48 \times \displaystyle\frac{{50}}{{60}}} \right)$ km = 40 km
Required Speed = $\left( {\displaystyle\frac{{40}}{{40/60}}} \right)$ km/hr = $\left( {\displaystyle\frac{{40 \times 60}}{{40}}} \right)$ km/hr=60 km/hr

11.  A man goes uphill with an average speed of 35 km/hr and comes down with an average speed of 45 km/hr. The distance travelled in both the cases being the same, the average speed for the entire journey is :
a. $38\displaystyle\frac{3}{8}$ km/hr
b. $39\displaystyle\frac{3}{8}$ km/hr
c. 40 km/hr
d. None of these

Answer: B

Explanation:
Average speed = $\left( {\displaystyle\frac{{2 \times 35 \times 45}}{{35 + 45}}} \right)$ km/hr = $39\displaystyle\frac{3}{8}$ km/hr.

12.  P and Q are two stations. A train goes from P to Q at 64 km/hr and returns to P at a slower speed. If its average speed for the whole journey is 56 km/hr, at what speed did it return?
a. 48 km/hr
b. 49.77 km/hr
c. 52 km/hr
d. 47 km/hr

Answer: B

Explanation:
Let the required speed be x km/hr.
Then, $\displaystyle\frac{{2 \times 64 \times x}}{{64 + x}} = 56$
$ \Rightarrow 128x = 64 \times 56 + 56x$
x = $\displaystyle\frac{{64 \times 56}}{{72}} = 49.77$ km/hr

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