## Divisibility Rules: Quantitative Aptitude Section

Posted by Jasleen Behl
There are some specific rules by which we can determine the divisor of the given number. Today I will discuss divisibility rules from 2 to 19. Using these rules you can easily determine a divisor of given number, however large it may be. Let me tell you the rules of divisibility from 2 to 19.

## Divisibility Rules

#### Divisibility by 2

Number which ends with even number or 0 is always divisible by 2.
For Example: 44, 120, 56, 70 etc are divisible by 2.
23, 57, 79 etc are not divisible by 3.

## Combinations

Posted by Ramandeep Singh
Unlike a permutation, in a combination the order of the objects selected does not matter. For example, if you were choosing a team of three students from a class of ten, the order you said "Joe, Sally, John" would not matter to who is in the group- saying "Sally, Joe, John" instead doesn't change anything.

## How to solve a combination

The number of ways you can make a combination of r objects out of a set of n objects is made from the formula below:

C=n!/r!(n-r)!

## An Explanation

You may be asking (like I did when I learned this material), "What? That formula doesn't make sense!".- However, it is based on the formula for a permutation. If you think about it, the n! makes perfect sense- thats how many options you'd have if you picked in order from all of them. Removing the (n-r)! gets it down to just the numbers you want (n-r equals the last number you arn't picking. If you pick 3 from 5, it'll be 5!/(5-3)!, so 5*4*3*2*1/2*1, which, after cancelling it out to 5*4*3, accomplishes the "pick x of 5" part.) The extra r! removes all the redundant options. For example, say you have 6 different balls, labeled A through F. If you pick three, you can pick ABC, ABD, ABE, ABF, etc. You'll end up with six of the same- ABC, ACB, BAC, BCA, CAB, CBA- all of which only count as one in a combination. As you'll notice, since we picked 3, 3!=3*2*1=6- the same number we have to divide out!

## An Example of a Combination

Lets take the example above. Out of a class of ten students, how many ways could you make a team of three students? We know n is ten because the set (the students) has ten objects in it. We know r is 3 because the team will have three students on it. To solve, first we set up the equation:

C=n!/r!(n-r)!
Then we fill in the values:
C=10!/3!(10-3)!
C=10!/3!7!

By taking apart the factorials, we can simplify the 10! and 7!. Alternativley, just start at 10 and multiply down, stopping right before 7.

C=10*9*8*7*6*5*4*3*2*1/3*2*1*7*6*5*4*3*2*1
C=10*9*8/3*2*1
C=720/6
C= 120

There you have it, there are 120 ways to pick a team of 3 people out of ten.

## Permutations

Posted by Ramandeep Singh
Permutations are very important in Quantitative Aptitude. They are most often used for finding probabilities or the number of ways something can be arranged. In fact, a permutation is simply the number of different ways a set can be arranged.

## An Example of a Permutation

One of the most used examples of a permutation is the number of ways to arrange a batting order for a baseball team.

Example1:  A baseball team has 9 players, each of which go up and bat before repeating the order. How many different ways can you organize 9 players in an order?

Solution: For the first position, there would be 9 options, for the second position 8 options, for the third 7, and so on, until for the 9th position there would be only one option (Since no player can go twice before all of the players have gone). Hence, the number of positions would be

9*8*7*6*5*4*3*2*1= 362,880

From that, we know that there are 362,880 ways to order a batting order for a baseball team. Writing 9*8*7*6... can be very tedious, though. Hence, we have what we call a "factorial"

## Factorials

Whenever you multiply a number by the number right before it, then by the number right before that and so on until you multiply it by one, we simply write it as "n!" where n is the number. So 9 factorial, like in the example above, would be written 9!

9!= 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1

## Permutations where the entire set isn't used

Most of the time, unlike in the baseball example, you will be selecting a few objects out of the set, not every object. The number of objects you take is called r. The size of the set is n.

It is expressed as P(n,r)

Example2: Choosing the 1st, 2nd and 3rd places out of a field of 9 players would be expressed as P(9,3), since there are 9 objects and three are being selected. How many different ways are there?

Solution: To solve this, you use the formula P(n,r) = n!/(n-r)!
So for the top three out of nine, it would be

P(n,r) = n!/(n-r)!
P(9,3) = 9!/(9-3)!
P= 9!/6!

Now if you take that back into your old form, you get this:

P=9*8*7*6*5*4*3*2*1/6*5*4*3*2*1

Since they are all being multiplied, you can simply cancel out the same numbers on the top and bottom. Hence, you may remove the 6, 5, 4, 3, 2 and 1 from both top and bottom.

P=9*8*7
P= 504

So there are 504 ways to choose 1st 2nd and 3rd place out of 9 players.

## Wait, that seems harder than it should be...

It is. Really, you can skip the whole P=n!/(n-r)! and just multiply the first r objects. so instead of that, you just do P=9*8*7. It's much simpler. Even so, you should remember the formula for tests.

