Question

10. A student starting from his house walks at a
speed of 2 1/2 km/hour and reaches his school 6
minutes late. Next day starting at the same time
he increases his speed by 1 km/hour and reaches
6 minutes early. The distance between the school
and his house is
(a) 4 km
(b) 3 1/2 km
(c) 1 3/4 km
(d) 6 km​

Answer 1

• To find : The distance between the school and his house

Solution :

Speed of a student = 2 ½ km/h = 5/2km/h

★ A student reaches his school 6 minutes late.

Consider the time be x

• 60min = 1 hour

Time = (x + 6) min = (x + 6/60) = (x + 1/10)h

• As we know that

★ Speed = distance/time

• Consider the distance be y

$\implies \sf \dfrac{5}{2} = \dfrac{y}{ x + \dfrac{1}{10}}$

$\implies \sf \dfrac{5}{2} = \dfrac{y}{ \dfrac{10x + 1}{10} }$

$\implies \sf \dfrac{5}{2} = y \times \dfrac{10}{10x +1}$

$\implies \sf \dfrac{5}{2} = \dfrac{10y}{10x +1}$

$\implies \sf 5(10x +1) = 20y$

$\implies \sf 50x +5=20y$

$\implies \sf 50x - 20y = - 5 \: \: \: \bf(Equation \: 1)$

★ Next day starting at the same time, he increases his speed by 1 km/hour and reaches 6 minutes early.

$\implies \sf \dfrac{5}{2} + 1 = \dfrac{y}{ x - \dfrac{1}{10}}$

$\implies \sf \dfrac{5 + 2}{2} = \dfrac{y}{\dfrac{10x - 1}{10}}$

$\implies \sf \dfrac{7}{2} = \dfrac{10y}{10x - 1}$

$\implies \sf 7(10x - 1) = 20y$

$\implies \sf 70x - 20y = 7 \: \: \: \bf(Equation \: 2)$

• Subtract both the equations

→ 50x - 20y - (70x - 20y) = -5 - 7

→ 50x - 20y - 70x + 20y = - 12

→ - 20x = - 12

→ x = 12/20 = 3/5

• Put the value of x in eqⁿ (1)

→ 50x - 20y = - 5

→ 50 × 3/5 - 20y = - 5

→ 30 - 20y = - 5

→ 30 + 5 = 20y

→ 35 = 20y

→ y = 35/20 = 7/4

→ y = 1 3/4 km

•°• The distance between school and house is 1 3/4 km.

Answer 2

Answer:

Given :-

• A student starting from his house walks at a speed of 2 ½ km/h and reaches his school 6 minutes late.
• Next day starting at the same time he increases his speed by 1 km/h and reaches 6 minutes early.

To Find :-

• What is the distance between the school and his house.

Solution :-

Let,

$\mapsto$ The time be x

$\mapsto$ The distance be y

Now,

$\implies \sf Speed_{(Student)} =\: 2\dfrac{1}{2}\: km/h$

$\implies \sf Speed_{(Student)} =\: \dfrac{4 + 1}{2}\: km/h$

$\implies \sf \bold{\green{Speed_{(Student)} =\: \dfrac{5}{2}\: km/h}}$

Now, let's find the time :

$\implies \sf Time =\: (x + 6)\: minutes$

$\implies \sf Time =\: \bigg(x + \dfrac{\cancel{6}}{\cancel{6}0}\bigg)\: hours\: \: \bigg\lgroup \sf\bold{1\: minutes =\: \dfrac{1}{60}\: hours}\bigg\rgroup$

$\implies \sf \bold{\green{Time =\: \bigg(x + \dfrac{1}{10}\bigg)\: hours}}$

Now, as we know that :

$\clubsuit$ Speed Formula :

$\mapsto \sf\boxed{\bold{\pink{Speed =\: \dfrac{Distance}{Time}}}}$

According to the question by using the formula we get,

$\implies \sf \dfrac{5}{2} =\: \dfrac{y}{\bigg(x + \dfrac{1}{10}\bigg)}$

$\implies \sf \dfrac{5}{2} =\: \dfrac{y}{\bigg(\dfrac{10x + 1}{10}\bigg)}$

$\implies \sf \dfrac{5}{2} =\: \dfrac{y}{1} \times \bigg(\dfrac{10}{10x + 1}\bigg)$

$\implies \sf \dfrac{5}{2} =\: \dfrac{10y}{10x + 1}$

By doing cross multiplication we get,

$\implies \sf 5(10x + 1) =\: 2(10y)$

$\implies \sf 50x + 5 =\: 20y$

$\implies \sf 50x - 20y =\: - 5$

$\implies \sf\bold{\purple{50x - 20y =\: - 5\: ------\: (Equation\: No\: 1)}}$

Again,

$\mapsto$ Next day starting at the same time he increases his speed by 1 km/h and reaches 6 minutes early.

$\implies \sf \dfrac{5}{2} + 1 =\: \dfrac{y}{\bigg(x - \dfrac{1}{10}\bigg)}$

$\implies \sf \dfrac{7}{2} =\: \dfrac{y}{\bigg(\dfrac{10x - 1}{10}\bigg)}$

$\implies \sf \dfrac{7}{2} =\: \dfrac{y}{1} \times \dfrac{10}{10x - 1}$

$\implies \sf \dfrac{7}{2} =\: \dfrac{10y}{10x - 1}$

By doing cross multiplication we get,

$\implies \sf 7(10x - 1) =\: 2(10y)$

$\implies \sf 70x - 7 =\: 20y$

$\implies \sf 70x - 20y =\: 7$

$\implies \sf\bold{\purple{70x - 20y =\: 7\: ------\: (Equation\: No\: 2)}}$

Now, by subtracting the equation no we get,

$\implies \sf 50x - 20y - (70x - 20y) =\: - 5 - 7$

$\implies \sf 50x {\cancel{- 20y}} - 70x {\cancel{+ 20y}} =\: - 12$

$\implies \sf 50x - 70x =\: - 12$

$\implies \sf {\cancel{-}} 20x =\: {\cancel{-}} 12$

$\implies \sf 20x =\: 12$

$\implies \sf\bold{\red{x =\: \dfrac{12}{20}}}$

By putting the value of x in the equation no 2 we get,

$\implies \sf 70x - 20y =\: 7$

$\implies \sf 70\bigg(\dfrac{12}{20}\bigg) - 20y =\: 7$

$\implies \sf \dfrac{84\cancel{0}}{2\cancel{0}} - 20y =\: 7$

$\implies \sf 42 - 20y =\: 7$

$\implies \sf - 20y =\: 7 - 42$

$\implies \sf {\cancel{-}} 20y =\: {\cancel{-}} 35$

$\implies \sf 20y =\: 35$

$\implies \sf y =\: \dfrac{\cancel{35}}{\cancel{20}}$

$\implies \sf y =\: \dfrac{\cancel{7}}{\cancel{4}}$

$\implies \sf\bold{\red{y =\: 1\dfrac{3}{4}\: km}}$

${\small{\bold{\underline{\therefore\: The\: distance\: between\: the\: school\: and\: house\: is\: 1\dfrac{3}{4}\: km\: .}}}}$

Hence, the correct options is option no (c) 1 ¾ km .