March 18th, 2014 by Ravi Handa

We believe we do a brilliant job when it comes to teaching students or creating content for it. But beyond that, we are pretty useless. As you can see, we do not even know how to get the word out about a job opening. The best that we can do, as of now, is to write a blog post about it. We need someone who can do this well. No – we don’t need someone in HR. We need someone who can get the word out better than we can. So, we are hiring for an **online marketer position**.

Following are the key responsibilities for this role:

- Implementing SEO activities.

- Implementing & tracking e-Mail campaigns.

- Setting up AdWords/Facebook Ad campaigns, optimizing and tracking results.

- Managing Social Media presence.

- Google Analytics reporting.

Experience: 0 to 2 years.

Timings: 10 AM to 6 PM (5 days a week)

Salary: 15000 – 20000 Rs. per month

Location: Baner-Balewadi, Pune

We are looking for self-driven person who is web savvy and is passionate for online marketing. This is a fulltime position. To get in touch with us, drop in an email on ravihanda@gmail.com or call us on 09765142632.

Posted in Uncategorized

January 30th, 2014 by Ravi Handa

If you have written CAT earlier, then probably you should skip this post. This is mainly directed towards those who are planning to write CAT and do not know what all gets covered in the exam.

CAT is divided into two sections:

1. Quantitative Aptitude and Data Interpretation

2. Logical Reasoning and Verbal Ability

Looking at the data of past few years, both sections have **30 questions** each. A student is given **70 minutes** for each section.

**Quantitative Aptitude (20/21 questions) **is basically questions which test your mathematical prowess. It can be further classified into 5 broad categories:

- Number System – Deals with properties of numbers, HCF, LCM, remainder, factorial, etc.
- Arithmetic – Deals with percentage, average, profit & loss, SI & CI, Time Speed Distance, Time & Work, etc.
- Algebra – Linear equations, Quadratic equations, functions, etc.
- Geometry – Lines, angles, triangles, circles, polygons, mensuration, etc.
- Modern Maths – Permutation and Combination, Set Theory, Probability, etc.

Maximum number of questions are asked from **Algebra and Geometry.**

**Data Interpretation (9/10 questions)** consists of questions based on pie charts, tables, etc. In these kind of questions, you would be given a large amount of data and questions will be based upon that data. I agree that there are no formulas which are required for stuff like this but a little bit of practice with different type of questions would take your prep a long way.

**Logical Reasoning (9/10 questions)** consists of questions based on puzzle and logic. Once again, it is not something that you can really prepare for in terms of concepts. But there are a variety of questions that you can practice on it. Looking at past year papers is a good idea.

**Verbal Ability (20/21 questions) **checks your mastery over the English language. It can be broadly classified into two categories:

**Reading Comprehension (9/10 questions) **- 3 passages will be given to you and then you will have to answer questions based on those passages.
**Verbal Usage (9/10/11 questions) **- It will consist of questions on Parajumbles, Statements & Assumption, Fact Inference Judgement, Syllogisms, Fill in the blanks, etc.

If you like this post, please share it with friends.

Do check out my online course on CAT 2014 coaching : http://www.wiziq.com/course/9277-lr-vr-di-ds-speed-calculations-quant-general-awareness

Posted in General Funda

January 22nd, 2014 by Ravi Handa

**A team of miner planned to mine 1800 tonnes in a certain number of days.Due to some difficulties in one third of the planned days, the team was able to achieve an output of 20 tons of ore less than the planned output.To make up for this, the team overachieved for the rest of the days by 20 tons.The end result for this that they completed the one day ahead of time.How many tone of ore did the team initially plan to ore per day?**

a. 50

**b. 100 **

**c. 150 **

**d. 200 **

**e. 250**

**Answer : **

Let us assume the no. of days as ’3d’ and the output per day as ‘x’

Then,** 3d*x = 1800 …(1)**

For the first ‘d’ days, the output was (x-20).

For the next ’2d-1′ days, the output was (x+20).

