Reena reaches a birthday party 20 minutes late if she walks at 3 km/h from her house. If she increases her speed to 4 km/h, she would reach 30 minutes early. What is the distance between her house and the party venue?

Explanation

Let the distance between Reena's house and the party venue be ddd km. When she walks at 3 km/h: Time taken to reach the venue =d3= \frac{d}{3}=3d​hours. She arrives 20 minutes late, so the actual time to reach on time =d3−2060= \frac{d}{3} - \frac{20}{60}=3d​−6020​. When she walks at 4 km/h: Time taken to reach the venue =d4= \frac{d}{4}=4d​hours. She arrives 30 minutes early, so the actual time to reach on time =d4+3060= \frac{d}{4} + \frac{30}{60}=4d​+6030​. Since both times are equal, we can set up the equation: d3−13=d4+12\frac{d}{3} - \frac{1}{3} = \frac{d}{4} + \frac{1}{2}3d​−31​=4d​+21​ Let's solve for ddd. The distance between Reena's house and the party venue is 10 km. ​