15 cats meet 3 cats each, and all those cats meets 25 cats and they all go for a curry
the ans is not
85,
70,
1125
1500,
1569,
2006-12-15 14:12:02 · 9 answers · asked by owais_cnwl 1 in Science & Mathematics ➔ Mathematics
I assume the ultimate question here is, "How many cats were there having a curry in the end?"
It will depend on how one interprets the words "each" and "all" (in other word, what the assumptions are associated with these connections).
If each of the first 15 cats meet 3 different additional cats, all met cats being exclusive to the one cat who met them, then this is 15 original cats + 45 met cats = 60 cats who then go on to meet 25 cats. If this means that as a group the 60 cats met another 25 cats, then the answer is 85 cats. But you have ruled that out.
If each of the 60 cats meets an exclusive set of 25 cats, then you have 60 cats meeting (60*25 = 1500 cats), for a total final group of 1560 cats. You ruled out 1569 -- if you meant 1560 and "1569" was just a typo, then we have another dead-end answer otherwise I would say this is the probable solution.
If the first group of 15 cats meets a second group of 3 cats (each of the 15 cats will meet the 3 new cats if they are in a group), then you have 18 cats. They go on to meet either 25 total cats in a third group for a final total of 43 cats having a curry, or else each of the 18 cats meets a unique set of 25 new cats for a total of 18 cats + (18*25 cats) = a total of 468 cats out for a curry.
Pretty much however you stack it, that's too damned many cats. And I'm not sure they would even LIKE curry. ;)
Oh yes, I suppose one could get really imaginative with their answer and say that the 15 cats each met three cats FROM WITHIN THE ORIGINAL GROUP OF 15 -- so you then have a total of 15 cats going on to meet the 25 cats. But if the first 11 of those 25 were the remaining cats in the first group, you only have 14 additional cats for a grand total of 29 cats out having a curry.
You culd probably play such games of semantics until the cows came home, at which time the cats could have milk which I am sure they would prefer over curry.
2006-12-15 14:31:07 · answer #1 · answered by Mustela Frenata 5 · 2⤊ 0⤋
each of 15 meets 3 so 15+15*3=60 if each of these 60 meets 25, 60+25*60=1560 cats total.
2006-12-15 15:21:37 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋
Silly answer : it must be cats reunion and they like clubbing.
Serious answer :
15 + (15*3) = 60
60 + (60*25) = 1,560
It must be 1,560 cats (assumption : every cat did not meet the same cat as they friend meet).
Are you from Cat society (hope you 're a cat boy or cat woman family :))
2006-12-15 14:34:55 · answer #3 · answered by btc_smd 2 · 0⤊ 0⤋
15*3=45
45+25=70
2006-12-15 14:20:59 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
15*3=45
45+25=70
is the answer 70
2006-12-16 01:16:49 · answer #5 · answered by Miks 1 · 0⤊ 0⤋
think its 43( 15+3+25=43)
2006-12-15 18:37:47 · answer #6 · answered by Anonymous · 0⤊ 0⤋
15x3=45
45+15=60
60x25= 1500
1500+60=1560
1560 all together
2006-12-15 14:18:09 · answer #7 · answered by Jamie ♥ 2 · 0⤊ 0⤋
The variety of answers doesn't mean that it's a difficult problem. It means that it is poorly-posed!!!
2006-12-15 14:23:31 · answer #8 · answered by modulo_function 7 · 2⤊ 0⤋
990
Solution:
15x3=45
45x25=990
2006-12-15 14:17:25 · answer #9 · answered by Donald Duck 2 · 0⤊ 0⤋