While solving the problem

Half the people on a bus get off at each stop after the first and no one gets on after the first stop.If only one person gets off at stop number 7 .How many people got on at the first stop ?

Here is how I am solving it but I guess I am wrong

Let no of people who got on = x

so $x + \frac{x}{2} + \frac{x}{4}+ \frac{x}{8}+ \frac{x}{16}+ \frac{x}{32}+ 1 = 0 $

Any suggestions ?

  • 2
    $\begingroup$ Another way is to just work backwards iteratively. $\endgroup$ – AnonymousCoward Jul 25 '12 at 5:20
  • $\begingroup$ I wonder how many people get off the bus at the stop no. 8. And I'm really affraid to try to express my guess... $\endgroup$ – CiaPan Sep 19 '16 at 14:02

Let us do it in your style. Say that there are $x$ people on the bus after the first stop.

At the second stop, $\frac{x}{2}$ get off, so $\frac{x}{2}$ are left.

At the third stop, $\frac{x}{4}$ get off, and $\frac{x}{4}$ are left.

At the fourth stop, $\frac{x}{8}$ get off, $\frac{x}{8}$ are left.

At the fifth stop, $\frac{x}{16}$ get off, $\frac{x}{16}$ are left.

At the sixth stop, $\frac{x}{32}$ get off, $\frac{x}{32}$ are left.

At the seventh stop, $\frac{x}{64}$ get off, and $\frac{x}{64}$ are left.

But $1$ person is left. So $\frac{x}{64}=1$, and therefore $x=64$.

Alternately, there were $2$ people on the bus just before the $7$th stop, so $4$ just before the sixth, so $8$ just before the fifth, so $16$just before the fourth, so $32$ just before the third, so $64$ just before the second stop.

Or else, (but this is too much work, $\frac{x}{2}$ got off at the second, and so on, leaving $1$ person on the bus after the seventh. This gives us the equation $$x=\frac{x}{2}+\frac{x}{4}+\frac{x}{8}+\frac{x}{16}+\frac{x}{32}+\frac{x}{64}+1.$$ Solve for $x$. Maybe to make life easier first multiply both sides by $64$.

Remark: Does this question involve series? Well, if we solve it my third way, which is fairly inefficient, then indeed we end up finding the sum of a finite geometric series. But there are more efficient ways that just involve a sequence.

  • $\begingroup$ Thanks.. Looks like I messed up at the last part. I actually summed up the equation and tried solving for x $\endgroup$ – Rajeshwar Jul 25 '12 at 5:32
  • $\begingroup$ Since all the people got off I equated a $0$. Thats where I was wrong $\endgroup$ – Rajeshwar Jul 25 '12 at 5:33
  • $\begingroup$ Another way to solve for $x$ is to factor $x$ and calculate the sum of this geometric progression $\endgroup$ – Belgi Jul 25 '12 at 5:37

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