# IIM CAT Preparation Tips

IIM CAT Preparation Tips: October 2014

## Oct 31, 2014

### CAT Preparation Online - Permutation and Combination

Question
Product of the distinct digits of a natural number is 60. How many such numbers are possible?

Explanation
60 = 2^2 * 3 * 5

60 cannot be written as a product of two single digit numbers. SO, the number in question should either have 3 or more digits.

Three-digit numbers
The digits could be 345 or 265
Digits being 345 - there are 3! such numbers
There are six numbers for each of these outlines. So, there are 3! + 3! = 12 three-digit numbers

Four-digit numbers
The digits could be 1345, 1265 or 2235.
Digits being 1345 - there are 4! such numbers
Digits being 1265 - there are 4! such numbers
Digits being 2235 - this is not possible as digits have to be distinct.

So, there are totally 24 + 24 four-digit numbers possible. 48 four-digit numbers.

Total number of numbers = 12 + 48 = 60.

(this post has been modified to plug error)

## Oct 23, 2014

### CAT preparation - Ratio and Proportion (Tough)

Question

John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If there are more chocolates of type C than of type B, and more of type B than of type A, what is the minimum possible number of chocolates overall?

Explanation

Again, big thanks to Mukund Sukumar for excellent solution.

Let john have 3x chocolates of type A and 7x of type B
Let Mike have 5y chocolates of type B and 4y of type C
Let john have 3z chocolates of type C and 5z of type A

So in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since C>B we get solving y<3z-7x --->(1)
Since B>A we get solving 5y>5z-4x ---> (2)

What gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only y=1 as choice, for which second condition doesnt satisfy.

So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second condition. so which gives choice the only y=4.

Hence x=1,y=4 and z=4 works and is the best possible answer.

For these values, we get A=23,B=27,C=28.

Minimum possible number of chocolates overall is 76.

### CAT Preparation Online - Tough one from Permutation and Combination

Question

A wonderful, but very tough question from Permutation and Combinations.

In how many ways, can we rearrange the word MONSOON such that no two adjacent positions are taken by the same letter? (Tough one. Tougher than what we will see in CAT)

Explanation

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2 N’s.
Let us focus on the three O’s.

We can place the three O’s in some blanks around the other letters. Or, three O’s can be placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in 5C3, or 10 ways.

Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1, 5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7}  - Whew.

Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4 letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be adjacent in these arrangements. We will need to eliminate these.

O’s in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2
words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 4, 6} or O__ __O__O__  - Ns could be in the slots {2, 3}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {2, 4, 6} or __O__O__O__  - Tehre are no possible slots for N. So, we count all 12 words on this list.

O’s in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10.

O’s in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.
Phew!.

There is a far more elegant method for accounting for the words where the 2 N’s appear together. This one came from Mukund Sukumar.

We need to account for the number of possibilities of N,N being together. So to subtract that part, consider 'NN' being together as one letter and place O's. In 3 out of the 4 slots in _M_NN_S_

The O’s can be selected in 4C3 ways. The MNNS can be rearranged in 3! Ways if the N’s have to appear together.

Or, we get 4C3 * 3! = 24 ways. So, we have 120-24 = 96 ways totally.

## Oct 15, 2014

### CAT Preparation Online - A simple question from Quadratic Equations

Questions

x^2 - 17x + |p| = 0 has integer solutions. How many values can p take?

Explanation

To begin with, if we assume roots to be a and b, sum of the roots is 17 and product of the roots is |p|. Product of the roots is positive and so is the sum  of the two roots.

So, both roots need to be positive.

So, we are effectively solving for

Number of positive integer solutions for a + b = 17.

We could have (1,16), (2, 15), (3, 14)......(8, 9) - There are 8 sets of pairs of roots. Each of these will yield a different product.

So, |p| can take 8 different values. Or, p can take 16 different values.

is that it? Or, are we missing something? Can p be zero? What are the roots of x^2 - 17x = 0. This equation also has integer solutions.

So, p can also be zero.

So, number of possible values of p = 16 + 1 = 17.

Wonderful question - chiefly because there are two really good wrong answers one can get. 8 and 16. So, pay attention to detail. No point telling yourself "Just missed, I just didnt think of zero. I deserve this mark". Being just wrong, will earn us a -1 instead of the honourable 0.

## Oct 14, 2014

### CAT Preparation Online - Simple one from Quadratic Equations

Question

How many integer solutions exist for the equation x2 - 8|x| - 48 =0?

Explanation

One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly simpler method.

Note that x2 is the same as |x|2, so we can treat the equation as a quadratic in |x|.

Or, |x|2 – 8|x| - 48 = 0
(|x| - 12|) (|x| + 4) = 0
|x| could be 12. |x| cannot be -4.

If |x| could be 12, x can be -12 and 12.

