Question

${\rm{A}}$and ${\rm{B}}$ run a kilometre and ${\rm{A}}$ wins by $25\,{\rm{seconds}}$. ${\rm{A}}$ and ${\rm{C}}$ run a kilometre and ${\rm{A}}$ wins by $275\,{\rm{m}}$.${\rm{B}}$ and ${\rm{C}}$ run the same distance and ${\rm{B}}$ wins by $30\,{\rm{sec}}$. What is the time taken by ${\rm{A}}$ to run a kilometre?

A.$2\,\min \,25\,\sec $
B.$2\,\min \,50\,\sec $
C.$3\,\min \,20\,\sec $
D.$3\,\min \,30\,\sec $

Answer 1

Hint:
In the solution, first we consider the time taken by $A$ to run ${\rm{1}}\,{\rm{km}}$. After that we have to find the different relations between $A,{\rm{ }}B$ and $C$ according to the question. This will give three expressions for $A,{\rm{ }}B$ and $C$. By solving the expressions, we have to calculate the time taken by $A$ to run ${\rm{1}}\,{\rm{km}}$.

Complete step by step solution:
Given,
${{\rm{1}}^{{\rm{st}}}}$Statement - $A$ and $B$ run a kilometre and $A$ wins by $25\,{\rm{seconds}}$. The winning time is given in this statement between $A$ and $B$.
${2^{{\rm{nd}}}}$Statement - $A$ and $C$ run a kilometre and $A$ wins by $275\,{\rm{m}}$. The winning distance is given in this statement between $A$ and $C$.
${3^{{\rm{rd}}}}$ Statement - $B$ and $C$ run the same distance and $B$ wins by$30\,{\rm{sec}}$. The winning time is given in this statement between $B$ and $C$.
Let the time taken by $A$ to run $1\,{\rm{km}}$ be ${\rm{y}}$.
Then the time taken by $B$ to run $1\,{\rm{km}}$ is $ = (y + 25)\;second\,$
Let us come to the ${2^{{\rm{nd}}}} $Statement - When $B$ and $C$ runs the same distance and $B$ wins by $30\,{\rm{sec}}$.
We know time taken by $B$ to run $1\,{\rm{km}}$ $ = (y + 25)\;second\,$
If $C$ loses by $30\,\sec $, then time taken by $C$ $ = (y + 55)\,{\rm{second}}$
$\therefore A = {\rm{ }}y$ ...... (1)
$B = {\rm{ }}y + 25$ ...... (2)
$C = {\rm{ }}y + 55$ ...... (3)
Now, let us come to the ${3^{{\rm{rd}}}}$ Statement - ${\rm{B}}$ and ${\rm{C}}$ run the same distance and ${\rm{B}}$ wins by $30\,{\rm{sec}}$.
It takes $C$ $55\,\sec $ to run ${\rm{275}}\,{\rm{m}}$.
So, in $1\,\sec $$C$ can run ${\rm{ = }}\dfrac{{{\rm{275}}}}{{{\rm{55}}}}{\rm{ = 5}}\,{\rm{m}}$.
Time taken by $C$ to run ${\rm{5}}\,{\rm{m = 1}}\,{\rm{sec}}$
So, time taken by $C$ to run ${\rm{1}}\,{\rm{km = }}\dfrac{1}{{0.005}} = 200\,\sec $
From equation (3),
$\begin{array}{l}C = {\rm{ }}y + 55\\ \Rightarrow 200{\rm{ }} = {\rm{ }}y + 55\\ \Rightarrow y{\rm{ }} = 145\;\sec \end{array}$

From equation (1),
$A = {\rm{ }}y{\rm{ }} = {\rm{ }}145{\rm{ sec }}$
Converting 145 second into minute.
${\rm{145 sec = 2}}\,{\rm{min 25 sec}}$
Thus, time taken by $A$ to run a kilometre $ = 2\,\min 25\sec $.
Hence option A is correct.


Note:
Here we have to find the time taken by $A$ to run a kilometre. Since their relative time and the distances are given, so, by solving their relations we can easily get the required final answer.