@krum said:don't have OA , actually got this 1 in litmus today
Ok. I got 12. Will tell approach. wait.
@krum said:don't have OA , actually got this 1 in litmus today
You did it using AP right?? This should actually be approached as GP if I am right...
@krum said:don't have OA , actually got this 1 in litmus today . I was under the impression that it would be a cake walk, but attempted only 16/20 in qa and got trolled in lr
litmus is pretty hard as I hear :mg:
@freebull said:tell me the approach puys !!
3/11??
.3*.008/(.3*.008+.5*.01+.2*.007)=.0024/.0088=3/11
@ayushnasa said:litmus is pretty hard as I hear
lr was on level with aimcats , rest were normal stuff
@krum
P.S. Figure not to scale...
N hope its readable...
4,k1,k2,k3,9 are radius on 5 levels. & x is extra part say.
Hence
4/(4+x)= k1/(k1 + 4*2 + x)
From here find k1.
U will get k1= ((8+x)/x) *4
Similarly for level 3 U will get k2= ((8+x)/x)^2 *4
Thus its a GP. U can check further if wanted.
So on 5th level we get
9=((8+x)/x)^4 *4
From here U can find k2.
k2=6. So diameter = 12.
P.S. Figure not to scale...
N hope its readable...
4,k1,k2,k3,9 are radius on 5 levels. & x is extra part say.
Hence
4/(4+x)= k1/(k1 + 4*2 + x)
From here find k1.
U will get k1= ((8+x)/x) *4
Similarly for level 3 U will get k2= ((8+x)/x)^2 *4
Thus its a GP. U can check further if wanted.
So on 5th level we get
9=((8+x)/x)^4 *4
From here U can find k2.
k2=6. So diameter = 12.
@adwaitjw said:@krumP.S. Figure not to scale...N hope its readable...4,k1,k2,k3,9 are radius on 5 levels. & x is extra part say.Hence4/(4+x)= k1/(k1 + 4*2 + x)From here find k1.U will get k1= ((8+x)/x) *4Similarly for level 3 U will get k2= ((8+x)/x)^2 *4Thus its a GP. U can check further if wanted.So on 5th level we get9=((8+x)/x)^4 *4From here U can find k2.k2=6. So diameter = 12.
OR as 3rd layer is the GM...calcutate rt(4*9)=rt(36)=6....Hence dia 12, I did this way :P
@ayushnasa said:OR as 3rd layer is the GM...calcutate rt(4*9)=rt(36)=6....Hence dia 12, I did this way
Yup... But wasnt sure at the start if it is actually GP. Attempted this kind for first time... So went by long method...
Next time will use GM method...
Next time will use GM method...
@freebull said:tell me the approach puys !!
assume 200 people..
standard -----Preferred----Ultra-preferred
100--------------60----------- 40
number die: 1------12/25-----14/50
multiply by 50: 50--------24------14
required probability = 24/(50+24+14) = 24/88 = 3/11 
Each user on a computer system has a password with 6 to 8 character long having upper case letters and a digit. Each password must contain atleast one digit. How many possible passwords are there??
@Shray14 said:Each user on a computer system has a password with 6 to 8 character long having upper case letters and a digit. Each password must contain atleast one digit. How many possible passwords are there??
Total upper case letters = 26
Total digits = 10
=> Total characters = 36
If pasword contains 6 characters, we can have 36P6 possibilties.
We need to remove cases when only letters are used. So, remove 26P6 cases.
Similarly for 7 and 8 character password.
So, total ways = (36P6 + 36P7 + 36P8 ) - (26P6 + 26P7 + 26P8 )
Total digits = 10
=> Total characters = 36
If pasword contains 6 characters, we can have 36P6 possibilties.
We need to remove cases when only letters are used. So, remove 26P6 cases.
Similarly for 7 and 8 character password.
So, total ways = (36P6 + 36P7 + 36P8 ) - (26P6 + 26P7 + 26P8 )
@krum said:spherical balls are put layer by layer in a conical flask inverted upside down such that each one touches the sides and the layer below it. If the diameter of bottom most ball is 8cm and that at the 5th level is 18cm. Find the diameter of the ball at the 3rd layer.1)12 cm2)13 cm3)8root3 cm4)none
a=4
ar,ar^2,ar^3,ar^4
ar^4 = 9
r^4=9/4
4.3/2 = 6 is the radius of ball at 3rd level
so,12 is the diameter.
@soumitrabengeri said:Total upper case letters = 26Total digits = 10=> Total characters = 36If pasword contains 6 characters, we can have 36P6 possibilties.We need to remove cases when only letters are used. So, remove 26P6 cases.Similarly for 7 and 8 character password.So, total ways = (36P6 + 36P7 + 36P8 ) - (26P6 + 26P7 + 26P8 )
@Shray14 said:Each user on a computer system has a password with 6 to 8 character long having upper case letters and a digit. Each password must contain atleast one digit. How many possible passwords are there??
(36^6 -26^6) + (36^7 -26^7) + (36^8 -26^8)
@Shray14 said:Each user on a computer system has a password with 6 to 8 character long having upper case letters and a digit. Each password must contain atleast one digit. How many possible passwords are there??
26-Alphabets
10-digits
Total-when password contains no digit
36C6+36C7+36C8-(26C6+26C7+26C8)