Two possible solutions exist.

### CAT - Question from Pipes and Cisterns

Question

There are n pipes that fill a tank. Pipe 1 can fill the tank in 2 hours, Pipe 2 in 3 hours, Pipe 3 in 4 hours and so on. Pipe 1 is kept open for 1 hour, pipe 2 for 1 hour, then pipe 3 and so on. In how many hours will the tank get filled completely?

Explanation

Almost all questions can be approached well by asking the question "What happens in 1 hour" (or 1 day, or 1 minutes)

In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first hour, the tank will not be filled

In 1 hour, pipe 2 fills 1/3 of the tank. So, in two hours we would have filled 1/2 + 1/3 of the tank, or 5/6 of the tank. So, by the end of the second hour, the tank would still not be filled.
Let us move to the third hour. In 1 hour, pipe 3 fills 1/4 of the tank. So, by the end of the third hour, we should have filled 5/6 + 1/4 = 13/12 of the tank.

Oops, one cannot fill 13/12 of a tank. What this tells us is that the tank gets filled in the 3rd hour.

When exactly during the third hour?

At the beginning of the third hour, we still have 1/6 of the tank still to fill. Pipe 3 can fill at the rate of 1/4 of the tank per hour. Or, pipe 3 will take 2/3 hours to fill the tank.

Or, the tank will be filled in 2 hours and 40 minutes.

This is an example of a sequence that is a harmonic progression. The formulae for HP are needlessly confusing. So, simple step-by-step approach works best.

## Oct 10, 2014

### 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.

## Oct 8, 2014

### CAT Number Systems - Fun question based on Armstrong numbers

Question

A 3-digit Armstrong number is a three-digit number where the number is equal to the sum of the cubes of the three digits. Give a few Armstrong numbers.

Explanation

To start with, be clear that CAT does not ask questions like these. You do not need to know what Armstrong number are; nor do you need to know how to get these to crack these exam. This is just a fun question to do some trial and error with.

The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and error (although a lil scientifically)

### CAT Number Theory - Interesting Question from Factorial

Question

How many trailing zeroes will be present in the base 12 representation of 55!?

Explanation

The question can be restated as follows - "What is the highest power of 12 that divides 55!"

Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.

So, in order to find the highest power of 12 that divides 55!, we need to look at the highest powers of 2 and 3 that divide 55!.

Successive division by 2 will give us the highest power of 2 that divides 55!:

55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.

Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 =  50

Successive division by 3 will give us the highest power of  3 that divides 55!:

55/3 = 18, 18/3 = 6, 6/3 = 2.

Highest power of 3 that divides 55! = 18 + 6 + 2 = 26.

So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide 55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!

We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's.

Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.

### Question on Number Theory

Question

A bus service runs from Chennai to Blr every third day with the first bus starting on Jan 1st. Another bus service runs from Chennai to Mumbai every 5th day starting from Jan 2nd. A third bus service from Chennai to Cochin runs on every 7th day starting from Jan 4th. In that year (which is not a leap year), on how many different days will a bus run from Chennai to all three cities?

Explanation

Let us call Jan 1st as day 1, jan 2nd as day 2 and so on.

So,
The Chennai-Blr services runs on days             1, 4, 7, 10, 13, 16, 19,......
The Chennai-Mumbai services runs on days    2, 7, 12, 17, 22, 27,......
The Chennai-Cochin services runs on days     4, 11, 18, 25, ......

First up, let us see if we can find one day where all three buses ply.

Chennai Blr runs on days of the form 3a + 1, Chennai_Mumbai on 5b + 2, and Chennai-Cochin on 7c + 4. If N can be written as 3a + 1, 5b + 2 and 7c + 4, then we should be able to get n as 105d + ___ . (105 is the LCM of 3, 5 and 7)

This is a very important idea. Get lots of practice on this idea. Wrap your head around this idea very clearly.

First let us combine 3a + 1 and 5b + 2.

A number of the form 3a + 1, on division by 15 can have one of the following remainders { 1, 4, 7, 10, 13}
A number of the form 5b + 2, on division by 15 can have one of the following remainders { 2, 7, 12}

Or, if a number is of the form 3a + 1 and 5b + 2, it has to be of the form 15k + 7.

Now, N = 15K + 7 and 7c + 4.

Number 15k + 7, on division by 105 can have one of the following remainders { 7, 22, 37, 52, 67, 82, 97}}
Number 7c + 4, on division by 105 can have one of the following remainders { 4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95, 102, 109}

Or, the number is of the form 105d + 67.

So, all three services will run on days 67, 67 + 105, 67 + 105*2 and so on.

Or, on days 67, 172  and 277. (Any other day would go beyond 365)

So, all three services would run together 3 days of the year.

Wonderful question testing concepts of LCM and Remainders.