A person has to select one of the three capsules.One capsule makes his weight increase by 1 Kg.Another one ensures that the weight doesnt change.The third one makes him gain 2 kg and also compels him to select one of the three capsules once again-all similar to the first set of three capsules.the person has no way to identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
@albiesriram said:if N is a 4 digit number such that if we remove the leftmost digit then resulting 3 digit number is 1/9th of the N. How many possible values of N?
1000a + 100b + 10c + d = 9(100b+10c+d)
=>1000a = 800b + 80c + 8d
=>125a = 100b + 10c + d
=> all the three digit multiples of 125 will give us the desired result
hence 1125, 2250, 3375, 4500, 5625, 6750, 7825 are 7 possible values.
ATDH.
@mani0303 said:
2)I think one formula is there for this...guess it's 4a^2 + 1 = 901...
no of solutions of |x-a| + |y-b| =k are 4k
and no of solutions for |x-a|+|y-b|+z-c|=k are 4k^2+2
so in this ques, answer would be 4*15^2+2= 902 :)
Team BV-- Pratik Gauri
@pakkapagal said:A person has to select one of the three capsules.One capsule makes his weight increase by 1 Kg.Another one ensures that the weight doesnt change.The third one makes him gain 2 kg and also compels him to select one of the three capsules once again-all similar to the first set of three capsules.the person has no way to identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
change because of 1st tablet = 1/3*1
change because of 2nd tablet = 1/3 *0
change because of 3rd tablet= 1/3 (2) + 1/3 * (1/3*1 + 1/3*0... )
total wt. gain G = 1/3*1 + 1/3*0 + 1/3 (2) + 1/3 * (1/3*1 + 1/3*0... )
G = 1/3*1 + 1/3 *0 + 1/3 (2 + (1/3*1 + 1/3*0...))
G= 1/3 + 1/3 (2+G)
2/3G= 1
gain = 3/2??
change because of 2nd tablet = 1/3 *0
change because of 3rd tablet= 1/3 (2) + 1/3 * (1/3*1 + 1/3*0... )
total wt. gain G = 1/3*1 + 1/3*0 + 1/3 (2) + 1/3 * (1/3*1 + 1/3*0... )
G = 1/3*1 + 1/3 *0 + 1/3 (2 + (1/3*1 + 1/3*0...))
G= 1/3 + 1/3 (2+G)
2/3G= 1
gain = 3/2??
@ChirpiBird said:change because of 1st tablet = 1/3*1 change because of 2nd tablet = 1/3 *0 change because of 3rd tablet= 1/3 (2) + 1/3 * (1/3*1 + 1/3*0... )total wt. gain G = 1/3*1 + 1/3*0 + 1/3 (2) + 1/3 * (1/3*1 + 1/3*0... )G = 1/3*1 + 1/3 *0 + 1/3 (2 + (1/3*1 + 1/3*0...))G= 1/3 + 1/3 (2+G)2/3G= 1gain = 3/2??
ji... 
@pakkapagal said:A person has to select one of the three capsules.One capsule makes his weight increase by 1 Kg.Another one ensures that the weight doesnt change.The third one makes him gain 2 kg and also compels him to select one of the three capsules once again-all similar to the first set of three capsules.the person has no way to identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
@pakkapagal said:A person has to select one of the three capsules.One capsule makes his weight increase by 1 Kg.Another one ensures that the weight doesnt change.The third one makes him gain 2 kg and also compels him to select one of the three capsules once again-all similar to the first set of three capsules.the person has no way to identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
1/3 + 1/3•(2 + 1/3 • ( 1/3 + 1/3•(2 + 1/3•(1/3 + ....
= 1 + 1/3 + 1/9... = 3/2
@pakkapagal said:A person has to select one of the three capsules.One capsule makes his weight increase by 1 Kg.Another one ensures that the weight doesnt change.The third one makes him gain 2 kg and also compels him to select one of the three capsules once again-all similar to the first set of three capsules.the person has no way to identifying the capsules beforehand and is equally likely to take any of the three capsules at any stage.Find the expected gain in weight.
