Question

Rakesh Yadav starts in Honda city from Delhi towards Goa. After sometime he realises that he will cover only $75\% $ of the distance in the scheduled time and he therefore doubles his speed immediately and thus manages to reach Goa exactly on time. Find the time after which Rakesh Yadav changed his speed, given that he could have been late by $3$ hours if he had not changed his speed.

Answer 1

Hint: Only value given here is the three hours difference between actual journey and expected journey. So we can interpret these three hours is used for the remaining $25\% $ of the journey. Using this idea and the speed and distance equations we get the expected time and subtracting three from this we get the actual time taken. Then splitting the distance equation before and after speed doubling, we can get to the answer.

Formula used:
Distance traveled is the product of speed and the time taken.
If a body travels $d$ distance in $t$ time, the speed of the body, $v = \dfrac{d}{t}$

Complete step by step answer:
Given that Rakesh Yadav travelled at a particular speed, let it be $v$ for some time that is, say $x$ hours, and realising he could only complete only $75\% $ of the journey in scheduled time, then doubled the speed.
Also, the time difference of his actual journey and his expected journey is three hours.
We are asked to find the time up to when he maintained the initial speed. That is to find the value of $x$.
So we can express these facts mathematically.
Let the total distance be $d$.
Also, let the initial speed be \[v\]. So if he travelled the whole journey in that speed, the time he would have taken was $t = \dfrac{d}{v}$.
But the time taken here actually is $t - 3$ hours.
If he travelled in initial speed $v$, he could only have completed $75\% = {\dfrac{3}{4}^{th}}$ of the whole journey, that is $\dfrac{3}{4}d$.
This means the extra $3$ hours will be used for the remaining $25\% = {\dfrac{1}{4}^{th}}$of the whole journey, that is for $\dfrac{1}{4}d$.
Distance travelled is the product of speed and the time taken.
$ \Rightarrow v \times 3 = \dfrac{1}{4}d$
Simplifying we get,
$\dfrac{d}{v} = 3 \times 4 = 12 - - - (i)$
Also $\dfrac{d}{v} = t$, the time should have taken (expected time).
Equating we get,
$ \Rightarrow t = 12$
That is, his expected time of journey if he maintained the initial speed was $12$ hours.
But due to doubling the speed he took only $t - 3 = 12 - 3 = 9$ hours.
If he doubled the speed after $x$hours, then the distance travelled by him in this $x$ hours is ${d_1} = speed \times time = vx$.
This means for the remaining $9 - x$ hours he used the double speed that is $2v$.
So, the distance travelled in $9 - x$ hours is ${d_2} = speed \times time = 2v(9 - x)$.
Then the total distance $d = {d_1} + {d_2}$
Substituting we get,
$ \Rightarrow d = vx + 2v(9 - x)$
Using $(i)$ we get $d = 12v$
$ \Rightarrow 12v = vx + 2v(9 - x)$
Cancelling $v$ from every term we have,
$ \Rightarrow 12 = x + 2(9 - x)$
$ \Rightarrow 12 = x + 18 - 2x = 18 - x$
Simplifying we get,
$ \Rightarrow x = 18 - 12 = 6$

$\therefore $ He doubled the speed after $6$ hours.

Note:
While doing this problem students can make a mistake that he covers three-fourth of the journey in initial speed and the remaining one-fourth in doubled speed. But this is wrong. We have to analyze the question carefully and understand what we are given and what we have to find.