Math, asked by Harishhfhjb9730, 1 year ago

Ram and shyam are hunting a wild boar at the start they are on a straight road at a distance of 1 kilometre from each other at point a and b respectively when ram was about to catch the ball it start running in the opposite direction towards shyam when it reaches shyam it turns back and start running towards ram

Answers

Answered by Anonymous
16

The question is incomplete.

I'll solve it superficially as u can understand and can solve the problem urself.

Let's assume Ram has a speed of y metre/min. while the boar has a speed of x metre/min.

Now let's assume that after t₁ min. the boar reaches Shayam.

Thus, t₁x = 1 km.

As the boar started running from Ram towards Shayam.

Now after reaching Shayam the boar runs in opposite direction at it's same speed of x metre/min. and after t₂ min. it finally faces Ram.

Then, an equation can be formed as follows;

y(t₁ + t₂) = 1000 + t₂x

If any three variables are known other can be solved.

Answered by amitnrw
0

Given : they are on a straight road at a distance of 1  km from each other at point A and B respectively.

Ram's speed is 4 times the speed of Shyam.

Boar can  only be caught by Ram and Shyam together.

To Find :  At what distance (in meters) from A do they catch the boar

Solution:

Ram and  Shyam are on a straight road at a distance of 1  km from each other .

Boar can  only be caught by Ram and Shyam together.

Hence Distance covere by Ram and shyam together = 1 km as they are 1 km Apart.

Shyam Speed =  A km/hr

Ram speed = 4A  km/hr

Let say After T hr they catch Boar

Distance covered by Shyam  = AT km

Distance covered by Ram  = 4AT   km

AT + 4AT  =  1 km

=> 5AT = 1 km

=> AT = 0.2 km

 distance from A = Distance covered by Ram  = 4AT = 4(0.2) = 0.8 km

 0.8 x 1000 = 800 m

distance (in meters) from A   they catch the boar = 800 m

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