visionIIM-ACL SaysMy take : remainder is 1.
My Take...1 too....
My Take...1 too....
hey guys... got an interesting one...
There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.
Cheers!
Nice Question...
let a,b,c,d,e be 5 numbers
here 4(a+b+c+d+e) = 120
(a+b+c+d+e) = 30
let say a+b = 23...sum of max. two numbers
and c+d = 0..sum of min two numbers..
so e = 7...
now...e+a could be any of the numbers from above question....be it 15,8,7,3,0....,19
so value of a could be (8,1,-4,-7,9,11,4,16,12)
from this given set...7 would be added to only 4 numbers...
and from these set of 8 values...rest of the numbers can be extracted...
also two numbers be such that sum of min of them is 0..and max of them is 23...
only -4,4,11,12..satisfy the given conditions...
hence second highest number is 4...
hey guys... got an interesting one...
There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.
Cheers!
ans are -4,4,12,11,7 --the trik is with 0 ( it can come wen two numbers are same but of diff magnitude)
ans 11
Nice Question...
let a,b,c,d,e be 5 numbers
here 4(a+b+c+d+e) = 120
(a+b+c+d+e) = 30
let say a+b = 23...sum of max. two numbers
and c+d = 0..sum of min two numbers..
so e = 7...
now...e+a could be any of the numbers from above question....be it 15,8,7,3,0....,19
so value of a could be (8,1,-4,-7,9,11,4,16,12)
from this given set...7 would be added to only 4 numbers...
and from these set of 8 values...rest of the numbers can be extracted...
also two numbers be such that sum of min of them is 0..and max of them is 23...
only -4,4,11,12..satisfy the given conditions...
hence second highest number is 4...
-4,4,11,7,12...
11 ans
i m geting 7 ..
for max value a+b+c+d+e=30 ..we need distinct numbers..
in case of same numbers it can be a=6,b=6,c=6,d=6,e=6..
for distinct we can change it by 4,5,6,7,8..making 7 as the second highest number......
plz crosscheck with the qstn--accrdn to qstn
"yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer."
in tht case u will not be getting zero
hey guys... got an interesting one...
There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.
Cheers!
Answer should be 11.
The numbers are -4,4,7,11,12.
Not clear.. !!
can you explain it further.. ?
sorry i made a calculation mistake
but the process was more o0r less correct,
see let us take the numbers as c1>c2>c3>c4>c5
now the sum given are combination of all this numbers taken two at a time
like c1+c2,c1+c3,c2+c3 etc
so if we add them up we will get all of them 4 times (as from the 5 numbers if we take 2 at a time we will get 10 combinations and all the numbers will be present in 4 pairs i.e. once with each of the other numbers)
by adding we get
4(c1+c2+c3+c4+c5)=120
or, c1+c2+c3+c4+c5=30
now its clear that ci+c2=23,c1+c3=19 and also c4+c5=0
surely so if we calculate we get c3=7,c1=12,c2=11
answer c2=11
i hope it helps
divishth SaysMy Take...1 too....
has to b 1
some1 please answer my question posted in the previous page
hey friends i have few doubts..i will post some of them 2day. any help will be appreciated ...thanks in advance!!!
1)Find the number of negative integral solutions of
2)Amrita hosted a birthday party and invited all her friends and asked them
to invite their friends.There are n people in the party.Only Satyam is not
known to Amrita.Each pair that does not include Amrita or Satyam has exactly
2 common friends.Also,Satyam knows everyone except Amrita.If only
2 friends can dance at a time,how many dance numbers will be there at the
party?
a)3n 7 b)2n + 5 c)4n 3 d)none of these.
it the ans a? 3n 7
PS-Other questions are not clear....
Well I am confused about the question.
I am getting the following ..
(t1+t2)+7*(t1+t2)* (1+8+8*8+8*8*8+8*8*8*8+--- inf)
here the common ration is 8
since 8>1 ..the series is divergent .. then how do we get any answer for this?
We cant calculate the sum of infinite GP series when common ratio is >=1
correction invited..
Sum of infinite series S=a/(1-r).
What is the restriction for 'r' here, Will r
Sum of infinite series S=a/(1-r).
What is the restriction for 'r' here, Will r
Yes..the summation of an infinite GP is possible only if r
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
E(17) = 16.
(38^16!)^1777 mod17 = (38^16k)^1777 mod 17 = 1
some questions from arun sharma...
1.There is a natural no tht become equal to the squire of a natural no when 100 is added to it and squire of another no when 169 is added to it. find the no.
2.define a no: k such thtit is the sum of the squires of the first m natural nos where m
wrong post
0 is the answer
since, 0 + 100 = 10^2
0 + 169 = 13^2
vc , 0 is a whole number and not natural number
0 is the answer
since, 0 + 100 = 10^2
0 + 169 = 13^2
but, question says natural number..
0 is not a natural number!!
some questions from arun sharma...
1.There is a natural no tht become equal to the squire of a natural no when 100 is added to it and squire of another no when 169 is added to it. find the no.
.
the number is 1056
naga25french Saysvc , 0 is a whole number and not natural number
ya got that dint read the question properly snipped it
some questions from arun sharma...
1.There is a natural no tht become equal to the squire of a natural no when 100 is added to it and squire of another no when 169 is added to it. find the no.
2.define a no: k such thtit is the sum of the squires of the first m natural nos where m
The no. is 1056. Here's how I did it:
Let x^2 = n + 100
and y^2 = n + 169, where n is the number which we have to find
Now, y^2 - x^2 = 69
(y-x)(y+x) = 69
Now, 69 can be got either by 1*69 or 3*23.
Using 1*69, let's assume (y+x) = 69 and (y-x) = 1. Then, we can get x = 34 and y = 35
Thus, n + 100 = 34^2 = 1156
n + 169 = 35^2 = 1225
Either way, we can find out that n = 1056 😃