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**Greetings Aspirants!**

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.

**So, Aspirants… Let’s begin here! **

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.**

* Note: The Cut off marks for Quantitative Aptitude Section will be decided on the basis of number of questions which were answered similarly by the aspirants.*

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.

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

For more Number System Short Tricks Methods, please visit - __How to Solve Number System Questions in Exams__

**Product of two numbers a and b **

**(a*b) = Their HCF * Their LCM.**

**But a*b*c ≠ HCF*LCM**

__Note__:

**HCF of two or more numbers is the greatest number which divide all of them without any remainder. **

**LCM of two or more numbers is the smallest number which is divisible by all the given numbers.**

**HCF of given fractions = (HCF of Numerator)/(LCM of Denominator) **

**HCF of given fractions ****= (LCM of Denominator)/(HCF of Numerator) **

**If d = HCF of a and b, then there exist unique integer m and n, such that d = am + bn.**

** **

**Some important HCF and LCM Rules- Factors and Multiples**

**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.**

** **

**Co-primes**

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

**HCF of a given number always divides its LCM.**

** **

**Methods of finding HCF of two or more numbers **

**Method 1: Prime Factors Method**

**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.**

**Example **

**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.**

** **

**Method 2: Division Method**

**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.**

**Method 3: HCF of Large Numbers**

**Find the obvious common factor from both the numbers and remove it. Also remove the prime number (if any found). Now perform division method to remaining numbers and find the HCF. Check out the example for better understanding.**

**Methods of finding LCM of two or more numbers **

**Method 1: Prime Factors Method**

**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.**

**Example**

**LCM of 8,12,15 and 21.**

**Now 8 = 2*2*2 = 23**

**12 = 2*2*3 = 22*3**

**15 = 3*5**

**21 = 3*7**

**So highest power factors occurred are – 23, 3, 5 and 7**

**LCM = 23*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-**

B – Bracket

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

O – Of

D - Division,

M - Multiplication,

A - Addition and

S - Subtraction

** **

**Modulus of a Real Number**

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.

** **

**Vinculum (or Bar):**

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.

** **a = D = (a*a)** **

**ab = D = 2*(a*b) **

**abc = D = 2*(a*c)+(b)2 **

**abcd = D = 2*(a*d) + 2*(b*c) **

**abcde = D = 2*(a*e) + 2*(b*d) + (c)2 **

**abcdef = D = 2*(a*f)2 + 2*(b*e)2 + 2*(c*d)2**

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

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) + 02 / 2*(0*7) / 7*7

(207)2 = 4/0/28/0/49

(207)2 = 4/0/28/0/49

(207)2 = 4 / (0+2) / 8 / (0+4) / 9

(207)2 = 4 / 2 / 8 / 4 / 9

(207)2 = 42849.

**Easy Method to calculate Cube**

We 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 0 9 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.

** **

** Quicker Methods – **To find out son’s age, use this formula

If t1 years earlier the father’s age was x times that of his son. At present the father’s age is y times that of his son. Then the present age of son will be?

**Son’s Age = t1 (x-1) / (x-y)**

** **

If present age of the father is y times the age of his son. After t2 years the father’s age become z times the age of his son. Then the present age of son will be?

**Son’s Age = (z-1)t1 / (y-z)**

t_{1} years earlier, the age of the father was x times the age of his son. After t_{2} years, the father’s age becomes x times the age of his son. Then the present age of son will be?

**Son’s Age = [(z-1)t _{2 } + (x-1)t_{1}] / (x-z)**

Son’s or Daughter’s Age = [Total ages + No. of years ago (Times – 1)] / (Times+1)

Son’s or Daughter’s Age = [Total ages – No. of years ago (Times – 1)] / (Times+1)

Father : Son

Present Age = x : y

T years before = a : b

Then, **Son’s age = y * [ T(a-b) / Difference of cross product ]**

And **Father’s age = x * [ T(a-b) / Difference of cross product ]**

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

Age of New Entrant = New Average + No. of Old Members * Increase

Weight of New Person = Weight of Removed Person + No. Of Persons * Increase In Average

Number of Passed Candidates = Total Candidates * (Total Average – Failed Average) / (Passed Average – Failed Average)

Number of Failed Candidates = Total Candidates * (Passed Average – Total Average) / (Passed Average – Failed Average)

Age of New Person = Age of Removed Person – No. of Persons * Decrease in Average Age

Average after x innings = Total Score – Increment in Average * y innings

If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by – 2xy / (x+y) km/hr.

