[Q] Geometry Tricks - 1 [Important Formulae]

Many students asked me to share some geometry tricks. So here is the Part-1 :

(1) Be it algebra or geometry, such questions are always there that don't deserve your rough space.

Like the above questions asks you the value of x. Now value of x can't be negative in this context. Because CO = x - 3 and any negative value of 'x' will make CO negative, like if x = -8, then CO = -11. We know that side can never be negative. Hence all the options that have negative value for x are wrong.
Answer : (B)

In the above question a certain area is asked. Now see the options and observe that in options A, C and D if we take r=7, 2 and 1 respectively, then the area will become zero, and area can never be zero in this case, although value of 'r' can be 7, 2 or 1.
Answer : (B)
Note : Such questions are rare in geometry and in most of the questions you will have to pick up your pen, but still I shared these questions just to unleash the jugaad within you. Moreover, if you are able to solve even a single question with such approach, you will save at least 1 crucial minute in the examination hall.

(2) In geometry too, my beloved concept of 'symmetry' plays an important role. If in any question you find that some symmetrical expressions/equations are given, you can assume the triangle to be equilateral.

Q. 1 

In the above question you can assume that the triangle is equilateral. Then angles A, B and C will be 60. Hence sin²A + sin²B + sin²C = (√3/2)² + (√3/2)² + (√3/2)² = 9/4

Answer : (B)
Note : If in any question, the sides of a triangle are given (like a, b and c), then you can assume a = b = c, but make sure any additional detail is not given. Like in the above question that I solved, some additional equations were given, but then too I supposed the triangle to be equilateral only because the equation was symmetrical. Had the equation been unsymmetrical, I could not have been able to assume a =b =c

Q. 2

Here the lengths of perpendiculars are given to be a, b and c. Note that no additional information is given, hence it is safe to assume a = b = c. Let the side of the triangle be 's'. The figure will look like this - 

We have to establish a relation between 'a' and 's'. In an equilateral triangle, the incentre, orthocentre, circumcentre and centroid, all coincide. So you can calculate 'a', by which ever method you like.

a = inradius = s/2√3 [The inradius of an equilateral triangle is s/2√3 and the circumradius is s/√3]

Hence s = 2√3a

We know the formula for calculating the area of an equilateral triangle = (√3/4)s²

= (√3/4) (2√3a)²

= 3√3a²

Now put a = b = c in all the options and check which one will give 3√3a²
A) √3a

B) 3√2a²

C) √2a

D) 3√3a²

Answer : (D)

Moving on...

I have found that SSC has some real love with 'Area of a triangle' and it keeps on asking it again and again under different contexts. Like in Tier 2 (2014) around 5 questions asked area of triangles.
Therefore it is very important that you memorize all the possible formulas to calculate it. This will save you a lot of time.
Note : All the below questions are taken from a single paper [Tier-2, 2014].

Formula 1 : (Applicable only for Right-Angled Triangle)

Q. 3

In the above question you may struggle to calculate the area. You can try this question yourself (with a timer), to see how much time you are taking...

There is a direct formula for such questions -

Apply this formula, area = (100^2 * sin30)/4 = 1250
Answer : (D)
Note : You can choose any of the angles of the triangle (except the 90 degree one) and you will get the same result. Like in the above question, the 3 angles of the triangle are 15, 75 and 90.
sin2*15 = sin30
sin2*75 = sin150
And we know that sin150 = sin30
Hence it doesn't matter which angle you take. But to avoid any confusion, always take the smaller angle (15 in this case).

Next question -

Q. 4

Again apply the same formula
Answer : (C)

Formula 2 :    ½*b*c*sinϴ

This formula is only applicable when ϴ lies between the sides 'b' and 'c'.

Q. 5

Apply the formula, area = 1/2 * 10 * 10 * sin45
Answer : (D)

Formula 3 : Area of a triangle = r * S
where r = inradius
S = semi-perimeter

Q. 6

Given, perimeter = 50, hence semi-perimeter = 25
Area = 6 * 25 = 150
Answer : (C)

Now when you know this formula, in Q. 2 above, where we had to establish relation between 's' and 'a', you can establish it by applying this formula too.
Area of an equilateral triangle = (√3/4)s²
Area of a triangle = inradius * semi-perimeter = a * (3s)/2

Now,

a * (3s)/2 = (√3/4)s²

or s = 2√3a    [Same result]

Geometry Part-2 to follow soon...

