
A car travels from A to B at \[40km/h\] . $1\dfrac{3}{4}$ hours later, another car starts from A and travels at \[48km/h\] reaches B, 15 minutes before the first car. The distance(in km) from A to B
A.420
B.480
C.500
D.450
Answer
587.1k+ views
Hint: At first we will assume the distance from A to B to be x, then using this distance we will find the time taken by both the cars to travel from A to B.
Now using the condition given i.e. $1\dfrac{3}{4}$ hours later, the second car starts from A reaches B, 15 minutes before the first car, we will form an equation in the variable of distance, and solving this equation we will get the required distance.
Complete step-by-step answer:
Given data: Speed of the first car $ = 40km/h$
Speed of the second car $ = 48km/h$
$1\dfrac{3}{4}$ hours later, the second car starts from A reaches B, 15 minutes before the first car
Now let the distance between A and B is x km.
Now we know that $time = \dfrac{{dis\tan ce}}{{speed}}$
Therefore, Time taken by the first car to travel from A to B $ = \dfrac{x}{{40}}$
Similarly, Time taken by the first car to travel from A to B $ = \dfrac{x}{{48}}$
Now it is given that $1\dfrac{3}{4}$ hours later, the second car starts from A reaches B ,15 minutes before the first car
$ \Rightarrow \dfrac{x}{{40}} = 1\dfrac{3}{4} + \dfrac{x}{{48}} + \dfrac{{15}}{{60}}$
Solving by taking like terms to one side we get,
$ \Rightarrow \dfrac{x}{{40}} - \dfrac{x}{{48}} = \dfrac{7}{4} + \dfrac{{15}}{{60}}$
Simplifying by taking LCM we get,
$ \Rightarrow \dfrac{{6x}}{{240}} - \dfrac{{5x}}{{240}} = \dfrac{7}{4} + \dfrac{1}{4}$
$ \Rightarrow \dfrac{x}{{240}} = \dfrac{8}{4}$
On cross multiplication we get,
$ \Rightarrow x = \dfrac{8}{4}\left( {240} \right)$
On simplification we get,
$\therefore x = 480$
Therefore the distance from A to B is 480 km
Hence, Option (B) is correct.
Note: While calculating for the given 15 minutes most of the students just substitute this value as it is but we can see all the other measurements are in kilometer and hours so it should also be converted into hours to get the correct answer otherwise the calculation will be incorrect and we will get the wrong answer.
Now using the condition given i.e. $1\dfrac{3}{4}$ hours later, the second car starts from A reaches B, 15 minutes before the first car, we will form an equation in the variable of distance, and solving this equation we will get the required distance.
Complete step-by-step answer:
Given data: Speed of the first car $ = 40km/h$
Speed of the second car $ = 48km/h$
$1\dfrac{3}{4}$ hours later, the second car starts from A reaches B, 15 minutes before the first car
Now let the distance between A and B is x km.
Now we know that $time = \dfrac{{dis\tan ce}}{{speed}}$
Therefore, Time taken by the first car to travel from A to B $ = \dfrac{x}{{40}}$
Similarly, Time taken by the first car to travel from A to B $ = \dfrac{x}{{48}}$
Now it is given that $1\dfrac{3}{4}$ hours later, the second car starts from A reaches B ,15 minutes before the first car
$ \Rightarrow \dfrac{x}{{40}} = 1\dfrac{3}{4} + \dfrac{x}{{48}} + \dfrac{{15}}{{60}}$
Solving by taking like terms to one side we get,
$ \Rightarrow \dfrac{x}{{40}} - \dfrac{x}{{48}} = \dfrac{7}{4} + \dfrac{{15}}{{60}}$
Simplifying by taking LCM we get,
$ \Rightarrow \dfrac{{6x}}{{240}} - \dfrac{{5x}}{{240}} = \dfrac{7}{4} + \dfrac{1}{4}$
$ \Rightarrow \dfrac{x}{{240}} = \dfrac{8}{4}$
On cross multiplication we get,
$ \Rightarrow x = \dfrac{8}{4}\left( {240} \right)$
On simplification we get,
$\therefore x = 480$
Therefore the distance from A to B is 480 km
Hence, Option (B) is correct.
Note: While calculating for the given 15 minutes most of the students just substitute this value as it is but we can see all the other measurements are in kilometer and hours so it should also be converted into hours to get the correct answer otherwise the calculation will be incorrect and we will get the wrong answer.
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