@techgeek2050 said:Find the maximum value of abc(b + c) if ab + bc + ca = 2
abc(b+c) = bc(ab + ac) = bc(2-bc) = 2bc - bc^2
max. value for bc = 1 is 1.... max value = 1?
@pratskool said:abc(b+c) = bc(ab + ac) = bc(2-bc) = 2bc - bc^2max. value for bc = 1 is 1.... max value = 1?
don't have the OA. just have a quesion. Can't i take ab = bc = ca = 2/3 for max [ab(bc) + bc(ca)] ??
plz help me out with this concept builder problem....
from each corners of the square of unit side, right angle triangles are cut off to form a regular octagon. what is the side of each octagon formed?
@DivineSeeker said:plz help me out with this concept builder problem....from each corners of the square of unit side, right angle triangles are cut off to form a regular octagon. what is the side of each octagon formed?
rt2 - 1/2 ?
let the side of triangle be = x
so side of octagon = rt(x^2 + x^2) = x*rt2
now x + rt2*x + x = 1
x = 1/(2 + rt2)
side of octagon = rt2/(2 + rt2) = (rt2 - 1)/2
@DivineSeeker said:@mailtoankit how come x + rt2*x + x = 1. could u plz elaborate it????thanks
wats the answer ?
Pythagorean triplets - an amazing trick. Great for saving time in a horrid quant section! 😃
the ans is (rt2 - 1)/2.
How many of the first 1200 natural numbers are either prime to 6 or to 15?
OA given is 400 bt m getting 720 
@pakkapagal said:How many of the first 1200 natural numbers are either prime to 6 or to 15?OA given is 400 bt m getting 720
1200*(1/2)*(2/3) + 1200*(2/3)*(4/5) - 2*1200*(1/2)*(2/3)*(4/5) = 400
@Logrhythm said:1200*(1/2)*(2/3) + 1200*(2/3)*(4/5) - 2*1200*(1/2)*(2/3)*(4/5) = 400
bhai 2 se kyu multiply kiya yha???
@pakkapagal said:bhai 2 se kyu multiply kiya yha???
urmm....let me explain with a smaller number
How many of the first 15 numbers are coprime to either 3 or 5
the thing here to notice is Either prime to 3 or 5
Let's do it manually first:
Coprime to 3: 10 numbers
Coprime to 5: 12 numbers
Coprime to 15 (both): 8 numbers
so number coprime to either 3 or 5 are 5,10,3,6,9 and 12 => 6 numbers
or 15*(2/3) + 15*(4/5) - 2*15*(2/3)*(4/5) = 10+12-16 = 6
Why does this occur is simply because of Either Or...and when we apply set theory, we are double counting it...so that is why we need to subtract it twice
Hope it is clear :)
@gautam22 Please don't call me Sir, I am a female. Call me Sukhada/Sukhi.
Hope you found the video useful though :)
Q.> The odds against an event are 4 to 5 and the odds in favor of another independent event are 3 to 7. Find the probability that BOTH of them occur
a. 2/18
b. 5/18
c. 9/18
d. 3/18
OA is 3 /18 but i am getting it as 31/45. Plz let me know if any one is able to get the OA :)
@ravi6389 said:Q.> The odds against an event are 4 to 5 and the odds in favor of another independent event are 3 to 7. Find the probability that BOTH of them occura. 2/18b. 5/18 c. 9/18d. 3/18OA is 3 /18 but i am getting it as 31/45. Plz let me know if any one is able to get the OA
probability that event A do not occur = 4/(4 + 5) = 4/9
probability that event A occur = 5/9
probability that event B occur = 3/(3 + 7) = 3/10
probability that event B do not occur = 7/10
probability that both of them occur = 5/9*3/10 = 3/18
@mailtoankit said:probability that event A do not occur = 4/(4 + 5) = 4/9probability that event A occur = 5/9probability that event B occur = 3/(3 + 7) = 3/10probability that event B do not occur = 7/10probability that both of them occur = 5/9*3/10 = 3/18
@mailtoankit .why dont we solve it like this ;->
p(A U B ) = P(A) + P(B) - P(A) * P(B)
so putting the values i get it as 31/45. Can u plz tell this?
@ravi6389 said:@mailtoankit .why dont we solve it like this ;->p(A U B ) = P(A) + P(B) - P(A) * P(B)so putting the values i get it as 31/45. Can u plz tell this?
we have to find P(A)*P(B)..(When both of them are occurring)
P(A U B)----> this is used,when we have to find either A or B and not both A and B