## MockaMania : Mocks on 14th Sept

*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

2. E none of these

3 B 64

4 B 1

tyroSeptember 14, 2008 at 5:09 pm

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 LifeSeptember 14, 2008 at 5:56 pm

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 LifeSeptember 14, 2008 at 6:06 pm

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

tyroSeptember 15, 2008 at 12:19 am

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)

milindSeptember 15, 2008 at 11:28 am

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

Celebrating LifeSeptember 15, 2008 at 1:32 pm

@ bhaskar

I did exactly the same way

as milind and i think it is good enough !!

outtimedSeptember 15, 2008 at 1:54 pm

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 LifeSeptember 15, 2008 at 2:02 pm

@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) 🙂

milindSeptember 15, 2008 at 3:56 pm

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 LifeSeptember 16, 2008 at 7:01 am

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

Celebrating LifeSeptember 16, 2008 at 7:04 am

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 LifeSeptember 16, 2008 at 7:10 am

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.

milindSeptember 16, 2008 at 7:23 am

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

Celebrating LifeSeptember 16, 2008 at 7:24 am

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 LifeSeptember 16, 2008 at 7:34 am

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

Celebrating LifeSeptember 16, 2008 at 7:36 am

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

Celebrating LifeSeptember 16, 2008 at 7:39 am

@Bhaskar

Wont (27,28) and (44,45) be included too?

milindSeptember 16, 2008 at 7:49 am

Bhaskar Question 5 you have solved wrong !

outtimedSeptember 16, 2008 at 7:52 am

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 LifeSeptember 16, 2008 at 8:42 am

Q7 E

Q8 E

Q9 A

tyroSeptember 18, 2008 at 3:16 pm

well done tyro !!!

outtimedSeptember 18, 2008 at 3:48 pm

Q6 C 39

tyroSeptember 18, 2008 at 5:10 pm

whats the answer to q.6 .

I am getting E.41

CATastrophe08October 2, 2008 at 9:07 pm

answer is 39

outtimedOctober 2, 2008 at 9:13 pm

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.

yodhaOctober 13, 2008 at 5:26 pm

ok i got now,

i did not count (49,50). hence ,E)

yodhaOctober 14, 2008 at 11:49 am