CAT Permutation Combination and Functions

Question

How many functions can be defined from Set A -- {1, 2, 3, 4} to Set B = {a, b, c, d} that are neither one-one nor onto? 

Explanation

To start with, if you do not know the meaning of one-one or onto, look these up. For good measure know the meanings of the terms surjective, injective etc also. CAT tests these terms.

Let us start by answering a far simpler question. How many functions are possible from Set A to Set B. This is equal to 4^4 = 256. 

Note that any function from Set A to Set B that is one-one will also be onto and vice versa. How? Why? - Think about this. Remember, tutors can only take the horse to the pond. :-)

So, we need to subtract only those functions that are one-one AND onto. Or, effectively we have to eliminate those functions where {1, 2, 3, 4} are mapped to {a, b, c, d} such that each is mapped to a different element. This is effectively same as the number of ways of rearranging 4 elements. Or, number of ways of doing this is 4! .

So, total number of functions that are neither one-one nor onto = 256 - 24 = 232. 

CAT Permutation and Combinations

Question

If we list all the words that can be formed by rearranging the letters of the word SLEEPLESS in alphabetical order, what would be the rank of SLEEPLESS?

Explanation

First let us think about the number of possible rearrangements. SLEEPLESS can be rearranged in 9!/ (2! 3! 3!) = 5040. 

Now, if rearrange these, we would have words starting with E, L, P and S.

Now, SLEEPLESS starts with S. So, let us think about how many words start with S. Number of words starting with S = 8!/ (2! 2! 3!) = 1680.

So, there would have been 5040 - 1680 words that have gone by before the first word starting with S. Or, 3360 words start with E, L or P.

The first word with S is SEEELLPSS. This would have a rank of 3361.

Now, let us go step by step. 

Words starting with SE____ - Number of such words = 7!/(2!*2!*2!) = 630 words

Next we have words starting with SL, but our word also starts with SL, so let us go deeper

Words starting with SLE would be the next step, but our word starts with SLE as well.SO, let us go one further step deeper

Words starting with SLEE. The first such word would be SLEEEPLSS

So, let us think about words starting with SLEEE - there would be 4!/(2!) words like this = 12 words like this

words starting with SLEEL - there would be 4!/(2!) words like this = 12 words like this

So, we have accounted for 3360 + 630 + 12 + 12 = 4014 words thus far.

Now, on to words starting with SLEEP - there are again 4!/(2!) words like this = 12 words like this

Our word is within these 12 words

Words starting with SLEEPE - there are 3!/2! words like this, or 3 words like this
Our word comes in the next bunch.

Words starting with SLEEPL -  SLEEPLESS is the first such word. Or, the rank fo SLEEPLESS = 4014 + 3 + 1 = 4018.

A very good question to get lots of practice on letters rearrangement. However, it is unlikely that you will face a question this time-consuming in CAT. Wonderful question to practice though.

Algebra Question and Solution

Question

|x| + |2y| + |3z| = 13, x * y * z is non-zero; x, y, z are all integers. How many sets of values are possible?

Answer
64 sets of values

Explanation

Inequalities - Interesting Question

Question
How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0?

Answer: 30

Explanation

Let us try out a few values to see if that gives us anything.

When x = 8, 10, 12, ....100 this goes to zero. So, these cannot be counted.
When x = 101, 102 or beyond, all the terms are positive, so the product will be positive. 

So, straight-away we are down to numbers 1, 2, 3, ...7 and then odd numbers from there to 99.

Let us substitute x =1,

All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3, 4, 5, 6, and 7.

Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question.

When x = 9, there is one positive terms and 46 negative terms. So, the product will be positive. 
When x = 11, there are two positive terms and 45 negative terms. So, the product will be negative. 
When x = 13, there are three positive terms and 44 negative terms. So, the product will be positive. 

and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7...and then 11, 15, 19, 23, 27, 31,..... and so.

What will be the last term on this list? 
99, because when x = 99, there are 46 positive terms and 1 negative term. 

So, we need to figure out how many terms are there in the list 11, 15, 19,....99. These can be written as 
4 * 2 + 3, 
4 * 3 + 3, 
4 * 4 + 3
4 * 5 + 3
4 * 6 + 3
...
4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of x exist satisfying the condition. 

CAT - Percents

Question: 

 

P is x% more than Q. Q is (x - 10)% less than R. If P > R, what is the range of values x can take?

A.  10% to 28%

B.  10% to 25%

C.  10% to 37%

D.  10% to 43% 

Correct Answer: (C)

Explanation:

P = Q

Q = R

R =

P > R

(100 + x) (110 – x) > 100 x 100

11,000 + 110x – 100x – x² > 10000

1000 + 10x – x² > 0

x² – 10x – 1000 < 0

x² – 10x + 25 < 1000 + 25

(x – 5)² < 1025

x – 5 < 32

x < 37

x could range from 10% to 37%

Answer Choice (C)

Difficulty level II

CAT - Inequalities

Question:

Solve the inequality x³ – 5x² + 8x – 4 > 0.

A.  (2, )

B.  (1, 2) U (2, )

C.  (-, 1) U (2, )

D.  (-, 1)

Correct Answer :  (B)

Explanation:

Let a, b, c be the roots of this cubic equation

a + b + c = 5

ab + bc + ca = 8

abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations}

The other approach is to use polynomial remainder theorem

If you notice, sum of the coefficients = 0

=>  P(1) = 0

=>  (x - 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic equation.

(x - 1) (x - 2) (x - 2) > 0

Let us call the product (x - 1)(x - 2)(x - 2) as a black box.

If x is less than 1, the black box is a –ve number

If x is between 1 and 2, the black box is a +ve number

If x is greater than 2, the black box is a +ve number

Since we are searching for the regions where black box is a +ve number, the solution is as follows:

1 < x < 2 OR x > 2

Answer Choice (B)

Difficulty level 2