SOLUTION: At some stops, a certain bus picks up 5 people. At other stops it picks up 2 and at the same time, let’s off 5. There are no other stops than these. It starts it’s run empty an

Algebra ->  Inequalities -> SOLUTION: At some stops, a certain bus picks up 5 people. At other stops it picks up 2 and at the same time, let’s off 5. There are no other stops than these. It starts it’s run empty an      Log On


   

Question 1194075: At some stops, a certain bus picks up 5 people. At other stops it picks up 2 and at the same time, let’s off 5. There are no other stops than these. It starts it’s run empty and picks up 5 people. At the end, it has 11 people aboard. If the number of stops is greater than 17, what is the least number of stops the bus makes?
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


x = number of stops where the bus picks up 5 people
y = number of stops where the bus picks up 2 people and lets off 5

At each of the x stops, the number of people on the bus increases by 5.
At each of the y stops, the number of people on the bus decreases by 3.

Then after (x+y) stops the number of people on the bus is 5x-3y.

We are to find the least number of total stops (x+y) greater than 17 for which the number of people left on the bus is 11:

5x-3y=11 ; subject to the constraint that x+y>17

This is a linear Diophantine equation -- a single equation with two variables, in which the values of the variables are integers.

Here is a standard formal way for solving this kind of equation.

(1) Solve the equation for one of the variables (it doesn't matter which)

5x-3y=11
5x=3y%2B11

Divide by 5, writing the result on the right as quotient plus remainder:

x+=+%283y%2B11%29%2F5+=+%2810%2B%283y%2B1%29%29%2F5+=+2%2B%283y%2B1%29%2F5

(2) Find the values of y that make that expression an integer; find the corresponding values of x, and find the value of x+y. We are looking for the smallest value of x+y that is greater than 17.
    y   x=2+(3y+1)/5  x+y
  ------------------------
    3   2+2 = 4        7
    8   2+5 = 7       15
   13   2+8 = 10      23

ANSWER: 23

A note about solving linear Diophantine equations by this method....

Note that in the table above, the values of x and y (and x+y) form arithmetic sequences. Specifically, the possible y values have a common difference of 5 and the possible x values have a common difference of 3. The "5" and "3" are because of the equation x=(3y+1)/5.

If you are solving a problem like this and your list of x and/or y values does not form an arithmetic sequence, then some of your calculations are incorrect.

At the same time, knowing that these lists of values form arithmetic sequences allows you to find other solutions without searching for x or y values. For example, in this problem, seeing that the first two possible y values are 3 and 8, you know that the common difference is 5, so the next possible y values are 13, 18, 23, ...; and since the first two possible x values are 4 and 7, you now that the next possible corresponding x values are 10, 13, 16....