Should be 27 Mins..
Here, A= 25/9 m/sec..B= 25/3 m/sec n C = 125/9 m/sec..
Thus, their ratio = 1:3:5..Their 1st meeting point will be at the staring point..which will happen after 540 sec..Thus, their 3rd meet will happen, 1620 sec or 27 mins..
@pirateiim478
@pirateiim478 said:Find whole no. of solutions for |x|+|y|+|z|=36?
maalik itna kyo le rahe ho..
ans = 38C2..
@pirateiim478
Mod[x] + Mod[y] + Mod[z] = a...The above Eqn will have 4a^2 + 2 Soln..
Further to add, Mod[x] + Mod[y]= a...The above Eqn will have 2a^2 Soln..
PS: Didn't saw Whole No. Solns.. 
@rayus said:maalik itna kyo le rahe ho.. ans = 38C2..
😛 Maalik!! 38C2 is for x+y+z=36 Here modulus is involved we need to consider 2 scenarios. One there will be sign changes for each variable and other has 3 cases.
1. One of them is 0
2. two of them is 0
3. None of them is 0
We need to find solutions for each case and multiply them.
@pyashraj said:@pirateiim478Should be 5186..Mod[x] + Mod[y] + Mod[z] = a...The above Eqn will have 4a^2 + 2 Soln..Further to add, Mod[x] + Mod[y]= a...The above Eqn will have 4a^2 Soln..
These are formulas to calculate area formed where as i asked for integral solutions. Also for 2nd one area formula will be 2a^2.
@pirateiim478 said:Maalik!! 38C2 is for x+y+z=36 Here modulus is involved we need to consider 2 scenarios. One there will be sign changes for each variable and other has 3 cases.1. One of them is 02. two of them is 03. None of them is 0We need to find solutions for each case and multiply them.These are formulas to calculate area formed where as i asked for integral solutions. Also for 2nd one area formula will be 2a^2.
bhai.. for all whole numbers |x| = x.. same for y and z.. toh modulus ka toh kaam hi nahi raha.. it becomes a simple x+y+z=36
@pirateiim478 OA kya hain iska? is it different from 38C2?
Find range of x^2-x+1/x^2+x+1?
@rayus said:bhai.. for all whole numbers |x| = x.. same for y and z.. toh modulus ka toh kaam hi nahi raha.. it becomes a simple x+y+z=36@pirateiim478 OA kya hain iska? is it different from 38C
Yes it is different!!
Yes it is different!!
Last one :A fair coin is tossed 10 times . Find the probability that two heads do not occur consecutively ??(A) 1/(2^4)(B) 1/(2^3)(C) 1/(2^5)(D) none
Good night @rayus
@pirateiim478 said:1.Find whole no. of solutions for |x|+|y|+|z|=36?2. Find no of +ve solutions for xyz=72?
1) For whole number solutions - x+y+z = 36. So 38C3 solutions
2) xyz = 72.
72 = 9*8 = 3^2*2^3
x = 3^a1*2^b1
y = 3^a2*2^b2
z = 3^a3*2^b3
a1+a2+a3 = 2. So 4c2 = 6 solutions
b1+b2+b3 = 3. So 5c2 = 10 solutions
Hence, total +ve integer solutions 60.
@pirateiim478 said:Last one :A fair coin is tossed 10 times . Find the probability that two heads do not occur consecutively ??(A) 1/(2^4)(B) 1/(2^3)(C) 1/(2^5)(D) noneGood night @rayus
2/2^5 = 1/2^4
@pirateiim478 said:Find range of x^2-x+1/x^2+x+1?
bhai ye question samajh nahi aaya.. eqn mein bracket laga do please.. confusion ho rahi hain..
@pirateiim478 said:@rayus said:bhai.. for all whole numbers |x| = x.. same for y and z.. toh modulus ka toh kaam hi nahi raha.. it becomes a simple x+y+z=36@pirateiim478 OA kya hain iska? is it different from 38CYes it is different!!
bhai.. isme if we have to find no. of whole no solution then it will be (should be) 38C2.. if we dont have to find whole number solution then cases banenge.. with and without 0... as you have mentioned in your post..
@pirateiim478 said:Last one :A fair coin is tossed 10 times . Find the probability that two heads do not occur consecutively ??(A) 1/(2^4)(B) 1/(2^3)(C) 1/(2^5)(D) noneGood night @rayus
Is it None of there??
Case0 - 10T = 1 case.
Case1 - 9T 1 H = 10 cases.
Case2 - 8T 2H = 9c2 = 36 cases.
Case3 - 7T 3H = 8c3 = 56 cases.
Case4 - 6T 4H = 7c4 = 35 cases.
Case5 - 5T 5H = 6c5 = 6 cases.
Total = 144.
Hence, probability = 144/2^10
Consider a regular polygon of n sides .The number of values of n for which the polygon will have angles whose degree values is an integer ?
@pirateiim478 said:1.Find whole no. of solutions for |x|+|y|+|z|=36?2. Find no of +ve solutions for xyz=72?
X=2^a1*3^b1
Y=2^a2*3^b2
Z=2^a3*3^b3
a1+a2+a3=3
5C2=10
b1+b2+b3=2
4C2=6
ordered ways=60
unordered ways=
a1=a2 and b1=b2=>4 ways
a2=a3 and b2=b3=>4 ways
a1=a3 and b1=b3=>4 ways
Total ways =12
60-12/6=8 ways
Unordered ways=6+8=14
@naga25french said:Consider a regular polygon of n sides .The number of values of n for which the polygon will have angles whose degree values is an integer ?
Is it 22??
The internal angle of an n sided regular polygon is given by 180 - 360/n
360 = 9*4*10 = 3^2*2^2*5*2 = 2^3*3^2*5
(3+1)(2+1)(1+1) = 4*3*2 = 24.
but we need to take out cases where n = 1 and 2 (as these are not polygons)
total = 24-2 = 22
Some one suggest me if I should study mass point geometry for CAT 2013. Is it beneficial and does not it take too much time to understand? 
@Logrhythm @gnehagarg @naga25french
@naga25french said:Consider a regular polygon of n sides .The number of values of n for which the polygon will have angles whose degree values is an integer ?
interior angle is (n-2)*180/n
180=3^2*2^2*5
Number of factors are 3*3*2=18
n=1,2 not possible
16 solutions
@naga25french
Should be 22 sir..
Let n be the Number of sides of the polygon..Then, Exterior angle= 360/n
Now, For each value of n, Ext. Angle shld be an Integer, coz each Interior angle shld be an Integer..
Thus, 360= 2^3*3^2*5^1..Number of Factors= 24..
However, 1 n 2 cannot form a Closd plygon..Hence, 24-2 = 22..
@Logrhythm @pyashraj
Right
N = 444(upto 50 digits) except for the nth digit(where n ranges from 1-50). If N is divisible by 13 for nth digit, then find how many value can n take ?
Right
N = 444(upto 50 digits) except for the nth digit(where n ranges from 1-50). If N is divisible by 13 for nth digit, then find how many value can n take ?
@gnehagarg said:X=2^a1*3^b1Y=2^a2*3^b2Z=2^a3*3^b3a1+a2+a3=35C2=10b1+b2+b3=24C2=6ordered ways=60unordered ways=a1=a2 and b1=b2=>4 waysa2=a3 and b2=b3=>4 waysa1=a3 and b1=b3=>4 waysTotal ways =1260-12/6=8 waysUnordered ways=6+8=14
can you please teach me how to calculate the unordered ways? i have never been able to wrap my head around that concept. thanks.