Question:
Ten ants are on the real number line. At time $t=0$, the $k-th$ ant starts at the point $k^2$ and travelling at uniform speed, reaches the point $(11-k)^2$ at time $t=1$. The number of distinct times at which at least two ants are at the same location is:
Options:
1. $45$
2. $11$
3. $17$
4. $9$
My method:- i found out the velovity of the k th particle and it turned out to be $11(11-2k)$
so the first ant move 99 m second moves 11 m and so on so the first ant crosses 9 other ants second crosses 8 and so on therefore the answer is $9+8+7...+1=45$
but the given answer is $17$
Note:-this question was asked in the exam and i want to challenge its key.So can anybody tell what the correct answer is?
A hint:
The timetable of ant $A_k$ $(1\leq k\leq 10)$ is given by $$x_k(t)=(1-t)k^2+t(11-k)^2\ .$$ Two ants $A_k$ and $A_l$ with $l>k$ could in principle meet at a certain time $t$ which is the solution of the equation $$x_k(t)=x_l(t)\ \tag{1}$$ A priori this solution could lay in the exterior of the $t$-interval $[0,1]$. In this case the ants would not meet during the experiment.
Now analyze how many different admissible values for $t$ you can get by solving equations of the type $(1)$.