
An aeroplane covers a certain distance at a speed of 240 km/hr in 5 hours. To covers the same distance in $1\dfrac{2}{3}$ hours, it must travel at a speed of:
A. 300 km/hr
B. 360 km/hr
C. 600 km/hr
D. 720 km/hr
Answer
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Hint: In this question, the time taken by the aeroplane is changing but the distance travelled by the aeroplane is fixed. We know that speed =$\dfrac{{{\text{distance travelled}}}}{{{\text{time taken}}}}$. So use the relation that speed is inversely proportional to the time taken.
Complete step-by-step answer:
In the question we have given:
Time taken is changing from 5 hours to $1\dfrac{2}{3}$ hours. And distance travelled by the aeroplane is fixed. We have to calculate the speed for which time taken is $1\dfrac{2}{3}$ hours.
We know the relationship between speed, distance travelled and time taken is given by:
${\text{Speed(S) = }}\dfrac{{{\text{Distance travelled(D)}}}}{{{\text{Time taken(t)}}}}.$
And we know that distance is constant. So we can write:
$
{\text{speed = }}\dfrac{{\text{k}}}{{{\text{time taken}}}} \\
\Rightarrow \dfrac{{{{\text{s}}_1}}}{{{{\text{s}}_2}}} = \dfrac{{{{\text{t}}_2}}}{{{{\text{t}}_1}}} \\
$ (1)
Where
k is a constant.
${t_1}$ is the time taken to travel the distance at a speed of 240 km/hr = 5 hours.
${t_2}$ is the time taken to travel the distance at a new speed = $1\dfrac{2}{3} = \dfrac{5}{3}$ hours.
Let the new speed be x km/hr.
Putting the values of time and speed for the two cases in equation 1, we get:
$
\dfrac{{{{\text{s}}_1}}}{{{{\text{s}}_2}}} = \dfrac{{{t_2}}}{{{t_1}}} \\
\Rightarrow \dfrac{{240}}{{\text{x}}} = \dfrac{{\dfrac{5}{3}}}{5} = \dfrac{1}{3} \\
\Rightarrow {\text{x = 240}} \times {\text{3 = 720}} \\
$
Therefore the new speed is 720 km/hr.
So option D is correct.
Note: One thing is clear that previously the aeroplane was travelling a specific distance in 5 hours, but now the same distance needs to be covered in a lesser time so applying general sense we know that it must boost its speed, so any option depicting speed lesser than the given initial speed can directly be discarded.
Complete step-by-step answer:
In the question we have given:
Time taken is changing from 5 hours to $1\dfrac{2}{3}$ hours. And distance travelled by the aeroplane is fixed. We have to calculate the speed for which time taken is $1\dfrac{2}{3}$ hours.
We know the relationship between speed, distance travelled and time taken is given by:
${\text{Speed(S) = }}\dfrac{{{\text{Distance travelled(D)}}}}{{{\text{Time taken(t)}}}}.$
And we know that distance is constant. So we can write:
$
{\text{speed = }}\dfrac{{\text{k}}}{{{\text{time taken}}}} \\
\Rightarrow \dfrac{{{{\text{s}}_1}}}{{{{\text{s}}_2}}} = \dfrac{{{{\text{t}}_2}}}{{{{\text{t}}_1}}} \\
$ (1)
Where
k is a constant.
${t_1}$ is the time taken to travel the distance at a speed of 240 km/hr = 5 hours.
${t_2}$ is the time taken to travel the distance at a new speed = $1\dfrac{2}{3} = \dfrac{5}{3}$ hours.
Let the new speed be x km/hr.
Putting the values of time and speed for the two cases in equation 1, we get:
$
\dfrac{{{{\text{s}}_1}}}{{{{\text{s}}_2}}} = \dfrac{{{t_2}}}{{{t_1}}} \\
\Rightarrow \dfrac{{240}}{{\text{x}}} = \dfrac{{\dfrac{5}{3}}}{5} = \dfrac{1}{3} \\
\Rightarrow {\text{x = 240}} \times {\text{3 = 720}} \\
$
Therefore the new speed is 720 km/hr.
So option D is correct.
Note: One thing is clear that previously the aeroplane was travelling a specific distance in 5 hours, but now the same distance needs to be covered in a lesser time so applying general sense we know that it must boost its speed, so any option depicting speed lesser than the given initial speed can directly be discarded.
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