Can you solve this problem without using algebra?

You have a typo. 24 tickets at $17.50 will be more than $59.00!

Thanks for pointing that out. It should be 24 tickets at $17.50 is less than $459.

My solution is to flirt with the ticket seller and ask them how their day went.

It’s not algebra, but it is science.

^^ Yes, pheromones and such, as well as biology, birds and bees

Sorry @LostInParadise…couldn’t resist.

I see I am not getting any takers. Here is the solution. 24 tickets at $17.50 each costs 24x$17.50 = $420, which is $39 less than $459. Substituting a $20.50 ticket for a $17.50 ticket costs an additional $20.50 – $17.50 = $3 per ticket. $39 divided by $3 per ticket is 13 higher priced tickets, and 24 tickets – 13 tickets gives 24 – 13 = 11 lower priced tickets.

What is interesting is that if you set the problem up algebraically, x+y=24 and 17.5x+20.5y= 259, and initially solve for y, the arithmetic calculations are identical to the above non-algebraic method.

Well, in a sense. When I read:

“A place sells two different types of ticket, one for $17.50 and one for $20.50. It sells 24 tickets and brings in $459.00. How many of each type of ticket does it sell?”
– I see this as a non-problem. They should be able to count the tickets sold. Or if they didn’t count, but somehow care to have a count after the fact, and do somehow have a total count of the money, they could solve it with math, algebra, or logic. But “without using algebra”? The right answer to that request sounds like, “don’t bother me with that annoying problem and vague restriction.”

But well, let’s see: It’s a trivial math problem using logic and not algebra per se. That is, you know there were 24 tickets sold, If they’d sold 24 of the expensive tickets, that’d be 410 + 82 = 492.

So they sold 492 – 459 = 33 less than that. Each cheaper ticket costs 3 dollars less, so they must have sold 11 cheaper tickets, and 13 more expensive tickets.

Though:

1) That’s essentially what algebra is, just with logic and natural language.

2) I feel dirty having given in and solved the stupid story problem for the lazy imaginary bullies with such strange accounting questions they insist others answer for them.

The arithmetic you did is identical with what you would have done if you started out solving for x in the two equations I showed for an algebraic solution. I am sorry if you did not find the word problem interesting enough, but isn’t it interesting that the non-algebraic, what you call logical, solution provides a meaning to each of the mechanical operations of the algebraic solution?

I didn’t mean you personally did anything wrong in asking this question.

I just mean to communicate my life-long reaction to 90% of math story problems, especially the ones that conjure surreal worlds where incompetent strangers must know the answer to some question, for no specified reason, and though there’s zero reason why they would know the data presented, but not the information they want. And no reason why they shouldn’t figure it out for themselves. So the pretext that this is a great example of how you might practically want to use math in the real world, is really terribly represented in such problems.

Which is a shame, because math really is the language of facts about the universe, and can be extremely relevant and helpful. But most story problems tend to present some plane of hell where incompetent strangers have unexplained accounting needs that they require grade school students to solve for them. And other surreal contrived nonsense that does the opposite of convince me I ought to do math about it.

And yes, it is interesting, that reasoning about a problem using natural language and logic, ends up having a natural algebraic-language equivalent, and is in fact a direct algebraic solution to the problem. THAT is the kind of observation that CAN potentially make algebra “click” as being useful . . . especially if there were a story problem that was more relatable.

Examples of more relatable story problems:

Chuck is crazy excited to buy a certain indie video game, which costs $33.99. His allowance is $12 every Monday when he gets back from school. He can mow lawns every Saturday or Sunday, for $5. It’s Wednesday now. What’s the soonest he could afford to buy the game, with the least amount of lawn-mowing?

Bart also wants the game, is only willing to mow a lawn once, and gets $15 allowance every Wednesday. When’s the earliest Bart will be able to afford the game, and what’s the latest day he could put off mowing a lawn?

If Chuck and Bart pool their money to buy the game, what’s the earliest they could get it?

Crazy Dan is speeding at 100 MPH when he zooms past stopped Officer Cletus. Officer Cletus takes 10 seconds to get his car moving, and then races in pursuit. It takes Cletus 10 more seconds to accelerate to 60 MPH, covering 1/12 of a mile. In the next 20 seconds, he accelerates to 140 MPH, covering another ⅔ of a mile. The county line is 5 miles from where Dan sped past Cletus. What is the slowest Dan can drive, to cross the county line before Cletus catches up to him at 140 MPH?

I could do it through trial and error. That’s part of the problem with teachers wanting you to show your work, if you’re good and also fast at thinking about a math problem logically without setting up the algebra, you only get half the points. I say fast, because trial and error usually takes more time.

Plus, as problems get harder the student won’t be able to set up or do the algebra, which is the biggest problem.