@albiesriram said:Some geometry..
x/1 = 1/CB
CB = 1/x
EB = CB-CE
= (1/x)-1
= (1-x)/x
@grkkrg said:the summation is repeated twice
which summation b=a+c??
but iske sol to b ke cases leke wud be 2(1+2+3+4.......1003) tak hi jayega.???
dusra kaun sa repetition ho raha hai..m kuch miss kar rah hu...point karao yaar..
@nick_baba said:which summation b=a+c??but iske sol to b ke cases leke wud be 2(1+2+3+4.......1003) tak hi jayega.???dusra kaun sa repetition ho raha hai..m kuch miss kar rah hu...point karao yaar..
2 * (1 + 1 + 2 + 2 + 3 + 3 + ... + 1003 + 1003) aa rha hai :)
@Brooklyn said:dekho:we knw x=-1 is a soln of the qthis gives us: a - b + c=0ie a+c=bnow b can be any value from 1 to 2008and a+c=bso we get :a+c =adding a variable x that takes care of extra value, we get:a+c+x=2008now a,c >0so a+c+x=2006total int soln=2008c2now nos such as 2008 can also be written as 1004 + 1004but we have constratint a,b,c differentso take these out , ie all nos which are multiple of 2 => 2008/2 = 1004Hope this helps
..there shud have been a hats-off smiley too...thanks.waise ye a+c+x=2006 ka soln wahi hai na p+q+r=n ka soln which is (n+3)c3 form hai na??
@grkkrg said:2 * (1 + 1 + 2 + 2 + 3 + 3 + ... + 1003 + 1003) aa rha hai
aiyyyoooo....
..silly mistake.
..silly mistake.@nick_baba said:..there shud have been a hats-off smiley too...thanks.waise ye a+c+x=2006 ka soln wahi hai na p+q+r=n ka soln which is (n+3)c3 form hai na??
nai !! p+q+r=n has total soln = (n+3-1)C(3-1)=(n+2)C 2 soln
Let H be a set of 2000 nonzero real numbers. How many negative
elements should H have in order to maximize the number of four-element
subsets of H with a negative product of elements?
elements should H have in order to maximize the number of four-element
subsets of H with a negative product of elements?
a) 1039 b) 961 c) either a or b d) neither a nor b
bhai logo zara is tarah k ques ko approach karne ka tareeka batao..meri bht phatati hai jahan set theory a jata hai..
@Brooklyn said:nai !! p+q+r=n has total soln = (n+3-1)C(3-1)=(n+2)C 2 soln
oh yess...jaldi jaldi me slip kar gaya..
...jaldi jaldi me slip kar gaya..let f be function such that f(313)=3+1+3=7,then find,f(100)+f(101)+f(102)+.......f(200)
a)900 b)902 c)1000 d)1002
@Sandy0303 said:let f be function such that f(313)=3+1+3=7,then find,f(100)+f(101)+f(102)+.......f(200)a)900 b)902 c)1000 d)1002
1002??
see..the series reduces like this:
for f(100) to f(109)----> 1-10
for f(110) to f(119)----> 2-11
for f(120) to f(123)----> 3-12
. .
.
.
.
like this f(190) to f(199)---> 10-19
Now the sum of an AP with first and last term given is n(a+l)/2
so the sum of the given nos. is 10(11+13+15....29)/2
--->1000
now we have left f(200)=2
hence the total shud be 1002
@Sandy0303 said:let f be function such that f(313)=3+1+3=7,then find,f(100)+f(101)+f(102)+.......f(200)a)900 b)902 c)1000 d)1002
1002
100*1+450+450+2=1002
@Sandy0303 said:let f be function such that f(313)=3+1+3=7,then find,f(100)+f(101)+f(102)+.......f(200)a)900 b)902 c)1000 d)1002
(1 * 100 + 2) + (45 * 10) + (45 * 10) = 102 + 450 + 450 = 1002
@nick_baba said:Let H be a set of 2000 nonzero real numbers. How many negativeelements should H have in order to maximize the number of four-elementsubsets of H with a negative product of elements?a) 1039 b) 961 c) either a or b d) neither a nor bbhai logo zara is tarah k ques ko approach karne ka tareeka batao..meri bht phatati hai jahan set theory a jata hai..
I figured this out from the options..either the answer is c or d
I cant find a way to solve the problem, but the answer comes out to be the same with options 1 and 2
1039C1*961C3 + 1039C3+961C1..considering 1039 -ve elements
1039C3*961C1 + 1039C1*961C3..considering 961 -ve elements
@nick_baba said:Let H be a set of 2000 nonzero real numbers. How many negativeelements should H have in order to maximize the number of four-elementsubsets of H with a negative product of elements?a) 1039 b) 961 c) either a or b d) neither a nor bbhai logo zara is tarah k ques ko approach karne ka tareeka batao..meri bht phatati hai jahan set theory a jata hai..
What is the answer to this ? I couldn't solve it, but I tried it with this approach
Let the number of negative elements be x
-> four element subsets where product is negative = xC1*(2000-x)C3 + xC3*(2000-x)C1
This is a fourth degree function in x, so I used wolfram method and noticed that the maxima appears at two places : 1039, 961 approx.
So, I guess it should be either a or b ?
My first post...
@Sandy0303 said:let f be function such that f(313)=3+1+3=7,then find,f(100)+f(101)+f(102)+.......f(200)a)900 b)902 c)1000 d)1002
ans shd be 1002
will form and AP for every 10 term.
just calculate = [10(11+13+15+...+29)/2]+2[for 200]=1002
@grkkrg said:(1 * 100 + 2) + (45 * 10) + (45 * 10) = 102 + 450 + 450 = 1002
@Sandy0303 said:@grkkrg can u explain it plse?
The first term is the sum of the hundreds digit of all the numbers.
Second term is the sum of the tens digit of all the numbers.
Third term is the sum of the units digit of all the numbers.
PS: 45 = 0 + 1 + 2 + 3 + ... + 9
Second term is the sum of the tens digit of all the numbers.
Third term is the sum of the units digit of all the numbers.
PS: 45 = 0 + 1 + 2 + 3 + ... + 9
@soham2208 said:What is the answer to this ? I couldn't solve it, but I tried it with this approachLet the number of negative elements be x-> four element subsets where product is negative = xC1*(2000-x)C3 + xC3*(2000-x)C1 This is a fourth degree function in x, so I used wolfram method and noticed that the maxima appears at two places : 1039, 961 approx.www.wolframalpha.com/input/?i=...So, I guess it should be either a or b ?
ans given is either a or b..so 