Bags, Marbles, and Scales
Math brain teasers require computations to solve.
In front of me, I have level scales. On each side, there are 3 bags. In each bag, there are some marbles. There are 150 Marbles in total. Using the clues below, work out how many marbles are in each bag.
Bag 1 + Bag 2 = 56 marbles
Bag 1 + Bag 3 = 54 marbles
Bag 2 + Bag 3 = 40 marbles

Bag 4 + Bag 5 = 61 marbles
Bag 4 + Bag 6 = 37 marbles
Bag 5 + Bag 6 = 52 marbles
Answer
Bag 1 = 35 marbles
Bag 2 = 21 marbles
Bag 3 = 19 marbles
Bag 4 = 23 marbles
Bag 5 = 38 marbles
Bag 6 = 14 marbles
To work this out, since there are 150 marbles, and the scales are level, both sides of the scales must have 75 marbles. If Bag 1 + Bag 2 = 56 marbles then Bag 3 = (7556) 19 marbles. So Bag 1 = 35 marbles and Bag 2 = 21 marbles. The same is done with Bags 4,5, and 6. Bag 4 + Bag 5 = 61 marbles. So Bag 6 = (7561) 14 marbles. So Bag 4 = 23 marbles and Bag 5 = 38 marbles.
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Comments
redraptor50
Oct 27, 2005
 I don't believe it, I got it right! chance? Very clever and well done 
joez11
Oct 27, 2005
 Good one, although the part about the scales and the 150 total need not even be part of the teaser. If we know there are 2 more marbles in bags 1 and 2 than in bags 1 and 3, then 2 and 3 MUST be 21 and 19 to equal 40. Same for 4,5, and 6. 
Samrat
Oct 27, 2005
 verrrryyyy eazzzzzeeee!!! 
scihead
Oct 27, 2005
 Yes, the part about having 150 marbles and level scales is not necessary. I didn't even use that part at all. Fun, but way easy! 
Swordoffury1392
Oct 27, 2005
 easy but fun. 
Skoolkid
Oct 28, 2005
 I didn't understand the idea behind the scale, but I used a system of equations.
A=Bag 1
B=Bag 2
C=Bag 3
1. A + B = 56
2. A + C = 54
3. B + C = 40
Subtracting equation 3 from 1, you get
4. A  C = 16
Adding equation 4 and 2 you get
A + A = 70 = 2A
A = 35.
And that checks with the answer. 
realm2346
Oct 29, 2005
 I went about this completely wrong, but it was very intriguing. . I like the guy with the algebra system....I was thinking of something else, which is why I didn't get the teaser. I know it had me going, because my head hurts now. 
maximo3491
Oct 30, 2005
 very easy for me, cause im a math wizz . i did algebra using variables. worked pretty well nice job, keep em comin 
karma48
Oct 30, 2005
 Great Numerous ways to solve; but half the fun was just that 
magicumbreon
Dec 22, 2005
 Very easy. I used the algebra method as well! 
magicumbreon
Dec 22, 2005
 Very easy. I used the algebra method as well! 
magicumbreon
Dec 22, 2005
 Very easy. I used the algebra method as well! 
mmmcla01
Jun 28, 2006
 Very Simple, but fun!
I just used the fact that there were a total of 150 marbles on level scales to check my solution, not to solve the problem. It reminded me of the problem that you posted about the minieral water, sandwich, crisps, and fizzy juice. I used the same method in this as I used in that. Anyway, very fun and easy!! 
(user deleted)
Mar 09, 2008
 You can solve these systems of equations very readily by placing them into a matrix and transforming it into reducedrowechelonform. I typed the matrix into matlab as follows, one column for each bag#, and the last column for number of marbles:
x =
1 1 0 0 0 0 56
1 0 1 0 0 0 54
0 1 1 0 0 0 40
0 0 0 1 1 0 61
0 0 0 1 0 1 37
0 0 0 0 1 1 52
for example, the second row reads:
1 of bag1, 0 of bag2, 1 of bag3, 0 of bag 4, 0 of bag5, 0 of bag6, 40 marbles.
Placing into reduced row echelon form:
rref(x)=
1 0 0 0 0 0 35
0 1 0 0 0 0 21
0 0 1 0 0 0 19
0 0 0 1 0 0 23
0 0 0 0 1 0 38
0 0 0 0 0 1 14
the second row implies that 1 of bag2 has 21 marbles. the result is read from the last column.
Scilab is an free alternative to matlab, and can help you solve many of these brain teases with ease.
The fun part is delving into linear albegra and learning WHY rref works like this =) 
javaguru
Jan 04, 2009
 As several people have pointed out, the information about the scales and the 150 total isn't needed. The rest is very easily determined from the three independent systems of equations.
About 20 seconds in my heada little too easy. 
Gexxo
Dec 29, 2009
 Wow the way I did this was really a lot more complicated than it should have been, haha. Oh well I still got the answer 
opqpop
Sep 11, 2010
 I like the comments regarding solving it w/o the given info.
Even w/o the given info, I usually solve these
a+b = x
b+c = y
c+a = z
where x, y, z are constants by adding all 3 equations, finding what a+b+c equals, and subtract each equation from the result to figure out a, b, c.
But using logic based on the difference between 2 of the variables and using the 3rd equation can be faster, as I've noticed from trying it out!
Thanks for the good learning. 
spikethru4
Feb 01, 2013
 It's true, the information about the scales is not necessary, but it does get you to the answer quicker without solving simultaneous equations (or putting a matrix into reduced row echelon form). 
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