nimisha try to solve 2nd qs as follows:
take 2 as common 2(11+111+....49 times 1)/9 nowwe will will find series for remainder
11/9=2 111/9=3 u will find a series of remainders thn u can solve it....i coudlnt find the ans..as u typed worng 😐 now will get back to u wd solution for 2^2+22^2+......
and besides the actualt question i wanted to ask was'
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2.. remainder from 9
(earlier there was a typing error)..plz let me know if u solve this..
nimisha i got a series for remainders 4+7+0+1+1+0+7+4+0+4 and soo on upto 49terms...now add all as u got a series toh it wont to solve..u will get final ans after dividing the sum of reaminderts by 9 agian
is the remainder 6?
confirm please....i will post the method soon....m in a bit of hurry
yes remainder is 6
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2..
find digit sum
N=2^2 + 4^2+ 6^2+..............98^2
=2^2(1^2+2^2+......49^2)
=4 * 49 * 50 * 99/6
=4*49*25*33
N mod 9
=4*4*7*6 mod 9
= -2*-3 mod 9
=6 mod 9
remainder 9
find digit sum
N=2^2 + 4^2+ 6^2+..............98^2
=2^2(1^2+2^2+......49^2)
=4 * 49 * 50 * 99/6
=4*49*25*33
N mod 9
=4*4*7*6 mod 9
= -2*-3 mod 9
=6 mod 9
remainder 9
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2..
find digit sum
N=2^2 + 4^2+ 6^2+..............98^2
=2^2(1^2+2^2+......49^2)
=4 * 49 * 50 * 99/6
=4*49*25*33
N mod 9
=4*4*7*6 mod 9
= -2*-3 mod 9
=6 mod 9
remainder 9
m nt getitng ur solution 😐 it seem worng to me
N=2^2 + 4^2+ 6^2+..............98^2
=2^2(1^2+2^2+......49^2)
could u explain it how did u come to this???
taking 2^2 common.....
so is der ne alternate 4 dis solutn?
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2.. remainder from 9
Take 2^2 common. Remaining series will be-
1^2+11^2+111^2+.......
The divisibility by 9 depends on the sum of digits of the number. Hence,
1%9=1
11%9=2
111%9=3
|
111.....9times%9=0
111....10 times%9=1
and so on
So 49 numbers can be divided into 5 groups of 9 no each and 4 seperate terms.
Now try to find the remainnder when ssquare of it is divided by 9
1^2%9=1
11^2%9=4
111^2%9=0
1111^2%9=7
11111^2%9=7
Other remainders are 0,4 and 1
Hence sum of remainders for 49 Nos = 5*24+12=132
Hence remainder= 4*132 % 9 = 4*6%9 = 6
prince_chittu Saystaking 2^2 common.....
okie probably u have done typo 😐 somewhere coz ur getitng 9 rem while it should b 6...anyways my method is correct..and some had explain it in detail
2^2 + 22^2+ 222^2+ 2222^2 + .+ 222(49 times)^2..
find digit sum
See we can write the above equation as
N=2^2 + 4^2+ 6^2+..............98^2
now taking 2^2 common we get
=2^2(1^2+2^2+......49^2)
applying sigma n^2=n(n+1)(2n+1)/6 we get
as 4* (49 * 50 * 99)/6
we need to find the N mod 9
remainder 6
anybody , Q6 pg 127 (Block review tests, block 2)
2nd edition of the book.
p.s. this is my first post.. do tell me if i need to post the entire ques here..
So 49 numbers can be divided into 5 groups of 9 no each and 4 seperate terms.
Now try to find the remainnder when ssquare of it is divided by 9
1^2%9=1
11^2%9=4
111^2%9=0
1111^2%9=7
11111^2%9=7
Other remainders are 0,4 and 1
Hence sum of remainders for 49 Nos = 5*24+12=132
Hence remainder= 4*132 % 9 = 4*6%9 = 6
hey..whhy have we divided into ggroups of 9?? arent the remainders repeating after every 8 terms?? and later, why have you multiplied by 4??
nimisha_2009 Sayshey..whhy have we divided into ggroups of 9?? arent the remainders repeating after every 8 terms?? and later, why have you multiplied by 4??
We have divided into groups of 9 because remainder of nos whose sum of digits is 1,10,19,28,37 (when divided by 9) is same i.e. 1.
Similarly for nos whose sum of digits is 2,11,20,29,38 i.e 2
and so on for other nos.
Remember we have taken 2^2(4) common in starting. So, in order to compensate that we have multiplied it by 4.
@nidhisoni- The solution given by prince and shivam are also correct. The trick here is that the divisibility by 9 is dependent on sum of the digits. Think abt the solution given by them. Its absolutely correct and infact the better one :-).
Puys,I think our aim should be here to find a better solution to the problem.
@tarun.varshney- please post the whole question also. Few ppl like me dont have the Arun Sharma. So, we can't help you unless we get to know the problem 😃
hi all pgs!
i visit the pg almost every day n read cat 2008 thread religiously.i am happy to see arun sharma thread.though started prep very late n a first timer,i would like to ask u guys that r you all able to solve LOD1 N LOD 2 with ease?plz tell me while calculating %age Q how v calculate Q likethis?
when 60%of a numberA is added to another number B,b BECOMES 175% OF ITS PREVIOUS VALUE.then which is true-
a--A>B
b--B>A
c--either can be true depending upon value of Aand B.
d--nothing can be said
just let me know how the thought should go?
thanks
Hi...anjali...
this is wat i thout of....
0.6A + B =1.75B
=> 0.6A = 0.75B
=> A =(0.75 / 0.6)B
Thus A is greater than B
I was solving alligation in arun sarma....
had pblm in solving 1 qs in alligation method...though cud solve in orthodox fashion.....
A mixture of alchohol and water has 10% water.....
Hw much more water will i add separately to make the conc of water as 12.5% ??BY METHOD OF ALLIGATION ONLY...plss...
Hi...anjali...
this is wat i thout of....
0.6A + B =1.75B
=> 0.6A = 0.75B
=> A =(0.75 / 0.6)B
Thus A is greater than B
you needn't even go that far..
try to do such simple problems mentally as much as possible.
follow the logic that there is an increase of ?? 75% in b and that is caused by 60% of A so clearly 75% of A would be an even bigger number so A is bigger.
I was solving alligation in arun sarma....
had pblm in solving 1 qs in alligation method...though cud solve in orthodox fashion.....
A mixture of alchohol and water has 10% water.....
Hw much more water will i add separately to make the conc of water as 12.5% ??BY METHOD OF ALLIGATION ONLY...plss...
Don't insist on using something just for its sake when you don't follow the logic.
use the method u r using..
but just for your info by alligation you would do it like:
water:alcohol
10:90(1:9) to 12.5:87.5(1:7)
difference in water=1/7-1/9=2/63
so water you would add=2/63*90=2.85 i think..somewhere there.
btw this is a very strong drink you are making!!! lol
generally its alcohol thats 12.5% by volume in drinks. check your heineken can.
there you go some general knowledge for you too!! who knows- might help in your interview. 😛
this is a qn from ch-1 number systems page 20.
Q.16- which of the following represents the largest 4 digit number which can be added to 7249 in order to make the derived number divisible by each of 12, 14, 21, 33 and 54?
a) 9123
b) 9383
c) 8727
d) none of these
i dint get the method used in the book to solve this question.
can we come up with an alternative?
A simple one, but my answer doesnt match with the one provided in the options
Ch-1 , num sys, page 21
Q24- The LCM of 4.5 , 0.0009 and 0.18 is?
a) 4.5
b) 45
c) 0.225
d) 2.25