To buy the super-hit SSC Hack Book, follow the below link-
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[Q] Sequence and Series Tricks

There are basically two types of questions that are asked by SSC from this part-
1. Finite sequence where the denominator is of the type 1x3, 3x5, 5x7, etc. (can be solved with direct formula)
2. An infinite sequence where you have to find the series (the sum of infinite sequence)

First of all let's see questions of type 1.

Q.1)     1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42
First break it into the required form-

Now apply the direct formula-

Difference is the gap between the two numbers in the denominator. So here difference = 2-1 = 1

Answer :  6/7 

Next question

Q . 2

Again break it into the required form-

1/3 + 1/15 + 1/35 + ... + 1/399

 = 1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/(19*21)

Apply the formula

Answer : 10/21

Now let us move to the second type of questions where we will be applying the real tricks-

Q . 3

To begin with, I have taken the most basic question. Although many of you might be knowing that the sum of first n natural numbers is n(n+1)/2 and hence the arithmetic mean of first n natural numbers will come out to be (n+1)/2

But let's say you forget the formula (exam pressure) and you can't recall it. What will your approach be?

The trick is simple- suppose a value of 'n'.

'n' defines how many terms from the sequence you are taking

Let us take n = 2

Arithmetic mean of first 2 natural numbers = (1+2)/2 = 3/2

Now put n = 2 in all the four options and check which option is giving 3/2 as the output

A) 3/2

B) 6

C) 6

D) 3

Answer : (A)

But till now we have killed only a mouse, let us go and catch some lions.

Q . 4

I will take n = 1 (you can take n = 2 as well but that will make the calculations a little complex in this question). But I will solve this question by taking n = 1 and n = 2 both, so that you become well-versed with the concept.

n = 1

Sequence will reduce into  (1 - 1/(n + 1))

Put n = 1

Value = 1/2

Now put n = 1 in all the four options and check which one of them is giving 1/2 as the output

A) 1

B) 1/2

C) 2

D) 1

Answer : (B)

Now n = 2

Sequence will reduce into  (1 - 1/(n + 1)) + (1 - 2/(n + 1))

Put n = 2

Value = 2/3 + 1/3 = 1

Now put n = 2 in all the four options and check which one of them is giving 1 as the output

A) 2

B) 1

C) 3

D) 3/2

Answer : (B)

Remember in the exam you will take either n = 1 or n = 2, I have taken both just to explain you the concept. But in some questions you may find that after taking n = 1, you are getting two options with the same value. In such cases you will have to take n = 2.

Let me take some questions of 'Averages' in this article itself. Firstly an easy one - 

                                 Q. 5.  

Assume the numbers as 1, 2, 3, 4 and 5

Average of these numbers = 3

Include the next two numbers (6 and 7)

Average of 1, 2, 3, 4, 5, 6, 7 = 4

Average has increased from 3 to 4

Answer : (A)

Now a tougher one - 

Q . 6

This question appeared in Tier 2 (2015). It talks about average and hence it would be wise to at least take n = 2 or n = 3. Let us take n = 2.

It is given that 'Average of n numbers is a' and no other information is given, so you can assume the numbers as well. Since you have assumed n = 2, so you will have to assume two numbers. Let me take 1 and 3 as the two numbers (note : you can take any two numbers you want).

Average of 1 and 3 = 2

Hence a = 2

First number is increased by 2 and second number is increased by 4. Hence 1 and 3 will become 3 and 7.

Average of 3 and 7 = (3 + 7)/2 = 5

Now put a = 2 and n = 2 in all the four options and check which one of them is giving 5 as the output.

A) 5

B) 3.5

C) 7

D) 5.5

Answer : (A)

I hope the concept is clear now...

To buy the super-hit SSC Hack Book, follow the below link-

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[Q] Fast Calculation Tricks

There are some fast calculation hacks that can come in handy. Let's see them one by one -


1. Squaring a number with lightning fast speed

       Well you might have come across various tricks to find the square of a number, but I can bet you that my trick is the simplest and the fastest :)
Firstly memorize squares of numbers upto 25. I think many of you might be knowing it already, but for those who don't, please mug it. Now this trick will help you in squaring numbers upto 125 and trust me, for SSC you need this much only.

