## John, James, and Harry: a math brain teaser

John, James, and Harry have $4.80 which they wish to divide equally among them. To do this, John, who has the most, gives to James and Harry as much as they already have. Then, James divides by giving John and Harry as much as they have after John’s division. Harry then divides with John and James in the same way, and it is found that they have equal sums.

How much had each at first?

*As always, if you can figure this out, you may enter your answer in the ‘comment’ section below. Will reveal answers later on this week!*

mimi112| Profile June 24th, 2009 - 4:56 amJohn had $2.60, James had $1.40 and Harry had $0.80.

fuzzy| Profile June 24th, 2009 - 10:24 amJohn had $2.60, James had $1.40, and Harry had 80 cents.

After 1st division John was left with 40 cents, James had $2.80 and Harry had $1.60.

After 2nd division James and John both had 80 cents each, and Harry had $3.20. Finally, after the 3rd division they each had $1.60.

TheMadEgyptian| Profile June 24th, 2009 - 11:11 amWorking backwards:

John James Harry

$1.60 $1.60 $1.60 Harry splits

$ .80 $ .80 $3.20 James splits

$ .40 $2.80 $1.60 John splits

$2.60 $1.40 $ .80 Beginning

tad| Profile June 24th, 2009 - 1:06 pmJohn = 2.60

James = 1.40

Harry = 0.80

This one is easiest worked out in reverse.

hex| PUZZLE MASTER | Profile June 24th, 2009 - 1:22 pmAssuming that “gives to James and Harry as much as they already have” implies he gives James as much as James has, and Harry as much as Harry has (not the sum of James and Harry divided by 2),

John has $2.60

James has $1.40

Harry has $0.80

Mashplum| PUZZLE MASTER | Profile June 24th, 2009 - 6:24 pmJohn = $2.60

James = $1.40

Harry = $0.80

Easy if you work backwards.

michaelc| Profile June 24th, 2009 - 6:51 pmJohn had $2.60.

James had $1.40.

Harry had $0.80.

Then they all had a $1.60.

So which one of the guys is John?

Shawn| PUZZLE GRANDMASTER | Profile June 25th, 2009 - 7:13 amHey, I can log-in again!!!

John: $2.60

James: $1.40

Harry: $0.80

bizarette18| PUZZLE MASTER | Profile June 25th, 2009 - 3:00 pm$2.60, $1.40, $0.80

Hendy| Profile June 25th, 2009 - 10:09 pmJohn James Harry

rnd 1 $2.60 $1.40 $0.80

rnd 2 $0.40 $2.80 $1.60

rnd 3 $0.80 $0.80 $3.20

rnd 4 $1.60 $1.60 $1.60

rachel| Profile June 25th, 2009 - 11:32 pmjohn starts with $2.60, james with $1.40 and Harry with .80c

then john gives james $1.40 and harry .80c

so..

john: .40c james: $2.80 harry: $1.60

then james gives john .40c and harry $1.60

john: .80c james: .80c harry: $3.20

then harry gives john .80c and james .80c

john: $1.60 james: $1.60 harry: $1.60

so they are all left with $1.60 each..

APEX.JP| Profile June 27th, 2009 - 9:00 amJohn 260

James 140

Harry 80

RK| Founder | Profile June 28th, 2009 - 11:59 pmvery good, answers above are correct!

pravinkumar1942@gmail.com| Profile July 16th, 2009 - 9:59 amanswer is

John had $2.60,James had $1.40 and harry had $0.80 at the start.

makada| Profile August 1st, 2009 - 8:54 amI really enjoyed solving this. Reminded me of my days in school.

Below I post the way I solved it

Initial money with john = x

Initial money with james = y

Initial money with henry = z

x + y+z = 480

x>y

x>z

After John’s division, each has the following amount

x-y-z, 2y , 2z

After James’ division, each has the following amount

2(x-y-z),2y-2z-(x-y-z),4z

After Henry’s division, each has the following amount

4(x-y-z),4y-4z-2(x-y-z),4z-[2(x-y-z)+2y-2z-(x-y-z)]

As they now have equal amounts then

4(x-y-z) = 160

4x-4y-4z=160

x-y-z=40……..(1)

x=40+y+z………(1a)

4y-4z-2(x-y-z) = 160

4y-4z-2x+2y+2z=160

6y-2x-2z=160

3y-x-z=80……..(2)

Substituting for x from Eq 1a into Eq 2

3y-40-y-z-z=80

3y-40-y-2z=80

2y-2z=120

y-z=60

y=60+z …..(3)

Substituting y in Eq 1a

x=40+60+z+z

x=100+2z…….(4)

4z-[2(x-y-z)+2y-2z-(x-y-z)]=160

4z-[2x-2y-2z+2y-2z-x+y+z]=160

4z-2x+2y+2z-2y+2z+x-y-z=160

7z-x-y=160………(5)

7z-100-2z-60-z=160

4z-160=160

4z=320

z=320/4=80………….(ans)

y=60+80=140….(ans)

x=100+2*80=100+160=260……ans

RK| Founder | Profile August 1st, 2009 - 10:50 amgood job makada, it was a difficult problem

makada| Profile August 1st, 2009 - 12:00 pmMy math teacher in school always used to tell me that I would some how find the most longest and cumbersome way to solve any problem…but he always said that as long as the answer is right, he will not stop my ‘creative’ thinking!!!

;-)