=> d(x-20) + (2d-1)(x+20) = 1800

=> dx – 20d + 2dx + 40d – x – 20 = 3dx {Replacing 1800 with 3dx from equation (1)}

=> 20d = x + 20

=> **d = (x+20)/20 …(2)**

=> 3 [(x+20)/20] x = 1800

=> x^2 + 20x = (1800/3)*20

=> x^2 + 20x – 12000 = 0

=> (x+120)(x-100) = 0

=> x = -120 or 100

**Since x is the output it cannot be negative.**

So, the initial planned output is **100 tonnes.Thus Option B**

Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

**Question : a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??**

**Answer :** Roots of 3x^2 + 4x + 5 = 0 are complex i.e. they are of the form** (p + iq) & (p – iq)** where i is iota = sqrt(-1)

For a quadratic equation, complex roots occur in conjugate pairs if the coefficients are real.

In the equation, ax^2 + bx + c = 0, the coefficients are sides of a triangle and hence real. So, if one of roots is common with the other equation say (p+ iq) then the other root will also be compulsorily common. This implies that **both equations have the same roots**.

This implies that **ax^2 + bx + c = 0 , is nothing else but the same equation**

**i.e. 3x^2 + 4x + 5 = 0**

Now, we know that the sides of the triangle are 3,4 & 5 where c = 5.

3,4 & 5 form a Pythagorean triplet making the triangle a right angled triangle.

** The angle opposite to the biggest side is 90 degrees in a right angled triangle**, which in this case is C.

Hence, **C = 90 degrees**.

Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

**Question : A person lent out some money for 1year at 6% per annum simple interest and after 18months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly?**

**a. Rs 4000**

**b. Rs 4200 **

**c. Rs 4400 **

**d. Rs 3600**

**Answer :**

In the first case, the rate of interest is 3% per half-year. The money will become

1.03x after 6 months from t = 0

1.06x after 12 months from t = 0

1.09x after 18 months from t = 0

**1.12x after 24 months from t = 0**** **and so on ….

In the second case, the rate of interest is 6% per half-year. The money is given out at t = 18.

The money will become:

**1.12x** after 6 months from t = 18, which is **24 months from t = 0**.

So after 2 years the returns from both the investments will be same. They will be 1.12x

This was the point at which he gets back 4704 Rs.

=> 1.12x = 4704

=> **x = 4200 Rs.**

Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

**Question : Six persons try to swim across a wide river.Its known that on an average only three persons out of 10 are successful in crossing the river.Whats the probability that at most four of the 6 persons will cross the river safely?**

**Answer : **

The probability distribution of the random variable X is called a **binomial distribution**, and is given by the formula:

* P(X)= nCr * p^r * q^(n-r)*

where

n = the number of trials

r = 0, 1, 2, … n

p = the probability of success in a single trial

q = the probability of failure in a single trial

(i.e. q = 1 âˆ’ p)

Keeping the above mentioned idea in mind,

p is given to us as 3/10 or 0.3

=> q = 1 – 0.3 = 0.7

Also, n = 6 {There are 6 people trying to cross the river}

P(0) = 6C0 * (0.3)^0 * (0.7)^6 = 1 * 1 * 0.117649 = 0.117649

P(1) = 6C1 * (0.3)^1 * (0.7)^5 = 6 * 0.3 * 0.16807 = 0.302526

P(2) = 6C2 * (0.3)^2 * (0.7)^4 = 15 * 0.09 * 0.2401 = 0.324135

P(3) = 6C3 * (0.3)^3 * (0.7)^3 = 20 * 0.027 * 0.343 = 0.18522

P(4) = 6C4 * (0.3)^4 * (0.7)^2 = 15 * 0.0081 * 0.49 = 0.059535

P(5) = 6C5 * (0.3)^5 * (0.7)^1 = 6 * 0.00243 * 0.7 = 0.010206

P(6) = 6C6 * (0.3)^6 * (0.7)^0 = 1 * 0.000729 * 1 = 0.000729

As you can see from above, total is = 0.117649 + 0.302526 + 0.324135 + 0.18522 +0.059535 + 0.010206 + 0.000729 = 1 {Add it up in calculator if you don’t believe me }

The question is asking us for at most 4 crosses, so that would be

P(0) + P(1) + P(2) + P(3) + P(4) = 1 – P(5) + P(6)

=> 0.117649 + 0.302526 + 0.324135 + 0.18522 + 0.059535 = 1 – (0.010206 + 0.000729)

