tricky math problem?

Question

At some stops, a certain bus picks up 5 people. At other stops, it picks up 2 and, at the same thime, lets off 5. 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. If the nmber of stops is greater than 17, what is the least number of stops the bus makes?

If anyone can figure this out, can you please help me step by step to figue this out? thank you verrry much! =D

Answer 1

You can think of the stops as +5 and -3.

My method would be to assume you pick up passengers until you are above 11, then subtract 3 until you are below, etc. In the end you should hit 11 exactly.

Stop 1: Add --> 5
Stop 2: Add --> 10
Stop 3: Add --> 15
Stop 4: Subtract --> 12
Stop 5: Subtract --> 9
Stop 6: Add --> 14
Stop 7: Subtract --> 11
Stop 8: Add --> 16
Stop 9: Subtract --> 13
Stop 10: Subtract --> 10
Stop 11: Add --> 15
Stop 12: Subtract --> 12
Stop 13: Subtract --> 9
Stop 14: Add --> 14
Stop 15: Subtract --> 11
Stop 16: Add --> 16
Stop 17: Subtract --> 13
Stop 18: Subtract --> 10
Stop 19: Add --> 15
Stop 20: Subtract --> 12
Stop 21: Subtract --> 9
Stop 22: Add --> 14
Stop 23: Subtract --> 11

I'm thinking 23 stops.

It takes 7 stops to get to 11 (4*5 - 3*3 = 11)
It then takes 8 stops to get back to 11 (3*5 - 5*3 = 0)
It will take another 8 stops to get back to 11 (3+5 - 5*3 = 0)

7 + 8 + 8 = 23

All together you had 10 stops picking up (net addition of 50 people) and 13 stops dropping off (net loss of 39 people). That results in 11 people after 23 stops.

I don't think you can do better.

Another way to think of this is:
Let A be number of stops picking up
Let B be number of stops dropping off
5A - 3B = 11

Now let B = n - A, where n is 18, 19, 20, etc.

5A - 3(n - A) = 11
5A - 3n + 3A = 11
8A - 3n = 11
A = (11 + 3n)/8

Now just try values of n until you get a perfect multiple of 8 for 3n + 11

3(18) + 11 = 54 + 11 = 65 nope
3(19) + 11 = 68 nope
3(20) + 11 = 71 nope
3(21) + 11 = 74 nope
3(22) + 11 = 77 nope
3(23) + 11 = 80 YES!

So we have proven that the minimum number of stops (above 17) is 23 stops.

A = 80/8 = 10
B = 23 - 10 = 13

10 stops picking up
13 stops dropping off
-------
23 stops minimum

Answer 2

x+y>17
5x+2y-5y=11
5x-3y=11
minimize x+y
x>0, y>0

5x=3y+11
y's unit digit=3 or 8
y=13 is the least
5x=3*13+11
=50
x=10

x+y=23