Search found 34 matches
- by gamemaster
Tue Dec 05, 2006 2:37 pm- Forum: Problem Solving
- Topic: Difficult Math Problem #70 - Permutations
- Replies: 5
- Views: 4212
if 10 have the same pace, and 10 of them do it in 16 hr, then each alone will do it in 160hr
now the question is how much time will it take to 8 machines with rate of 160 hr to do the job
- by gamemaster
Sun Nov 19, 2006 7:30 am- Forum: Problem Solving
- Topic: help....
- Replies: 3
- Views: 2595
the distance she has to walk is 2*pi*r = 2pi =~ a little more than 6 miles
she walk 3 miles in 1 hr ...
- by gamemaster
Sun Nov 19, 2006 7:23 am- Forum: Problem Solving
- Topic: help with this question....
- Replies: 2
- Views: 3346
- by gamemaster
Sun Nov 19, 2006 7:20 am- Forum: Problem Solving
- Topic: help...
- Replies: 2
- Views: 3516
- by gamemaster
Thu Nov 16, 2006 6:40 am- Forum: Problem Solving
- Topic: Percentage problem
- Replies: 1
- Views: 3216
note that they say MUST true, not possibly true
and because we get:
xy + z = xy + xz
xz = z
only 5. must be true
- by gamemaster
Thu Nov 16, 2006 6:36 am- Forum: Problem Solving
- Topic: All answers are satisfying.. which is correct?
- Replies: 6
- Views: 3792
.
well in case its 3 and 5, the standard deviation would be
sqrt[42/7] = sqrt[6]
in case its 2 and 6 it will be:
sqrt[48/7] which is bigger than sqrt[6] but smaller than sqrt[8] so closer to sqrt[8] than sqrt[6] is
so i still think its 2 and 6
- by gamemaster
Mon Nov 13, 2006 11:12 am- Forum: Problem Solving
- Topic: Difficult Math Question #50 - Standard Deviation
- Replies: 16
- Views: 9525
answer
the average of the series is 4
the standard deviation is: sqrt [(40)/5] = sqrt [8]
we would like the new sum to be ~56 to get sqrt[8] so the closest will be:
2 and 6 then the standard deviation would be: sqrt[48/7]
D
- by gamemaster
Mon Nov 13, 2006 3:22 am- Forum: Problem Solving
- Topic: Difficult Math Question #50 - Standard Deviation
- Replies: 16
- Views: 9525
- by gamemaster
Mon Oct 16, 2006 6:34 am- Forum: Problem Solving
- Topic: Difficult Math Problems #28 - Consecutive integers
- Replies: 6
- Views: 14083
- by gamemaster
Tue Oct 10, 2006 4:13 am- Forum: Problem Solving
- Topic: Difficult Math Question #25 - Sets
- Replies: 3
- Views: 4009
- by gamemaster
Fri Oct 06, 2006 8:29 am- Forum: Problem Solving
- Topic: Difficult Math Question #23 - Algebra
- Replies: 5
- Views: 4471
x = 13!
a = x^16 - x^8 / x^8 + x ^ 4 =
(x^8+x^4)(x^8-x^4)/(x^8+x^4) =
x^8 - x^4
a/x^4 =
x^4 - 1
digit number of 13! must be 0 because its like (1*2*3...*11*12*13) * 10
so the final answer would be 1
- by gamemaster
Thu Oct 05, 2006 12:29 am- Forum: Problem Solving
- Topic: Difficult Math Question #23 - Algebra
- Replies: 5
- Views: 4471
i think i got it ...
the solution applies only for cases where the order of picking matters, so in this case:
{1,2,3,4} {5,6}
is different than
{2,2,3,4} {3,1}
This should be clarified in the question!, otherwise the solution should be different ... (i think )
- by gamemaster
Thu Oct 05, 2006 12:16 am- Forum: Problem Solving
- Topic: Difficult Math Question #20 - Combinations
- Replies: 5
- Views: 4846
but arent you repeating on options like this? this is what's confusing, i dont think you can just multiply (13^4) by 48*47 lets take 1,2,3 of type hart and 4,5,6 of 3 different types, so: {1,4,5,6} {2,3} {2,4,5,6} {1,3} {3,4,5,6} {2,1} are the same choises, you have here a very large number of repet...
- by gamemaster
Thu Oct 05, 2006 12:13 am- Forum: Problem Solving
- Topic: Difficult Math Question #20 - Combinations
- Replies: 5
- Views: 4846
i think its about time to see the answer
i think it should be < 13^4 but i didnt find an easy way to solve it
can you tell us how to solve this?
Thanks
- by gamemaster
Wed Oct 04, 2006 4:58 am- Forum: Problem Solving
- Topic: Difficult Math Question #20 - Combinations
- Replies: 5
- Views: 4846