100 people board a bus. Passengers include Men, Women and Children (note that there is at least 1 man/woman/child). Each passenger is entitled to cakes. Each man gets 10 cakes, each woman gets 3 cakes and each child gets 1/2 a cake (these are not minimums but absolutes). Total available cakes equals 100.
Find the breakdown of men, women and children who boarded the bus.
(Working out an analytical solution is preferred)
Saturday, January 12, 2008
Teaser
Posted by Mad Max at 11:14 PM
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5 comments:
I just used 2 linear equations to solve it...it comes to
man = 5
woman= 1
kids = 94
@ Mindframes: Can you give your reasoning also please.
I just solved it in the good old algebraic way. I wrote down the following 2 equations (from the given data)
10x + 3y + 0.5z = 100
x + y + z = 100
Solving the above (multiply 2nd eqn by 0.5 and subtract from 1st eqn)
9.5x + 2.5y = 50 or
19x + 5y = 100
By looking at the above eqn, I could see that x=5 and y=1 is atleast 1 solution.
Substituting those values in eqn 2, you get the value for z=94.
5M+1W+94C
I initially was solving with
10A+3B+.5C = 100 & A+B+C = 100
But it was getting difficult, so did a smaller value assumption for A since that cannot be a huge number.
Tried 3 and then 5 and it worked for 5.
@ Mindframes and Brainwaves: Awesome job folks. The answer is correct. BTW, Mindframes, this is the only solution to the problem. My thinking was pretty much inline with yours. The key is there is no need to guess the value on X.
19X + 5Y = 100. This implies that since Y has to be an integer, X has to be a multiple of 5. Since X cannot be zero or greater than 10 (since this means Y and Z are zero which violates the constraint), X has to be 5. This reduces the problem to a simultaneous equation set up with 2 equations and 2 unknowns. The rest is easy. This is a unique solution given the constraints.
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