Please help Ben, Jack and Emma were playing a game with 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 to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben.

They all passed more than one counter. after this they all had the same number of counters. How many could each of them have started with

( if it's not too much plz try to solve this algebraically ).

Emma started with 10

Jack with 8

and Ben with 12

(30 total)

Step-by-step explanation:

Emma started with 10

Jack with 8

and Ben with 12

(30 total)

after the swap, they each end up with 10:

Ben kept 8 and got 2 (a fifth of 10)

Jack kept 6 and got 4 (a third of 12)

Emma kept 8 and got 2 (a quarter of 8)

The constraints that the total number of counters is < 40, and that they each pass more than one counter makes this a unique solution.  Without these constraints, any combination in the ratio of 4:5:6 (J:E:B) would work...

Now for the work:

If Ben passed 1/3, he kept 2/3

If Jack passed 1/4, he kept 3/4

If Emma passed 1/5, she kept 4/5

so we know:

2/3B + 1/5E = 3/4J + 1/3B = 4/5E + 1/4J

you can rearrange these in pairs to find the ratio between B,E, and J, the original number of counters each had.

I actually multiplied everything by 60 first to get rid of the fraction, even though they came back quickly:

40B + 12E = 20B + 45J = 48E + 15J

From the first two parts:

20B = 45J - 12E (**)

and from the last two:

20B = 48E - 30J (***)

giving: 45J - 12E = 48E - 30J

or 75J = 60E, or J = 4/5E

now plugging back into  **  

20B = 45(4/5 E) - 12E

20B = 36E - 12E = 24E

B = 6/5 E

so now J:E:B are in the ratio of (4/5E):E:(6/5)E, or

4:5:6

we can say J = 4x, E = 5x, and B = 6x, and find values of x that satisfy the given conditions

if x = 1, J = 4, E = 5, and B = 6, but Emma and Jack have to pass more than one, so this doesn't work

if x = 2, J = 8, E = 10, and B = 12 (the solution that works)

if x = 3, J = 12, E = 15, and B = 18, giving a total of 45 counters (too many)