If half of the journey is travelled at a speed of x km/hr and the next half at a speed of y km/hr, then average speed during the whole journey is 2xy / (x+y) km/hr.

If a man goes to a certain place at a speed of x km/hr and returns to the original place at a speed of y km/hr, then the average speed during up and down journey is 2xy / (x+y) km/hr.

If a person travels 3 equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz / (xy+yz+zx) km/hr.

**Percentage = [Value / Total Value * 100]**

If two values are respectively x% and y% more than a third value, then the first is the **(100+x) / (100+y) *100%** of the second.

If A is x% of C and B is y% of C, then A is **x/y*100% of B**.

x% of quantity is taken by the first, y% of the remaining is taken by the second and z% of the remaining is taken by third person. Now is A is left in the fund then there was (**A*100*100*100) / (100-x) (100-y) (100-z)** in the beginning.

x% of quantity is added. Again y% of increased quantity is added. Again z% of the increased quantity is added. Now, it becomes A, then the initial amount is given by** (A*100*100*100) / (100+x) (100+y) (100+z)**

If the original population of a town is P and the annual increase is r%, then the population in n years will be –** ****P + P*r/100 = P*(1+r/100)**

The population of a town is P. It increases by x% during the 1^{st} year, increases by y% during the 2^{nd} year and again increases by z% during the third year. Then, the population after 3 years will be –** ****P*(100+x)(100+y)(100+z) / 100*100*100**

When the population decreases by y% during the 2nd year, while for the 1st and 3rd years, it follows the same, the population after 3 years will be – **P*100+x)(100-y)(100+z) / 100*100*100**

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

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

** **** Gain or Loss % = **(Loss or Gain / CP) * 100 %

Gain % = [Error / (True Value – Error)] * 100 %

Gain % = [(True Weight – False Weight) / False Weight] * 100 %

Total % Profit = [(% Profit + % Less in wt) / (100 – % Less in wt)] * 100 %

If CP of x articles is = SP of y articles, then **Profit % =** **[(x –y) / y] * 100**

Cost Price = (100 * More Charge) / (% Diff in Profit)

Selling Price = More Charge * (100+ First Profit%) / (% Diff in Profit).

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

If we add Time for both the groups T1 and T2 respectively, then the formula will become – M1 * D1 * T1 * W1 = M2 * D2 * T2 * W2

And if we add efficiency for both the groups E1 and E2 respectively, then the formula becomes – M1 * D1 * T1 * E1 * W1 = M2 * D2 * T2 * E2 * W2

If A can do a piece of work in x days and B can do it in y days, then A and B working together will do the same work in **[(x*y)/(x+y)]**

If A, B and C can do a work in x, y and z days respectively, then all of them working together can finish the work in **[(x*y*z) / (xy + yz + zx)]**

If A and B together can do a piece of work in x days and A can do it in y days, then B alone can do the work in** (x*y) / (x-y)**

Original Number of Workers = (No. of more workers * No. of days taken by the second group) / No. of less days

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

If a pipe can empty a tank in y hours, then the part of the full tank emptied in 1 hour = 1/y.

If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, then the net part filled in 1 hour, when both the pipes are opened = (1/x) – (1/y).

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

If a pipe can fill a tank in x hours and another can fill the same tank in y hours, then the net part filled in 1 hour, when both the pipes are opened = (1/x) + (1/y).

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

If a pipe can fill a tank in x hours and another can fill the same tank in y hours, but a third one empties the full tank in z hours and all of thm are opened together, then the net part filled in 1 hour = (1/x) + (1/y) + (1/z).

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

A pipe can fill a tank in x hours. Due to a leak in the bottom, it is filled in y hours. If the tank is full, then the time taken by the leak to empty the tank = xy / (y-x) hours.

Speed = Distance / Time

If the speed of a body is changed in the ratio a:b, then the ratio of the time taken changes in the ration b:a.

If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, then the average speed during the whole journey is 2xy / (x+y) km/hr.

Meeting point’s distance from starting point = (S1 * S2 * Difference in time) / (Difference in speed)

Distance travelled by A = 2 * Distance of two points (a / a+b)

Distance = [(Multiplication of speeds) / (Difference of Speeds)] * (Difference in time to cover the distance)

Meeting Time = (First’s starting time) + [(Time taken by first) * (2nd’s arrival time – 1st’s starting time)] / (Sum of time taken by both)

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

When x and y trains are moving in same direction, then their relative speed = Speed of x - Speed of y

When a train passes a platform, it should travel the length equal to the sum of the lengths of train & platform both.

