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

Correct option is C

Given:

1. Reena is 20 minutes late if she walks at 3 km/h.

2. Reena is 30 minutes early if she walks at 4 km/h.

We need to find the distance between her house and the party venue.

Formula Used:

Let d be the distance between Reena's house and the venue.

Time taken at $$3 km/h: \frac{d}{3} hours.$$​​

Time taken at $$4 km/h: \frac{d}{4} hours.$$​​

The difference in time between the two speeds (since one results in being 20 min late and the other 30 min early) is 50 minutes $$(or \frac{50}{60} = \frac{5}{6} hours).$$​​

Solution:

1. Set up the equation based on the time difference:

$$\frac{d}{3} - \frac{d}{4} = \frac{5}{6}$$​

- Find a common denominator for the fractions:

$$\frac{4d - 3d}{12} = \frac{5}{6}$$​

​$$\frac{d}{12} = \frac{5}{6}$$​

- Multiply both sides by 12 to isolated:

​$$d = \frac{5}{6} \times 12 = 10$$ 

Therefore, the distance between Reena's house and the venue is 10 km.