(a) For numbers close to 50

      (i) Numbers less than 50

        (ii) Numbers greater than 50
                  The process is almost same -  Compute (N - 50)
                  Add this difference to 25
                  E.g. 56*56=
                   1. 56 - 50 = 6
                   2. Last two digits = 6 * 6 = 36
                   3. First two digits = 25 + 6 = 31
                 Therefore, 56 * 56 = 3136

(b) Numbers close to 100

     (i) Numbers less than 100

      (ii) Numbers greater than 100
              The process is almost same -  Compute (N - 100)
              Add this difference to N
              E.g. 107*107=
              1. 107 - 100 = 7
              2. Last two digits = 7 * 7 = 49
              3. First three digits = 107 + 7 = 114
          Therefore, 107 * 107 = 11449

2. Squaring a number containing only '1'. E.g. 111, 1111, 1111111, etc.        Just count the number of ones. Like in 1111, there are 4 ones. So the square of 1111 is 1234321 (go from 1 to 4 and then back). Similarly, the square of 111111 = 12345654321

3. Finding product of two numbers of the form ab and ac, such that b+c=10. E.g. product of 43 and 47Last two digits = 3*7 = 21
First two digits = 4*(4+1) = 20
Hence 43*47 = 2021

4. Finding product of two numbers close to 100

(a) 103*109

Last two digits = 3*9 = 27

First three digits = (103 + 9) or (109 + 3) = 112

So 103*109 = 11227

(b) 92*97

Last two digits = 8*3 = 24  [100 - 92 = 8 and 100 - 97 = 3]

First two digits = (92 - 3) or (97 - 8) = 89

Hence 92*97 = 8924

Bonus Hack :

A very common question if SSC - 

I will tell you a formula that no 'paramount' will tell you :P

Note : Always take the difference between numerator and denominator as positive.

Answer : 3/(4-3) * 20 = 60 minutes

Next question - 

Rakesh runs at 5/4 of his usual speed and reaches the playground 5 minutes earlier. What is his usual time? Answer : 5/(5-4) * 5 = 25 minutes To buy the super-hit SSC Hack-Book, follow the below link- Buy SSC Hack-Book

[R] Reasoning Tricks - 1

The moment you receive your question paper, just write A=1, B=2, C=3...Z=26 at the back of it. This is better than memorizing the values because exam pressure is a bitch :) you may end up recalling wrong values. Moreover, it is easier to decipher the pattern if you have written down the alphabets on a paper.

Since the reasoning questions are pre-defined, I will try to give tricks for few of them.

1. 2-3 questions in reasoning section are vocabulary based. Many students leave these questions because they are confusing/difficult. But these questions are the easiest.

            (1)

 (2)

These questions may pose difficult if you don't know the meaning of the words giddiness, expunge, etc. But don't worry SSC itself solves these questions for you. How? Have a look at the hindi translation of these questions(which is usually given on the adjacent page)...

    (1)

(2)

Looking at this, you can easily decipher that the answers for the above questions are (A) and (C)

Always refer the hindi translation when ever you face any difficulty in solving vocabulary based questions of Reasoning section.

2. Then there are questions, where you have to arrange some words alphabetically. Don't commit the mistake of forming the entire sequence. Learn to solve such questions with options.

Like in the above question it is clear that the first 2 words will be Recollect and Remember. So the sequence will start with 1, 2, ...Stop here only and look at the options. Only option D starts with 1, 2...Don't go on arranging the rest of the words because it is not required and will only waste your time. Similarly for the below question

Smallest organism among these is the mosquito. Hence the sequence will start with 3. Option C is the only option starting with C. 

3. Now let us see a very famous question of CGL - Paper cutting. There are two rules to solve such questions - 

a) Note the position at which the cut is made(most of the questions can be solved with this rule only)

b) Count the layers

This will make sense if we take any example - 

To solve this question, just look at the position at which the cut is made (let's call it position X)

Now match the pattern of X with all the 4 options and see which option has exactly the same pattern.

Answer : C

Again you can see that option A matches the pattern

Answer : A

Both the questions above were solved with rule (a). Now let's see some questions where rule (b) is helpful - 

First of all apply rule (a)

Rule (a) only eliminates Option (2).

So here we will apply rule (b), i.e. counting the number of layers on which the cut is made.

When you are holding a piece of paper, you are actually holding 1 layer. When you fold it once, you will get 2 layers. Making a third fold will result in 4 layers and fourth fold will result in 8 layers.

In this question, the paper is folded twice and hence total no. of layers formed are 4.

There are total 4 layers and two holes are made on them.

