Q 11 12
@Subhashdec2 well my thought process went like this.....
1. First i realized that the sum would be unsolvably convoluted,(not CAT level) if the entire string of variables did not fit into some expansion like ab+a+b=(a+1)(b+1)-1
2. So i tried extrapolating the above logic for three variables (a+1)(b+1)(c+1) -1 gives you abc+ab +ac+bc+a+b+c. So basically going by this school of thought,and having identified the pattern which is, for X different variables the equation(x1+1)(x2+1)......(xn+1)-1 would give you an equation which has NcN number of terms which are product of all variables from x1 to xn, Nc(n-1) terms which are product of (n-1) terms taken at a time,...and so on...
ANd hence the expansion of the entire question's equation would be (a+1)(b+1)(c+1)(d+1)= 2010...factorize it and since all variables are positive factorize it as 67*3*2*5
hence a+b+c+d= 73
(P.S. i forgot to subtract 4 in the previous answer and wrote 77, but ans is 73)
1. First i realized that the sum would be unsolvably convoluted,(not CAT level) if the entire string of variables did not fit into some expansion like ab+a+b=(a+1)(b+1)-1
2. So i tried extrapolating the above logic for three variables (a+1)(b+1)(c+1) -1 gives you abc+ab +ac+bc+a+b+c. So basically going by this school of thought,and having identified the pattern which is, for X different variables the equation(x1+1)(x2+1)......(xn+1)-1 would give you an equation which has NcN number of terms which are product of all variables from x1 to xn, Nc(n-1) terms which are product of (n-1) terms taken at a time,...and so on...
ANd hence the expansion of the entire question's equation would be (a+1)(b+1)(c+1)(d+1)= 2010...factorize it and since all variables are positive factorize it as 67*3*2*5
hence a+b+c+d= 73
(P.S. i forgot to subtract 4 in the previous answer and wrote 77, but ans is 73)
@Tusharrr said:Two points are chosen uniformly at random on the unit circle and joined to make a chord C1. This process is repeated 3 more times to get chords C2,C3,C4. The probability that no pair of chords intersect can be expressed as a/b where a and b are coprime positive integers. What is the value of a+b?
I'm not sure but I guess we can proceed like this :-
There are 4 chords means 8 random points are choosen. Now using these 8 points we can form 4 chords in C(8, 2)*C(6, 2)*C(4, 2) = 8!/(2!2!2!2!) = 28*15*6 ways
Now, we have to find those cases when none of these 4 chords intersect. We can find these cases using CATALAN Numbers (for more info check wikipedia, they have explained it very well)
So, no of such cases = C(8, 4)/5 = 14
required probability = 14/(28*15*6) = 1/180
___________________________________________________________________________
""probability of getting 5 heads = (1/10)(9*1/32 + 1) = 41/320Probability that a fair coin is selected and we get 5 heads = (1/10){9*(1/32)} = 9/320required prob = (9/320)/(41/320) = 9/41
cud u plz explain it???"""
we are getting 5 heads, there are two ways of getting 5 heads:-
i) we choose a fair coin and we got head in all 5 flips
ii) we chose the biased coin, in this case we will always get a head
case i) - there are 9 such coins, so probability of getting a fair coin is 9/10 and in all 5 flips we will get a head
=> total probability = (9/10)(1/2)(1/2)(1/2)(1/2)(1/2) = 9/320
case ii) - there is only one biased coin, probability of choosing it is 1/10 and in all 5 flips we will get a head
=> total probability = (1/10)(1)(1)(1)(1)(1) = 1/10
=> total probability = (1/10)(1)(1)(1)(1)(1) = 1/10
required probability = (i)/{(i) + (ii)}
@chillfactor said:I'm not sure but I guess we can proceed like this :-There are 4 chords means 8 random points are choosen. Now using these 8 points we can form 4 chords in C(8, 2)*C(6, 2)*C(4, 2) = 8!/(2!2!2!2!) = 28*15*6 waysNow, we have to find those cases when none of these 4 chords intersect. We can find these cases using CATALAN Numbers (for more info check wikipedia, they have explained it very well)So, no of such cases = C(8, 4)/5 = 14required probability = 14/(28*15*6) = 1/180@simplydevesh said:probability of getting 5 heads = (1/10)(9*1/32 + 1) = 41/320Probability that a fair coin is selected and we get 5 heads = (1/10){9*(1/32)} = 9/320required prob = (9/320)/(41/320) = 9/41cud u plz explain it???we are getting 5 heads, there are two ways of getting 5 heads:-i) we choose a fair coin and we got head in all 5 flipsii) we chose the biased coin, in this case we will always get a headcase i) - there are 9 such coins, so probability of getting a fair coin is 9/10 and in all 5 flips we will get a head=> total probability = (9/10)(1/2)(1/2)(1/2)(1/2)(1/2) = 9/320case ii) - there is only one biased coin, probability of choosing it is 1/10 and in all 5 flips we will get a head=> total probability = (1/10)(1)(1)(1)(1)(1) = 1/10required probability = (i)/{(i) + (ii)}
In a nutshell Baye's theorem
@iLoveTorres said:bhai approach would be very long and difficult to be posted there is only 1 way the Sudoku can be completed
How come dost? 2 na? Symmetrical...
