Thursday, February 3, 2011

Cookies and Milk

This puzzle comes with a prize! First person to submit a correct solution will get a Mastermind game from the Pressman Toys. To read about the history of the Mastermind game, click here.

The new school principal thought she'd get the children to drink the milk at recess more cheerfully by offering cookies with it. She ordered a box of cookies delivered to school daily. A lot of children had cold and were out on different days. One day the principal noted that if there were five fewer children the next day, each kid would get two more cookies. However, the absentee list was low the next day, and she had four more children instead of five fewer. As a result each child received one cookie less than the day before. How many cookies did the children get the second day?

This puzzle is from the "The Mensa Genius Quiz-a-day book" and every ordinary genius like you can solve it.

Image above is by michelleTNS distributed under CCL.

Answers accepted all day long on Friday, on our Family Puzzle Marathon. They will be hidden until Saturday morning (EST) and everyone who contributed something reasonable will get a puzzle point. Please, explain your answer.


Kim said...

5 cookies on Day 2

On Day 1 there were 20 kids eating 6 cookies each
On Day 2 there were 24 kids eating 5 cookies each

Here's how we get it:

Let x=number of cookies/kid on Day 1
let k=number of kids on Day 1
let c=total number of cookies

On day 1: xk=c
and we know (x+2)(k-5)=c
and (x-1)(k+4)c

so xk-5x+2-10=c

and since xk=c


multiply the bottom equation by 2 and add them


so 3x=18

On day 1 we have x

Bean said...

Is this where I post the answer? On the first day, there were 20 kids with 6 cookies each. On the second day, there were 24 kids, so they got 5 cookies each.

Anonymous said...

answer: 5 cookies the 2nd day. Do you want me to show the algebra, too?


Lynnet said...

The hard here is representing the situation algebraically and getting it to make sense.

C=# of cookies per box
K=# of kids on day 1
K-5=# of kids on hypothetical day 2
K+4=# of kids on actual day 2

C/K=#of cookies each kid gets on day 1
C/(K-5)=# of cookies each kid gets on hypothetical day 2
C/(K+4)=# of cookies each kid gets on actual day 2

(C/K)+2=# of cookies each kid got on hypothetical day 2
(C/K)-1=# of cookies each kid gets on actual day 2


At this point, it's just a matter of algebra, but the algebra is kind of messy. My mom helped me solve this problem by substitution because I'm only 11 years old. I can do simple algebra but these are really messy.

The # of cookies each kid gets on day 2 is 5 cookies.

It took over 2 pages densely covered to solve this problem!
My mom used to play cows and bulls with her class but she always heard it called pico-fermi-bagel. A pico is a cow and fermi is a bull.

Lynnet's mom said...

By the way, I wanted to comment that Lynnet, mom, dad, and 5yo brother all talked about the menu challenge for a great while last week -- we just got home very late so she didn't post. Still a terrific challenge! I like it when you mix in problems where there isn't a clear "right" answer.

Dennis (of Dennis and Katrina) said...

On the second day, each child got 5 cookies.

On day 1, there were 20 students and each got 6 cookies. (total 120)

The principal noticed that if she had 15 students, each would get 8 cookies, or 2 more.

As it turned out, less were sick, so there were 24 students, and each got 5 cookies.


Anonymous said...

Barry G via email:
Hi Maria,
Each kid would receive 5 cookies the second day... 6 cookies the prev day.
I based it on 120 cookies and 20 kids.
120/20 = 6 cookies ea
120/(20+4) = 5 cookies ea
If there were 5 fewer kids, each kid would receive 8 cookies (120/15), where 8 is 2 more than today, and 3 less than the 2nd day.

Anonymous said...

The answer is 5. If 'n' is the normal number of children and 'c' is the normal number of cookies each receives, then the total number of cookies is nc. If there had been 5 fewer children each would have received 2 more cookies, so (n-5)(c+2)=nc, the total # of cookies nc being the same. There were 4 additional children and each got one less cookie, so (n+4)(c-1)=nc. With those two formulas you can reduce them and determine that n=20 and c=6. The normal # of cookies is 6, so when there were 4 more children each got 6-1=5.


SteveGoodman18 said...

Each child got 5 cookies the second day.

If we let x = the number of cookies and y = the number of kids on the given day, then the number of cookies per child is given by x/y.

Since if there were 5 fewer children, each child would get two more cookies, we have x/y = x/(y-5) - 2.

Since when there were 4 more children, each child got one fewer cookie, we have x/y = x/(y+4) + 1.

The solution to this system of equations is x = 120 cookies, and y = 20 students present on the given day.

There were y + 4 = 24 students present the second day, giving each child 120/24 = 5 cookies.

The principal is lucky that the number of students present was always a factor of 120.

anne-marie said...

If x is the number of kids in the class
Y being the number of kids absent on the given day
Z being the number of cookies per kids.

I have three equations and three unknown
X_y= z+(y/x) z
X -(y-5) = z+ (y-5/(x-(y-5)) z which is = z+(y/x)z+2
x-y+4= z+( ( y+4)/( x-y+4))z which is z+( y/x)z-1

Donna said...

I used Algebra to solve this, and don't really feel like typing all the equations in, but here goes. I let x = the # of cookies each kid got on day 1, while y = the # of kids on day 1. So xy would be the total number of cookies on any given day, since that does not change. Then I came up with two equations using x and y: One for the scenario where there were 5 fewer students, with each student subsequently getting 2 more cookies. That equation looked like this: (y+2)(x-5) = xy. The second equation was based on the actual event of 4 more students and each student getting 1 less cookie: (y-1)(x+4) = xy. I then multiplied everything out, and cancelled the xy terms from both sides, and got the following equations: 2x-5y=10, and 4y-x=4. I then solved for y by multiplying the second equation by 2 and adding to the first. I got y = 6.....meaning that on day 1 each student got 6 cookies. Then I substituted that back in for y in the first equation and solved for x: x=20, meaning there were 20 kids the first day. So on day 2, there would've been 24 children (4 more) and each would've gotten 5 cookies (1 less). BTW,the total # of cookies on each day was 120. [6x20, and 5x24]

kj said...

For simplicity, we assume that all cookies are consumed each day (i.e. there are no leftover cookies on hand to give out on future days)

Let n = # of children on day one
c = # of cookies per child on day one

then (n - 5)*(c + 2) = n*c
and (n + 4)*(c - 1) = n*c

multiplying out the first equation we have
n*c + 2*n - 5*c - 10 = n*c
2*n - 5*c = 10

and the second equation gives us
n*c - n + 4*c - 4 = n*c
n = 4*c - 4

plugging this into the first equation gives
8*c - 8 - 5*c = 10
3*c = 18
c = 6

n = 4*6 - 4 = 20

so there are 20 children at school the first day, and they each receive 6 cookies.

(The principal noted that if only 15 children come the second day, they will each receive 8 cookies -- 20*6 = 120 = 15*8. Actually, 24 children attend on the second day, and they each receive 5 cookies -- 20*6 = 120 = 24*5.)

Maria said...

Wow. You all cracked it. Cookies and milk for everyone!

Kim was the first one and will receive the Mastermind game. Kim - please email me your mail address and Pressman toys will ship the game directly to you.

Kim, Bean, Allen, Lynnet, Dennis, Barry G., SteveGoodman18, anne-marie, Donna, kj - all get a puzzle point. You all are geniuses.

Lynnet's family - you are doing an amazing thing by making puzzle brainstorming a family affair. And I agree with Lynnet's mom - puzzles with no clear answer that tickle our imagination and creativity are so much fun.

Next week - a very special Valentines puzzle.

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