GO Classes 2023 | IIITH Mock Test 1 | Question: 91

Answer 1

Answer: C

Let the original speed be $x$, time taken at $x$ speed = $t$, Distance = $d$

Originally, $d = x*t$

At $x$ speed, $d  = \frac{5}{7}x*(t+16)$ 

Equating both of the equation

$x*t = \frac{5}{7}x(t+16) \implies 7t = 5t+80 \implies2t = 80 \implies t = 40$ min

Answer 2

Let the original speed $ = S$ and the original time $ = T$ minutes.
 
 $\text{Speed(S)} = \dfrac{\text{Distance(D)}}{\text{Time(T)}}$
 
 $\implies S \propto \dfrac{1}{T}$
 
 $\implies \dfrac{S_{1}}{S_{2}} = \dfrac{T_{2}}{T_{1}}$
 
 $\implies \dfrac{S}{\left(\dfrac{5}{7}\right)S} = \dfrac{T + 16}{T}$
 
  $\implies \dfrac{7}{5} = \dfrac{T + 16}{T}$
  
  $\implies 7T = 5T + 80$
  
  $\implies 2T = 80$
  
  $\implies T = 40\,\text{minutes}.$
  
  $$\textbf{(OR)}$$
  
  Let usual speed $ = S$
  
  and, new speed $ = \left(\dfrac{5}{7}\right)S$
  
  If usual time $ = T$ minutes, then new time $ = \left(\dfrac{7}{5}\right)T$ minutes.
  
  $\therefore \left(\dfrac{7}{5}\right)T - T = 16$
  
  $ \implies T = 40$ minutes.
  
  So, the correct answer is $(C).$