=> 0.989065 = 1 – 0.010935

=> **Ans is ~ 0.99**

Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

**Question :** The L.C.M of 5/2, 8/9, 11/14 ?

**Answer :**

**LCM = LCM of Numerator / HCF of Denominator**

**HCF = HCF of Numerator / LCM of Denominator**

** **

So,in this case;

LCM (5/2, 8/9, 11/14) is :

LCM (5,8,11) / HCF (2,9,14) = 440 / 1 = **440**

Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

The concept of ‘divide and conquer’, derived from the Latin phrase ‘Divide et impera’, was put into use effectively by everyone from Caesar to Napoleon to The British in India. Even Gaddafi tried using the same but as current events show us – he wasn’t very effective. Dividing rather divisibility rules to be specific can come in really handy at times in solving problems based on Number Systems.

The standard rules which nearly all of us are very comfortable with are the ones for 2^{n} and 5^{n}. For these all that one needs to do is look at the last ‘n’ digits of the number. If the last ‘n’ digits of a number are divisible by 2^{n} or 5^{n}, then the number is divisible by 2^{n} or 5^{n} and vice versa. For details about other numbers, I suggest that you read on.

**Funda 1:**

** **For checking divisibility by ‘p’, which is of the format of 10^{n} – 1, sum of blocks of size ‘n’ needs to be checked (Blocks should be considered from the least significant digit ie the right side). If the sum is divisible by p, then the number is divisible by p.

Eg 1.1 Check if a number (N = abcdefgh) is divisible by 9

- 9 is 10
^{1} – 1
- Sum of digits is done 1 at a time = a + b + c + d + e + f + g + h = X
- If X is divisible by 9, N is divisible by 9
- Also, N is divisible by all factors of 9. Hence the same test works for 3.

Eg 1.2 Check if a number (N = abcdefgh) is divisible by 99

- 99 is 10
^{2} – 1
- Sum of digits is done 2 at a time = ab + cd + ef + gh = X
- If X is divisible by 99, N is divisible by 99
- Also, N is divisible by all factors of 99. Hence the same test works for 9, 11 and others.

Eg 1.3 Check if a number (N = abcdefgh) is divisible by 999

- 999 is 10
^{3} – 1
- Sum of digits is done 3 at a time = ab + cde + fgh = X
- If X is divisible by 999, N is divisible by 999
- Also, N is divisible by all factors of 999. Hence the same test works for 27, 37 and others.

**Funda 2:**

** **For checking divisibility by ‘p’, which is of the format of 10^{n} + 1, *alternating sum* of blocks of size ‘n’ needs to be checked (Blocks should be considered from the least significant digit ie the right side). If the alternating sum is divisible by p, then the number is divisible by p.

*(***Alternating Sum**: Sum of a given set of numbers with alternating + and – signs. Since we are using it to just check the divisibility, the order in which + and – signs are used is of no importance.)

Eg 1.1 Check if a number (N = abcdefgh) is divisible by 11

- 11 is 10
^{1} + 1
- Alternating sum of digits is done 1 at a time = a – b + c – d + e – f + g – h = X
- If X is divisible by 11, N is divisible by 11

Eg 1.2 Check if a number (N = abcdefgh) is divisible by 101

- 101 is 10
^{2} + 1
- Alternating sum of digits is done 2 at a time = ab – cd + ef – gh = X
- If X is divisible by 101, N is divisible by 101

Eg 1.3 Check if a number (N = abcdefgh) is divisible by 1001

- 1001 is 10
^{3} + 1
- Sum of digits is done 3 at a time = ab – cde + fgh = X
- If X is divisible by 1001, N is divisible by 1001
- Also, N is divisible by all factors of 1001. Hence the same test works for 7, 11, 13 and others.

** **

**Funda 3: ****Osculator / seed number method**

*For checking divisibility by ‘p’,*

*Step 1*: Figure out an equation such that

If we have this equation, the osculator / seed number for ‘p’ will be . *(-m in case of 10m+1 and +m in case of 10m – 1)*

*Step 2*: Remove the last digit and multiply it with the seed number.

*Step 3*: Add the product with the number that is left after removing the last digit.

*Step 4*: Repeat Steps 2 and 3 till you get to a number which you can easily check that whether or not it is divisible by p.

Eg: Check whether 131537 is divisible by 19 or not.

