@jaspunit said:3 circles touching each other, find possible radius of circle circumscribing all these, if radius of each 3 circle is 'r'?
(2r/rt3)+r
@Asfakul said:How many ordered pairs (a, b) exist such that L.C.M. of a and b is (2^3)(5^7)(11^13) (a, b ˆˆN)?a) 2460b) 2835c) 2645d) 2840
7*15*27=2835
@catahead said:From QE: http://www.quantexpert.co.in/questionoftheday.htmlQ1 and Q2
Q1 3000
Q2 (9/2)^8/3
Q2 (9/2)^8/3
@mani0303 said:It's 2835 ie (4^2 - 3^2)*(8^2 - 7^2)*(14^2 - 13^2) = 7*15*27 = 2835
@Subhashdec2 said:7*15*27=2835
ye kaise kiya???
@mani0303 said:It's 2835 ie (4^2 - 3^2)*(8^2 - 7^2)*(14^2 - 13^2) = 7*15*27 = 2835
Sir.. I'm horribly out of touch :embarrassed:
Can you please explain the formula you have used? :neutral:
@mani0303 said:It's 2835 ie (4^2 - 3^2)*(8^2 - 7^2)*(14^2 - 13^2) = 7*15*27 = 2835
Sir I'm in horribly out of touch :embarrassed:
Can you please explain the formula you have used? :neutral:
@jaspunit said:@Dexian3 circles touching each other, find side of triangle circumscribing all these, if radius of each 3 circle is 'r'?
(2+rt3)r
@vijay_chandola said:Sir I'm in horribly out of touch Can you please explain the formula you have used?
hey,- good to see you old fella
Let's say a = (2^x1)(5^y1)(11^z1) and b = (2^x2)(5^y2)(11^z2)
the pair x1 and x2 can take 4 different values including 0 - so the possible pairs = 16,but we should exclude pairs that don't have 3
Hence,4^2 - 3^2 and similarly for other 2
@AsihekAdhvaryu said:my ans is 2r[1+root(3)], can you please explain?
circum radius is (2r/rt3) + r........
i think yahin par kuchh cal mistake hoga...
wat have u got as circum radius...?
i think yahin par kuchh cal mistake hoga...
wat have u got as circum radius...?
@Dexian said:circum radius is (2r/rt3) + r........i think yahin par kuchh cal mistake hoga...wat have u got as circum radius...?
I am getting as 2r/root(3)+2r
@AsihekAdhvaryu said:I am getting as 2r/root(3)+2r
i think it shud b +r and not +2r .........
if u c ur diagram... only r needs to be added and not 2r
if u c ur diagram... only r needs to be added and not 2r
@priyank333333 said:@Dexian ..@AsihekAdhvaryu ..approach plz ?
it becomes a equi triangle wid side 2r.....
now find the circumradius of this triangle and add r
now find the circumradius of this triangle and add r
in a quadrilateral ABCD, BC=10, CD=14, AD=12 and /_CBD = /_BAD =60. IF AB=a+root b. where a and b are a positive integers then a+b = Options
1) 193
2) 201
3) 204
4) 207
5) none of the above
1) 193
2) 201
3) 204
4) 207
5) none of the above
How many ordered triples (x,y,z) of integer solutions are there to the system of equations
x2+y2+z2=194
x2z2+y2z2=194=4225?
Here x2 means x sq and y2 means y sq and z2=z sq