A bus goes to 3 bus stops, at each stop 3/4 of the people on the bus get off and 10 get on. what is the minimum number of people to start on the bus?

Mathematics Asked by Mathie102 on December 1, 2020

A bus goes to $3$ bus stops, at each stop $3/4$ of the people on the bus get off and $10$ get on. what is the minimum number of people to start on the bus?

I think the number would need to be divisible by $4$ and an integer, since you can’t have a non "full person".

I assumed that the people on the bus do NOT include the driver.

What I have so far:
let $”n” = #$ of people on the bus.

First stop: $n/4 + 10$

Second stop: $(n/16+10/4) +10$

Third stop: $(n/64 + 50/16) +10= (840+n)/64$

Don’t know how to move on from here to solve… and how do I account for the amount of people who left the bus?

Please help! Thanks!

4 Answers

Let the bus have x passengers in the beginning

At first stop: No. of people left in the bus= x/4 + 10

At second stop: No. of people left on the bus= x/16 + 50/4

At third stop: No. of people left on the bus= x/64 + 210/16

Hence, the number of people left on the bus after the bus stops three times is (840 + x)/64

Obviously, the number of people can neither be fractional, nor negative

Hence, 840+x has to be a multiple of 64.

The minimum possible value of x for which 840+x is a multiple of 64 is 64*14-840

Hence x=56

Answered by Smriti Sivakumar on December 1, 2020

You found out that when $n_0geq1$ passengers are on the bus at the start then after three stops there are $$n_3={840+n_0over64}$$ people on the bus. As $n_3$ has to be an integer the smallest $n_0$ that would qualify is $n_0=56$, making $n_3=14$. In order to make sure we have to check that for this $n_0$ the intermediate numbers $n_1$ and $n_2$ are integers as well.

By the way: When $x_k$ is the number of passengers after $k$ stops then we have the recursion $$x_{k+1}={1over4} x_k+10 .$$ The "Master Theorem" gives the general solution $$x_k=ccdot 4^{-k}+{40over3}qquad(kgeq0) ,$$ but this expression does not care about integerness. Therefore we really have to go through the cases.

Answered by Christian Blatter on December 1, 2020

Let $R equiv {0, 1, 2, cdots }, ;S equiv {1, 2, 3, cdots }.$

Initially, there are $x_0$ people.
After the 1st stop, there are $x_1$ people.
After the 2nd stop, there are $x_2$ people.
After the 3rd stop, there are $x_3$ people.

(1) $;x_0$ goes to $(1/4)x_0 + 10 = x_1.$
(2) $;x_1$ goes to $(1/4)x_1 + 10 = x_2.$
(3) $;x_2$ goes to $(1/4)x_2 + 10 = x_3.$

It is immediate that $x_0, x_1, x_2$ are all multiples of 4 $;Rightarrow$
$exists ;a,b,c ,in ,S ;ni $ $; x_0 = 4a, ; x_1 = 4b, ; x_2 = 4c.$

(4) By (2), $;b + 10 = 4c ;Rightarrow; c geq 3 ;Rightarrow; exists ;k ,in ,R ;ni c = 3 + k ;Rightarrow $
$x_2 = (12 + 4k) ;Rightarrow$

[by (2)] $;b = (x_2 - 10) = 2 + 4k ;Rightarrow $
$x_1 = (8 + 16k) ;Rightarrow$
[by (1)] $;(1/4) x_0 = a = (x_1 - 10) = [(8 + 16k) - 10] = 16k - 2 ;Rightarrow$

$k geq 1;$ and $;x_0 = 4a = 64k - 8 ;Rightarrow$ The min value for $x_0$ is 56.

Addendum Originally, I thought that the answer was 40.
Then I realized that I was misreading the query. That is, at each stop, 3/4 of the people get off (before 10 get on), not 1/4 of the people.

Addendum-1 A fair criticism of my answer is that I didn't try to focus on the OP's work, and guide his work to a solution. I neglected to try, because with a problem like this I'm only comfortable taking baby steps, so I'm uncomfortable trying to critique a sophisticated approach.

Answered by user2661923 on December 1, 2020

Now expand your third expression to put everything over a common denominator. For example at the third stop you have $frac{stuff}4+7=frac {stuff+28}7$ $stuff$ still has fractions in it, so unpack them. See what the denominator comes out to be and see what the smallest $n$ is to make the fraction an integer.

Answered by Ross Millikan on December 1, 2020

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