A,B and C start running on a circular track from the same point. A and B run in a clcokwise fashion. C runs anti clockwise. When A and C meet for the first time, C is at a distance which is equal to quarter of circumference of circular track. It is also known that A runs faster than C. The ration of speeds of A,C and B cannot be. \(\large \color{black}{ a.)\quad 5:1:2 \quad \hspace{.33em}\\~\\ b.)\quad 3:1:1 \quad\hspace{.33em}\\~\\ c.)\quad 4:2:1 \quad \hspace{.33em}\\~\\ d.)\quad 3:2:1 \quad \hspace{.33em}\\~\\ }\)
well we know that B is only 1/4 of the way done running the track and that A went 3/4 around the track, and since C runs slower that A, it will be less than A but greater than B.
how u concluded that " A went 3/4 around the track".
because it said that "When A and C meet for the first time, C is at a distance which is equal to quarter of..." so A would have to meet him at 3/4 of the distance, because they both went opposite ways
ok, so how to move further
what do you mean?
have u got the answer
yes i have
which one
my answer is 3:2:1
how u got it
because A goes 3 times as fast as B, BUT C is slower than a but faster than b
how u get "A goes 3 times as fast as B".
because a is already 3/4 of the way around the track when b is 1/4 the way around
but it is C who is 1/4 th way around , read the question.
ohh it said which cannot be the answer.. ok. then The answer is b. i didnt read the question good enough
@iGreen what would you say?? or @Adi3
@phi
The question is a bit confusing. first, I think the ratio of speeds of A:C is 3:1 if the ratios are given as speeds of A:B:C (and not A:C:B as implied) then I would expect 3x: y: x where x is an integer thus 3: y : 1 works or 6:y:2 works so B and D "work" they want the ratios that don't work, so I would pick A and C
i have checked the question several times , the ratio of speeds of \(A:C:B\) is asked.
and the answer given in book is \(\large \color{black}{d.)\quad 3:2:1\\~\\ }\)
@Michele_Laino
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