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MockaMania : Mocks on 14th Sept

with 27 comments


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!

Written by Implex

September 14, 2008 at 9:01 am

27 Responses

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  1. 1. A only 1
    2. E none of these
    3 B 64
    4 B 1

    tyro

    September 14, 2008 at 5:09 pm

  2. My take on 2 and 3..
    2. 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🙂

    Celebrating Life

    September 14, 2008 at 5:56 pm

  3. 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

    Celebrating Life

    September 14, 2008 at 6:06 pm

  4. Q 3 is such a simple problem, messed up that one as well. I am not sure about the second one though.

    tyro

    September 15, 2008 at 12:19 am

  5. 3)the only positions for the 4 ones are 2/3/4/5.
    Baaki 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

    September 15, 2008 at 11:28 am

  6. @Milind, I think or at least hope, that there must a better solution than mere substitution.

    Celebrating Life

    September 15, 2008 at 1:32 pm

  7. @ bhaskar
    I did exactly the same way
    as milind and i think it is good enough !!

    outtimed

    September 15, 2008 at 1:54 pm

  8. 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.

    Celebrating Life

    September 15, 2008 at 2:02 pm

  9. @Bhaskar
    Of 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)🙂

    milind

    September 15, 2008 at 3:56 pm

  10. 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.:)

    Celebrating Life

    September 16, 2008 at 7:01 am

  11. 5.Zero solutions. X can take only +values of 1 and 2 since, y can’t be negative or zero.

    Celebrating Life

    September 16, 2008 at 7:04 am

  12. 8.The first digit is always 1. So number of 3 digit numbers are 1.
    Number 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)

    Celebrating Life

    September 16, 2008 at 7:10 am

  13. 8)First(least) tri-1 number is 111
    for 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.

    milind

    September 16, 2008 at 7:23 am

  14. 9.64+32+4,64+4+2,32+2+1,32+16+2,16+8+1. So total 5 possibilities.Answer choice A)

    Celebrating Life

    September 16, 2008 at 7:24 am

  15. 6.Multiples of 4 from 1 to 25–>25 values
    Multiples 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)

    Celebrating Life

    September 16, 2008 at 7:34 am

  16. I guess only only 40 values for now..Still working:o

    Celebrating Life

    September 16, 2008 at 7:36 am

  17. 7.(8,9),(24,25),(48,49),(49,50)..Total 4 possibilities.Choice c)

    Celebrating Life

    September 16, 2008 at 7:39 am

  18. @Bhaskar
    Wont (27,28) and (44,45) be included too?

    milind

    September 16, 2008 at 7:49 am

  19. Bhaskar Question 5 you have solved wrong !

    outtimed

    September 16, 2008 at 7:52 am

  20. ya 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]..

    Celebrating Life

    September 16, 2008 at 8:42 am

  21. Q7 E
    Q8 E
    Q9 A

    tyro

    September 18, 2008 at 3:16 pm

  22. well done tyro !!!

    outtimed

    September 18, 2008 at 3:48 pm

  23. Q6 C 39

    tyro

    September 18, 2008 at 5:10 pm

  24. whats the answer to q.6 .
    I am getting E.41

    CATastrophe08

    October 2, 2008 at 9:07 pm

  25. answer is 39

    outtimed

    October 2, 2008 at 9:13 pm

  26. answers to Q5 : E)4
    Q7 : D)5 (8,9) (24,25) (27,28) (44,45) (48,49)

    are these correct. correct me, if not.

    yodha

    October 13, 2008 at 5:26 pm

  27. ok i got now,
    i did not count (49,50). hence ,E)

    yodha

    October 14, 2008 at 11:49 am


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