Question

A boy walking at a speed of 10 km/hour reaches his school 15 minutes late. Next time he increases his speed by 2 km/hr , but still, he is late by 5 minutes. Find the distance of his school from his house.​

Answer 1

$\rule{200}3$

$\large{\underline{\boxed{\gray{ \textbf{Diagram\::-}}}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{2 mm}\put(2,2){\line(0,1){5}}\qbezier(-0.8,9)(-1,9)(1.5,7)\qbezier(-2.3,8)(-2.4,6)(-2.5,4)\qbezier(-2.5,4)(-2.8,2)(-3.2,0)\linethickness{4 mm}\put(-1.6,8){\line(0,1){2}}\put(-1.5,18){\circle*{4}}\linethickness{1.5 mm}\qbezier(-1.7,15)(-4.1,12)(-4.1,12)\put(-4.1,12){\line(0,-1){3}}\qbezier(0,15)(1.5,12)(1.5,12)\qbezier(1.5,12)(3,11.8)(4,11.6)\linethickness{0.2mm}\put(-2.95,10){\dashbox{0.1}(2.6,6)[l]}\linethickness{2mm}\put(-4,30){\line(1,0){45}}\put(-4,-15){\line(1,0){45}}\put(2,-1){\line(0,1){3}}\linethickness{0.6mm}\put(-4.6,-1){\line(1,0){45}}\linethickness{0.4mm}\put(8,20){\dashbox{0.1}(19,4)[l]}\put(2.5,15){\vector(1,1){5}}\put(8.5,21){speed = 12km/hr}\linethickness{0.2mm}\put(6.5,-6){\dashbox{0.1}(19,4)[l]}\put(5.5,-5){\vector(-1,0){10}}\put(8.5,-5){Distance = x km}\put(27,-5){\vector(1,0){9}}\end{picture}$

$\rule{200}3$

$\large{\underline{\boxed{\gray{ \textbf{Given\::-}}}}}$

$\sf The \:diffrence\: in\: time\: = 15 - 5$

$\Longrightarrow$ $\sf 10 min$

$\Longrightarrow$ $\sf \dfrac{10}{60}\: hr$

$\Longrightarrow$ $\sf \dfrac{1}{6} \:hr$

$\sf Speed = 10 + 2 = 12\: km\:hr^{-1}$

$\large{\underline{\boxed{\gray{ \textbf{To\:find:-}}}}}$

The distance of his school from the house.

$\large{\underline{\boxed{\gray{ \textbf{Solution\::-}}}}}$

Let the distance of his school from the house be x.

$\dagger\:$ $\underline{ According\: to\: Question\::-}$

$\dashrightarrow$ $\sf \dfrac{x}{10} - \dfrac{x}{12} = \dfrac{1}{6}$

$\dashrightarrow$ $\sf \dfrac{x}{5} - \dfrac{x}{6} = \dfrac{1}{3}$

$\dashrightarrow$ $\sf \dfrac{6x - 5x}{30} = \dfrac{1}{3}$

$\dashrightarrow$ $\sf \dfrac{x}{30} = \dfrac{1}{3}$

$\dashrightarrow$ $\sf x = \dfrac{30}{3}$

$\pink\dashrightarrow$ $\underline{\boxed{\pink{\sf x = 10\:km}}}$ $\orange\bigstar$

Therefore, the distance of his school from the house is 10 km.

$\rule{200}3$

Answer 2

Given :-

Case 1

Speed of boy = 10 km/hr

Time taken to reach school = 15 minutes late.

Case 2

Speed of boy = 10 + 2 = 12 km/hr

Time taken to reach school = 5 minutes late.

⠀⠀⠀⠀⠀⠀⠀

To find :-

Distance of school from his house.

⠀⠀⠀⠀⠀⠀⠀

Solution :-

Let,

Distance = x

⠀⠀⠀⠀⠀⠀⠀

Now,

Difference between time

$\sf = Time \: taken \:in \:case\: 1 \:-\: Time \:taken \:in \:case \:2$

= 15 - 5

= 10 minutes.

⠀⠀⠀⠀⠀⠀⠀

Converting minutes to hours :-

10 minutes

$= \sf \: \frac{10}{60} hours$

$\sf = \frac{1}{6}\: hours$

⠀⠀⠀⠀⠀⠀⠀

New speed = 12 km/hr

⠀⠀⠀⠀⠀⠀⠀

We know that,

Time = Dist./Speed

$\sf T^{1} \: = \frac{x}{10} \:$

$\sf T^{2} \: = \frac{x}{12} \:$

⠀⠀⠀⠀⠀⠀⠀

t1 - t2 = 1/6

$\sf \frac{x}{10} \: - \frac{x}{12} =\frac{1}{6} \\\frac{x}{60} =\:\frac{1}{6} \\6x=60\\x=10$

⠀⠀⠀⠀⠀⠀⠀

Distance between his school and home

= x

= 10km