Question

1. A man has to reach his office at 9:30 am. If he walks 3 km/hr, he is 10 minutes
late and if he walks 4 km/hr, he is 15 minutes early. What is the distance (in km)
between his house and office?


please let me know the answer​

Answer 1

Given :

A man has to reach his office at 9:30 am

If he walks 3 km/hr, he is 10 minutes  late

If he walks 4 km/hr, he is 15 minutes early

To Find :

The distance  between his house and office

Solution :

Let The distance = d km

Let the time taken to cover d distance = t hours

∵  Distance = Speed × Time

At 3 km/h speed , late by 10 min

So,   $D_1$ = 3 km/h × ( t + $\dfrac{10}{60}$ )  hours                    ........1

Again

At 4 km/h speed , early by 15 min

So,   $D_2$ = 4 km/h × ( t - $\dfrac{15}{60}$ )  hours                   ........2

Now, As both side distance cover is same

So,    $D_1 = D_2$

i.e  From eq 1 and eq 2  , we get

3 km/h × ( t + $\dfrac{10}{60}$ )  hours  =  4 km/h × ( t - $\dfrac{15}{60}$ )  hours  

Or,  3  × ( t + $\dfrac{1}{6}$ )  =  4  × ( t - $\dfrac{1}{4}$ )

Or,   3 t + 3 × $\dfrac{1}{6}$ = 4 t - 4 × $\dfrac{1}{4}$

Or,   3 t + $\dfrac{1}{2}$ = 4 t - 1

Or,   4 t - 3 t = 1 +  $\dfrac{1}{2}$

Or,             t = $\dfrac{2+1}{2}$

i.e              t = $\dfrac{3}{2}$  hours

∴  Total time taken =  $\dfrac{3}{2}$  hours

Again

∵ $D_1$ = 3 km/h × ( t + $\dfrac{10}{60}$ )  hours    

Put the value of t into eq 1

So, $D_1$ = 3 km/h × ( $\dfrac{3}{2}$ + $\dfrac{1}{6}$ )  hours    

           = 3 ×  $\dfrac{10}{6}$

           = 5 km

Hence, The distance  between Man's house and office is 5 km   Answer

Answer 2

Answer:

How does this kid all the questions i have????????????????????????????????????????????

Step-by-step explanation: