Ims Simcat 9( Some good Problems from Quant)
The paper had cat2007 pattern, only difference was it was +2, and -0.5
1) If a^k has k^4 divisors, where k is a natural number, then which of the following is true?
I a=k=1 II a^k>=210 III a^k>=2^k; k>=2
A) Only I B) only II C) Only III D) I or II E) I or III
2) f(n)=2g(n)+f(-n); for all non-zero integers n
g(n)= n*g(n-1) for all n>0 and g(0)=1
then Find g(-10)+g(-9)+....+g(0)+...g(9)+ g(10)
A) 10!+1 B)2*10!+1 C) 2*10! D) 1 E) None of these
3) Distinct two digit numbers are written one after the other to form a six-digit number. How many six-digit numbers thus formed have four consecutive 1s in them?
A) 90 B) 64 C) 65 D) 56 E )72
4) x=3m-1 and y=5n-1 where m,n,x,y are natural numbers less than 16. Find the number of pairs (m,n) satisfying the equations x^2=2y^2-7
A) 0 B) 1 C) 2 D) 3 E ) 4
5) FInd the number of real solutions of the system of equations
y=|x-1|+|x-2| and y+1=x(3-x)
A) 0 B) 1 C) 2 D ) 3 E) 4
6) A natural number ( greater than one) is called squareful number if its prime factorisation contains at least one square. How many squareful numbers below 101 are there
A) 63 B) 61 C) 39 D) 67 E) 41
7) How many pair of consecutive natural numbers less than 51 are squareful numbers as defined above
A) 2 B) 3 C) 4 D) 5 E ) 6
8) A binary number is called tri-one if it has exactly three 1s. If all tri-ones are arranged in ascending order, what is the rank of the least 8 digit tri-one number?
A) 35 B) 32 C) 34 D) 40 E) 36
9) How many tri-ones as defined above , less than 110 in decimal, when converted to decimal is divsible by 5 in base 10?
A) 5 B) 6 C) 7 D) 8 E )10
updated!
1. A only 1
ReplyDelete2. E none of these
3 B 64
4 B 1
My take on 2 and 3..
ReplyDelete2. Now consider f(-n) = 2g(-n)+f(n)..Now g(-n)=-g(n)..So the only remaining term of the summation is g(0) = 1..Answer choice D)
3.Now when the numbers are distinct, the only possibility of having 4 consecutive 1's is having 2,3,4,5 positions as 1..Now the 1st position can have any number from 2 to 9(Both inclusive)..And the last digit can have digits from 0 to 9 except 1. So totally 9*8 = 72 possibilities, answer choice E).
Good to see this blog resurrected :)
1.answer choice 1)..K>1 is not valid since for any k>1..The number of factors would be of the form (kx1+1)(kx2+1)..which can never divide k^4
ReplyDeleteQ 3 is such a simple problem, messed up that one as well. I am not sure about the second one though.
ReplyDelete3)the only positions for the 4 ones are 2/3/4/5.
ReplyDeleteBaaki bachi 1st aur 6th
First can have 2/3/4/5/6/7/8/9=8
sixth can have 2/3/4/5/6/7/8/9/0=9
so, 72 values
4)y=5n-1
so, possible values of y are 4/9/14
on putting these in eqn x^2=2y^2-7 we get only one that fits in such that root(x) is natural
So, x=5=3m-1
(m,n)=(2,1)
@Milind, I think or at least hope, that there must a better solution than mere substitution.
ReplyDelete@ bhaskar
ReplyDeleteI did exactly the same way
as milind and i think it is good enough !!
Ya. Three is not a big deal. I did the same, but was waiting if someone would better that in a way by generalizing the equation.
ReplyDelete@Bhaskar
ReplyDeleteOf late I have realized the importance of substitution in mocks :)
Till now I was gng by the conventional approach. But having a re-look at the papers, I feel more than 50% of the problems can be done that way (and pretty fast) :)
Ya subst works..But not always..So at times you end up wasting a lot of time without much headway..So a balance with constraints is required b/w conventional and subst approaches.:)
ReplyDelete5.Zero solutions. X can take only +values of 1 and 2 since, y can't be negative or zero.
ReplyDelete8.The first digit is always 1. So number of 3 digit numbers are 1.
ReplyDeleteNumber of 4 digit numbers are 3
Number of 5 digit numbers are 6
Number of 6 digit numbers are 10
Number of 7 digit numbers are 15
So the rank of least 8 digit number is 15+10+6+3+1+1 that is 36th
Answer choice E)
9.64+32+4,64+4+2,32+2+1,32+16+2,16+8+1. So total 5 possibilities.Answer choice A)
ReplyDelete8)First(least) tri-1 number is 111
ReplyDeletefor 4 digit nos. first digit will be 1. remaining 3 can have 3C1 perm(arrange a single 0 and 2 ones)=3
5 digit nos. 1st digit is 1. Remaining have 2 accomodate 2 1s and 2 0s. So, 4C2=6
5C2=10, 6C2=15. In all 34 numbers of upto 7 digits. So the no. with 8 digits is ranked 35
9)we have tri ones only .. so the number is of the form 2^x+2^y+2^z=5k
Now, there can be 2 ways.
First is when 1 is at the units digit. In such case 2^y+2^z=K4
2 such numbers 25/35
Second is when 2^x is not 0:)
numbers are 50/70/100
So, in all 5 nos.
This could be done better with a more general approach. Trying to form that.
6.Multiples of 4 from 1 to 25-->25 values
ReplyDeleteMultiples of 9 from 1 to 11 except 4, 10 values
Multiples of 25 from 1 to 3, 3 values
Multiples of 49 1 and 2, 2 values
and 1
So totally 41 values option E)
I guess only only 40 values for now..Still working:o
ReplyDelete7.(8,9),(24,25),(48,49),(49,50)..Total 4 possibilities.Choice c)
ReplyDelete@Bhaskar
ReplyDeleteWont (27,28) and (44,45) be included too?
Bhaskar Question 5 you have solved wrong !
ReplyDeleteya question 5 is two solutions 1 and 2..Why did I type zero solutions[:o]..And ya (27,28) I noted and then missed typing..Didn't count (44,45) at all[:o]..
ReplyDeleteQ7 E
ReplyDeleteQ8 E
Q9 A
well done tyro !!!
ReplyDeleteQ6 C 39
ReplyDeletewhats the answer to q.6 .
ReplyDeleteI am getting E.41
answer is 39
ReplyDeleteanswers to Q5 : E)4
ReplyDeleteQ7 : D)5 (8,9) (24,25) (27,28) (44,45) (48,49)
are these correct. correct me, if not.
ok i got now,
ReplyDeletei did not count (49,50). hence ,E)