SOLUTION: A bus driver who starts with a large empty bus picks us 11 riders at every odd- numbered stop and drops off 5 riders at every even-numbered stop. The stop after which the bus will

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Question 1199189: A bus driver who starts with a large empty bus picks us 11 riders at every odd- numbered stop and drops off 5 riders at every even-numbered stop. The stop after which the bus will first contain more than 165 passengers is stop number
a) 27 b) 28 c) 52 d) 53 e) 55
Found 4 solutions by ikleyn, math_tutor2020, Edwin McCravy, greenestamps:
Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
A bus driver who starts with a large empty bus
picks us 11 riders at every odd-numbered stop and drops off 5 riders at every even-numbered stop.
The stop after which the bus will first contain more than 165 passengers is stop number
a) 27 b) 28 c) 52 d) 53 e) 55
~~~~~~~~~~~~~~ 

After the stop #2, there are   0 + (11-5)        =  6 riders in the bus.


After the stop #4, there are   6 + (11-5) =  6+6 = 12 riders in the bus.


After the stop #6, there are  12 + (11-5) = 12+6 = 18 riders in the bus.



         As you can notice, there is a PATTERN: 
    after the stop #(2n), there are 6n riders in the bus.



Now,               165/6 = 27.5.


So, we expect the number n from this relation  n <~~~> 165/6 = 27.5,  
which gives the number of the stop somewhere around  2n ~ 55.


Based on this estimation, let's look what happens  
in vicinity of n= 27 (in vicinity of the stop #55).  


             See the table below.


    stop #      n       number of passengers 
                          after this stop 
  ---------------------------------------------

      50       25         6*25     = 150             

      51                  150 + 11 = 161             

      52       26         6*26     = 156             

      53                  156 + 11 = 167     <<<---=== here the number of passengers first time exceeded 165

      54       27         6*27     = 162

      55                  162 + 11 = 173   

      56       28         6*28     = 168


From the table, you see that for the first time, the number of passengers will exceed 165 after stop #53.


ANSWER.  For the first time, the number of passengers will exceed 165 after stop #53.

Solved.

------------------

This problem and my solution teaches you three key thing:


        (a)   To organize your thoughts and calculations;

        (b)   To find a pattern;

        (c)   To make your logical conclusions based on estimations.




Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

The sequence
11,6,17,12,23,18,...
represents the number of riders on the bus for any particular stop.

We start with 11 riders on stop 1 (odd number).
Then the bus drops off 5 riders on stop 2 (even number) to have 11-5 = 6 people remaining.

For the next odd numbered stop, stop 3, there are 11 more riders to have 6+11 = 17 people.
For the next even numbered stop, stop 4, exactly 5 people leave to have 17-5 = 12 people remaining.

This process continues until the bus is full of course, or when there aren't any other stops to make.
We'll assume that the bus can hold 165 people or a bit more.


-----------------------------------------

Going back to this sequence:
11,6,17,12,23,18,...

Let's highlight the even numbered terms.
11,6,17,12,23,18,...
This is the set of multiples of 6.
Since 2 is a factor of 6, each item is even.
Therefore, it is impossible to have 165 people on the bus for any even numbered stop since 165 is odd.

Let's highlight the odd numbered terms.
11,6,17,12,23,18,...
Notice that we're increasing by 6 each time.
This is exactly because 11-5 = 6 is the net change in the number of people on the bus.

Therefore, the odd numbered terms of this sequence are arithmetic
a = 11 = first term
d = 6 = common difference
a(n) = a+d*(n-1) = nth term
a(n) = 11+6*(n-1)

Let's see if we get 165 in this sequence
a(n) = 11+6*(n-1)
165 = 11+6*(n-1)
165 = 11+6n-6
165 = 6n+5
6n+5 = 165
6n = 165-5
6n = 160
n = 160/6
n = 26.667
We don't have 165 in the odd numbered terms since n must be a positive integer.
But it gives us an idea where the min number over 165 could be located.

Let's revisit these highlighted terms:
11,6,17,12,23,18,...
The "11" is the 1st odd number and 1st term overall.
The "17" is the 2nd odd number and 3rd term overall.
The "23" is the 3rd odd number and 5th term overall.
And so on.
For the kth odd numbered term, it refers to the (2k-1)th term overall, where k is a positive integer.

If n = 26, then,
a(n) = 11+6*(n-1)
a(26) = 11+6*(26-1)
a(26) = 161
The 26th odd numbered bus stop will have 161 people on the bus.
We haven't gone over 165, but we're really close.

If n = 27, then,
a(n) = 11+6*(n-1)
a(27) = 11+6*(27-1)
a(27) = 167
We now have gone over 165.
This is the smallest such n to achieve the goal.

