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# Ram walks to school from home every morning and it takes him 30 minutes. Once on his way he realized that he had forgotten his physics book at home. If he continued walking to school at the same speed, he would be there 12 minutes before the bell. But he went back home for the physics book and reached school 6 minutes after the bell. Assuming he had walked all the way with his usual speed, the fraction of the way to his school he covered till the moment he turned back is _______.

Last updated date: 13th Jun 2024
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Hint: The time required to go back home and make it to school is the time of the bell plus 6 minute. The extra distance he would have travelled had he not gone back would have been total distance minus distance already covered and the time taken to travel the distance would be time of the bell minus 12 minutes.

Formula used: In this solution we will be using the following formulae;
\[v = \dfrac{d}{t}\] where \[v\] is the speed of an object, \[d\] is the distance travelled, and \[t\] is the time taken to cover such distance.

Complete Step-by-Step Solution:
Let the distance Ram travel be a distance \[x\], but the total distance be \[d\].
Now at point \[x\] if Ram had continued, he would have gotten to school 12 minute before the bell. Let the time of the bell be \[{t_1}\], then the time he would have taken to get to school will be \[{t_1} - 12\]. But the distance remaining was \[d - x\]. Hence,
\[{t_1} - 12 = \dfrac{{d - x}}{v}\] where \[v\] is the speed of his movement.
\[ \Rightarrow {t_1} = \dfrac{{d - x}}{v} + 12\]
However, after going home, he finally got to school 6 minutes after the bell, i.e. \[{t_1} + 6\]. But the total distance travelled is \[x + d\] (because he went back by \[x\] and travelled the full distance \[d\]). Hence,
\[{t_1} + 6 = \dfrac{{x + d}}{v}\]
\[ \Rightarrow {t_1} = \dfrac{{x + d}}{v} - 6\]
Hence, by equation
\[\dfrac{{d - x}}{v} + 12 = \dfrac{{x + d}}{v} - 6\]
By collecting like terms, we have
\[\dfrac{{d - x}}{v} - \dfrac{{x + d}}{v} = - 6 - 12\]
Hence, by performing subtraction on both sides, we get
\[ - \dfrac{{2x}}{v} = - 18\]
\[ \Rightarrow x = 9v\]
Since, usually it takes him 30 minutes to travel the distance \[d\], then,
\[d = 30v\]
Then the ratio of the distance travelled till he came back to the total distance is
\[\dfrac{x}{d} = \dfrac{{9v}}{{30v}} = \dfrac{3}{{10}}\]

Note: For understanding, observe how the quantity used was speed and not velocity. The velocity in general would have been inappropriate to find the distance since it neglects any interval of space travelled twice. Hence, the velocity will neglect the distance \[x\] and only consider the distance \[d - x\].