Question

Q- A school boy ,travelling at 3km/h reaches his school 32 min late but if he travels at 5 km/h then he reaches his school 28 min early . Find the distance b/w school and his home.

Answer 1

Let the Distance between School and Home be : D

When Boys Travels with a Speed of 3 km/hr, He Reaches his School in $\bf{: (\frac{D}{3})hr}$

When Boys Travels with a Speed of 5 km/hr, He Reaches his School in $\bf{: (\frac{D}{5})hr}$

Given : If the Boys Travels at 3 km/h, He reaches his School 32 Minutes Late and if he Travels at 5 km/h, He reaches his School 28 Minutes Early.

It means : The Difference between the Time Taken by the Boy while travelling at Speed 3 km/hr and while travelling at Speed 5 km/hr should be Equal to (32 + 28) = 60 Minutes.

$\bf{\implies (\frac{D}{3} - \frac{D}{5}) = 60\;Minutes}$

$\bf{\implies (\frac{D}{3} - \frac{D}{5}) = 1\;Hour}$

$\bf{\implies (\frac{5D - 3D}{15}) = 1}$

$\bf{\implies (\frac{2D}{15}) = 1}$

$\bf{\implies 2D = 15}$

$\bf{\implies D = 7.5}$

Distance Between School and His Home is 7.5 km

Answer 2

Answer:

Let the Distance between School and Home be : D

When Boys Travels with a Speed of 3 km/hr, He Reaches his School in \bf{: (\frac{D}{3})hr}:(

3

D

)hr

When Boys Travels with a Speed of 5 km/hr, He Reaches his School in \bf{: (\frac{D}{5})hr}:(

5

D

)hr

Given : If the Boys Travels at 3 km/h, He reaches his School 32 Minutes Late and if he Travels at 5 km/h, He reaches his School 28 Minutes Early.

It means : The Difference between the Time Taken by the Boy while travelling at Speed 3 km/hr and while travelling at Speed 5 km/hr should be Equal to (32 + 28) = 60 Minutes.

\bf{\implies (\frac{D}{3} - \frac{D}{5}) = 60\;Minutes}⟹(

3

D

−

5

D

)=60Minutes

\bf{\implies (\frac{D}{3} - \frac{D}{5}) = 1\;Hour}⟹(

3

D

−

5

D

)=1Hour

\bf{\implies (\frac{5D - 3D}{15}) = 1}⟹(

15

5D−3D

)=1

\bf{\implies (\frac{2D}{15}) = 1}⟹(

15

2D

)=1

\bf{\implies 2D = 15}⟹2D=15

\bf{\implies D = 7.5}⟹D=7.5

Distance Between School and His Home is 7.5 km