@amresh_maverick said:f(x)=x^2/(x^2-1) then 31f(30)*f(29).............*f(3)*f(2) equals to ??
60 ?
Folks, since functions is the current theme, try solving this one
Let f: R -> R and f(x+2) = 1/2 + (f(x) - (f(x))^2)^1/2. Then which among the following is always true?
(a) f(2) = f(4) (b) f(3) = f(7) (c) f(4) = f(10) (d) Atleast 2 of the foregoing
@bodhi_vriksha Please tell us the solution P&C; question wherein 4couples were to be arranged.. I got 21 ways? What's the answer??
@ScareCrow28 : what are u doing on this thread now!?
..ur seasons over na?! noi 
P.S: sry for spamming!
..ur seasons over na?! noi P.S: sry for spamming!
Thirty six numbers are filled in the cells of a matrix as shown in the figure given below. Six numbers are chosen from the matrix such that no two numbers belong to the same row or the same column. In how many ways can the numbers be chosen?
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
a. (6^2 Ć 5^2 Ć 4^2 Ć 3^2 Ć 2^2 Ć 1^2) b. (6 Ć 6 Ć 6 Ć 6 Ć 6 Ć 6)
c. 720 d. None of these
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
a. (6^2 Ć 5^2 Ć 4^2 Ć 3^2 Ć 2^2 Ć 1^2) b. (6 Ć 6 Ć 6 Ć 6 Ć 6 Ć 6)
c. 720 d. None of these
@Faruq said:Thirty six numbers are filled in the cells of a matrix as shown in the figure given below. Six numbers are chosen from the matrix such that no two numbers belong to the same row or the same column. In how many ways can the numbers be chosen?1 2 3 4 5 67 8 9 10 11 1213 14 15 16 17 1819 20 21 22 23 2425 26 27 28 29 3031 32 33 34 35 36a. (6^2 Ć 5^2 Ć 4^2 Ć 3^2 Ć 2^2 Ć 1^2) b. (6 Ć 6 Ć 6 Ć 6 Ć 6 Ć 6)c. 720 d. None of these
Is the answer 720?
Approach i followed which m pretty sure is wrong..
if we select 1 from the first row then
we can select
5 no from the 2nd row
4 no from the 3rd row
3 no from the 4th row
2 no from the 5th row
1 no from the 6th row
so similarly for 2,3,4,5,6
hence total no of ways = 6! = 720 .
:(
Approach i followed which m pretty sure is wrong..
if we select 1 from the first row then
we can select
5 no from the 2nd row
4 no from the 3rd row
3 no from the 4th row
2 no from the 5th row
1 no from the 6th row
so similarly for 2,3,4,5,6
hence total no of ways = 6! = 720 .
:(
@pathetic said:AC is the diameter of a circle. B and D are points on its circumference such that ?BAD = 60 Ā° and AB = AD. The perimeter of the quadrilateral ABCD is 20 cm. Find the radius of the circle. (in cm)
radius is 5(rt3-1)
@sonamaries7 said:@ScareCrow28 : what are u doing on this thread now!? ..ur seasons over na?! noi P.S: sry for spamming!
Kya karu.. š Just love to come here! Even if I dnt want to 
On giving 3 pencils free with every 5 pens bought, a shopkeeper makes a profit of 20% and on giving 6 pencils free with every 2 pens bought, he suffers a loss of 25%. Find the approximate profit percent made by the shopkeeper when he gives 4 pencils free with every 6 pens bought. (Assume that the pencils are identical and the same applies to the pens.)
a. 18% b. 20% c. 24% d. 16%
a. 18% b. 20% c. 24% d. 16%
@Faruq said:On giving 3 pencils free with every 5 pens bought, a shopkeeper makes a profit of 20% and on giving 6 pencils free with every 2 pens bought, he suffers a loss of 25%. Find the approximate profit percent made by the shopkeeper when he gives 4 pencils free with every 6 pens bought. (Assume that the pencils are identical and the same applies to the pens.)a. 18% b. 20% c. 24% d. 16%
x and y be SP and CP of pen
a and b be SP and CP of pencil
5x-5y-3b=1/5(5y+3b)
5x=6y+18b/5
25x=30y+18b
6b+2y-2x=1/4(6b+2y)
2x=9b/2 +3y/2
4x=9b+3y
17x/24=y
72b=15x
b=5x/24
6x-6y-4b/(6y+4b)
6/(17/4+5/6) - 1
6*12/61 - 1
0.18=18%
a and b be SP and CP of pencil
5x-5y-3b=1/5(5y+3b)
5x=6y+18b/5
25x=30y+18b
6b+2y-2x=1/4(6b+2y)
2x=9b/2 +3y/2
4x=9b+3y
17x/24=y
72b=15x
b=5x/24
6x-6y-4b/(6y+4b)
6/(17/4+5/6) - 1
6*12/61 - 1
0.18=18%
@Faruq said:On giving 3 pencils free with every 5 pens bought, a shopkeeper makes a profit of 20% and on giving 6 pencils free with every 2 pens bought, he suffers a loss of 25%. Find the approximate profit percent made by the shopkeeper when he gives 4 pencils free with every 6 pens bought. (Assume that the pencils are identical and the same applies to the pens.)a. 18% b. 20% c. 24% d. 16%
Approx soln....
Consider a base of 30 (LCM of 2, 5, 6) pens. On this
18 pencils => +20%
90 pencils => -25%
so 20 pencils = 18.xx
So 18 (out of given answer choices).
(Not a mathematically rigorous solution! Just trying something different)
regards
scrabbler
Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
@Faruq said:Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
AK=p (probability of winning)
SS=2p
K=4p
p=1/7
hence 6/7??
@Faruq said:Three horses : Kanishka, Silver Streak and Arabian Knight are the only horses competing in a race and only one of these three can win the race. If Kanishka is twice as likely to win as Silver Streak and Sliver Streak is twice as likely to win as Arabian Knight, then what is the probability of Arabian Knight losing this race?
6/7 ?
4 : 2 : 1 ---> odds in favor
probability of arabian knights winning = 1/7
probability of losing = 6/7
There are ānā necklaces in a safe box (n > 1). Every necklace has the same number of diamonds. Each necklace has at least 2 diamonds. The total number of diamonds in these ānā necklaces is between 500 and 600. If this data is sufficient to find the value of n, then what is the value of ānā?
a. 19 b. 23 c. 29 d. None of these
a. 19 b. 23 c. 29 d. None of these