Question

A person goes to his office from his home daily at a fixed speed. But today he decided to reduce his speed by $10\% $ as a result he reached his office $16$ minute late. The actual time he takes to reach his office daily is:
A. $90$ minute
B. $180$ minute
C. $45$ minute
D. $64$ minute

Answer 1

Hint: Speed is distances per unit time.
By making necessary equations for the given problem , it can be solved.

Complete step by step answer:
Let the fixed speed with which the person goes to his office daily be $ = v$
The actual time takes is t and the distance between office and home be x
So, as we know that speed $ = \dfrac{{dis\tan ce}}{{time}}$
So, $v = \dfrac{x}{t}$…. (i)
Now, according to question,
When speed $ = v - 20\% $ of v
$
   = v - \dfrac{{20}}{{100}}v \\
   = \dfrac{{80}}{{100}}v \\
 $
Then, time taken $ = t + 16$
As speed $ = \dfrac{{dis\tan ce}}{{time}}$
So, $\dfrac{{80v}}{{100}} = \dfrac{x}{{t + 16}}$… (ii)
Dividing (ii) by (i), we get
\[\dfrac{{80v}}{{\dfrac{{100}}{v}}} = \dfrac{x}{{\dfrac{{t + 16}}{{\dfrac{x}{t}}}}}\]
\[
   \Rightarrow \dfrac{{80v}}{{100}} \times \dfrac{1}{v} = \dfrac{x}{{t + 16}} \times \dfrac{t}{x} \\
   \Rightarrow \dfrac{8}{{10}} = \dfrac{t}{{t + 20}} \\
   \Rightarrow 8\left( {t + 16} \right) = 10t \\
   \Rightarrow 8t + 128 = 10t \\
   \Rightarrow 2t = 128 \\
   \Rightarrow t = \dfrac{{128}}{2} = 64 \ minutes \\
 \]

So, the correct answer is “Option D”.

Additional Information:
As the time increases by an amount of $16$ minutes. So, the final answer is also in minutes. The actual time with daily speed is $64$ minutes. When the speed is decreased by $20\% $ then time increases by $16$ minutes that is it becomes $64 + 16 = 80$ minutes, so $80$ minutes is the final time when the speed is decreased.

Note:
Remember that the velocity decreases by $20\% $ so, the final velocity is $100 - 20 = 80\% $ not $20\% $ that’s why the final velocity is $\dfrac{{80v}}{{100}} = 0.8v$ not$0.2v$.