So total holes in the final figure  = No of layers * No of holes made = 4*2 = 8

This means in the answer, there will be 8 holes in total

Options (1) and (2) have 8 holes, but option (2) was eliminated with rule (a). Hence 

Answer : (1)

Bonus Trick :

In SSC CGL, there is a very common question of nested square roots and there are total 3 cases for it (addition, subtraction and multiplication)

Case 1

In such questions, you should avoid forming quadratic equations because that will be time consuming. Here is the trick : Break the number written inside the square root into n*(n + 1) form. So 12 = 3*4

Answer is (n+1), i.e., 4

Similarly, for 72 (written as 8*9), the the answer will be 9; for 20(written as 4*5) the answer will be 5, and so on. In CGL questions, the number will always break into n*(n + 1) form.

Case 2

If instead of '+' you have '-' (above image), then

72 = 8*9

The answer is n, instead of n+1. So the above expression evaluates to 8.

If instead of + you have * (multiplication), then the answer is the number itself. E.g. - 

Case 3

Simple, isn't it? :)

I will cover more Reasoning topics in Part-2.

To buy the super-hit SSC Hack-Book, follow the below link-

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[Q] Trigonometry Tricks - 1

Trigonometry is yet another scoring section of CGL. Most of the questions can be solved with jugaad.

Just like Algebra, where we assumed the values of variables, in trigonometry we will assume the value of 'theta'. And like algebra, make sure the value you are assuming for theta will not make the denominator zero.
How to assume value -
1. When you don't have to deal with fractions you can assume ϴ = 90 or 0
E.g. (acosϴ - bsinϴ)
2. When fraction is given and putting ϴ = 90 or 0 is making denominator zero, then you can go with ϴ = 45.
3. Don't assume a value for theta at which the trigonometric function is not defined. E.g. When tanϴ is given, you can't assume ϴ = 90.
4. When you are assuming two angles, go with A = 60 and B = 30
Note : These are not hard and fast rules and you can assume any value you like, but make sure denominator ≠ 0. Sometimes when you assume ϴ, you may end up with two options(say A and B) that are giving similar results (but two options will still get eliminated, i.e., C and D). Now change the value of theta and check only A and B.

You only need to memorize the values of sin, cos and tan (for ϴ = 0, 30, 45 and 60). The values of cosec, sec and cot can be obtained by reciprocating sin, cos and tan respectively.
Now let's solve CGL questions

Q . 1. 

Put ϴ = 90

x = a(1 + 0) = a

y = b(1 - 0) = b

So x²/a² + y²/b² = 2

Answer : C

Some complex questions can be solved by assuming the value of theta

Q . 2.

Again assume ϴ = 90
Then a*1 + b*0 = cor a = c
You have to find the value of acosϴ - bsinϴ
acosϴ - bsinϴ = -b (Since ϴ = 90)
Now put a = c in all the 4 options to check which one can give '-b' as the output

Answer : D

Q . 3

Assume ϴ = 45

a (tan45 + cot45) = 1

So a = 1/2

sin45 + cos45 = b

So b = √2

Put a = 1/2 and b = √2 in all the options and check which among the 4 equations is right (i.e. LHS should be equal to RHS)

Answer : A (both LHS and RHS are equal to 1)

Assume ϴ = 45
(tanA - secA - 1)/(tanA + secA + 1) = -√2/(2 + √2) = -1/(√2 + 1) = 1 - √2 (rationalize)
Put ϴ = 45 in all the 4 options and check which one will give (1 - √2) as the output
A) √2  - 1
B) √2 + 1
C) 1 - √2
D) √2 - 1
Answer : C

Sometimes you would need to assume two angles

Q . 4

Let A = 60 and B = 30
Then n = 3 and m = √3
cos²A = 1/4 (since A = 60)
Now put n = 3 and m = √3 in all the options and check which one will give 1/4
Answer : B


There is one type of question which is frequently asked by SSC - 

Q . 5.

