@vbhvgupta said:If the ratio of Harmonic mean to the gm of 2 nos is 12:13, find the ratio of the no.4/9 or 9/4 2/3 or 3/22/5 or 5/2 3/4 or 4/33/4 or 4/5
2ab/(a+b)/rt(ab)=12/13
put options
4/9 or 9/4 satisfies
if x, y,z are in gp and a^x and b^y and c^z are equal thn a b c are in???
hello puys.......i have scored 55% in my graduation which i completed in 2012......i gave cat this year and got 78 percentile.... i plan to give cat again next year.........will the IIM's not give me a call even if i get the requisite percentile as my graduation marks are below 60%?????? please reply
Find the sum of all possible whole no divisors of 720??
concept??
@cattaker2012 said:hello puys.......i have scored 55% in my graduation which i completed in 2012......i gave cat this year and got 78 percentile.... i plan to give cat again next year.........will the IIM's not give me a call even if i get the requisite percentile as my graduation marks are below 60%?????? please reply
Post here.!! :)
@vbhvgupta said:Find the all possible whole no divisors of 720??concept??
720=2^4.3^2.5
So it has 5.3.2 = 30 factors
So it has 5.3.2 = 30 factors
@catahead said:720=2^4.3^2.5So it has 5.3.2 = 30 factors
@warrior2012 said:is it 30?
how to find sum of these?
@vbhvgupta said:Find the sum of all possible whole no divisors of 720??concept??
720=2^4*3^2*5=5*3*2=30?
@vbhvgupta said:if x, y,z are in gp and a^x and b^y and c^z are equal thn a b c are in???
a^x=b^y=c^z = K(say) so a,b,c are in GP
Now xz=y^2 => logaK . logcK = (logbK)^2 => loga.logc = (logb)^2
So loga,logb,logc are in GP
Plz confirm if the questin was to find for a,b,c or loga,logb,logc
PS: Edited
Now xz=y^2 => logaK . logcK = (logbK)^2 => loga.logc = (logb)^2
So loga,logb,logc are in GP
Plz confirm if the questin was to find for a,b,c or loga,logb,logc
PS: Edited
@vbhvgupta said:how to find sum of these?
There are some factor theories that you need to know
Read some here http://quantexpert.co.in/qenotes/quantnotes/numbers/47-factor-theories.html
Read some here http://quantexpert.co.in/qenotes/quantnotes/numbers/47-factor-theories.html
@catahead said:a^x=b^y=c^z = K(say) so a,b,c are in GPSee ac = k^(1/x).k^(1/z) = k^(1/xz) = k^(1/y^2) = b^2
k^(1/x)*k(1/z)=k(1/x+1/z) hoga na??
@vbhvgupta said:how to find sum of these?
the sum i got is 2418 using N = { (ap+1 - 1)(bq+1 - 1)(cr+1 - 1) } / { (a-1)(b-1)(c-1) }
@vbhvgupta said:if x, y,z are in gp and a^x and b^y and c^z are equal thn a b c are in???
getting loga,logb,logc in GP??
@vbhvgupta said:if x, y,z are in gp and a^x and b^y and c^z are equal thn a b c are in???
niether i am getting it AP, GP or HP
whats the OA
whats the OA
Hi,
This type of question always comes ! Still a little confusing for me.
Q - Find the number of numbers between 300 and 400 ( both included ), that are NOT divisible by 2,3,4 and 5.
A - 26.
I am using the counting principle
( last - First ) / difference + 1 (im little confused and not sure how to use )
@anytomdickandhary Sir jee Is concept pe bhi thoda prakash daliye...
@ananyboss said:Hi,
Nos. divisible by 2 = 51.
Nos. divisible by 3 = 34, but remove even multiples so we get 17.
Nos. divisible by 5 = 21, but remove multiple of 3, so we get 7.
Nos. divisible by 3 = 34, but remove even multiples so we get 17.
Nos. divisible by 5 = 21, but remove multiple of 3, so we get 7.
we are not considering 4 as they are included in 2
total nos. divisible by 2,3,4,5 = 75.
total nos. not divisible by 2,3,4,5 = 26. ....Is it cleared
total nos. divisible by 2,3,4,5 = 75.
total nos. not divisible by 2,3,4,5 = 26. ....Is it cleared
@ananyboss said:Hi,This type of question always comes ! Still a little confusing for me.Q - Find the number of numbers between 300 and 400 ( both included ), that are NOT divisible by 2,3,4 and 5.A - 26.I am using the counting principle( last - First ) / difference + 1 (im little confused and not sure how to use )@anytomdickandhary Sir jee Is concept pe bhi thoda prakash daliye...
U need to erase all multiples of 2,3,5 (4 is not needed as when u eliminate multiples of 2 , multiples of 4 are also elimintaed)
Now, during this process , u eliminated the multiples of 6,10,15 more than once.
So you need to calculate them and add.
and lastly, you again added the multiples of 30 more than once.
This is little iterative, but can be done with little patience
Now, during this process , u eliminated the multiples of 6,10,15 more than once.
So you need to calculate them and add.
and lastly, you again added the multiples of 30 more than once.
This is little iterative, but can be done with little patience
@Logrhythm said:to find out days against dates....use zeller's rule10 + [129/5] + 89 + [89/4] + [19/4] - 38 = 10 + 25 + 89 + 22 + 4 - 38 = 112 112 is 7k form...hence it was a sunday..i have attached a txt file explaining zeller's rule with this post...
plz calculate it for 19 jan 1964 .. dnt know why i am getting it wrong ..