- 19*1 = 10*2 – 1 (Seed number is +2)
- 13157-> 13153+7*2 = 13167 -> 1316+7*2 = 1330 -> 133+0*2 = 133
- 133 is divisible by 19
- 131537 is divisible by 19

I hope that these divisibility rules will enable you to *divide and conquer* few of the Number Systems problems that you encounter during your preparation.

Posted in Quant Funda

January 22nd, 2014 by Ravi Handa

**Factor Theory **

I understand that the title might be a little misleading but at least I could not come up with something better. So before you end up getting disappointed, let me first list down the topics that I am going to cover in this particular blogpost:

- Number of factors of a given number
- Number of
*even* factors or *odd* factors of a given number
- Sum of all factors of a given number
- Sum of all
*even* factors or *odd* factors of a given number

We know that a number N can be written as a product of its factors as given below

- N = a
^{p} x b^{q }x c^{r} …

Here a,b,c… are prime factors of N

& p,q,r … are the powers of the prime factors of N.

**In such a case the number of factors of N are given by the formula**

The obvious question which arises is, why this formula in particular. It is actually a game of choices and options available. Let us see how.

Suppose you are going on a camping trip and you have packed all the necessary items. Now you are wondering how many movie DVDs, music CDs & MP3 players you should take. (*Please do not judge me by the example. I have no experience about camping trips but I guess that is obvious by the example mentioned*). In your collection, you have 5 movie DVDs, 4 music CDs and 2 MP3 players.

Try to answer these questions:

- How many choices / options do you have for the number of movie DVDs that you can take?
- How many choices / options do you have for the number of music CDs that you can take?
- How many choices / options do you have for the number of MP3 players that you can take?
- Are the answers to all the above questions independent of each other?

Your answers for the above questions should be:

- 6 (
*You can take either 1 or 2 or 3 or 4 or 5 Movie DVDs. You also have the option of not taking a movie DVD at all.)*
- 5 (
*You can take either 1 or 2 or 3 or 4 Music CDs. You also have the option of not taking a music CD at all.)*
- 3 (
*You can take either 1 or 2 or 3 or 4 or 5 MP3 players. You also have the option of not taking a MP3 player at all.)*
- Yes. All answers are independent of each other.

So, what is the total number of options / choices do you have while packing for the camping trip?

Answer = 6 x 5 x 3 = 90

If you understood the above example, the number of factors formula should be a piece of cake.

Consider a number **N = 2**^{5 }x 3^{4} x 5^{2}

It means that you have: 2,2,2,2,2 && 3,3,3,3 && 5,5

Any combination of the above will make a factor. You have 6 choices for picking up the number of 2s in a factor, 5 choices for picking up the number of 3s in a factor and 3 choices for picking up the number of 5s in a factor.

So, the total number of factors = 6 x 5 x 3 = 90.

Now, for finding out the number of even factors consider the camping trip case.

If one of your friends insists that he wants to watch Twilight while on the camping trip and you should bring the movie DVD because you own a copy. (*Once again I will restrain myself and not pass a judgement on you or your friend or the choice of the movie)*

So, in this case the number of choices for picking up the movie DVD has reduced from 6 to 5 because you HAVE TO bring the Twilight DVD.

So, the total number of choices that you have now is 5 x 5 x 3 = 75

Same logic can be extended to finding out the number of even factors.

Number of even factors = 5 x 5 x 3 = 75 {Because your factor has to contain at least one 2, analogous to the twilight DVD}

Number of odd factors = 5 x 3 = 15 {In this case, your factor cannot contain any 2s, analogous to not being allowed to take a movie DVD}

As a matter of fact, if you have the total number of factors and the total number of even factors; their difference would directly give you the total number of odd factors.

Such logic can be extended to find out the number of factors divisible by a particular number.

Number of factors divisible by 12 = 4 x 4 x 3 = 48 {In this case, your factor will have to contain at least two 2s and one 3}

We know that a number N can be written as a product of its factors as given below

- N = a
^{p} x b^{q }x c^{r} …

Here a,b,c… are prime factors of N

& p,q,r … are the powers of the prime factors of N.

**In such a case the sum of factors of N are given by the formula**

The logic remains the same in this case also. If you expand the above expression, you will end up with all the factors.

