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!
John had $2.60, James had $1.40 and Harry had $0.80.
John 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.
Working 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
John = 2.60
James = 1.40
Harry = 0.80
This one is easiest worked out in reverse.
Assuming 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
John = $2.60
James = $1.40
Harry = $0.80
Easy if you work backwards.
John 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?
Hey, I can log-in again!!!
John: $2.60
James: $1.40
Harry: $0.80
$2.60, $1.40, $0.80
John 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
john 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..
John 260
James 140
Harry 80
very good, answers above are correct!
answer is
John had $2.60,James had $1.40 and harry had $0.80 at the start.
I 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
good job makada, it was a difficult problem
My 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!!!
;-)