2x + 3y = n has exactly 2011 non-negative integral solutions. Determine the SUM of the possible values of n.
@19rsb said:The lower corner of a page in a book is folded over so as to reach the inner edge of the page.What is the fraction of the width folded over when the area of the folded part is minimum.1)1/32)2/33)3/44)None of these
3/4?
solved using options, if its none then
solved using options, if its none then
@19rsb said:The lower corner of a page in a book is folded over so as to reach the inner edge of the page.What is the fraction of the width folded over when the area of the folded part is minimum.1)1/32)2/33)3/44)None of these
Suppose I fold it from a distance 'x' from the inner edge and at an angle A
Width = W, length = L , Also, the fold meets the length at distance y from top
Once, you make the diagram:
Area of triangle = 1/2*(L-y)*(W-x)
now, (W-x)CosA = x => cosA = (x/(W-x))
Also, (L-y)sinA = W
Area = 1/2*(W)*(W-x)/sinA = (1/2)*(W*(W-x)^2)/[root(W*(W-2x)]
Solved it using Wolfram's method, and it says that x = W/3 
..
.. => Ans = 2/3
@ziddiarmaan said:@hiteshkhurana82 : i am getting 21/55 ..
same i am getting. justed wanted to confirm bcz was not very sure for that.. lets see if masters have sth diff in store.
@jain4444 said:2x + 3y = n has exactly 2011 non-negative integral solutions. Determine the SUM of the possible values of n.
Edited:72375
@jain4444 said:
2x + 3y = n has exactly 2011 non-negative integral solutions. Determine the SUM of the possible values of n.
2x + 3y = n
Case 1 :
n = 3k
=> y = (3k - 2x )/3 = k - (2/3)x
So, x can take values from 3.0 to 3.2010 and therefore k can take values from 3.0 to 3.2010
=> k ranges from 0 to 6030
And hence, n comes out to be 2*6030 = 12060
Case 2:
n= 3k + 1
I suppose again 12060 cases and n = 12061
and 12060 cases in 3k+2 form also and n = 12062
Similarly n can also be 12064 and 12065..
So total = 12060*6 +15 = 72375 ? ....
@hiteshkhurana82 said:On a straight line of length 10 cm, two points A, B are selected at random. What is the probability that AB >4
Plot it on the graph
O is origin and represents one end point of line AB, P and Q are other end points (P on x - axis and Q on y - axis)
O is origin and represents one end point of line AB, P and Q are other end points (P on x - axis and Q on y - axis)
x - axis represents length of OA and y - axis represents length of OB
Sample space is 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10
Allowed area is |x - y| > 4
So probability = (100 - 6*6)/100 = 64/100 = 16/25
@jain4444 said:2x + 3y = n has exactly 2011 non-negative integral solutions. Determine the SUM of the possible values of n.
Least value of n is 2 + 3 + 2010*6 = 12065 and largest value of n is 2*3 + 3*2 + 2010*6 = 12072
Sum of possible values = 12065 + 12066 + ... + 12072 = 4*24137 = 96548
Edit:- Oops solved for positive x and y
For non-negative, it will be
12060 to 12067
Sum = 96508
@chillfactor said:
Least value of n is 2 + 3 + 2010*6 = 12065 and largest value of n is 2*3 + 3*2 + 2010*6 = 12072Sum of possible values = 12065 + 12066 + ... + 12072 = 4*24137 = 96548
Sir, for n = 12060, 2011 non-negative solutions are hai na?? Or am I making some mistake?
Please can you check my solution?? :(
@ScareCrow28 said:Sir, for n = 12060, 2011 non-negative solutions are hai na?? Or am I making some mistake?Please can you check my solution??
Sorry earlier I solved it for positive x and y, just edited my post
@ravihanda said:@sahil.oberoi Just keep working hard. There isn't much to it. My mail id is ravihanda[at]gmail[dot]com
Sir I have mailed you with an issue...
Please check
Thanks
Please check
Thanks
@hiteshkhurana82 said:On a straight line of length 10 cm, two points A, B are selected at random. What is the probability that AB >4
p(AB8/25*2/10+6/10*4/10+8/25*2/10=16/250+240/1000+16/250=(240+32*4)/1000=368/1000
p(AB>4)=.632
i have taken mean over here instead of integration, so answer might be off a bit
p(AB>4)=.632
i have taken mean over here instead of integration, so answer might be off a bit
@chillfactor said:Sorry earlier I solved it for positive x and y, just edited my post
Sir, Please can you tell me where I missed some possibilities of n in my solution?
@chillfactor said:
Least value of n is 2 + 3 + 2010*6 = 12065 and largest value of n is 2*3 + 3*2 + 2010*6 = 12072Sum of possible values = 12065 + 12066 + ... + 12072 = 4*24137 = 96548Edit:- Oops solved for positive x and yFor non-negative, it will be12060 to 12067Sum = 96508
sir I read this formula somewhere
ax+by = n ; If this eq has M non negative integral sol ---> then sum of all the possible values of M is (M-1)*a^2*b^2 + [a*b*(2ab-a-b)]/2
(2010) * 2^2 * 3^2 + [3*2*(2*3*2 - 3 - 2)]/2 = 72381
(2010) * 2^2 * 3^2 + [3*2*(2*3*2 - 3 - 2)]/2 = 72381
@gautam22 said:sir 72375 aana chahiye.....12060+12061.........+12065.....sir ye 6 extra kahan se aa raha hai??????
n can vary from 12060 to 12065 only. I have the same doubt :(
because 6 ke gap ke bad 1 solution badh jayega..
@chillfactor @jain4444 Sir jee, please clarify
@gautam22 said:air 66 aur 67 kyun count kiye hain unpe to 2012 values nahi ho jayengi?????????sir plz correct me if i m wrong
2x + 3y = n
x = 3k + a and y = b - 2k, here a is the least possible value of x and b is largest possible value of y
to get least value of n, a = 0 and b = 2010*2 = 4020
Least n = 0 + 3*4020 = 12060
To get largest n, a = 2 and b = 2010*2 + 1
Largest n = 4 + 12060 + 3 = 12067
@chillfactor said:2x + 3y = nx = 3k + a and y = b - 2k, here a is the least possible value of x and b is largest possible value of yto get least value of n, a = 0 and b = 2010*2 = 4020Least n = 0 + 3*4020 = 12060To get largest n, a = 2 and b = 2010*2 + 1Largest n = 4 + 12060 + 3 = 12067
Sir, for n= 12066
There will be 2012 non-negative solutions na??
@ScareCrow28 said:Sir, for n= 12066There will be 2012 non-negative solutions na??
2x + 3y = 12067
y is always odd
Least value of y = 1
Largest value of y = 4021
So, y can take {(4021 - 1)/2} + 1 = 2011 values
Yeah for n = 12066, we have 2012 solutions (didn't check for 12066)