Friday, October 28, 2011

Halloween dilemma: candy or tchotchkes

Halloween is approaching in a few days and you still have not bought any giveaways. You usually try to be thoughtful and buy candy as well as tchotchkes for those with allergies or strict parents.  Last year expecting 300 kids you bought 300 pieces of candy and 300 tcotchkies. You had only 45 candies and 75 tchotchkes left. Indeed 300 trick-o-treaters showed up according to someones count.

What percentage of the trick-o-treaters took both: candy and tchotchkes?
If you expect 400 kids this year, how many of each, candy and tchotchkes, you should buy to not have any left-overs assuming the same percentages of candy-toy preferences as last year?

Image by eszter, distributed under CCL.

Answer ideas accepted any time until midnight on Saturday October 29th (EST), on our Family Puzzle Marathon. They will be hidden till then and everyone who submitted something reasonable will get a puzzle point.


kj said...

First, I assume every no one walked away empty handed, so each trick-or-treater took either a piece of candy, a tchotchke, or both.

So 45 trick-or-treaters took only a tchotchke, and 75 trick-or-treaters took only a piece of candy.

That means that 300 - 75 -45 = 180 trick-or-treaters took both a piece of candy and a tchotchke.

300 - 75 = 225 kids took a tchotchke.
300 - 45 = 255 kids took a piece of candy.

So 180/300 = 3/5 => 60% of the kids took both.

You will need (225/300)*400 = 300 tchotchkes
and (255/300)*400 = 340 pieces of candy

and (300 + 340) - 400 = 240 = 0.6*400 trick-or-treaters will take both a piece of candy and a tchotchke.

Wang said...

300 - 45 = 255
300 - 75 = 225

255 + 225 - a = 300
480 - a = 300
a = 180

If 180 kids took both candies then 255 - 180 = 75 kids took only the candies and 225 - 180 = 45 kids took only the tchotchkes.

180 + 75 + 45 = 300 kids in total.

So 180 / 300 = 60% of the kids took both candies.

If 400 kids come and 60% of the kids take both candies then

240 kids will take both candies. 160 kids will take one candy each.

To be safe and let the kids take any candy they want then it is best to get at least 240 candies of each kind + 160 of each in case the 160 kids all take only the candy or only the tchotchkes.

Therefore 400 candies should be bought of each just to be safe.

Annie said...

This year 60% took both candy and tchotchkes:
x + y + z = 300 where x= # candy takers, y = # tchotchkes takers and z
= #takers of both.
x+z =255 (candy takers)
y + z= 225 (tchotchkes takers)

180 took both (60%), 75 took candy only (25%) and 45 took tchotchkes only (15%)

With 400 trick-or-treaters next year you should buy 300 tchotchkes and 340 candies. (240 will take both, 60 just the tchotchkes, and 100 just the candy. If I didn't get the problem correct at least I definitely know how to spell "tchotchkes!! My mother-in-law will be thrilled!)

Ilya said...

Assuming that each kid can take no more than one of each kind, there are 180 that took both, 75 that took only candies, 45 that took only toys. That is 25%, 60% and 15% respectively. Applying the same percentages to 400 kids, we have 240 kids taking both, 100 kids taking only candies and 60 kids taking only toys. Which means we should only prepare 340 candies and 300 toys. Now, since we still have leftovers from last year :-), perhaps we only have to buy 295 candies and 225 toys. The last bit is a throwback to this series of jokes:

Jerome said...

You have to assume that every one of the children must take at least one item. Otherwise, the problem is ambiguous and does not have a unique solution.

The total number of items taken was
Candy 300 - 45 =255
Toys 300 - 75 = 225

The total number of items taken was 255 + 225 = 480

Therefore 480 - 300 = 180 items that were taken in excess of 300 which means that 180 kids took both.

The % = 180/300 * 100 = 60% took more than 1 item.

Now to the second part of the question.

255 children took the candy. That's 255/300 = 85%

So if there are 400 kids wouldn't you have to get 85%*400 = 340 pieces of candy?

Similarly 225 kids took the toy. That means that 225/300 * 100 = 75% of every caller took a toy. So for 400 kids you would have to order 300 toys, I think.

Jerome said...

I have to clarify my restrictions. Each child must take one item but no more than two and each item (I think) must be different. I haven't worked out what would happen if each child took two candies or two toys but I think it goes into ambiguity.

Maria said...

You all are absolutely right: 60% of the kids took two items - a candy and a toy. 85% of the kids took a candy, 75% - a toy. Expecting same percentages this year, we should run and buy 340 pieces of candy and 300 tchotchkes.

However, it is freezing cold and snowing in the North East this year. I doubt that we will get more than 100 trick-o-treaters.

A puzzle point for kj, Wang, Annie, Ilya, Jerome.

One of the easiest ways to solve this puzzle was via an illustration. kj emailed me hers: Halloween Candy solution by kj

TracyZ said...

Algebra is so much fun and seeing this puzzle made my heart sing and my face smile, and my mind get to work ... though doing algebra requires less creativity than many of your other puzzles, some of which completely flummox me :)

I tried to get my answer in by the deadline but as I was responding last night, my electricity went down from the storm *sigh* and didn't come back. Estimates are that power will be restored in 24-48 hours. I hope that is the case. Maybe some future problem on your blog could consider power outages and the probabilities of power being restored within certain timeframes. Over 3 million people lost power in this storm -- at least I know my neighbors and I are not alone --- and some may not get power back from the storm for a week or longer.
Love your blog and weekly puzzle.

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