The are 25 HUNGRY fish, what is the max. number of full(NOT HUNGRY Fish) Left if each hungry fish should eat 3 fish inorder to become (NOT HUNGRY), No hungry fish are allowed to stay at the end and no fish is allowed to eat more than three.
2007-03-05 04:43:33 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You are all prob.correct because that what I thought of also but my teacher says its not 6, but its possible that she's trying to confuse us as always are you sure? no other way, she wants the MAX. numbers.
2007-03-05 04:57:07 · update #1
I got the answer from the teacher its 6.
2007-03-06 01:05:30 · update #2
well The catch here was That she wanted The MAX. Number, thats why 6 is the answer not 4 !
2007-03-08 16:19:32 · update #3
Stage 1:
6 (NOT HUNGRY) fish at the end of stage 1
Each has eaten 3 other (HUNGRY) fish - 18 fish have been eaten
18 + 6 = 24
1 (HUNGRY) fish is left with the 6 (NOT HUNGRY) fish.
Stage 2:
The 1 (HUNGRY) fish eats 3 of the (NOT HUNGRY) fish.
This leaves 4 fish, all (NOT HUNGRY).
2007-03-05 04:50:21 · answer #1 · answered by MamaMia © 7 · 0⤊ 0⤋
Several ways to number this but how about this. There are a total of 4 full fish and 0 hungry fish left at the end. Fish (fish eaten by this fish).
Fish 1 eats( fish2, 3, 4) ALIVE AT END
5 eats (6,7,8), EATEN BY 17
9 eats (10,11,12) EATEN BY 17
13 eats (14, 15, 16) EATEN BY 17
17 eats (5, 9, 13) ALIVE
18 eats (19, 20, 21) ALIVE
22 eats (23, 24,25) ALIVE
2007-03-05 04:57:29 · answer #2 · answered by Allison S 5 · 0⤊ 0⤋
There's 1 left, it ends by this one eating 3 of the full fish.
2007-03-05 04:54:59 · answer #3 · answered by Norrie 7 · 0⤊ 0⤋
6: Fish #4 eats #1-3. Fish #8 eats #5-7. Fish #12 eats #9-10.
....Fish #16 eats #13-15. Fish #20 eats #17-19.
....Fish #22 eats #21. Fish #25 eats #22-24.
One fish (#22 above) has to eat another before being eaten itself.
2007-03-05 04:57:07 · answer #4 · answered by gebobs 6 · 0⤊ 0⤋
enable Mrs. Thompon's age be 10x + y enable Larry's age be 10y + x x must be extra desirable than y different smart she is a organic and organic oddity to declare the least so ... 10x + y - (10y + x) = 3/11 (10x + y + 10y + x) Multiply the two sides by potential of 11 110x + 11y - 110y - 11x = 3(11x + 11y) 99x - 99y = 33x + 33y Now transter the x's to the left, the y's to the impressive. 99x - 33x = 33y + 99y 66x = 132y It seems such as you do not have sufficient suggestions, yet you do. The solutions could desire to be complete numbers. So what number will artwork below here situations: x could desire to be extra desirable than y The values that artwork are x = 2 y = a million x = 4 y = 2 x = 6 y = 3 x = 8 y = 4 It looks to me like the middle 2 may be ok. 40 two and 24 makes her 18 whilst he replaced into born. looks sturdy to me sixty 3 - 36 = 27 She replaced into 27 whilst he replaced into born. nonetheless sturdy. 80 4 - 40 8 = 36 Even it extremely is okay. the only one you are able to truly get rid of is the 1st one: she'd be too youthful, yet you're able to desire to examine it out for your self.
2016-10-17 07:59:47 · answer #5 · answered by Anonymous · 0⤊ 0⤋
6 fish will eat 6*3 = 18
now 1 hungry and 6 not hungry fishes are left
of that the hungry one will eat 3
so 3 not hungry fish are left finally
2007-03-05 04:51:28 · answer #6 · answered by ganesan 2 · 0⤊ 0⤋
ok, 1 fish eats 3 others so that is 4 down...21 left.
1 fish eats 3 more...4 more down.. 17 left.
1 fish eats 3.......13 left.
1 eats 3....9 left.
1 eats 3....5 left.
1 eats 3....1 left.....hmmm can't work? because 6 full and one left....can't have one left.....let me know if ya get an answer
2007-03-05 04:53:25 · answer #7 · answered by Brandiibaby 2 · 0⤊ 0⤋
x = full fish
3x = dead fishes
x + 3x = 25
=> 4x = 25
=> x = 6 (w/ 1 empty fish at the end which leaves the pond)
2007-03-05 04:51:49 · answer #8 · answered by milamtil 2 · 0⤊ 0⤋