Two knights Sir Lancelot and Arthur riding on their horses in a combat charge towards each other in a battling field from opposite ends at uniform speed , but
Sir Lancelot is faster that Arthur.They first pass at a point 18.5 m from Arthur's end.Completing one length each of them rests for 1 min on the edge.After a minute they charge at each other again and meet at 10.5 m from Sir Lancelot's end.How long is the battling field.

Question

Two knights Sir Lancelot and Arthur riding on their horses in a combat charge towards each other in a battling field from opposite ends  at uniform speed , but
Sir Lancelot  is faster that Arthur.They first pass at a point 18.5 m from Arthur's end.Completing one length each of them rests for 1 min on the edge.After a minute they charge at each other again and meet at 10.5 m from Sir Lancelot's end.How long is the battling field.

irishboy123 · Answer

i thought Arthur was the King.  can he also be a knight?  we need to get to the bottom or the heraldry before we can mess around with numbers.

mathmath333 · Answer

Arthur was the knight as well, what do u think

irishboy123 · Answer

seems you are right, sir.  William the Conqueror was knighted when he became king and he was already a Norman knight.  Henry VII was knighted too, *after* becoming King.  

I think your question stands!!

nnesha · Answer

$\large\color{white}{{\rm i~hate~}\rm word~problems}$

ganeshie8 · Answer

29m ?

mathmath333 · Answer

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mathmath333 · Answer

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mathmath333 · Answer

so $29$ m cannot be the total length

hartnn · Answer

if the length is not 29, then they are taking different times to bypass each other in the 2 rounds.

hartnn · Answer

s = speed, x = total distance
$\Large t_1 =\dfrac{x-18.5}{s_L} = \dfrac{18.5}{s_A}$
$\Large t_2 = \dfrac{x-10.5}{s_L} = \dfrac{10.5}{s_A}$

hartnn · Answer

atleast one more info. is required to solve that system

mathmath333 · Answer

"Sir Lancelot  is faster that Arthur".

as their speeds are different, so their times will be different too .

 i think u r close to solve this maybe

mathmath333 · Answer

$\large \color{black}{
a.)  55.5\quad m\hspace{.33em}\~\
b.) 45 \quad m\hspace{.33em}\~\
c.) 66 \quad m\hspace{.33em}\~\
d.) 49 \quad m\hspace{.33em}\~\
}$

hartnn · Answer

my initial thoughts were, 
since they are not changing their speeds,
they should again meet at 18.5m from Lancelot's end in the next round.

if not speed, what did they change? 
the ground is same... so how come time change?

mathmath333 · Answer

i think this has some relevance then 

"each of them rests for 1 min on the edge" or may be not

hartnn · Answer

i think that info is just redundant.

hartnn · Answer

since ground is same, the only thing that can change their meeting point is their speeds

mathmath333 · Answer

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