# IIM CAT Preparation Tips

IIM CAT Preparation Tips: November 2012

## Nov 29, 2012

### CAT Functions

Questions from functions have increased over the past few years. Given below is an interesting question from functions.

Question
Consider functions  f(x) = x2 + 2x,  g(x) = √(x +1) and h(x) = g(f(x)). What are the domain and range of h(x)?

Domain = ( - infinity, +infinity)
Range - [0, infinity]

h(x) = g(f(x)) = g(x2 + 2x) = Ö( x2 + 2x + 1) = Ö(x+1)2 = | x + 1|

This bit is very important, and often overlooked

sqrt(9) = 3, not +3
If x2 = 9, then x can be +3, but Sqrt(9) is only +3.

So, Sqrt(x2)= |x|, not +x, not + x

Domain of | x + 1| = ( -infinity, + infinity), x can take any value.
As far as the range is concerned, | x + 1| cannot be negative. So, range = [0, infinity)

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## Nov 23, 2012

### CAT Question - Logarithms

This is a reasonably tough question. But since the messrs conducting CAT seem to have upped the ante on Logarithm, this is probably par for the course

Question

log5x = a (This should be read as log X to the base 5 equals a)log20x = b
What is
logx10 ?

Correct Answer: (a + b)/2ab

Explanation

log5x = a
log20x = b
logx5 = 1/a

logx20 = 1/b
logx100/5 = 1/b
logx100 – logn5 = 1/b
2logx10 – 1/a = 1/b
logx10 = ½(1/a + 1/b)
logx10 = (a + b)/2ab

## Nov 19, 2012

### CAT Inequalities - Few questions

Have given three interesting questions from Inequalities.

Questions

1.  Find the range of x for which (x + 2) (x + 5) > 40.
2.  How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200
3.  Find the range of x where ||x - 3| - 4| > 3.

1.  x < -10 or x > 3
2.  9 different values
3.  (-infinity, -4) or (2,4) or ( 10, infinity)

1.  Find the range of x for which (x + 2) (x + 5) > 40.

There are two ways of trying this one. We can expand and simplify this algebraically. x^2 + 7x + 10 > 40 or
x^2 + 7x - 30 > 0
(x + 10) (x -3) > 0
The roots are -10 and +3.
=> x should lie outside the roots.

Now, what is this based on?
There is a simple thumb rule for solving quadratic inequality
For any quadratic inequality ax2 + bx + c < 0
Factorize it as a(x - p) ( x - q) < 0

Whenever a is greater than 0, the above inequality will hold good if x lies between p and q.
a(x - p) (x - q) will be greater than 0, whenever x does not lie between p and q. In other words x should lie in the range ( -infinity, p) or (q, infinity)

Now, coming back to the question (x+10) (x -3) > 0
Or,x < -10 or x > 3

Second method: 5 * 8 = 40, -8 * -5 = 40
So, if x + 2 > 5 this will hold good => x > 3
If x + 2 is less than -8 also, this will hold good => x < -10. The first method is far more robust.

2.  How many integer values of x satisfy the equation x( x + 2) (x + 4) (x + 6) < 200

To begin with - 0, -2, -4 and -6 work. These are the values for which the left-hand side goes to zero.

There are 4 terms in the product. If all 4 are positive or all 4 are negative the product will be positive
The product can be negative only if exactly 1 or exactly 3 are negative. When 1 or 3 terms are negative, the product is clearly less than 200.

When x = -1, one term is negative
When x = -5, three terms are negative

So, adding these two numbers also to the set of solutions {-6, -5, -4, -2, -1, 0} satisfy the inequality.

Beyond this it is just trial and error.

Let us try x = 1. Product is 1 * 3 *  5 * 7 = 105. This works
x = -7 gives the same product. So, that also works.

So, the solution set is now refined to  {-7, -6, -5, -4, -2, -1, 0, 1}

x = 2 => Product is 2 * 4 * 6 * 8 = 8 * 48. Not possible. Any x greater than 1 does not work
x = -8 is also not possible. Any value of x less than -7 does not work.

So, the solution set stays as {-7, -6, -5, -4, -2, -1, 0, 1}

The one missing value in this sequence is -3. When x = -3, product becomes -3 * -1 * 1 * 3. = 9. This also holds good.
So, values {-7, -6,-5, -4, -3, -2, -1, 0, 1} hold good. 9 different values satisfy this inequality

3.  Find the range of x where ||x - 3| - 4| > 3

If we have an inequality |y| > 3, this will be satisfied if => y > 3 or y < -3.

So, the above inequality simplifies to two inequalities

Inequality I: | x - 3| - 4 > 3
| x - 3| - 4 > 3 => | x - 3 | > 7
x - 3 > 7 or x - 3 < -7
Or, x > 10 or x < -4

Of, x lies outside of  -4 and 10. Or, x can lie in the range ( - infinity, -4) or ( 10, infinity)

Inequality II: |x -3| - 4 < -3
|x - 3 | - 4 < - 3
=> | x - 3 | < 1
=> -1 < x - 3 < 1 or x lies between 2 and 4

So, final solution is given by the range
( - infinity, -4) or (2,4) or ( 10, infinity)

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CAT weekend batches @ Chennai
Weekend batches @ Anna Nagar from 24th November, @ Velachery from 25th November. @ Mylapore from 2nd December.

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## Nov 10, 2012

### CAT 2013 - Simple Questions sitting under elaborate wording

Competitive exams often ask questions with a 'wrapper' around them. Its important to get to the right question quickly. Give the underlying questions for the following -

1. Give the smallest 4-digit number with an odd number of factors (easy one)

2. Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13, 420, ..........}, how many elements of Set S are integers?

3. f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such that we end up with a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2

Discussion:

The questions sitting underneath the above statements are as follows

1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is the smallest 4-digit perfect square?

2. Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13, 420, ..........}, how many elements of Set S are integers? => How many odd factors does 6300 have?

3. f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such that we end up with a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2? => How many 5-digit number exist that when divided by 9 leave a remainder of 2?

1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is the smallest 4-digit perfect square? 1024

2. Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13, 420, ..........}, how many elements of Set S are integers? => How many odd factors does 6300 have?

6300 = 2^2 * 3^2 * 5^2 * 7. Number of odd factors of this number = (2+1) * (2+1) * (1+1) = 18. For discussion on number of odd factors, look here

3. f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such that we end up with a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2? => How many 5-digit number exist that when divided by 9 leave a remainder of 2?

There are 90000 5-digit numbers. There will be 10000 numbers that leave a remainder of 2 on division by 9 within these. Answer = 10000.

g(n) mentioned above is identical to the remainder when a number is divided by 9. Once you pick that, this question become a sitter. Competitive exams are good at masking questions. As much as possible, learn from first principles. If one had thought about why the test of divisibility for 9 works, this bit would have been clear.

CAT 2013 Course offered by 2IIM @ ChennaiCourse handled by IIM Alumni, Batches @ Anna Nagar, Mylapore and Velachery.

Weekend batches @ Anna Nagar from 24th November, @ Mylapore from 17th November, @ Velachery from 18th November.