## What to remember:

• A permutation is the number of ways objects can be ordered in a set. Order DOES matter.

• When the whole set is used, it is simply n!

• n! means n factorial, or n*(n-1)*(n-2)... down to 1.

• When you only choose some of the set, simply take the number you are supposed to choose and use that many numbers from the start of the factorial. So P(9,3)=9*8*7 and P(121,4)=121*120*119*118

• Remember the formula P=n!/(n-r)! for tests.

## Simple or linear Equations: Tricks and Examples.

Posted by Jasleen Behl
I have already discussed a concept - Quadratic Equations. Today I will discuss some examples of simple equations which have been proved to be a very important topic for various competitive exams. The problems of linear equations can be easily solved by using simple tricks. Lets discuss how.

## Examples with solutions

Example1: If 3x + 6 = 4x - 2, then find the value of x?

## Geometry (Part-2) for SSC CGL Tier-I

Posted by Jasleen Behl
In my previous session of Geometry, I discussed some properties of triangles. Also, I shared some important questions for SSC CGL. Today I'm going to discuss congruency and similarity of triangles which are one of the most important properties of triangles. There are some rules to determine whether the given two triangles are congruent or similar or nothing. Lets discuss it one by one.

## Congruency of Triangles

Two triangles are said to congruent if
• they overlap each other when placed one over the other.
• they have same area.
• identical in all aspects.

### When two triangles are congruent?

Now, the question is how we can detect that whether the given two triangles are congruent or not.

There are 5 rules or I can say ways by which we decide whether the triangles are congruent are not.

#### 1. S-S-S (Side- Side-Side)

When two given triangles have same three sides,then both are said to be congruent triangle.

(MARKED WITH BLUE COLOR to EQUAL TERMS.)
For example, following two triangles have equal sides. Therefore, these are congruent triangles

#### 2. S-A-S (Side- Angle- Side)

When two sides and included angle is same (A is in between two S's), then also both triangles are congruent.
Following two triangles are congruent:

#### 3. A-S-A (Angle- Side- Angle)

When two angles and included side of both triangles are equal (S is in between two A), then triangles are congruent.
Following two triangles are similar:

#### 4. A-A-S (Angle- Angle- Side)

When two angles and non-included side (  not the included side) is equal, then both triangles are congruent.
Following two triangles are congruent:

#### 5. R-H-S ( Right angle- Hypotenuse- Side)

This rule is just for right angle triangles. For congruency, there must be one right angle and hypotenuse and any one side must be equal.

Following two right angle triangles are congruent:

### Special Case:

#### A-A-A (Angle- Angle- Angle)

Two triangle having all three same angles are NOT congruent because you cant assume that they will have same sides i.e. same angles does not mean sides will be same.

## Similarity of triangles

Two triangles are said to be similar if
• sides of both triangles are in same proportion.
• corresponding angle are same
• different size and area

### When two triangles are Similar?

To check whether two triangles are similar or not, there are only 3 rules. Discussed as follows:

#### 1. A-A ( Angle- Angle)

If any two angles of triangles are equal, then they are said to be similar. For example, following triangles are similar triangles:

#### 2. S-A-S (Side- Angle- Side)

If two side of triangles are proportionate and include angle is same, then both triangles are similar.
Consider the following triangle:
7/3.5 = 10/5 =2 i.e. proportionate sides. Also one angle is same. Therefore, these are similar triangles.

#### 3. S-S-S (Side- Side- Side)

If three sides of one triangle is proportionate to three side of other triangle, then they are said to be similar triangle.
All the three sides are proportional.

8/4 = (12)/6 = (10)/5 =2
Therefore, triangles are similar.

Remember that,
All congruent triangles are similar. But, all similar triangles are not congruent.

## IBPS-Specialist Officer result status out

Posted by Jasleen Behl
All candidates of IBPS- Specialist officer are informed that result status have been declared. You can check your results here: RESULT.

### Information for all Qualified candidates

Firstly, congratulations to all who have been qualified. Well Done! Now start preparing for your interviews. And the people who are not qualified, don't feel bad, be motivated and keep preparing for other bank exams.
• IBPS will not release final score before allotment. So, you have to wait for atleast 2-3 months.
• After this, start preparing for interviews.

## Study Plan for SSC CGL Tier-I

Posted by Jasleen Behl
Today I'm going to present a helpful study plan for all SSC CGL Tier-I aspirants. I will just share the complete pattern of exam and the weightage of all the sections. Before, you read this article, firstly check the syllabus of SSC CGL. Then start making a study plan using the following discussion. Suggested books are discussed here. As we all know SSC CGL includes four sections given below and each section  is consist of 50 questions.

Quantitative Aptitude: 50 Marks
General Intelligence and Reasoning: 50 Marks
General Knowledge: 50 Marks
English Language: 50 Marks