Suppose you have a train with 11 carriages. The sum of the number of people in any 3 consecutive carriages (i.e. 1-2-3, 2-3-4, 3-4-5, etc.) is 99. Given that there are 381 people on the train, how many people are on carriage 9? Provide an work used to find the answer.
there's 15 people in carriage 9
here's how I did it:
if you want to make any 3 consecutive numbers (number of people in the carriage) add up to the same number, you have 2 patterns to choose from:
1. X N N X N N X N N X N N 2. X N X X N X X N X X N (N and X represent different numbers)
then I just tried a bunch of numbers and both patterns
and I got N = 42 X = 15, pattern 1
P.S. sorry if it doesn't make any sense I'm really bad at explaining LOL
here's how I did it:
if you want to make any 3 consecutive numbers (number of people in the carriage) add up to the same number, you have 2 patterns to choose from:
1. X N N X N N X N N X N N 2. X N X X N X X N X X N (N and X represent different numbers)
then I just tried a bunch of numbers and both patterns
and I got N = 42 X = 15, pattern 1
P.S. sorry if it doesn't make any sense I'm really bad at explaining LOL
ABCDEFGHIJK= 1-11
A+B + C = 99
B+C+D = 99
C+D+E = 99
D+E+F =99
E+F+G = 99
F+G+H = 99
G+H+I = 99
H+I+J = 99
I+J+K = 99
Everything = 381
Adding (A+B+C) and (D+E+F) and (G+H+I) we get 297, so J+K = 84, or I = 15, so 15 yeah
A+B + C = 99
B+C+D = 99
C+D+E = 99
D+E+F =99
E+F+G = 99
F+G+H = 99
G+H+I = 99
H+I+J = 99
I+J+K = 99
Everything = 381
Adding (A+B+C) and (D+E+F) and (G+H+I) we get 297, so J+K = 84, or I = 15, so 15 yeah
:| bruh
@GloweyySquidd said in #2:
> there's 15 people in carriage 9
>
> here's how I did it:
> if you want to make any 3 consecutive numbers (number of people in the carriage) add up to the same number, you have 2 patterns to choose from:
If you forget about the total number of people in the train and in three consecutive carriage and you just want all the 3 consecutive carriages to have the same number of people, there are 8 equations (one per three consecutive carriages except for the first one) and 11 unknowns (one per carriage). So the space of solutions is a vector space of dimension at least 3 [actually exactly 3 because the equations are linearly independent]. But what you described is the union of two vector spaces each of dimension 2.
The "correct" pattern is
X Y Z X Y Z X Y Z X Y.
Once we know that, the solution of Calby can be rewritten as follows:
X+Y+Z=99 (eq1)
4X+4Y+3Z=381 (eq2)
4(eq1)-(eq2): Z=4*99-381=15.
The reason you were still able to find the correct solution is because the original system has two independant equations for three unkowns, so its space of solutions has dimension 1. Restricting to a certain pattern amounts to add an extra equation [in your case Y=Z or X=Z], that turns out to be independent from the other two, thus obtaining three independent equations for three unknown, hence a unique solution for (X, Y, Z). But since there was already a unique solution for Z, what you find for (X, Y, Z) must agree with it.
> there's 15 people in carriage 9
>
> here's how I did it:
> if you want to make any 3 consecutive numbers (number of people in the carriage) add up to the same number, you have 2 patterns to choose from:
If you forget about the total number of people in the train and in three consecutive carriage and you just want all the 3 consecutive carriages to have the same number of people, there are 8 equations (one per three consecutive carriages except for the first one) and 11 unknowns (one per carriage). So the space of solutions is a vector space of dimension at least 3 [actually exactly 3 because the equations are linearly independent]. But what you described is the union of two vector spaces each of dimension 2.
The "correct" pattern is
X Y Z X Y Z X Y Z X Y.
Once we know that, the solution of Calby can be rewritten as follows:
X+Y+Z=99 (eq1)
4X+4Y+3Z=381 (eq2)
4(eq1)-(eq2): Z=4*99-381=15.
The reason you were still able to find the correct solution is because the original system has two independant equations for three unkowns, so its space of solutions has dimension 1. Restricting to a certain pattern amounts to add an extra equation [in your case Y=Z or X=Z], that turns out to be independent from the other two, thus obtaining three independent equations for three unknown, hence a unique solution for (X, Y, Z). But since there was already a unique solution for Z, what you find for (X, Y, Z) must agree with it.
All right, you guys are getting good at this. I might need to pull out all the stocks for the next one. I also need something not directly of a YouTube video.
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