@ravi.theja said:In a convex polygon,interior angle is in AP with samllest angle 120 and common difernce 5.find numbr of sides of polygon?
15
17
18
9
9
@ravi.theja said:Define a number k such that it is the sum of the squares of the first M natural nos(i.e k=1^2+2^2+...+m^2), where ma) 10 b) 11
c) 12 d) 13
is it 12?
@ravi.theja said:Define a number k such that it is the sum of the squares of the first M natural nos(i.e k=1^2+2^2+...+m^2), where ma) 10 b) 11
c) 12 d) 13
@pyashraj said:Smallest Interior angle= 120...Thus, Largest Exterior angle=60Now, 2nd smallest interior angle=125..thus, 2nd largest exterior angle=55Thus, Ext angle forms an AP with a=60, and CD=-5Let, n be the Number of sides..Thus, 360=n/2*[2*60 + (n-1)*(-5)]=>720 = n(125 - 5n)=>n*(25-n) = 144Hence, n= 9..k = M(M + 1)(2M + 1)/6
For k to be divisible by 4, one of M or (M + 1) should be divisible by 8 (as we have an extra two in denominator), as (2M + 1) will always be odd.
For M = 8, 16, 24, 32, 40, 48; M will be divisible by 8 and
for M = 7, 15, 23, 31, 39, 47, 55; (M + 1) will be divisible by 8
So, for total 13 values of M 55, k will be divisible by 4..Courtesy: @chillfactor Sir [4 the 2nd ques]
pyashraj bhai..
even i thought the same at the 1st place but m
there are 28 identical coins,all of which except for one weigh the same,using a common balance what is the minimum number of weightings required to ensure that the coin with different weight is identified??
3
4
5
6
Aimcat question huva...solution confusing
@pavimai said:there are 28 identical coins,all of which except for one weigh the same,using a common balance what is the minimum number of weightings required to ensure that the coin with different weight is identified??3456Aimcat question huva...solution confusing
3??
half of the coins.. each time till we have 7 coins eft:
28--> 14-->7
now as they speak of MIN no of weighting required:
out of those 7 weights that we have,we hand pick the coin with diff weight in the very 1st go and weight the remaining 6 coins after dividing them in half i.e 3-3.
:-/
@pavimai said:there are 28 identical coins,all of which except for one weigh the same,using a common balance what is the minimum number of weightings required to ensure that the coin with different weight is identified??3456Aimcat question huva...solution confusing
4??? 😃 28 ==> 14 ===> 7 ===> 3 ==> get the result
wen we get final 7 keep one aside..and if its balanced then the left out one is the diff weightd one..else make the 6 as 3 n 3 n weigh it...take the 3 with heigher weight n place 1 aside..n do for two 😃 so minimum is 4 ???
wen we get final 7 keep one aside..and if its balanced then the left out one is the diff weightd one..else make the 6 as 3 n 3 n weigh it...take the 3 with heigher weight n place 1 aside..n do for two 😃 so minimum is 4 ???
no its 4
1st weighing- 9 9 10(g1)
2nd weighing-from 10- 3 3 4 (g2)
3rd weighing-from 4- 1 1 2(g3)
4th weighing-from 2 -1 1
so minimum 4 weighings
@Zedai said:3??half of the coins.. each time till we have 7 coins eft:28--> 14-->7now as they speak of MIN no of weighting required:out of those 7 weights that we have,we hand pick the coin with diff weight in the very 1st go and weight the remaining 6 coins after dividing them in half i.e 3-3.:-/
@ravi.theja said:4??? 28 ==> 14 ===> 7 ===> 3 ==> get the resultwen we get final 7 keep one aside..and if its balanced then the left out one is the diff weightd one..else make the 6 as 3 n 3 n weigh it...take the 3 with heigher weight n place 1 aside..n do for two so minimum is 4 ???
super
@Zedai said:So, apparently i have witnessed GOD committing mistake!
in that case please be prepared to witness many such acts... :D
ATDH.
Que) If 5x+12y+13z=25,What could be the minimum value of x^2+y^2+z^2?Guys this is a Cauchy inequality type problem.........
Kindly post the ans as well as the solution......
I am totally confused in this one
Kindly post the ans as well as the solution......
