Question

A student starts from his house with a speed 2kmph and reaches his college 3min late.next day he increases his speed by 1kmph and reaches college 3 min earlier the distance between his house and college is

Answer 1

Answer:

Nine Kilometers is the answer

Answer 2

Answer:

The distance between his house and college is 0.6 km.

Explanation:

Given : A student starts from his house with a speed 2 kmph and reaches his college 3 min late. Next day he increases his speed by 1 kmph and reaches college 3 min earlier.

To find : The distance between his house and college?

Solution :

Let x be the distance between house and college.

Let t be the time to reach from house to college.

A student starts from his house with a speed 2 kmph and reaches his college 3 min late.

Speed = 2 kmph

Time = $t+\frac{3}{60}$

Distance =  Speed × Time

$x=2(t+\frac{3}{60})$ ....(1)

Next day he increases his speed by 1 kmph and reaches college 3 min earlier.

Speed = 2+1=3 kmph

Time = $t-\frac{3}{60}$

Distance =  Speed × Time

$x=3(t-\frac{3}{60})$ ......(2)

As the distance is same equate (1) and (2),

$2(t+\frac{3}{60})=3(t-\frac{3}{60})$

$2t+\frac{6}{60}=3t-\frac{9}{60}$

$3t-2t=\frac{6}{60}+\frac{9}{60}$

$t=\frac{6+9}{60}$

$t=\frac{15}{60}$

$t=\frac{1}{4}$

The time taken to reach house to college is $\frac{1}{4}$ hours.

Substitute in equation (1),

$x=2(\frac{1}{4}+\frac{3}{60})$

$x=\frac{1}{2}+\frac{1}{10}$

$x=\frac{10+2}{20}$

$x=\frac{12}{20}$

$x=\frac{3}{5}$

$x=0.6km$

Therefore, The distance between his house and college is 0.6 km.