OT: Stuck on a very simple algebra problem

[Buccaneer]

I'm being braindead. If I bought $840 in marbles and 1/4 of them I paid $10 each, another 1/4 $40 each and the remaining 1/2 I paid $20 each - what is the formula to determine how marbles I bought?

[Passacaglia]

0.25 * x* 10 + 0.25 * x * 40 + 0.5 * x * 20 = 840

[Passacaglia]

Simplifying:

2.5 * x + 10 * x + 10 * x = 840

22.5 * x = 840

x = 840/22.5

x = 1680/45

[Passacaglia]

Do the math.

[finkenst]

How do you buy 37.3333333333333333333 marbles?

[Buccaneer]

Thanks Pass, I was dividing by the wrong number for some stupid reason.

[QuikSand]

Originally posted by finkenst
How do you buy 37.3333333333333333333 marbles?

Good question. Not many puzzle-makers would pick a seemingly pre-constructed total of $840 just to result in some repeating fractional number (37.3333...) of marbles.

Sure we don't have a typo somewhere in here?

[QuikSand]

Might be a puzzle in itself -- see if we can deconstruct what the puzzle-maker was thinking, or where is the typo (if there is one).

Theory: 840 is a convenient multiple of 35. An errant quick glance at the two "halves" costing $10+40 (yes, I know this is flawed thinking) and $20 would yield an "average" cost of $35... which would yield a nice, round figure of 24 total marbles. Might that have been the originally-intended solution? (Makes more sense to me than 37 1/3)

[finkenst]

Originally posted by QuikSand
Might be a puzzle in itself -- see if we can deconstruct what the puzzle-maker was thinking, or where is the typo (if there is one).

Theory: 840 is a convenient multiple of 35. An errant quick glance at the two "halves" costing $10+40 (yes, I know this is flawed thinking) and $20 would yield an "average" cost of $35... which would yield a nice, round figure of 24 total marbles. Might that have been the originally-intended solution? (Makes more sense to me than 37 1/3)

Any whole number makes more sense than 37 1/3 marbles.

The two halves costing 10 and 40 gives an average of $25/marble. Which means the average price per marble should be $22.50. And of course 840/22.5 = 37 1/3 .

I had to get some paper and a pencil out to work this problem. I came up with
10x + 40x + 20(2x) = 840
90x = 840
x = 840/90
x = 9 1/3.

Multiply by 4 since x is 1/4 of the marbles and came out with 37 1/3.

best regards,

tim

[finkenst]

dola,

thinking about this a little more...
if it's 1/4 @ $10 and 1/4 @ $30 and 1/2 @ $20...

Using Pass's formula:
.25 * x* 10 + 0.25 * x * 30 + 0.5 * x * 20 = 840
2.5x + 7.5x +10x = 840
20x = 840
x=42 marbles.

.
.
.

which looks like a good round number of marbles, but 1/4 of 42 is 10.5 which clearly can't be right.

Algebra teacher should give a question that makes sense.

--tim

[Chief Rum]

What algebra teacher? Bucc is like 80 years old. Where did you dig this thing up, Stevo?

CR

[Phoenix]

this is 7th grade math

[QuikSand]

Originally posted by finkenst
The two halves costing 10 and 40 gives an average of $25/marble. Which means the average price per marble should be $22.50.

I'm well aware that the average of 10 and 40 is 25, and not 50. Perhaps you missed my statement that "yes, I know this is flawed thinking." I recognize that my logic was incorrect - I was just trying to see how someone could have set up that problem on purpose. My theory (that someone forgot to divide the 50 by two in soe way) was a semi-logical way to get to a nice, round answer with a total cost of $840, which was the point.

However, I'm guessing we'll never know, unless Bucc goes back and sees that perhaps he made a typo of some sort.

[Buccaneer]

No typo, just mixing metaphors.

[finkenst]

QS,

Sorry, I knew you knew the average of 10,40 is 25...

of course, the teacher could have just been throwing numbers up to see what the equation to solve it would be.

tim