For decorating accents in her living room, Lisa bought three bags of marbles, colored red, blue, and green.
Each bag held the same number of marbles.
She emptied the red marbles into the first bowl, the blue marbles into the second bowl, and the green marbles into the third bowl.
To mix them up, she scooped up some (but not all) of the marbles in the first bowl and mixed them into the marbles that were in the second bowl.
Then she scooped up some of the marbles in the second bowl and mixed them into the marbles that were in the third bowl.
Finally, she scooped up some of the marbles in the third bowl and mixed them into the marbles that remained in the first bowl.
She scooped up the same number of marbles each time.
When she was finished, each bowl contained a pretty blend of colors in which no color present in the mixture constituted more than half
or less than a fourth of the marbles in the bowl.
When she finished there were 12 red marbles in third bowl.
What was the total number of marbles, how were the colors divided among the bowls, and how many marbles did the scoop hold?
Solution to the Problem:
Here is the solution:There were 144 marbles total.
The first and third bowls had 12 red marbles, 12 blue marbles, and 24 green marbles.
The second bowl had 24 red and 24 blue marbles.
The scoop held 36 marbles.
Here is a breakdown of what happened each time Lisa scooped marbles from one bowl and put them in another:
Initially,
bowl 1: 48 red marbles
bowl 2: 48 blue marbles
bowl 3: 48 green marbles
Lisa scoops 36 red marbles from bowl 1 and mixes them in bowl 2:
bowl 1: 12 red marbles
bowl 2: 48 blue marbles 36 red marbles
bowl 3: 48 green marbles
Lisa scoops 12 red marbles and 24 blue marbles from bowl 2 and mixes them in bowl 3:
bowl 1: 12 red marbles
bowl 2: 24 blue marbles 24 red marbles
bowl 3: 48 green marbles, 12 red marbles, 24 blue marbles
Lisa scoops 12 blue marbles and 24 green marbles from bowl 3 and mixes them in bowl 1:
bowl 1: 12 red marbles, 12 blue marbles, 24 green marbles
bowl 2: 24 blue marbles 24 red marbles
bowl 3: 12 red marbles, 12 blue marbles, 24 green marbles
Click here for Dr. Kishan's excellent solution
Colin Bowey gets extra credit because he found that the information in the problem does not appear to determine a unique scoop size. Any whole-number scoop size from 36 to 47 marbles produces the same final distribution.
Here are his results:
Simulation of every possible scoop size
Every scoop size from 36 to 47 works:
S=36 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 36R→ After scoop 1 12R 36R+48B 48G Scoop 2 12R+24B→ After scoop 2 12R 24R+24B 12R+24B+48G Scoop 3 ←0R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=37 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 37R→ After scoop 1 11R 37R+48B 48G Scoop 2 13R+24B→ After scoop 2 11R 24R+24B 13R+24B+48G Scoop 3 ←1R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=38 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 38R→ After scoop 1 10R 38R+48B 48G Scoop 2 14R+24B→ After scoop 2 10R 24R+24B 14R+24B+48G Scoop 3 ←2R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=39 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 39R→ After scoop 1 9R 39R+48B 48G Scoop 2 15R+24B→ After scoop 2 9R 24R+24B 15R+24B+48G Scoop 3 ←3R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=40 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 40R→ After scoop 1 8R 40R+48B 48G Scoop 2 16R+24B→ After scoop 2 8R 24R+24B 16R+24B+48G Scoop 3 ←4R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=41 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 41R→ After scoop 1 7R 41R+48B 48G Scoop 2 17R+24B→ After scoop 2 7R 24R+24B 17R+24B+48G Scoop 3 ←5R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=42 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 42R→ After scoop 1 6R 42R+48B 48G Scoop 2 18R+24B→ After scoop 2 6R 24R+24B 18R+24B+48G Scoop 3 ←6R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=43 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 43R→ After scoop 1 5R 43R+48B 48G Scoop 2 19R+24B→ After scoop 2 5R 24R+24B 19R+24B+48G Scoop 3 ←7R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=44 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 44R→ After scoop 1 4R 44R+48B 48G Scoop 2 20R+24B→ After scoop 2 4R 24R+24B 20R+24B+48G Scoop 3 ←8R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=45 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 45R→ After scoop 1 3R 45R+48B 48G Scoop 2 21R+24B→ After scoop 2 3R 24R+24B 21R+24B+48G Scoop 3 ←9R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=46 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 46R→ After scoop 1 2R 46R+48B 48G Scoop 2 22R+24B→ After scoop 2 2R 24R+24B 22R+24B+48G Scoop 3 ←10R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G S=47 Stage Bowl 1 Bowl 2 Bowl 3 Start 48R 48B 48G Scoop 1 47R→ After scoop 1 1R 47R+48B 48G Scoop 2 23R+24B→ After scoop 2 1R 24R+24B 23R+24B+48G Scoop 3 ←11R+12B+24G Final 12R+12B+24G 24R+24B 12R+12B+24G
Correctly solved by:
| 1. Dr. Hari Kishan |
D.N. College, Meerut, Uttar Pradesh, India |
| 2. Seth Cohen | Concord, New Hampshire, USA |
| 3. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |