Math, asked by intelligentnepali, 1 year ago

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

Answered by amitnrw
8

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

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