@rnishant231 said:if x = 45 , then we ll have base side of square = double the other side as per the figure..pls see...
Yes, that was the point left unnoticed...a plain question otherwise..
@AbhiFMS said:Hi,I have just started the prep for 2013.This question has been nagging my mind since i came across it.Find the area of largest square that can be fit into a semi-circle of area 50 pi.I browsed a bit and got to know the approach,thou initially was not able to solve it all by myself..The approach which was explained in the yahoo answers was using the circle equation x^2 + y^2 = r^2.I would like to know the other approach that can be used?can we use similar triangles concept here?cos,i started it off like that but couldnt get to the right ans...Thanks in Advance!
80.
method same as scrabbler :)
area = pi*r^2/2 = 50pi
r^2=100
r=10
x^2 + (x/2)^2 = 100
x= rt80
area of sq = x^2=80.
method same as scrabbler :)
area = pi*r^2/2 = 50pi
r^2=100
r=10
x^2 + (x/2)^2 = 100
x= rt80
area of sq = x^2=80.
forgot basics , pl help 
generic formula for
no. of solution of a1+a2+a3+a4...+an = k ???
@techgeek2050 said:can u explain the 8C3 ?
@techgeek2050 said:can u explain the 8C3 ?
a+b+c+k=6
to subtract a=0
a+1+b+c+k=6
a+b+c+k=5
8C3
to subtract a=0
a+1+b+c+k=6
a+b+c+k=5
8C3
@rnishant231 said:forgot basics , pl help generic formula for no. of solution of a1+a2+a3+a4...+an = k ???
(k+n-1)C(n-1)
@rnishant231 said:forgot basics , pl help generic formula for no. of solution of a1+a2+a3+a4...+an = k ???
Equivalent to distributing k identical objects (1s) into n distinct groups - we can use partitioning to get (k+n-1)C(n-1)
regards
scrabbler
regards
scrabbler
@gnehagarg said:a+b+c+k=6to subtract a=0a+1+b+c+k=6a+b+c+k=58C3
when a=0
b + c + k = 16 => 18C2
Keeping track of all concepts this time 
the thread is lively again after the GODS have started pouring wisdom...
@heylady said:Please see the attached.nowBD=R sin xOB=R cosxArea of ABCD: 2Rcos x* R sin x= R^2 sin2xMaximum when sin2x=1i.e. 2x=90x=45Area(Max)=R^2R^2=25*pi/ pi/2R^2=50Hence Area(Max)=50...@scrabbler
This will give you the area of the maximum rectangle in the semicircle. But not the maximum square? you have not taken into consideration that the sides should be equal i.e. 2Rcosx = Rsinx.
regards
scrabbler
regards
scrabbler
@scrabbler said:This will give you the area of the maximum rectangle in the semicircle. But not the maximum square? you have not taken into consideration that the sides should be equal i.e. 2Rcosx = Rsinx.regardsscrabbler
I overlooked the square part...
@rnishant231 said:forgot basics , pl help generic formula for no. of solution of a1+a2+a3+a4...+an = k ???
( k + n - 1) C n-1 ===> for non negative integer solutions
(k - 1) C n - 1 ===> positive integer solution
@jain4444 said:how many three digit numbers are there whose sum is not greater than 16 ?
The idea of my this post is to walk you through an observation and pattern.
Suppose x is the sum of digits of a 3 digit number and N(x) denotes the number of ways of getting x as the sum. Then, N(x) =N(28-x), where x -> [1,27] For this reason maximum possible sum corresponds to N(14)= 45 +9 + 7 +5 +3 +1
If the three digit number is abc, then for x = 9 to 19 we have 9 values of a. However, for remaining values of x, there are x or 28 窶度 (which ever is single digit) values of a.
