The Answer key is as
1) C 2) E 3) B 4) A 5) E 6) D 7) B 8) E 9) C 10) E
Question 1 is easy just use a-1/a= 1 and 1/a-a=1 will give two roots sum them
Question 2 10^j-1o^i=10^i(10^(j-1)-1)
1001=7.11.13=10^3+1
so clearly 10^3+1 divides 10^6-1 and therefore 10^6k-1
hence j-i=6k, k is a natural number
applying other constraints we get option e
Question 3) 1+2+3..30=30.31/2=31.15=465
[465/2]=232 now suppose a subset A of S doe not have sum more than 232 then A' must have sum more than 232 hence 1/2 of the subsets of S will have sum more than 232
so 2^30/2=2^29
question 4 use the concept of reflection we will get min distance sum as 5root(2)
Question 5) Function is not correctly defined as 0 is not a natural number
so option e)
Question 6)
let the radius be r
and the point of tangency be P and Q and triangle be ABC. P lies on AB and Q on BC
let AP = m and BQ = n
m^2 = 15^2-x^2
n^2 = 20^2-x^2
m = 9 and n = 16 x =12 arcPQ = 6pi
question 7) see n numbers product is n and sum is zero
if n is odd then sum can't be zero
similalry check for other cases it will easily come out n =4k
question 8 toughest problem of the test
let g(n)=p(n)-n
then g(17)=-7=g(24)
let a,b be integers such that p(n)=n+3
then a-17 divides g(a)-g(17)=3-(-7)=10
similarly for 24
hence we find a-17 and a-24 both divide 10 this means k=a-17 and k+7 both divide 10
this means k=-5 or-2
a=19 b=22
hence ab=418
question 9 can be easily done
question 10) tricky enough problem
look for a series and solve it you will get E
Didn't expect such a disaster of a result!!!! But I enjoyed doing these problems. More I do, more will I learn.
ReplyDeleteIn Q 2 took j-i as 6 rather than 6k.
ReplyDeleteKeep working, Problems are simpler than they appear to be !!
ReplyDeleteHow is Q4 to be done using the concept of reflection?
ReplyDeleteCould you please explain Q1 in a little detail
ReplyDeleteI tried much but didn't understand your approach at all
why u did a-l/a ??