Quantitative Aptitude Section in **SSC CGL Exam** is considered to be the toughest one. The candidates need more accuracy and speed to attempt all the questions of this section.

**Take online mock test papers and solve previous years question papers to boost up your speed and accuracy in SSC CGL Tier-1. Grab them for free now!**

As in the previous blog of **SSC CGL Tier-1 30 Days Study Plan**, we have discussed the time allocation strategy during exam.

According to this strategy the aspirants are required to solve 25 questions in 30 minutes only. You can’t spend more than 30 minutes on Quantitative Aptitude Section as you’ll have to cover other sections as well.

In such situation when you have to solve a mathematics question in less than 1 minute duration, you need to practice really hard before appearing in the exam.

In this blog below, we will benefit you with the strategies and tips to attempt QA questions during exams. Also I’ll share a list of **SSC CGL Math formulas** that will be beneficial in quick revision.

The Quantitative Aptitude Section can benefit you with a total of 50 marks if you attempt all the 25 questions correctly. And if you answer the questions incorrectly, then you can lose 0.50 marks on each wrong answer.

So wisely analyze the question paper before you start attempting the questions.

** First of all, understand the nature of all the questions like- which question will take more time to solve and which question will take less time.**

** After that, calculate your marks according to the less time taking questions.**

** If you can score well within short questions only, then take them at first. And after solving them, if you have enough time left to solve lengthy questions, then you can attempt those otherwise leave.**

** Don’t try to attempt the questions by using long methods. Quantitative Aptitude questions can be solved within short tricks and methods.**

** Maintain speed and accuracy while solving the questions.**

In order to conquer the Quantitative Aptitude Section with good marks, you need to develop a great determination and confidence.

There’s nothing like reading and learning in Mathematics. The game is all about logical and mental abilities and presence of mind to make a quick response while solving questions.

** You need to understand the ****Math aptitude questions’** nature first and then solve them in minimal time duration with the smartest idea you can use.

** The most basic four things in Mathematics are – Addition, Subtraction, Division and Multiplication.**

These calculations are very basic in mathematics which are used in most of the questions. So if you perform them faster, you can save your valuable time in each question.

** Remember multiplication tables up to 20. If you remember the tables, the calculations would be easier and faster as well.**

Check the example here.

If you have to find out the result of 16*13, then you can simple solve this as –

16(10+3) = 16*10 + 16*3 = 160+48 = 208

And 13(10+6) = 130+78=208

So the answer of 16*13 is 208.

** Learn the short tricks, solve quicker examples and clear your fundamentals to attempt more questions in exam.**

** Solve ****SSC CGL Previous Years Question Papers and Sample Test Papers** to assess your performance and make more improvements.

Now find out the reasons of your low performance and try to resolve them.

** Revise with ****SSC Math quick notes**, containing formulas and short tricks.

**It’s better to take online mock tests or test series from 1 month before the exam and note down your each day performance.**

Follow the above strategies to start your preparation for Quantitative Aptitude Section in SSC CGL Tier-1.

“Success is the progressive realization of a worthy goal! – Earl Nightingale”

As per the **SSC CGL Tier-1 Syllabus** of Quantitative Aptitude, we have shortlisted the important topics that should be taken at the first while preparing for the QA section.

To reach your goal of SSC CGL qualification, you will have to prepare a perfect study plan and implement it as well.

You can check out the **SSC CGL daily study plan till 1 month** and follow it to boost up your learning techniques.

Here in this section, we are going to benefit you with the quick revision methods and formulas.

To ease out your revision anxiety, Eduncle has come up with a SSC CGL Tier-1 Maths Short Tricks and Formulas PDF.

Please check below:

** 1 + 2 + 3 + 4 + 5 + … + n = n(n + 1)/2**** (12 + 22 + 32 + ..... + n2) = n ( n + 1 ) (2n + 1) / 6**** (13 + 23 + 33 + ..... + n3) = (n(n + 1)/ 2)2**** Sum of first n odd numbers = n2**** Sum of first n even numbers = n (n + 1)**** (a + b)*(a - b) = (a2 - b2)**** (a + b)*2 = (a2 + b2 + 2ab)**** (a - b)*2 = (a2 + b2 - 2ab)**** (a + b + c)*2 = a2 + b2 + c2 + 2(ab + bc + ca)**** (a3 + b3) = (a + b)*(a2 - ab + b2)**** (a3 - b3) = (a - b)*(a2 + ab + b2)**** (a3 + b3 + c3 - 3abc) = (a + b + c)*(a2 + b2 + c2 - ab - bc - ac)**** When a + b + c = 0, then a3 + b3 + c3 = 3abc**** (a + b)*n = an + (nC1)*an-1b + (nC2)*an-2b2 + … + (nCn-1)*abn-1 + bn.**

For more Number System Short Tricks Methods, please visit - **How to Solve Number System Questions in Exams **

But a*b*c ≠ HCF*LCM

If number a, divided another number b exactly, we say that a is a factor of b.

In this case, b is called a multiple of a.

Two numbers are said to be co-primes if their H.C.F. is 1.

Break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers.

The product will be the required HCF.

If you have to find the HCF of 42 and 70.

Then 42 = 2*3*7

And 72 = 2*5*7

Common factors are – 2 and 7 so, HCF = 2*7 =14.

Divide the greater number by the smaller number, divide the divisor by the remainder, divide the remainder by the next remainder and so on until no remainder is left. The last divisor is required HCF.

