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A 66-seat bus picks up 1 passenger at the 1st bus stop, 2 passengers at the 2nd stop, 3 passengers at the 3rd bus stop and so on. When will all the seats in the bus be filled?

 Apr 6, 2021
 #1
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\(\boxed{11}\) stops will be taken

 Apr 6, 2021
 #3
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How did you get 11? See my solution for how I got 12. I think you might have factored the quadratic the wrong way as $(n+12)(n-11)=0$ or something.

RiemannIntegralzzz  Apr 6, 2021
 #4
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really? i honestly just used the long way and just added until i got 66, 1+2+3+4+5+6+7+8+9+10+11 is 66, i believe. i may have done it wrong, sorry!

Guest Apr 6, 2021
 #5
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Wait you were right, I just realized that I actually factored the quadratic the wrong way :)) the irony. Thank you for helping out!

RiemannIntegralzzz  Apr 6, 2021
edited by RiemannIntegralzzz  Apr 6, 2021
 #6
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ohh well we're all human right? laugh mistakes always happen :) good job though, your explanation is much more logical than mine lol

Guest Apr 6, 2021
 #7
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Thank you!

RiemannIntegralzzz  Apr 6, 2021
 #2
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You have the sequence $1+2+3+4+5+...+(n-1)+n$, so the sum of this will be $\frac{n(n+1)}{2}$.

 

So, you are looking for $\frac{n(n+1)}{2}=66$.

 

$n(n+1)=132$

$n^2+n-132=0$

$(n+12)(n-11)=0$.

 

This means that $n=11$ or $n=-12$. $n$ obviously cannot be negative, so the only solution is $n=11$, so at the $\boxed{11^{th}}$ stop.

 Apr 6, 2021
edited by RiemannIntegralzzz  Apr 6, 2021
 #8
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66  is  the  sum of the first  n positive integers....so  we  have

 

N ( N + 1)   / 2    = 66

 

N ( N + 1)  = 132

 

N^2  + N  - 132   =   0

 

(N + 12) ( N -11)  = 0

 

Since N  is positive....N - 11   = 0  .....N  =11

 

It will be full after  the 11th stop

 

 

cool cool cool

 Apr 6, 2021

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