Rock750 wrote:
vinaymimani wrote:
We know that the average speed is 2.8 mph. Thus,
as \(s<2.8<s+1\)
or \(s>1.8\) and \(s<2.8\). As s is an integer, the only value can be s=2. Thus,
Let the total length be d. Let the path where the sun shone on him be d1 = kd (0<k<1)
\(\frac{d}{2.8}\)=\(\frac{d1}{2}\)+\(\frac{d2}{3}\) =\(\frac{kd}{2}+\frac{(1-k)d}{3}\)
or \(\frac{10}{28}\)=\(\frac{k}{2}+\frac{(1-k)}{3}\)
By simple substitution, we can find that k = 1/7.
Hi vinaymimani
How did u get that : as \(s<2.8<s+1\)
or \(s>1.8\) and \(s<2.8\).
Regards
Hi Rock750
I will give a general proof for the above question :
From the given picture, let n1 and n2(n1,n2 are not equal to 0) be the speed for the length d1 and d2 respectively.
Also,
d=d1+d2 , d1 = kd, where 0<k<1
Now, average speed for this length is
\(\frac{d}{[d1/n1+d2/n2]}\) =\(\frac{d}{[kd/n1+d(1-k)/n2]}\) = let's call this value as AVG
Now, Considering n1>n2;lets assume that this average speed will be always between n1 & n2 or n2<AVG<n1\(n2<\frac{n1*n2}{[n2*k+n1*(1-k)]}\) ..... I
and
\(\frac{n1*n2}{[n2*k+n1*(1-k)]}<n1\).........II
Thus, from I, our assumption will be right iff
\([n2*k+n1*(1-k)]<n1\)
or (n1-n2)k>0.
As we had already assumed that, this stands true.
Similarly, from II, we have our assumption to be true iff
\((n2-n1)*(1-k)<0\); which is again true.
Thus for any positive value of n1,n2 the inequality
n2<AVG<n1 will always holds.
In the given sum, n2 = s, n1 = s+1 and AVG = 2.8 mph. Thus, 2.8 will always lie between s and (s+1). The question mentions that "s" is an
integer for this very purpose.
I hope it was clear enough.
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