I need a math wizard stat!!!

[Deleted]

Tickets were sold for a raffle $2 each or 3 for $5, 2100 tickets were sold and ~$4085 was raised

How many tickets were sold at each price? I think the equation is 1.67x + 2y = 4085, x+y = 2100

Can anyone help me?

[Deleted]

Virgil Showlion pretty please

[myrrh]

I think you are correct, and I get
y ($2 tickets) = ~1752 
x ($5 for 3) = ~348

[Deleted]

Is it 4805 or 4085?

[Deleted]

Jan 8, 2016 10:49:00 GMT -5 myrrh said:

I think you are correct, and I get
y ($2 tickets) = ~1752 
x ($5 for 3) = ~348

This = 5244$

[Deleted]

You could be right Rukh, I will have to keep figuring

[Regis]

If the $4,085 is correct, I get 1755 tickets at $2 each ($3,510.00) and 345 tickets (115 groups at $5 = $575.00).

[shanendoah]

Either laterbloomer's or @tbird's equations will work. The next step is to combine them.

I'm going to use later's. Since we know that x + y = 2100, we know that y = 2100-x. We use that in the first equation, so know we have:
1.67x + 2(2100-x) = 4085
This simplifies to .33x = 115
This tells us that 348 tickets were sold at x - the 3 for $5 amount, and 1752 were sold at $2 apiece.

We can double check this using rukh's equations
x = 2100-3y
2(2100-3y) + 5y = 4085
y = 115
since in this case, y is the number of 3 ticket packs sold, we need to multiply 115 x 3, which is exactly the same as the dividing it by .33 we did in the previous set of equations.

So we are confirmed 348 tickets were sold at the 3 for $5, and 1752 were sold at $2 each.

Now, these numbers are not exact. There's some decimals in there. But the take was only approximately $4085. I am certain that if we had the exact numbers, we'd get exact numbers.

[Deleted]

Not having exact numbers is making me a little shabby...

[ArchietheDragon]

You are not going to win. 

[shanendoah]

Jan 8, 2016 11:24:49 GMT -5 @tbird said:

Jan 8, 2016 11:18:35 GMT -5 shanendoah said:

Either laterbloomer 's or @tbird 's equations will work. The next step is to combine them.

I'm going to use later's. Since we know that x + y = 2100, we know that y = 2100-x. We use that in the first equation, so know we have:
1.67x + 2(2100-x) = 4085
This simplifies to .33x = 115
This tells us that 348 tickets were sold at x - the 3 for $5 amount, and 1752 were sold at $2 apiece.

We can double check this using rukh's equations
x = 2100-3y
2(2100-3y) + 5y = 4085
y = 115
since in this case, y is the number of 3 ticket packs sold, we need to multiply 115 x 3, which is exactly the same as the dividing it by .33 we did in the previous set of equations.

So we are confirmed 348 tickets were sold at the 3 for $5, and 1752 were sold at $2 each.

Now, these numbers are not exact. There's some decimals in there. But the take was only approximately $4085. I am certain that if we had the exact numbers, we'd get exact numbers.

One of the reasons that later got into trouble was her equations using non-integers and rounding. There just isn't any need to set these up with built in errors.

And I got different answers than you. When I plugged my x and y solutions into both equations, they came out as the perfectly round number that went in. 

You are right. If I had actually multiplied 115 x 3, I would have gotten 345, instead of 348, and that's because 1/3 is not the same as .33 - meaning setting up the equation with fractions (dreaded fractions) instead of decimals would have been much more accurate.
So, the answer is 345 tickets sold at 3 for $5 and and 1755 sold at $2 each.

[myrrh]

Thank you 
shanendoah!

[Deleted]

I got   5T + 2t = 4085     3T + t = 2100  so t = 2100 - 3T

so 

5T + 2 (2100-3T)  = 4085 

5 T + 4200 - 6T = 4085

T = 115  x 3 = 345 tickets     115 x 5 = $575
t                  = 1755 tickets          X 2= $3510 

$575 +  $3510 = $4085 if that is the number

[shanendoah]

Jan 8, 2016 11:34:24 GMT -5 @tbird said:

Jan 8, 2016 11:32:01 GMT -5 myrrh said:

Thank you 
shanendoah!

Perhaps for explaining why you were right and I was wrong?

Or perhaps for saying fractions are better than decimals? 

[Regis]

Jan 8, 2016 11:41:38 GMT -5 shanendoah said:

Jan 8, 2016 11:34:24 GMT -5 @tbird said:

Perhaps for explaining why you were right and I was wrong?

Or perhaps for saying fractions are better than decimals? 

Fractions are better than decimals when you round the decimals.

[Virgil Showlion]

laterbloomer: Both your construction and @tbird's are correct. Hers is the way I'd have done it, but be careful to note that her y variable represents the number of 3-ticket groups sold, hence 3y is the number of tickets sold at the discounted price in her setup.

Just as a helpful hint in solving such problems in future, there's no rule that states we have to use generic variable names like x and y. Using more descriptive names often helps clarify things in your own mind and for others. For example:

Let n_2$/1 be the number of tickets sold at $2 apiece, and n_$5/3 be the number of groups of 3 tickets sold at $5 per group. Then:

2n_2$/1 + 5n_$5/3 = 4,085
1n_2$/1 + 3n_$5/3 = 2,100

and the total number of tickets sold in the groups is given by 3n_$5/3.

[Deleted]

You guys are awesome. Except Virgil Showlion, you were supposed to dumb it down for me and give me the answer!

[Virgil Showlion]

Jan 8, 2016 13:09:17 GMT -5 laterbloomer said:

You guys are awesome. Except Virgil Showlion , you were supposed to dumb it down for me and give me the answer!

First of all: check whether the total sales are indeed supposed to be $4,085 (that's what the problem statement says, but your first equation has $4,805).

Multiply the second equation by -2 and add it to the first equation. This gives you:

-n_$5/3 = -115

hence n_$5/3 = 115 groups of 3 tickets (comprising 3n_$5/3 = 345 total tickets).

Substitute this into the second equation and solve for n_$2/1, which turns out to be 1,755.

It never hurts to check, hence we verify that indeed:

2·1755 + 5·115 = 4,085
1·1755 + 3·115 = 2,100

shanendoah's numbers are correct.

[Deleted]

Sorry guys, it was $4,085.

[Virgil Showlion]

www.wolframalpha.com/input/?i=2x%2B%285%2F3%29y%3D4085%2Cx%2By%3D2100

[shanendoah]

Virgil Showlion - Really? Referring them to Wolfram for this?

[Virgil Showlion]

Jan 8, 2016 16:20:25 GMT -5 shanendoah said:

Virgil Showlion - Really? Referring them to Wolfram for this?

Just to confirm the correctness of the solutions given, sure.

The link is to Wolfram Alpha, not wolfram.com. WA is a "computational knowledge engine". One of its principal uses is that you can give it a system of equations and ask it to find the solutions. It doesn't hurt as a verification tool.