So the total sum possibilities of any 3 digit number can be generated as follows:
For x= 1 to 9 and 19 to 27 -> Sum = Summation (Summation(n))
X=10 ->Summation(9) + 9 = 45 +9
X=11 -> 45 +9 + 7
X=12 -> 45 +9 + 7 +5 =66 = N(16) {This is the same value that we got in our previous problem on N(16). Remember?}
X=13 -> 45 +9 + 7 +5 +3
X=14 -> 45 +9 + 7 +5 +3 +1-> Maximum Sum
X=15 -> 45 +9 + 7 +5 +3 +1 -1 = 45+9+7+5+3 = N(13)
X=16 -> 45 +9 + 7 +5 +3 +1 -1 -3=45+9+7+5 = N(12)
X=17 -> 45 +9 + 7 +5 +3 +1 -1 -3-5= 45+9+7= N(11)
X=18-> 45 +9 + 7 +5 +3 +1 -1 -3-5-7= 45+9 = N(10)
X=19-> 45 +9 + 7 +5 +3 +1 -1 -3-5-7-9= 45= N(9)
For the purposes of the problem at hand, Summation(Summation(16)) = Summation (Summation (9)) + 45*7 + 9*7 +7*6+5*5+3*4+1*3-1*2-3*1
=>Summation(Summation(16))= 165 +420 +40-5=620
So the total sum possibilities of any 3 digit number can be generated as follows:
For x= 1 to 9 and 19 to 27 -> Sum = Summation (Summation(n))
X=10 ->Summation(9) + 9 = 45 +9
X=11 -> 45 +9 + 7
X=12 -> 45 +9 + 7 +5 =66 = N(16) {This is the same value that we got in our previous problem on N(16). Remember?}
X=13 -> 45 +9 + 7 +5 +3
X=14 -> 45 +9 + 7 +5 +3 +1-> Maximum Sum
X=15 -> 45 +9 + 7 +5 +3 +1 -1 = 45+9+7+5+3 = N(13)
X=16 -> 45 +9 + 7 +5 +3 +1 -1 -3=45+9+7+5 = N(12)
X=17 -> 45 +9 + 7 +5 +3 +1 -1 -3-5= 45+9+7= N(11)
X=18-> 45 +9 + 7 +5 +3 +1 -1 -3-5-7= 45+9 = N(10)
X=19-> 45 +9 + 7 +5 +3 +1 -1 -3-5-7-9= 45= N(9)
For the purposes of the problem at hand, Summation(Summation(16)) = Summation (Summation (9)) + 45*7 + 9*7 +7*6+5*5+3*4+1*3-1*2-3*1
=>Summation(Summation(16))= 165 +420 +40-5=620
Team BV - Vineet
@jain4444 said:how many three digit numbers are there whose sum is not greater than 16 ?
when 1 of the digits exceed 10, others cant. hence these are mutually exclusive...
1st let us find no. of solutions for
a + b + c
a + b + c + d = 16
x + b + c + d = 15 so no. of solutions = 18C3 = 816
now.., we cant have values x = 9,10,11... 15
for x = 9 we have b + c + d = 7, we have 21 solutions...
for x = 10 we have b + c + d = 6, we have 15 solutions... and so on we have 21 + 15 + 10 + 6 + 3 + 1 = 56 solutions , similarly applying same for b and c we have 168 solutions
so subtracting that from 816 we have 816-168 = 648
@pratskool said:when 1 of the digits exceed 10, others cant. hence these are mutually exclusive...1st let us find no. of solutions for a + b + c a + b + c + d = 16x + b + c + d = 15 so no. of solutions = 18C3 = 816now.., we cant have values x = 9,10,11... 15for x = 9 we have b + c + d = 7, we have 21 solutions...for x = 10 we have b + c + d = 6, we have 15 solutions... and so on we have 21 + 15 + 10 + 6 + 3 + 1 = 56 solutions , similarly applying same for b and c we have 168 solutions so subtracting that from 816 we have 816-168 = 648
can u plz see what's wrong with this
when one of a,b,c >=10 ,
a' + b + c + k = 6 => 3*9C3 = 252 ways
816 - 252 = 564 ??
@techgeek2050 said:can u plz see what's wrong with thiswhen one of a,b,c >=10 ,a' + b + c + k = 6 => 3*9C3 = 252 ways816 - 252 = 564 ??
sir, i am myself not sure about my method... neways 1stly no. of solutions of
a' + b + c + k = 6 is 9C3, but you do not have to multiply by 3 , since it takes into account each combination.... 2ndly i am not able to understand your method..
if one of a,b,c>=10, the solution is void, true...
but the rest of say when a >=10, b + c =6,5,4,3.... and so on.... hence many number of combinations...
In a right circular cone of radius 4cm and height 10cm, largest cube is inscribed whose base rests at base of the cone. A sphere is then inserted between top of cube and the vertex of cone. What can be the maximum radius of the sphere?
Team BV--Pratik Gauri
@pratskool said:sir, i am myself not sure about my method... neways 1stly no. of solutions of a' + b + c + k = 6 is 9C3, but you do not have to multiply by 3 , since it takes into account each combination.... 2ndly i am not able to understand your method..if one of a,b,c>=10, the solution is void, true...but the rest of say when a >=10, b + c =6,5,4,3.... and so on.... hence many number of combinations...
what i mean to say is this
if one of a,b,c>=10
let a = a' + 10 , the equation becomes
a' + b + c + k = 6
similarly for b and c, a + b' + c + k = 6 , and a + b + c' + k = 6
for each solution = 9C3
so total = 3*9C3 ways are there wherein one of a,b,c is >=10 , so that is what should be subracted.
P.S - bhai don't call me sir.. m just a noob. 
What is the probability that a path from (0,0) to (8,6) pass through (5,4), assuming that all paths move along grid lines and that movement must be either the positive x or the positive y direction?