## MockaMania Oct 5 2008

**Here are a few question from IMS simcat 11**

**Question 1) For what value of n>0 does the following pair of equations yield exactly three solutions for y?**

**|x|=|y|**

**x=y^2+n(n+3)-4**

**1) 0 2) 1 3) 4 5) no unique vale 6) none of these.**

**Question 2**

**The sum of 2k+1 consecutive natural numbers is 2n such that the sum of first k+1 natural numbers among these (2k+1) numbers is same as next k numbers. Which of the following cannot be the 7th least natural number among the 2k+1 natural numbers, if k>=3**

**1) 231 2) 175 3) 535 4) 325 5) none of these
**

**Question 3**

**if x=1+1/(x+1/(1+1/(x+…))) ; then which of the following bet represents x**

**1) 1<x<2 2) 1.5<x<2 3) 1<x<1.5 4) 1<x<1.2 5) 0<x<1**

**Question 4**

**let a(n) be q sequence such that n is an inetger and n>=1
**

**a(n)-a(n+1)=a(n+4)-a(n+5), find a(80)-a(84) if a(75)=103 and a(83)=205**

**1)- 51 2) -36 ) -17 4) 75 5) cannot be determined.**

**Question 5**

**Maya has six indentical pots, which she is planning to arrange in a straight line in her showcase. before that each of these pots is to be colored either red or yellow or green or blue, such that at least one pot is coloured with each of the four colours. In how many different ways can she arrange the pots in the showcase so that now two pots of the same colour are adjacent?**

**1) 120 2) 84 3) 840 4) 600 5) 936**

**Question 6**

**The second rightmost digit of (102)^33 is **

**1) 8 2) 2 3) 4 4) 6 5) none of these**

**7) A number is said to be crazy number is the product of the digits is equal to products of the distinct primes in its prime factorisation. How many crazy numbers less than 100 have less than 3 distinct prime factors in their prime factorisation?**

**1) 4 2_ 5 3)8 4) 6 5) none of these**

More to follow

Answer choice a)-51

Now in the equation keep substituting from a75 to a79 and add up all the equations. We get a79=(a75+a83)/2. Now its easy , a75-a79=a79-a83=a80-a84

Celebrating LifeOctober 6, 2008 at 5:19 am

The above solution was for Q.4

1)Answer choice B

Now consider two cases x=y and x=-y..Now in these two cases, there should be a common root.

So subtracting the equations, we get y=0 so n=1 or n=-4(not considered as its <0)

Celebrating LifeOctober 6, 2008 at 5:27 am

3.Answer choice 3)..Substituting the continuity equation by x and solving, we get x^3-x^2-1=0. So If we check for 1.2 it’s -ve and for 1.5, it’s positive..so the best fit would be 1<x<1.5

Celebrating LifeOctober 6, 2008 at 5:36 am

2. This is the easiest of lot if you can go overcome the initial hiccups..Now equating the sum of the first ‘k+1’ terms with the next ‘k’ terms, we get x=k^2 where x is the first term of the series..Now given the choices for the 7th term so the answer choice-6 should be a perfect square..Answer choice 4)

Celebrating LifeOctober 6, 2008 at 5:46 am

all of them correct 🙂

outtimedOctober 6, 2008 at 10:05 am

1.Took it as the equation of a parabola, equating to the value (0,0) of vertex .

we get the equation

n^2+3n-4=0

n=1

option 2

2. option 4

325

Q3 (3)

x^3-x^2-1=0

value would be just below 1.5

Q4.

AP

-4d=-4*51/4= -51

option 1

tyroOctober 6, 2008 at 11:45 am

Q 6

none of these 9

tyroOctober 6, 2008 at 4:40 pm

yupp right tyro, you are on a roll

outtimedOctober 6, 2008 at 5:07 pm

Question5: 936

riyaOctober 6, 2008 at 6:02 pm

Q.5

Now we can have 2 cases

1. where 3 are identical and 3 are of different colors..

2.2of one colour,2 of another another 1 each of 2 different colors

1st case: consider the positions as 123456

so the 3 identical can take positions(1,3,5)(1,3,6)(1,4,6)(2,4,6)

so total number of permutations in this case is 4*4c1*3! = 96

Now 2nd case is a bit bigger one. But lets try to limit them..lets consider the combination of the positions of the colors..when the 1st pair can take (1,3) the other pair can take (2,4)(2,5)(2,6)(4,6)..Now similarly there are 4 pairs each for (1,4)(1,5) and (1,6). So totally 16*2 combinations..So 32*4c2*2!= 384

Now totally 384+96 = 470. It’s 840/2..wheres the flaw?

Celebrating LifeOctober 6, 2008 at 6:37 pm

7.65,56,67,76. only 4.Answer choice 1)

Celebrating LifeOctober 6, 2008 at 6:47 pm

In the 10th..pairs should also be counted for (2,4)(2,5)(2,6)(not involving1) and also for (3,5) (not involving 1 and 2)..so total pairs are 16+6 = 22 so now for 2nd case, the possible combinations are 22*2*4c2*2 = 528. so, the total number is 528+96 = 624..Shall see if I can better this.

Celebrating LifeOctober 6, 2008 at 7:01 pm