Question 697720
At some stops, a certain bus picks up 5 people.  At other stops, it picks up 2 and, at the same time lets off 5.  
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Let x = the number of times the bus increases the no. of passengers by 5.

When the bus picks up 2 and lets off 5, it decreases the no. of passengers
by 3.

Let y = the number of times the bus decreases the no. of passengers by 3.

Then the number of passengers is 5x - 3y 
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There are no other stops but these.  It starts its run empty and picks up 5 people.  At the end, it has 11 people aboard.
<pre>
So we have
      5x - 3y = 11

The smallest coefficient in absolute value is |-3| = 3.

So we write 5 and ll in terms of their nearest multiple of 3.

The nearest multiple of 3 to 5 is 6, so we write 5 as 6-1,
The nearest multiple of 3 to 11 is 12, so we write 11 as 12-1, 

  (6-1)x - 3y = 12-1
  6x - x - 3y = 12 - 1

Divide every term through by 3

  2x - {{{x/3}}} - y = 4 - {{{1/3}}}

Isolate all the fractions on the right  

 2x - y - 4 = {{{x/3}}} - {{{1/3}}}

The left side is an integer, so the right side is also an
integer.  Let that integer be A.  Then

 2x - y - 4 = A  and  {{{x/3}}} - {{{1/3}}} = A
                      x - 1 = 3A
                          x = 3A + 1

Substitute 3A+1 for x in the left equation

2(3A+1) - y - 4 = A
 6A + 2 - y - 4 = A
     6A - 2 - y = A
         5A - 2 = y

So we have (x,y) = (3A+1,5A-2)

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       </pre>
a. If the number of stops is greater than 75, what is the least number of stops the bus makes?
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          x + y > 75

Substitute (x,y) = (3A+1,5A-2)

    3A+1 + 5A-2 > 75
           8A-1 > 75
             8A > 76
              A > {{{76/8}}}
              A > 9.5

x + y increases with increasing A, so the
minimum value of x+y (the number of stops, is when A
is the smallest integer greater than 9.5, which is 10.

Therefore (x,y) = (3A+1,5A-2) = (3·10+1,5·10-2) = (31,48)

So the smallest number of stops is the minimum value of x+y = 31+48 = 79

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  </pre> 
b. If the number of stops is greater than 703, what is the least number of stops the bus makes?
<pre>
          x + y > 703

Substitute (x,y) = (3A+1,5A-2)

    3A+1 + 5A-2 > 703
           8A-1 > 703
             8A > 704
              A > {{{704/8}}}
              A > 88

x + y increases with increasing A, so the
minimum value of x+y (the number of stops), is when A
is the smallest integer greater than 88, which is 89.

Therefore (x,y) = (3A+1,5A-2) = (3·89+1,5·89-2) = (268,443)

So the smallest number of stops is the minimum value of x+y = 268+443 = 711

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c. For part A, how many of the stops must be where the total gain in passengers is 5? 
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That's just the value of x for part a, which is 31.
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For part B?
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That's just the value of x for part b, which is 268.

Edwin</pre>