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Game: Math skill

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posted on Mar, 6 2014 @ 07:40 PM
After intense debate on the 2 vs 288 thread I think I need to "reactivate" my math skill a bit. I think why not you too ?. Some of you may already knew the answer, so please refrain from answering or try the algebra version, please let few others sharpen their math skill

So heres a simple question.

In verbose math:
I got \$100, bought 100 items and all money spent, the items are:
X priced at \$6, Y priced at \$3 and Z priced at 0.10cent,
how many X, Y and Z will I get ?

In algebra:
6x + 3y + 0.10z = 100 = x + y + z
What is x, y and z ?
Please show the working if possible, so we all can "relearn".

For math version, you know you get the answer correct when you get it, so no need to tell the answer I guess.
However I would like to see the answer solved in algebra, if possible.
And no googling , that would defeat the purpose
Thanks
Game:Chart identification
Game:Neat math riddle

posted on Mar, 6 2014 @ 07:58 PM
Hmmm, I have a weird way of doing this (to me its the easy way).

First I divide 100 by 6 , saw that I needed to do 15x6 to get 90.
That leaves room for three 3's, which is 9.
And so 10 dimes are left for the last whole 1.

posted on Mar, 6 2014 @ 07:59 PM
I guess I could go with 16 6's and then take 1 3 and 10 dime pieces too.

posted on Mar, 6 2014 @ 08:01 PM
Or maybe I took 30 of the 3's , a 6, and 40 dimes?

I don't know all of them add up to 100.

Or what about one 6, one 3, and 910 dimes?

posted on Mar, 6 2014 @ 08:05 PM
Tell me the answer I'm obviously not getting it.
Unless the joke is that there are tons of variable answers?

I just woke up recently and wasn't prepared for math riddles.
Pretty mind blowing actually how easily mistakes happen.

posted on Mar, 6 2014 @ 08:08 PM

Oh I missed the part where you said exactly how many items you bought, DOH!
That's embarrassing.

I need some coffee omg...

edit on 6-3-2014 by muzzleflash because: (no reason given)

posted on Mar, 6 2014 @ 08:08 PM

I'm drinking coffee so I have no excuse for botching it.
edit on 6-3-2014 by DenyObfuscation because: idiocy

posted on Mar, 6 2014 @ 08:10 PM
The answer is specific. Pure numbers.
The sum of the answers must be 100.

I'm really interested in the algebra solution.
6x + 3y + 0.10z = 100
x + y + z = 100

x = ?, y = ?, z = ?
edit on 6-3-2014 by NullVoid because: (no reason given)

posted on Mar, 6 2014 @ 08:10 PM

No he only bought exactly 100 items he said.

posted on Mar, 6 2014 @ 08:48 PM
I'm trying the algebra version using papamath and algebrahelp.com,
looks like dead end - the numbers either in notation or way off from the answer.

If you use the math version, you CAN find the answer, but when using the algebra, I'm stuck. Is it because there are 3 variables ? I gonna try again with less variable.

Ok, with only 1 variable, I get the answer, but that pretty easy algebra isnt it ?
edit on 6-3-2014 by NullVoid because: (no reason given)

posted on Mar, 6 2014 @ 08:51 PM
This can't be solved algebraically, 3 unknowns and 2 equations. You need one more independent equation.

posted on Mar, 6 2014 @ 09:27 PM

Yeah I even called my bro to check he agrees with what you said.

I've been playing with it and I'm thinking it's irrational.
Or I suck, haha!
edit on 6-3-2014 by muzzleflash because: (no reason given)

posted on Mar, 6 2014 @ 09:28 PM

saneguy
This can't be solved algebraically, 3 unknowns and 2 equations. You need one more independent equation.