To check the complete procedure in example format, then please visit here – **Number System Division and Remainder Rules**.

If you have to find out the HCF of 42237 and 75582.

Then first check out for the common factor –

42237 = 4693*9

75582 = 2*9*4199

Here we can remove 9 and 2 is a prime number so we can also extract this.

Now calculate with remaining numbers.

After performing Division method - 4693/4199, we get the remainder 494.

Now the remainder 494 is divided by 2 but divisor 4199 is not.

So it should be proceeded further after dividing 494 by 2 i.e. 247.

Now performing division method as – 4199/247, we get the remainder as 0.

Now by multiplying 247 with 9 we can have our required HCF i.e. 247*9=2223.

Resolve the given numbers into their Prime Factors and then find the product of the highest power of all the factors that occur in the given numbers. The product will be the LCM.

LCM of 8,12,15 and 21.

Now 8 = 2*2*2 = 2^{3}

12 = 2*2*3 = 2^{2}*3

15 = 3*5

21 = 3*7

So highest power factors occurred are – 2^{3}, 3, 5 and 7

LCM = 2^{3}*3*5*7 = 840.

**'BODMAS' Rule**

**Through this rule, you can understand the correct sequence in which the operations are to be executed and**

This rule depicts the correct sequence in which the operations are to be executed and the sequence can be evaluated.

Here are some rules of simplification given below-

(First of all remove all the brackets strictly in the order (), {} and || and after removing the brackets, you can follow the below sequence)

Modulus of a real number a is defined as

|a| = | a, if a > 0 |

-a, if a < 0 |

Thus, |5| = 5 and |-5| = -(-5) = 5.

When an expression contains Vinculum, before applying the 'BODMAS' rule, we simplify the expression under the Vinculum.

**Duplex Combination Method for Squaring**

**In this method either we simply calculate the square by multiplying the same digit twice or we have to perform cross multiplication.**

Here are the following Duplex rules and formulas, please check below.

Now I’ll make you understand the complete squaring procedure in a better way with the help of examples.

We have to find out the solution of (207)^{2} instantly.

Then,

(207)^{2} = D for 2 / D for 20 / D for 207 / D for 07 / D for 7

(207)^{2 }= 2*2 / 2*(2*0) / 2*(2*7) + 0^{2 }/ 2*(0*7) / 7*7

(207)^{2 }= 4/0/28/0/49

(207)^{2 }= 4/0/_{2}8/0/_{4}9

(207)^{2 }= 4 / (0+2) / 8 / (0+4) / 9

(207)^{2 }= 4 / 2 / 8 / 4 / 9

I would like to explain the cube method through example only.

Find out the result of (16)^{3}

Here we’ll write like this –

1 6 (6*6) (6*6*6) = 1 6 36 216

To find out the cube, it will be solved like this –

1 6 36 216

12 72

__________________

4 _{3}0 _{12}9 _{21}6 = 4096.

**Formulas -**** **If the current age is x, then n times the age is nx.** **If the current age is x, then age n years later/hence = x + n.** **If the current age is x, then age n years ago = x - n.** **The ages in a ratio a : b will be ax and bx.** **If the current age is x, then 1/n times the age is x/n.

** **

**Son’s Age = t _{1 }(x-1) / (x-y)**

**Son’s Age = (z-1)t _{1 }/ (y-z)**

**Son’s Age = **

** **Average = (Total of data) / (No. of data)

** **

Increase in expenditure = y

Increase in no. of students = z

And number of students (originally) = N, then

The original expenditure = N *

** Profit = Selling Price (SP) – Cost Price (CP)**

** **

** **If M1 persons can do W1 work in D1 days and M2 persons can do W2 work in D2 days then the formula will be -

**M1 * D1 * W1 = M2 * D2 * W2**

** **

**M1 * D1 * T1 * W1 = M2 * D2 * T2 * W2**

**M1 * D1 * T1 * E1 * W1 = M2 * D2 * T2 * E2 * W2**

** **If a pipe can fill a tank in x hours, then the part filled in 1 hour = 1/x.

** **

Time taken to fill the tank, when both the pipes are opened = xy / (y-x).

Time taken to fill the tank = xy / (x+y).

Time taken to fill the tank = xyz / (yz + xz - xy) hours.

** **Speed = Distance / Time

** **

** **When x and y trains are moving in opposite direction, then their **relative speed = Speed of x + Speed of y**

** **

** **If the speed of the boat is x and if the speed of the stream is y while upstream then the effective speed of the boat is = x - y

** **

Man’s rate with current = x + y

Man’s rate against current = x – y

** **If the gradients are mixed in a ratio, then

** **SI = p*t*r/100

** **

** **When Interest is compounded annually –

** Amount = P **

** **Area of Rectangle = Length * Breadth

** **

** **

**To get the above formulas in PDF Format, you can click here - SSC CGL Tier-1 Short Tricks and Formulas PDF.**

So, aspirants above, we have catered each and every topic and their important formulas from the SSC CGL Tier-1 Syllabus for **SSC CGL Math preparation**.

Hoping that the above formulas will be helpful to you while doing quick revision. If you want to share out any of your suggestions or feedbacks, then you are most welcomed to the comment section below.

We’ll try to get it touch with you as soon as possible.

To get the above formulas in PDF Format for free, you can visit to the Eduncle’s Free Downloads Section.

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