Can you solve problems of this type without using algebra?

If passing means 51 votes, then the total number of votes on each side would be 62 to 38. Take the total number of votes difference (24), split it in half add one half to the winning side, subtract the other half from the losing side.

That is exactly what I had in mind. I wonder how many others know how to do this.

I wonder if I just put an algebraic equation into words. I’ll have to look into this.

All computing is algebra! Boolean logic is algebra. I have a math question for you…
If I have twelve beers and thirteen friends then…

Which twelve friends have to go home?

Well, I can, sort of, but probably any technique that can be done “without algebra”, probably also has a formal algebraic representation, and algebra is just a language for describing mathematical reasoning, and math is “the language of numbers”, so in that sense, it’s hard to avoid algebra entirely when thinking of math. Just saying you have one brother is the same as “brothers = 1”, which is algebra.

Though I guess in another sense, some people may think of “using algebra” as “using algebra without thinking (much?) about the concepts the algebra represents”. Which is perhaps, ironically, harder to do, the more you understand what algebra represents.

Let me elaborate on the answer that @seawulf575 gave. Consider the voting problem. Start by dividing the senators into two groups of 50. The total number is 100 and difference between the groups is zero.

Think in terms of moving people from one group to the other. That way the total remains the same as we increase the difference. What happens if we move one person? One group gets larger by 1, from 50 to 51. The other group gets smaller by 1, from 50 to 49. The difference is 51 – 49 = 2, and in general the difference is twice as great as the number moved from one group to the other. That means we need to move half of 24= 12, to get a difference of 24, giving 50+12 = 62 in one group and 50 -12 = 38 in the other.

If I were teaching an algebra class, I would give this problem before teaching the algebraic method of solving it. I am sure you all know the algebraic approach. x+y=100 and x-y=24. Adding and subtracting these two equations will lead to the answer.

In fact, flow dynamics is the only engineering discipline that does require algebra.
a/c guys use it to size ducts. Plumbers use fluid mechanics to size pipes.
But really, who cares?

Algebra in simple terms means solve for x.

I took a different tack.
100–24=76
76/2=38
38+24=62

What was your reasoning behind those arithmetic operations?

There is a psychological test associated with this type of problem.
A plastic bat and ball together cost $1.10. If the bat costs a dollar more than the ball, what does the ball cost? If you are like most people, even those who are highly educated, your gut instinct for the solution will be incorrect. Link

@LostInParadise The ball costs $0.05.

Was that your initial reaction? For most people, ten cents is what instantly pops into their heads.

I never take my first answer blindly. I always check my math. Looking at the problem 10 cents seems to be the answer but you have to do the math first. As soon as you realize you will not meet the dollar difference the answer becomes clear.