Question

A car travels from $ P $ to $ Q $ at a constant speed. If its speed were increased by $ 10km/hr, $ it would have taken one hour less to cover the distance. It would have taken further $ 45 $ minutes less if the speed was further increased by $ 10km/hr $ . What is the distance between the two cities?
A. $ 420km $
B. $ 540km $
C. $ 600km $
D. $ 650km $

Answer 1

Hint: We have to know about the formula of kinematics of motion to proceed such types of problems. In this question we only use one formula which given by $ {\text{speed = }}\dfrac{{{\text{distance}}}}{{{\text{time}}}} $ . Also we should know that $ 1hr = 60\min $ .

Complete step-by-step answer:
Let us consider the distance between $ P $ & $ Q $ is $ x{\text{ }}km $ .
Which is covered by initial speed is $ u{\text{ }}km/hr $
Initial time $ = \dfrac{x}{u}hr $
When speed is increased by $ 10km/hr, $ then we have
Final time $ = \dfrac{x}{{u + 10}} $
If speed is increased by $ 10km/hr $ then the difference between their initial time and final time is $ 1hr $ then we have,
 $ \dfrac{x}{u} - \dfrac{x}{{u + 10}} = 1 $
 $ \dfrac{1}{u} - \dfrac{1}{{u + 10}} = \dfrac{1}{x} $ ….. \[\left( 1 \right)\]
Similarly if the speed is further increased by $ 10km/hr, $ then we have
Final time $ \Rightarrow = \dfrac{x}{{u + 20}} $
Then difference of time taken is $ 45\min = \dfrac{{45}}{{60}}hr = 0.75hr $ more from previous case time,
Now the difference between the final time and initial time will be $ 1 + \dfrac{3}{4} = \dfrac{7}{4}hr $
Now the equation will be,
 $ \Rightarrow \dfrac{x}{u} - \dfrac{x}{{u + 20}} = \dfrac{7}{4} $
Multiplying $ \dfrac{4}{{7x}} $ both side we get,
 $ \Rightarrow \dfrac{4}{{7u}} - \dfrac{4}{{7\left( {u + 20} \right)}} = \dfrac{1}{x} $ …. \[\left( 2 \right)\]
Equating both equation 1 & 2
 $ \Rightarrow \dfrac{1}{u} - \dfrac{1}{{u + 10}} = \dfrac{4}{{7u}} - \dfrac{4}{{7\left( {u + 20} \right)}} = \dfrac{1}{x} $
Equating first two terms
 $ \Rightarrow \dfrac{1}{u} - \dfrac{1}{{u + 10}} = \dfrac{4}{{7u}} - \dfrac{4}{{7\left( {u + 20} \right)}} $
Taking L.C.M both side we get,
 $ \Rightarrow \dfrac{{u + 10 - u}}{{u\left( {u + 10} \right)}} = \dfrac{{4\left( {u + 20 - u} \right)}}{{7u\left( {u + 20} \right)}} $
Further solving then we have,
 $ \dfrac{{10}}{{\left( {u + 10} \right)}} = \dfrac{{4 \times 20}}{{7\left( {u + 20} \right)}} $
After Cross multiplication we have
 $ 70\left( {u + 20} \right) = 80\left( {u + 10} \right) $
\[70u + 1400 = 80u + 800\]
Rearranging $ u $ terms at one side then we have,
\[10u = 600 \Rightarrow u = 60km/hr\]
Substituting the value of $ u $ in equation 1
 $
  \dfrac{1}{u} - \dfrac{1}{{u + 10}} = \dfrac{1}{x} \Rightarrow \dfrac{1}{{60}} - \dfrac{1}{{60 + 10}} = \dfrac{1}{x} \\
  \dfrac{{70 - 60}}{{60 \times 70}} = \dfrac{1}{x} \Rightarrow x = 420km \\
  $

So, the correct answer is “Option A”.

Note: These types of questions are conventionally very lengthy. The equation should be resolved very carefully where students can make mistakes. Unit of distance should be seen very carefully in options according to that we should answer.