When you see secA + tanA = something (let's say 'p')   ... (1)
you can write, secA - tanA = 1/p                           ... (2)
Now add (1) and (2)
2secA = p + 1/p
secA = (p² + 1)/2p [You can memorize this formula]
tanA = (p² - 1)/2p

In the above question, p=2
So secA = 5/4
Now we have to find sinA. The best way to determine the value of a trigonometric function when the value of other function is given, is by making a triangle.
secϴ = Hypotenuse/Base
Here secA = 5/4, hence hypotenuse = 5 and base = 4, which means perpendicular = 3
sinA = perpendicular/hypotenuse = 3/5 = 0.6
Answer : C



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How to approach Quantitative Aptitude section

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I would like to share some tips for Quantitative Aptitude section. The questions asked in CGL are very basic and anyone can solve them, but the difference lies in the fact that how much time you are taking in each one of them. Try to solve the questions just by looking at the options(without writing a single word). For e.g. let's take the questions that appeared in CGL 2015 tier-2 - 

Now this an easy question and everyone can solve it by making linear equations or any other method. But first see if this question really deserves your rough space. Since the monthly incomes of A and B are 8x and 5x, their difference will be 3x (or a multiple of 3). Rs. 42,000 is the only option possible because it is the only multiple of 3.

Next question -

Among the given options, 24 is the only number in which the units place digit is twice the tens place digit. No need to form equations.

Next-

HCF of the numbers is 21, hence both the numbers are divisible by 21. By looking at the options we find that 84 is the only number that satisfies this condition.

At least 15 questions can be solved by this method, and the time saved here can be utilized in solving other complex questions...

How to prepare for SSC CGL ? (Book-list and Tips)

Even one month of serious studies can prove fruitful for CGL. But you will have to use your days judiciously. Some tips from my side-

1. Your most important source is the previous year question papers because many questions are repeated as it is. I just purchased a single book(Kiran Prakashan) containing previous year papers and solved it 3-4 times. After completing that book, I got the feeling that 75% syllabus is al ready covered.

2. General Awareness - This section has the least benefit-cost ratio. You will invest more time, but your return will be low. I will recommend you not to waste much time here(at max 1 hour daily). Moreover I have found that Lucent's GK book/Manorama Year Book has the best compilation, especially for CGL. So instead of referring multiple sources and wasting your limited time, just go for them(either of the two).

3. Reasoning - Most scoring section of Tier-1, because the questions are pre-defined. Just practise the question papers, and you will definitely score well (you should at least get 40+ here, in less than 30 minutes).

4. Quantitative Aptitude - Most important part of the syllabus. This is the section where you should devote most of your time. Just purchase any good book (Kiran's Quickest Mathematics is fine) and practise thoroughly. Mugging shortcuts will help you to get an edge over others. Try solving every question within 1 minute.

5. English - Most tricky portion. Every section of this portion is unpredictable. You can't guess which idioms, antonyms, synonyms, phrasal verb, etc. will be asked.
   Error correction - at least prepare the rules used in previous year papers
   Idioms, One word substitution, Synonyms, etc. - Learn 10 words each, daily.
   Direct- Indirect and active-passive- Very scoring. Just mug the rules
   Comprehension - You can't do much here. Reading news paper can help.

I think I have touched every portion and I will re-iterate, don't ignore the previous year question papers.

Book-list

First of all, own publicity :)
I wrote an ebook containing all the shortcuts one needs to know for clearing CGL. It covers all the topics of quantitative aptitude, reasoning and some important topics of english. You can purchase it from here-

SSC Hack book

Moving on to other books-

1) Quantitative Aptitude

         (i)
             
             OR

              

          (ii)
                

2) English

           (i) Main reference book

                 

             OR

                 

              (ii) For vocabulary
                   Normal Lewis
                    
                 OR
                 MB Publication (good for Idioms, Phrasal Verbs, One Word Substitution, etc.)
                    

                (iii) For grammar
                    

3) General Awareness

              (i)

                   

               (ii)
                   

4) Previous Year Papers (a single book for both Tier-1 and Tier-2) - HIGHLY RECOMMENDED!

                    

5) Ultimate Guide to SSC CGL (covers entire syllabus including Reasoning)

                   

I would recommend you to buy only one book for each section (unless you can finish both). For English, I have mentioned two books : Plinth to Paramount covers grammar rules well and SP Bakshi is good for idioms, phrasal verbs, vocabulary, etc. You can buy both the books if you want. Moreover, I haven't explicitly mentioned any book for Reasoning section, as it will be covered by Ultimate Guide to SSC CGL (book no. 5 above) and previous year papers. But still if you think your reasoning in poor, you can go for either of the two books given below(although they are more suited for Banking exams)-

6) Reasoning

       (i)

                    
        (ii)

                    


Finally to give final touch to your preparation, solve mock tests, that will help you in time-management during the exams-

(7) Mock Tests

           (i)

                   

            (ii)

                    

Although both the books contain errors, but these are the only ones available in the market. So buy any one of them.

Best of Luck :)