Solve Fun Logic Puzzle: 111 People & 4 Jewels

[alane1994]

There are 111 people in a competition. The competition has 4 boxes and 4 jewels. Each box is identical and is completely opaque (i.e. you cannot see inside the box once it is closed). The jewels are all different: diamond, ruby, emerald and topaz. Everyone in the competition knows this. The host (who is NOT one of the 111 taking part), places one jewel in each box and then seals the boxes and writes a letter on each box: A, B, C and D - all done WITHOUT any of the 111 competitors watching. The competitors are then asked to guess which jewel is in which box.

-9 people get all 4 of their guesses wrong
-15 people guess exactly one jewel correctly
-25 people guess exactly 2 jewels correctly

How many people:
a) guess exactly 3 jewels correctly
b) guess exactly 4 jewels correctly

Bit of a hey, I'm back... again... puzzle!

 

[alane1994]

If you want a hint, feel free to ask! :)

 

[MarkFL]

My solution:

Spoiler

Since it is impossible to guess only 3 correctly, that leaves the remaining 62 to have guessed all 4 correctly.

 

[eddybob123]

My solution:

Spoiler

Since it is impossible to guess only 3 correctly, that leaves the remaining 62 to have guessed all 4 correctly.

[sp]Unless one or more people guessed twice of the same jewel, so they would have a slightly higher chance of guessing one of those right. But it[/sp]

 

[CellCoree]

a) Based on the given information, we can use the process of elimination to determine that 111 - 9 - 15 - 25 = 62 people did not guess exactly 3 jewels correctly. Therefore, the remaining 111 - 62 = 49 people must have guessed exactly 3 jewels correctly.

b) Similarly, 111 - 9 - 15 - 25 - 49 = 13 people did not guess exactly 4 jewels correctly. Therefore, the remaining 111 - 13 = 98 people must have guessed exactly 4 jewels correctly.