Let A be the gain in weight ..
A= 1/3*1 + 1/3*0 + 1/3*[2+A]
Here 1/3 denotes probability of event A,B and C .
Solving the above equation
We get A=3/2 :)
Team BV-- Pratik Gauri
@saurabhlumarrai said:The number of points with integer co-ordinates within the circle with radius 6 and center at origin is:a)24b)85c)109d)114
114?
EDIT.: sry, 109 ayega.
i counted 4 points on the circle and 0,0 twice.
so 114-5 = 109
EDIT.: sry, 109 ayega.
i counted 4 points on the circle and 0,0 twice.
so 114-5 = 109
@ajeetaryans said:ans is A = 280B= 280 c = 3so the maximum sum 563 of A + B + C
Although your answer is correct, but A,B , C must be distinct. check again :)
@iLoveTorres said:kindly post your approach.. mine is way too complicated..
22 points inside one quadrant(without points on axis), and 11 +10 on x,y axis.
so total 22*4 +21=109.
so total 22*4 +21=109.
@ChirpiBird said:22 points inside one quadrant(without points on axis), and 11 +10 on x,y axis.so total 22*4 +21=109.
kindly clarify why are you considering 21 points on x,y axis. you would have 12+12+1=25 points right?
@iLoveTorres said:kindly clarify why are you considering 21 points on x,y axis. you would have 12+12+1=25 points right?
see it like this -> 0 to 5 on +ve x-axis + (-1) to (-5) on -ve x-axis + 1 to 5 on +ve y-axis + (-1) to (-5) to -ve y-axi = 21 points
what you have done is counted the origin 4 times, so it is simply 5*4 + 1 (for origin) = 21 points or 6*4 - 3 (for extra counting) = 21 points...
abb sahi hai bhai??
@iLoveTorres said:kindly clarify why are you considering 21 points on x,y axis. you would have 12+12+1=25 points right?
(0,5) .. to (0,-5) .. 11 points
similarly (-5,0) to (5,0) ...again 11 points. but 0,0 will be counted twice.
so 11+10 = 21 points on x,y.
within likha h. so dont consider points on the circle. 6,0 .. 0,6... 0,-6, and -6,0 mat karo consider.
Q)find the remainder when (21!)^2 is divided by 43.
(don't have the OA,so with approach please)
@toshalimitra said:Q)find the remainder when (21!)^2 is divided by 43.(don't have the OA,so with approach please)
41! % 43 = 1 (wilson's th.)
21!(22*23*24*...41) % 43 = 1
21!(-21*-20*-19*...-2) % 43 = 1
21!*21! % 43 = 1
so 21!^2 % 43 = 1
@toshalimitra said:Q)find the remainder when (21!)^2 is divided by 43.(don't have the OA,so with approach please)
using Wilson's theorem, -42! mod 43 = 1..so the remainder is 1
@saurabhlumarrai said:The number of points with integer co-ordinates within the circle with radius 6 and center at origin is:a)24b)85c)109d)114
Use Pick's Theorem:
Area of region= I + B/2 -1
B= Boundary points
I=Integral points (on or within)
So, Area = 22/7 * 6^2 = 113.xx
Now Boundary Points: (6,0) , (-6,0) , (0,6) , (0-6) ---> Total 4.
So, B = 4
So I = Ceil [113.xx - 2 + 1] = 113
But we have to take within circle = 113 - B = 113 - 4 = 109 points
Area of region= I + B/2 -1
B= Boundary points
I=Integral points (on or within)
So, Area = 22/7 * 6^2 = 113.xx
Now Boundary Points: (6,0) , (-6,0) , (0,6) , (0-6) ---> Total 4.
So, B = 4
So I = Ceil [113.xx - 2 + 1] = 113
But we have to take within circle = 113 - B = 113 - 4 = 109 points