Some maths brain teasers qns
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huh....Originally posted by DailyFreeGames.com:Wah, this is a nice thread, I got a teaser question, sounds very simple but is actually very tricky lor...
There are 2 pair of couples sitting on a bench in a park, how many unique photos can they take assuming each unique photo show the 2 couples sitting in different position and the fact that the couples do not wish to be seperated from their boyfriend/girlfriend?
let each couple be 1 unit lor...
so for 2 units, you can get 4 different combination, as follows:
1st couple only
2nd couple only
1st couple left side 2nd couple right side
1st couple right side 2nd couple left side
then u know within each unit, the bf and gf can switch location, giving a further 2! more ways
Final answer = 4 * 2 = 8 -
Is the answer 8?
M1W1M2W2
Since the couples do not wish to be separated, M1W1 and M2W2. In this case, position is important.
Thus, no of ways = 2! x 2 x 2
= 8
Not sure if the ans is correct. -
Originally posted by shagadelic1892:
Here's my solution:
Another cheap-labour question but slightly more difficult
A and B working together can complete a job in 10 days. B and C working together can complete the job in 15 days. A and C working together can complete the job in 12 days. How many days would it take A, B and C working together to complete the job?
For A and B, fraction of job completed each day = 1/10 => A + B = 1/10
For B and C, fraction of job completed each day = 1/15 => B + C = 1/15
For A and C, fraction of job completed each day = 1/12 => A + C = 1/12
Next, add up the 3 equations and we have:
2A + 2B + 2C = 1/4 => A + B + C = 1/8
This equation implies that A, B and C working together can complete 1/8 of the job in one day.
Hence, no. of days required for A, B and C working together to complete the job = 8 -
wah, you guys are good, man! The correct answer for my question is 8.
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An interesting qn,
The infinite sequence
123456789101112131415161718192021.......
is obtained by writing the positive integers in order. The 2003rd digit in the sequence is
(A) 0
(B) 1
(C) 2
(D) 3
(E) None of the above. -
after first 9 digits, assume all remaining numbers are 2 digitsOriginally posted by tanjun:An interesting qn,
The infinite sequence
123456789101112131415161718192021.......
is obtained by writing the positive integers in order. The 2003rd digit in the sequence is
(A) 0
(B) 1
(C) 2
(D) 3
(E) None of the above.
(2003-9)/2 = 997 2-digits number
10 is 1st
11 is 2nd
99 is 90th < 997
90*2 + 9 = 189 digits
assume remaining number all 3-digits number
(2003 - 189)/3 = 604 2/3 3-digits number
100 is 1st
101 is 2nd
704 is 605th number
7 0 4 is (2002nd, 2003rd, 2004th digit)
2003rd digit is thus 0 (A) -
Correct. Answer is 0.