Saroj, Hari and Shyam started their journey form city A to city B at the same time. Shyam took Saroj on his moped, while Hari started walking towards city B. Shyam dropped Saroj somewhere on the way, asked him to continue walking towards City B, drove back and met Hari walking towards City B. All three of them reached City B at the same time. While the walking speed of Saroj and Hari was a uniform 5km/hr. The walking speed of the moped was a uniform 20km/hr. The Distance between two cities was 70 kms.
1. How much time did they take to reach City b from City A?
2. How long did Hari walk?
3. What is the total distance travelled by the moped?
4. At what distance form City A, was Hari when Shyam and Hari met?
Answers
Answer:
6.5 hr
20 km
130 km
20 km
Step-by-step explanation:
Let say Shyam dropped saroj after 20D km
Time taken by shayam = 20D/20 = D hr
Distance Left for Saroj = 70-20D km
Further time taken by Saroj = (70-20D)/5 = 14 - 4D hr
Total Travel time = D + 14 - 4D = 14 - 3D hr
After D hr Shayam Returned back from 20D km
at D hr Hari was at 5D km
Distance Between then = 20D - 5D = 15D km
time taken by both to cover 15 km = 15D/(5 + 20) = 15D/25 = 0.6D hr
Distance Travelled by hari in 0.6D = 5 * 0.6D = 3D km
Distance travelled by moped to meet Hari = 15D - 3D = 12D km
Hari was Picked at 5D + 3D = 8D km
Distance Left = 70 - 8D
Time taken on Moped = (70 -8D)/20 = 3.5 - 0.4D hr
Total time taken = D + 0.6D + 3.5 - 0.4D = 1.2D + 3.5 hr
1.2D + 3.5 = 14 - 3D
=> 4.2D = 10.5
=> D = 2.5
time they take to reach City b from City A = 14 - 3D = 14 - 3*2.5 = 6.5 hr
Hari Walked = 8D = 8*25 = 20 km
Hari walked for D + 0.6D = 1.6D = 1.6*2.5 = 4hr
total distance travelled by the moped = 20D + 12D + 70-8D
= 70 + 24D = 70+ 24 *2.5 = 130 km
distance form City A, Hari was when Shyam and Hari met = 8D = 8*2.5 = 20 km