
Walking \[6/{7^{th}}\] of his usual speed, Sanchit is 12 minutes late. The usual time taken by him to cover that distance is:
A. 2 hr
B. 1 hr 12 mins
C. 1 hr 15 mins
D. 45 mins
Answer
551.1k+ views
Hint:Speed of any object is given by the distance covered by that object in a particular interval of time. Therefore, speed is equal to the distance divided by the time. In the problem, it is given that Sanchit is 12 minutes late when he walks with \[6/{7^{th}}\] of his usual speed. The usual time taken by him to cover that distance can be calculated by applying the formula of speed-distance -time.
Formula Used:
The formula for speed is given as: \[s = \dfrac{d}{t}\]
where, \[s\] is the speed , \[d\] is the distance and \[t\] is the time
Complete step by step answer:
Let the usual time taken by Sanchit to cover the given distance \[d\] be \[t\]. Then his speed \[s\] is given by
\[s = \dfrac{d}{t}\]
Rearranging the equation in terms of \[t\]
\[t = \dfrac{d}{s}\] \[ \to (1)\]
This is the usual time taken.
If he walks with \[6/{7^{th}}\] of the original speed then the new speed will be;
\[s' = \dfrac{6}{7}s\]
where, \[s'\] is the new speed and \[s\] is the usual speed.
Then, the new time will be
\[t' = \dfrac{d}{{\left( {\dfrac{6}{7}} \right)s}} \\
\Rightarrow t' = \dfrac{{7d}}{{6s}}\] \[ \to (2)\]
With this new time, Sanchit is 12minutes late.
\[{\text{New time - Usual Time = 12}}\]
\[\dfrac{{7d}}{{6s}} - \dfrac{d}{s} = 12 \\
\Rightarrow \dfrac{{7d - 6d}}{{6s}} = 12 \\
\Rightarrow \dfrac{d}{{6s}} = 12 \\
\therefore \dfrac{d}{s} = 72\]
But from equation (1),\[\dfrac{d}{s} = t\]. Therefore, \[t = 72{\text{ minutes}}\]. Therefore, the usual time taken by him to cover that distance is 72 minutes or 1 hour 12 minutes.
Hence, option B is the correct answer.
Note: In the given problem, Sanchit is walking \[6/{7^{th}}\] of his usual speed.Therefore, he is 12 minutes late. This is because, speed is inversely proportional to time.When he is walking \[6/{7^{th}}\] of his usual speed, his speed decreases and therefore the time taken to cover the distance increases. Thus, he needed an extra 12 minutes to cover the distance.
Formula Used:
The formula for speed is given as: \[s = \dfrac{d}{t}\]
where, \[s\] is the speed , \[d\] is the distance and \[t\] is the time
Complete step by step answer:
Let the usual time taken by Sanchit to cover the given distance \[d\] be \[t\]. Then his speed \[s\] is given by
\[s = \dfrac{d}{t}\]
Rearranging the equation in terms of \[t\]
\[t = \dfrac{d}{s}\] \[ \to (1)\]
This is the usual time taken.
If he walks with \[6/{7^{th}}\] of the original speed then the new speed will be;
\[s' = \dfrac{6}{7}s\]
where, \[s'\] is the new speed and \[s\] is the usual speed.
Then, the new time will be
\[t' = \dfrac{d}{{\left( {\dfrac{6}{7}} \right)s}} \\
\Rightarrow t' = \dfrac{{7d}}{{6s}}\] \[ \to (2)\]
With this new time, Sanchit is 12minutes late.
\[{\text{New time - Usual Time = 12}}\]
\[\dfrac{{7d}}{{6s}} - \dfrac{d}{s} = 12 \\
\Rightarrow \dfrac{{7d - 6d}}{{6s}} = 12 \\
\Rightarrow \dfrac{d}{{6s}} = 12 \\
\therefore \dfrac{d}{s} = 72\]
But from equation (1),\[\dfrac{d}{s} = t\]. Therefore, \[t = 72{\text{ minutes}}\]. Therefore, the usual time taken by him to cover that distance is 72 minutes or 1 hour 12 minutes.
Hence, option B is the correct answer.
Note: In the given problem, Sanchit is walking \[6/{7^{th}}\] of his usual speed.Therefore, he is 12 minutes late. This is because, speed is inversely proportional to time.When he is walking \[6/{7^{th}}\] of his usual speed, his speed decreases and therefore the time taken to cover the distance increases. Thus, he needed an extra 12 minutes to cover the distance.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