1342
4213
2431
3124
and
1423
3241
4132
2314
regards
scrabbler
1342
4213
2431
3124
and
1423
3241
4132
2314
regards
scrabbler
@albiesriram said:
My take:
7 - 2 (x^2-y^2 = 241 so y^2 = 4)
8 - 2 (Mentioned in previous post)
9 - 73 (Someone else already posted a detailed solution 😞 )
regards
scrabbler
In a mixed doubles tennis tournament, a team always consists of one male and one female player. In a certain mixed double tennis tournament, four females and four males are participating. In every round, every possible pair will play against every other possible pair. After every round, one female and one male will be eliminated. The process continues until only one female and one male is left, and this pair is called the winning pair of the tournament. What is the total number of matches to be played to decide the winning pair?
(1) 92
(2) 82
(3) 72
(4) 62
(1) 92
(2) 82
(3) 72
(4) 62
There are five cities in a state and each of them is to be connected to exactly two other cities using telephone lines. In how many ways can this be done?(a) 12 (b) 24 (c) 36 (d) 9
how to solve this???
@Marchex said:In a mixed doubles tennis tournament, a team always consists of one male and one female player. In a certain mixed double tennis tournament, four females and four males are participating. In every round, every possible pair will play against every other possible pair. After every round, one female and one male will be eliminated. The process continues until only one female and one male is left, and this pair is called the winning pair of the tournament. What is the total number of matches to be played to decide the winning pair?(1) 92 (2) 82 (3) 72 (4) 62
92?
16*9 / 2 + 9*4 / 2 + 4*1 / 2 = 72 + 18 + 2 = 92
regards
scrabbler
16*9 / 2 + 9*4 / 2 + 4*1 / 2 = 72 + 18 + 2 = 92
regards
scrabbler
The numbers 1, 2, €Ś €Ś.. n are written in the natural order. Numbers in odd places are struck off to form a new sequence. This process is continued till only one number is left.
If n = 1997, the number left is
(1) 1996
(2) 1988
(3) 512
(4) 1024
If the number left is 512, the maximum possible value of n is
(1) 1025
(2) 1023
(3) 513
(4) 1024
If n = 1997, the number left is
(1) 1996
(2) 1988
(3) 512
(4) 1024
If the number left is 512, the maximum possible value of n is
(1) 1025
(2) 1023
(3) 513
(4) 1024
@hell_fire said:There are five cities in a state and each of them is to be connected to exactly two other cities using telephone lines. In how many ways can this be done?(a) 12 (b) 24 (c) 36 (d) 9how to solve this???
it is already solved by many people. check previous posts
@albiesriram said:
10 - 27 (3*ext angle = 40 so...)
11 - rt2/3 I think (not sure)
12 - trying....thoda bouncer jaa raha hai!
Edit 12 should be 1. The circle :splat:
regards
scrabbler
11 - rt2/3 I think (not sure)
Edit 12 should be 1. The circle :splat:
regards
scrabbler
@hell_fire said:There are five cities in a state and each of them is to be connected to exactly two other cities using telephone lines. In how many ways can this be done?(a) 12 (b) 24 (c) 36 (d) 9how to solve this???
place the cities in a pentagon's vertices. now number of ways of doing so are (5-1)! = 24 :-)
The duration of a railway journey varies as the distance and inversely as the velocity; the velocity varies directly as the square root of the quantity of coal used per km, and inversely as the no of carriages in the train. In a journey of 50 KM in half an hour with 18 carriages, 100 kg of coal is required. how much coal will be consumed in a journey of 42 KM in 28 minutes with 16 carriages.
Q1