For **N = 2**^{5 }x 3^{4} x 5^{2}

Sum of all factors

Sum of all the even factors

Sum of all the odd factors

Sum of all the factors divisible by 12

I hope that this post was helpful and you will not face any problems in finding out the number of factors and related stuff.

Posted in Quant Funda

January 22nd, 2014 by Ravi Handa

**Dealing with Factorials**

We all know what factorials (n!) are. They look friendly and helpful but looks can be deceiving, as many quant problems have taught us. Probably it is because that Factorials are simple looking creatures, most students prefer attempting questions based on them rather than on Permutation & Combination or Probability. I will cover P&C and Probability at a later date but in today’s post I would like to discuss some fundas related to factorials, which as a matter of fact form the basis of a large number of P&C and Probability problems.

Some of the factorials that might speed up your calculation are:

*0! = 1; 1! = 1; 2! = 2; 3! = 6; 4! = 24; 5! = 120; 6! = 720; 7! = 5040. *

**Funda 1: Rightmost non-zero digit of n! or R(n!)**

*R(n!) = Last Digit of [ 2*^{a} x R(a!) x R(b!) ]

*where n = 5a + b*

Eg 1.1: What is the rightmost non-zero digit of 37! ?

- R (37!) = Last Digit of [ 2
^{7} x R (7!) x R (2!) ]
- R (37!) = Last Digit of [ 8 x 4 x 2 ] = 4

Eg 1.2: What is the rightmost non-zero digit of 134! ?

- R (134!) = Last Digit of [ 2
^{26} x R (26!) x R (4!) ]
- R (134!) = Last Digit of [ 4 x R (26!) x 4 ]

We need to find out R (26!) = Last Digit of [ 2^{5} x R (5!) x R (1!) ] = Last digit of [ 2 x 2 x 1 ] = 4

- R (134!) = Last Digit of [ 4 x 4 x 4 ] = 4

** **

**Funda 2: Power of a prime ‘p’ in a factorial (n!)**

*The biggest power of a prime ‘p’ that divides n! (or in other words, the power of prime ‘p’ in n!) is given by the sum of quotients obtained by successive division of ‘n’ by p. *

Eg 2.1: What is the highest power of 7 that divides 1342!

- [1342 / 7] = 191
- [191 / 7] = 27
- [27 / 7] = 3
- Power of 7 = 191 + 27 + 3 = 221

Eg 2.2: What is the highest power of 6 that divides 134! ?

As 6 is not a prime number, we will divide it into its prime factors. 3 is the bigger prime, so its power will be the limiting factor. Hence, we need to find out the power of 3 in 134!

- [134/3] = 44
- [44/3] = 14
- [14/3] = 4
- [4/3] = 1
- Power of 3 in 134! = 44 + 14 + 4 + 1 = 63

Eg 2.3: What is the highest power of 9 that divides 134! ?

As 9 is not a prime number, we will divide it into its prime factors. 9 is actually 3^{2}. The number of 3s available is 63, so the number of 9s available will be [63/2] = 31.

Highest power of 9 that divides 134! is 31.

Highest power of 18 and 36 will also be 31. Highest power of 27 will be [63/3] = 21.

*Note: To find out the highest power of a composite number, always try and find out which number (or prime number) will become the limiting factor. Use that to calculate your answer. In most cases you can just look at a number and say that which one of its prime factors will be the limiting factor. If it is not obvious, then you may need to find it out for two of the prime factors. The above method can be used for doing the same. *

** **

**Funda 3: Number of ending zeroes in a factorial (n!)**

*Number of zeroes is given by the sum of the quotients obtained by successive division of ‘n’ by 5.*

This is actually an extension of Funda 1. Number of ending zeroes is nothing else but the number of times n! is divisible by 10 or in other words, the highest power of 10 that divides n!. 10 is not a prime number and its prime factors are 2 and 5. ‘5’ becomes the limiting factor and leads to the above-mentioned idea.

Eg 3.1: What is the number of ending zeroes in 134! ?

- [134/5] = 26
- [26/5] = 5
- [5/5] = 1
- Number of ending zeroes = 26 + 5 + 1 = 32

I hope that this gets you started with factorials and you might start singing this song.

Posted in Quant Funda