Ben, Jack and Emma were playing a game with a box of 40 counters - they were not using all of them.

They each had a small pile of counters in front of them.

All at the same time, Ben passed a third of his counters in Jack, Jack passed a quarter of hBen, Jack and Emma were playing a game with a box of 40 counters - they were not using all of them. They each had a small pile of counters in front of them. is counters in Emma, and Emma passed a fifth of her counters in Ben. They all passed on more than one counter. After this they all had the same number of counters. How many could each of them have started with

Guest Mar 12, 2017

#1**0 **

We have that

E + B + J < 40

And....when all the trading of counters has ceased..we find that

{E - E/5) = (B - B/3 + E/5 + J/4) = (J - J/4 + B/3)

(4/5)E = [ (2/3)B + E/5 + J/4 ] = [ (3/4)J + B/3 ]

Using the first and third parts of the equality, we have

(4/5)E = B/3 + (3/4)J multiply through by 60→ 48E = 20B + 45J (1)

Using the first two parts of this equality, we have

(4/5)E = [ (2/3)B + E/5 + J/4 ]

(3/5)E = (2/3)B + J/4 multiply through by 60 → 36E = 40B + 15J

Then multiply through by -3 → -108E = -120B - 45J (2)

Add (1) and (2)

-60E = -100B

B = (3/5)E

And using the first two parts of the equality with substitution for B we have that

(3/5)E = (2/3)(3/5)E + J/4

(3/5)E = (2/5)E + J/4

(1/5)E = J/4

J = (4/5)E

So

E + B + J < 40

E + (3/5)E + (4/5)E < 40

(12/5)E < 40

E < 16+2/3

So......since E is divisible by 5, the greatest that Emma could have started with is 15

Which means that Ben started with (3/5)(15) = 9

And Jack started with(4/5)(15) = 12

And

E + B + J =

15 + 9 + 12 =

36 < 40

So....they could have started with 36 counters

BTW.....there are other possibilites

E = 10 , B = 6 and J = 8 = 24 counters

E = 5, B = 3 and J = 4 = 12 counters

CPhill
Mar 12, 2017