@krum said:yaar ye sab to theek hai, J ko kaise B ke sath laga diya?
B ka shaadi ni hua tho.. L and F kaha se ayengee ? isliye marry older ones fst
@techsurge said:@krum if you put J with M how many sons are there who are married ??
B is the father of F who is his only married child, F's child can be married
@catter2011 said:B ka shaadi ni hua tho.. L and F kaha se ayengee ? isliye marry older ones fst
dadi unki swarg sidhar gayi hogi
chodo yaar, ye dusron ke family ke ladfde mien nai padna
chodo yaar, ye dusron ke family ke ladfde mien nai padna
@catter2011 said:tu avi ni samjha.. L is B's married chid, whoz unmarried
Got it..what a question!!!!!
The savings of Ramesh and Varun are in the ratio 3 : 2, while their expenditures are in the ratio 3 :4 in the same order. Also the ratio of their incomes in the given order is 4 : 5. What is the ratio of saving of Ramesh to his expenditure?
here's an easy one :
Find all positive integers n for which 1 + 2^n + 3^n + 4^n is divisible by 5.
| attitude is more important than facts |
@surajmenonv said:here's an easy one :Find all positive integers n for which 1 + 2^n + 3^n + 4^n is divisible by 5.| attitude is more important than facts |
all odd natural numbers
@catter2011 said:The savings of Ramesh and Varun are in the ratio 3 : 2, while their expenditures are in the ratio 3 :4 in the same order. Also the ratio of their incomes in the given order is 4 : 5. What is the ratio of saving of Ramesh to his expenditure?
is it 6:7??????
@catter2011 said:The savings of Ramesh and Varun are in the ratio 3 : 2, while their expenditures are in the ratio 3 :4 in the same order. Also the ratio of their incomes in the given order is 4 : 5. What is the ratio of saving of Ramesh to his expenditure?
ans : 1/7
@catter2011 said:The savings of Ramesh and Varun are in the ratio 3 : 2, while their expenditures are in the ratio 3 :4 in the same order. Also the ratio of their incomes in the given order is 4 : 5. What is the ratio of saving of Ramesh to his expenditure?
(4i-3e)/(5i-4e)=3/2
=>8i-6e=15i-12e
=>7i=6e
=>i/e=6/7
4i-3e=3s
=>24/7e-3e=3s
=>3/7e=3s
=>s/e=1/7
=>8i-6e=15i-12e
=>7i=6e
=>i/e=6/7
4i-3e=3s
=>24/7e-3e=3s
=>3/7e=3s
=>s/e=1/7
r v
savings 3s 2s
exp 3e 4e
income 4i 5i
now income = exp + sav
so,
4i = 3s + 3e
5i = 2s + 4e
solve to get s/e = 1/7
| attitude is more important than facts |
@krum said:all odd natural numbers
i am afraid not....think again
infact try for n = 2... its valid
| attitude is more important than facts |
@catter2011 said:@19rsb hmm.. donno the answer.. i am getting 1:7 , post ur approach @19rsb
yaa buddy it is 1/7....some calc mistake.....lost in d world of mistakes..but trying hard 2 come out of it........
@surajmenonv said:i am afraid not....think againinfact try for n = 2... its valid| attitude is more important than facts |
(1+2^n+3^n+4^n)mod5
==>(1+(2)^n+(-2)^n+(-1)^n)mod5
1+(2)^n+(-2)^n+(-1)^n =0 ; n=odd
1+(2)^n+(-2)^n+(-1)^n
==>2+2*2^n ; n=even
==>2+2*2^n=5k
as sum of two even numbers is always even, we need numbers of the form 10k
==>2*2^n=__8
==>2^n=__4
==>n=2,6,10,14,....
including all the odd numbers , we get all numbers except multiples of 4
==>(1+(2)^n+(-2)^n+(-1)^n)mod5
1+(2)^n+(-2)^n+(-1)^n =0 ; n=odd
1+(2)^n+(-2)^n+(-1)^n
==>2+2*2^n ; n=even
==>2+2*2^n=5k
as sum of two even numbers is always even, we need numbers of the form 10k
==>2*2^n=__8
==>2^n=__4
==>n=2,6,10,14,....
including all the odd numbers , we get all numbers except multiples of 4
@surajmenonv said:i am afraid not....think againinfact try for n = 2... its valid| attitude is more important than facts |
all natural no. except multiple of 4?
@krum said:all odd+ 2,6,10...
ya thats correct....in simpler words all nos which are not multiles of 4.
post ur approach too for others help
| attitude is more important than facts |
In a group of 130 players numbered 1 to 130, all even numbered players choose Cricket, whose numbers are divisible by 5 choose Football and those whose numbers are divisible by 7 choose Hockey. How many choose none of the three sports?
@catter2011 said:In a group of 130 players numbered 1 to 130, all even numbered players choose Cricket, whose numbers are divisible by 5 choose Football and those whose numbers are divisible by 7 choose Hockey. How many choose none of the three sports?
45?
EDITED
@catter2011 said:In a group of 130 players numbered 1 to 130, all even numbered players choose Cricket, whose numbers are divisible by 5 choose Football and those whose numbers are divisible by 7 choose Hockey. How many choose none of the three sports?
cricket - 130/2=65
football - 130/5=26
hockey - 130/7=18
cricket + football - 13
football + hockey - 3
cricket + hockey - 9
all 3=1
none=130-(65+26+18-13-3-9+1) = 45
football - 130/5=26
hockey - 130/7=18
cricket + football - 13
football + hockey - 3
cricket + hockey - 9
all 3=1
none=130-(65+26+18-13-3-9+1) = 45