I am totally confused in this one
@bhatkushal said:Que) If 5x+12y+13z=25,What could be the minimum value of x^2+y^2+z^2?Guys this is a Cauchy inequality type problem.........Kindly post the ans as well as the solution......I am totally confused in this one
Cauchy Inequality
(X^2 + Y^2 + z^2) (a^2 + b^2 + c^2) >= (ax + by + cz)^2
(X^2 + Y^2 + z^2) (5^2 +12^2 + 13^2) >= (5x+12y+13z)^2
(X^2 + Y^2 + z^2) >= 625/(25 + 144 + 169)
(X^2 + Y^2 + z^2) >= 25^2/ 13^2 *2
minimum value of (X^2 + Y^2 + z^2) is 1.535
(X^2 + Y^2 + z^2) (a^2 + b^2 + c^2) >= (ax + by + cz)^2
(X^2 + Y^2 + z^2) (5^2 +12^2 + 13^2) >= (5x+12y+13z)^2
(X^2 + Y^2 + z^2) >= 625/(25 + 144 + 169)
(X^2 + Y^2 + z^2) >= 25^2/ 13^2 *2
minimum value of (X^2 + Y^2 + z^2) is 1.535
@bhatkushal said:Que) If 5x+12y+13z=25,What could be the minimum value of x^2+y^2+z^2?Guys this is a Cauchy inequality type problem.........Kindly post the ans as well as the solution......I am totally confused in this one
considering the case X,Y,Z are integers.. then min of x^2+y^2+z^2 can be obtained only if values of x,y,z are minimum ..Maximise Z den Y den X ==> z=1 ,y=1,x=0 ==> sum minimum value=2
@bhatkushal said:Que) If 5x+12y+13z=25,What could be the minimum value of x^2+y^2+z^2?Guys this is a Cauchy inequality type problem.........Kindly post the ans as well as the solution......I am totally confused in this one
Those not interested in my verbose rant may please not read ahead
rather just providing the solution I would like to put certain points/observations regarding Cauchy's Inequality and problems related to it.
The inequality states
(a1^2 + a2^2......an^2)*(b1^2 + b2^2 +.......bn^2) >= (a1b1+ a2b2 + a3b3.......anbn)^2
more often than not we find it difficult to apply in problems even when we know the theorem by heart, so I would try to simplify it using some very kiddish observations
1. there are n pairs of numbers (a1,b1) (a2,b2).........(an,bn)
2. for each set of numbers cauchy did two things; he squared and multiplied
3. One side he first squared then then multiplied
4. On other side he first multiplied and then squared
5. And then he went on to say that Sqaure first multiply later >= Multiply first Sqaure Later
Now the bold terms help us identify if we can apply cauchy theorem in a given question or not.
For example in the given question there are squares (i.e x^2 + y^2 + z^2) and also pair of numbers multiplied..i.e (5 and x are multiplied, 12 and y are multiplied and 13 and z are multiplied... so we have pair of numbers (5,x) (12,y) and (13,z)
so try and figure which part of the inequality statement is missing.
So try and fit (x^+y^2+z^2) and (5x + 12y + 13z) in the chauchy's inequality
(x^2 + y^2 + z^)*(...........????.............) >= (5x + 12y + 13z)^2
now we know what needs to be filled in the ??? (it will be the sum of square of the numbers from each of the pair, i.e 5^2 , 12^2 and 13^2)
(x^2 + y^2 + z^)*(5^2 + 12^2 + 13^2) >= (5x + 12y + 13z)^2
Bingo!!
=>(x^2 + y^2 + z^)*(5^2 + 12^2 + 13^2) >= (25)^2 {replace the value as 25)
=>(x^2 + y^2 + z^) >= (25)^2/(5^2 + 12^2 + 13^2)
ATDH.
@bhatkushal said:@insane.vodka how did you get the value 625...
got it @anytomdickandhary explained it pretty well.......
@anytomdickandhary said:Those not interested in my verbose rant may please not read ahead rather just providing the solution I would like to put certain points/observations regarding Cauchy's Inequality and problems related to it.The inequality states (a1^2 + a2^2......an^2)*(b1^2 + b2^2 +.......bn^2) >= (a1b1+ a2b2 + a3b3.......anbn)^2more often than not we find it difficult to apply in problems even when we know the theorem by heart, so I would try to simplify it using some very kiddish observations1. there are n pairs of numbers (a1,b1) (a2,b2).........(an,bn)2. for each set of numbers cauchy did two things; he squared and multiplied3. One side he first squared then then multiplied4. On other side he first multiplied and then squared5. And then he went on to say that Sqaure first multiply later >= Multiply first Sqaure LaterNow the bold terms help us identify if we can apply cauchy theorem in a given question or not.For example in the given question there are squares (i.e x^2 + y^2 + z^2) and also pair of numbers multiplied..i.e (5 and x are multiplied, 12 and y are multiplied and 13 and z are multiplied... so we have pair of numbers (5,x) (12,y) and (13,z)so try and figure which part of the inequality statement is missing. So try and fit (x^+y^2+z^2) and (5x + 12y + 13z) in the chauchy's inequality(x^2 + y^2 + z^)*(...........????.............) >= (5x + 12y + 13z)^2now we know what needs to be filled in the ??? (it will be the sum of square of the numbers from each of the pair, i.e 5^2 , 12^2 and 13^2)(x^2 + y^2 + z^)*(5^2 + 12^2 + 13^2) >= (5x + 12y + 13z)^2Bingo!!=>(x^2 + y^2 + z^)*(5^2 + 12^2 + 13^2) >= (25)^2 {replace the value as 25)=>(x^2 + y^2 + z^) >= (25)^2/(5^2 + 12^2 + 13^2)ATDH.
Cauchy ke saath bana raha tha kya??? 
Find the least five digit number which when divided by 63, 56 and 42 leaves a remainder 1 in each case,
A.10082
B.10081
C.10061
D.10071
A.10082
B.10081
C.10061
D.10071