How ? Is this ok ?
pretty much the same, but its all I can make up.
6x + 3y + 0.1z + (x + y + z) = 200

posted on Mar, 6 2014 @ 09:36 PM

29 * \$3 = 87\$
1 * \$6 = 6 \$
70 *0.1 = 7 \$

both totals - 100

it took me about 1 min to figure it out - arithmetic / visualisation - not algebra

the solution [ for me ]

I have to spend 9 dollars - or less on the 10 cent item

then I spend the rest on the 3 dollar piece

that immeditatly offered a " partial solution of 70 10 centers and 31 3 dollar pieces = 100 bucks / 101 pieces

replace 1 3 dollar items with one at 6 bucks

voila

edit on 6-3-2014 by ignorant_ape because: (no reason given)

posted on Mar, 6 2014 @ 09:42 PM

Nice, lol, I went 6x 15 all the way down to about x6 and gave up there.
If only I had worked it to 6x1!

posted on Mar, 6 2014 @ 09:50 PM
Maybe there is, or could be, an equation that quantifies one of the qualifiers?

Like making one of the variables equal one?

I'm not good at math jargon, sorry.

Edit: I solved it the same way as ignorant ape -- I plugged away until I got the answer, but I think there must be some way to make a formula for solving these types of questions.

Maybe it is associative algebra?

Who knows a bunch of math formulas?
edit on 3/6/2014 by Bleeeeep because: (no reason given)

posted on Mar, 6 2014 @ 09:53 PM

Very nice indeed, took me 15 minutes.
Some people never able to solve it, probably they are using algebra/formula/whatever instead of basic arithmetic ?

6x + 3y + 0.10z = 100 = x + y + z

Yes, I think its associative algebra (whatever that is), see R modules in that Wikipedia.
Assuming we know z. Is it possible to get x and y ? I'm really no idea about this.
x + y + 70 = 6x + 3y + 70 = 100

-------------------------------------------------
Since the question solved before 2nd page lets not waste it. Another question
No google, it defeat the purpose.

Question:
There are 8 ball, equal size and everything except 1, its weight screwed, you are given a balancer/scale and only 3 times balancing act, determine which ball have wrong weight and is it heavier or lighter than others.

IMHO this one pretty easy compared to the 1 29 70 question above.
edit on 6-3-2014 by NullVoid because: (no reason given)

posted on Mar, 6 2014 @ 10:15 PM

I have to know for sure if the odd ball out is heavier or lighter. If you say which one then we could for sure answer with 3 chances at the scale. If you do not say if the odd ball is heavier or lighter then it will take more than 3 times, I think...

1. Start by dividing the balls 4 on one scale and 4 on the other. See which is heavier.

2. take heavier set and divide into 2 sets of 2.. see which is heavier.

3. take heavier set and divide into 2 sets of 1... you have answered.

But first you need to know if it is heavier or lighter, I think.

Edit: If step 2 shows the heavier set is actually equal when split into 2 sets of 2 then you need another turn... 3 will not be enough. At this point you split the lighter set into two sets of 2 (which is your third weighing) then you take the lighter of those two sets and find the lightest one with your forth weighing.

50/50 chance of finding it with 3 weighings and 100% with 4? Is this correct?
edit on 3/6/2014 by Bleeeeep because: (no reason given)

posted on Mar, 6 2014 @ 11:10 PM

Good try, but the answer is still 3 trial and you can tell its heavier/lighter.
If I told you its lighter/heavier then it should be as you said, but too bad, its not that way

This problem have many variations, sometime 12 balls 3 balancing etc etc, but I encountered just this one.
Its usage maybe for the computer bit parity checking, maybe, just my guess.
8 bit = 1byte. Lighter = 0, heavier = 1.

posted on Mar, 7 2014 @ 04:01 AM
(if not the same remove one ball each side for the answer) 2nd try

If the same add a ball to each side...if the same...2nd try
(if not the same you have the answer)

other scenario...

(remove a ball from each side for the answer in two tries
using simple logic)

If the same...
using simple logic

edit on 7-3-2014 by rival because: (no reason given)

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