The k = 27 odd numbered term refers to the 2k-1 = 2*27-1 = 54-1 = 53rd term overall of the sequence 11,6,17,12,23,18,...

------------------------------------------

Answer: 53 (choice D)
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

The correct answer is 53.

A bus driver who starts with a large empty bus picks us 11 riders at every odd-
numbered stop and drops off 5 riders at every even-numbered stop. The stop after
which the bus will first contain more than 165 passengers is stop number
a) 27 b) 28 c) 52 d) 53 e) 55
Each pair of stops gains +11-5 or 6 passengers.  So, immediately after every 
even-numbered stop, 6 more passengers are on the bus than were on the bus
at the previous even-numbered stop. 

We are looking for the largest even number of stops such that

6n < 165
 n < 27.5

The largest even number of stops less than 27.5 is 26.
There will have been 26 even-numbered stops on the 52nd stop, so there will be
26*6=156 passengers just after the 52nd stop.  The 53rd stop will increase the
number of passengers to 156+11=167.

So the number of stops asked for is 53, the first stop that puts the number of
passengers over 165.

The output of a brute force computer program to justify my answer:

Just after stop no. 1, there are 11 passengers.
Just after stop no. 2, there are 6 passengers.
Just after stop no. 3, there are 17 passengers.
Just after stop no. 4, there are 12 passengers.
Just after stop no. 5, there are 23 passengers.
Just after stop no. 6, there are 18 passengers.
Just after stop no. 7, there are 29 passengers.
Just after stop no. 8, there are 24 passengers.
Just after stop no. 9, there are 35 passengers.
Just after stop no. 10, there are 30 passengers.
Just after stop no. 11, there are 41 passengers.
Just after stop no. 12, there are 36 passengers.
Just after stop no. 13, there are 47 passengers.
Just after stop no. 14, there are 42 passengers.
Just after stop no. 15, there are 53 passengers.
Just after stop no. 16, there are 48 passengers.
Just after stop no. 17, there are 59 passengers.
Just after stop no. 18, there are 54 passengers.
Just after stop no. 19, there are 65 passengers.
Just after stop no. 20, there are 60 passengers.
Just after stop no. 21, there are 71 passengers.
Just after stop no. 22, there are 66 passengers.
Just after stop no. 23, there are 77 passengers.
Just after stop no. 24, there are 72 passengers.
Just after stop no. 25, there are 83 passengers.
Just after stop no. 26, there are 78 passengers.
Just after stop no. 27, there are 89 passengers.
Just after stop no. 28, there are 84 passengers.
Just after stop no. 29, there are 95 passengers.
Just after stop no. 30, there are 90 passengers.
Just after stop no. 31, there are 101 passengers.
Just after stop no. 32, there are 96 passengers.
Just after stop no. 33, there are 107 passengers.
Just after stop no. 34, there are 102 passengers.
Just after stop no. 35, there are 113 passengers.
Just after stop no. 36, there are 108 passengers.
Just after stop no. 37, there are 119 passengers.
Just after stop no. 38, there are 114 passengers.
Just after stop no. 39, there are 125 passengers.
Just after stop no. 40, there are 120 passengers.
Just after stop no. 41, there are 131 passengers.
Just after stop no. 42, there are 126 passengers.
Just after stop no. 43, there are 137 passengers.
Just after stop no. 44, there are 132 passengers.
Just after stop no. 45, there are 143 passengers.
Just after stop no. 46, there are 138 passengers.
Just after stop no. 47, there are 149 passengers.
Just after stop no. 48, there are 144 passengers.
Just after stop no. 49, there are 155 passengers.
Just after stop no. 50, there are 150 passengers.
Just after stop no. 51, there are 161 passengers.
Just after stop no. 52, there are 156 passengers.
Just after stop no. 53, there are 167 passengers.

So, when the bus starts after the 53rd stop, there 
will be 167 > 165 passengers.
 
Edwin
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


To start with, we can eliminate answer choices b and c because they are even numbers, and the number of riders on the bus DECREASES at each even-numbered stop.

The number of riders increases by 11 at every odd-numbered stop, so the number of riders will surpass 165 when the number of riders after an even-numbered stop is greater than 165-11 = 154.

So look for the pattern of the numbers of riders after the even-numbered stops.

After stop #2 (the 1st even-numbered stop): 11-5 = 6
After stop #4 (the 2nd even-numbered stop): 6+(11-5) = 12
After stop #6 (the 3rd even-numbered stop): 12+(11-5) = 18

The pattern is clear: after the n-th even-numbered stop, the number of riders is 6n.

Now find at which even-numbered stop the number of riders is greater than 154:

6n > 154
n > 154/6 = 25 2/3

It is after the 26th even-numbered stop that the number of riders is first greater than 154.

The 26th even numbered stop is stop number 52; so it is after the 53rd stop that the number of riders first surpasses 165.

ANSWER: d) 53
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