Distance = (Difference in Distance) * [(Sum of Speed) / (Diff in Speed)]

Length of Train = [(Length of Platform) / (Difference in Time)] * (Time taken to cross a stationary pole or man)

Speed of faster train = (Average length of two trains) * [(1/Opposite Direction’s Time) + (1/Same Direction’s Time)]

Speed of slower train = (Average length of two trains) * [(1/Opposite Direction’s Time) – (1/Same Direction’s Time)]

Length of the train = [(Difference in Speed of two men) * T1 * T2)] / (T2-T1)

Length of the train = [(Difference in Speed) * T1 * T2)] / (T1-T2)

Length of the train = [(Time to pass a pole) * (Length of the platform)] / (Diff in time to cross a pole and platform)

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

And if downstream then the speed of the boat = x + y

If x km/hr be the man’s rate in still water and y km/hr is the rate of the current. Then

Man’s rate with current = x + y

Man’s rate against current = x – y

A man can row x km/hr in still water. If in a stream which is flowing at y km/hr, it takes him z hrs to row to a place and back, the distance between the two places is = z * (x2 – y2) / 2x

A man rows a certain distance downstream in x hours and returns the same distance in y hours. If the stream flows at the rate of z km/hr, then the speed of the man in still water is given by – z* (x + y) / (y - x) km/hr.

Man’s rate against current = Man’s rate with current – 2 * rate of current

Distance = Total Time * [(Speed in still water)^{2} – (Speed of current)^{2}] / 2 * (Speed in still water)

Speed in Still Water = [(Rate of Stream) * (Sum of upstream and downstream time)] / (Diff of upstream and downstream time)

If the gradients are mixed in a ratio, then

[(Quantity of cheaper) / (Quantity of dearer)] = [(CP of dearer) – (Mean Price)] / (Mean price) – (CP of cheaper)]

Quantity of Sugar Added = [Solution * (Required% value – Present% value)] / (100 – required% value)

Required quantity of water to be added = [Solution * (Required Fractional Value – Present Fractional Value)] / 1 – (Required Fractional Value)

SI = p*t*r/100

The annual payment that will discharge a debt of INR A due in t years at the rate of interest r% per annum is = **(100 * A) / [(100 * t) + r*t* (t-1)]/2**

P = (Interest * 100) / [(t1*r1) + (t2*r2) + (t3*r3) + …..]

Rate = [100 * (Multiple number of principal – 1)] / Time

Sum = (More Interest * 100) / (Time * More Rate)

When Interest is compounded annually – **Amount = P [1 + (r/100)] ^{t}**

When Interest is compounded half-yearly – **Amount = P [1 + (r/200)] ^{2t}**

When Interest is compounded quarterly –** Amount = P [1 + (r/400)] ^{4t}**

When rate of Interest is r1%, r2% and r3% then –** Amount = P [1 + (r1/100)] * [1 + (r2/100)] * [1 + (r3/100)]**

Simple Interest for 2 years = 2*r = 2r% of capital

Compound Interest for 2 years = [2r +(r^{2}/100)]% of capital

Simple Interest for 3 years = 3*r = 3r% of capital

Compound Interest for 3 years = [3r +(3r^{2}/100) + (r^{3}/100^{2})]% of capital.

Area of Rectangle = Length * Breadth

(Diagonal of Rectangle)^{2} = (Length)^{2} * (Breadth)^{2}

Perimeter of Rectangle = 2 * (Length + Breadth)

Area of a Square = (Side)^{2} = 1/2 * (Diagonal)^{2}

Perimeter of Square = 4 * Side

Area of 4 walls of a room = 2 * (Length + Breadth) * Height

Area of a parallelogram = (Base * Height)

Area of a rhombus = 1/2 * (Product of Diagonals)

Area of a Equilateral Triangle = Root of (3) / 4 * (Side)^{2}

Perimeter of an Equilateral Triangle = 3 * Side

Area of an Isosceles Triangle = b/4 * root of 4a^{2} – b^{2}

Area of Triangle = 1/2 * Base * Height

Area of Triangle = root of [s(s – a) * (s – b) * (s-c)]

Area of Trapezium = 1/2 * (Sum of parallel sides * perpendicular distance between them)

Circumference of a circle = 2*(22/7)*r

Area of a circle = (22/7) * r^{2}

Area of a parallelogram = 2 * root of [s(s – a) * (s – b) * (s-d)]

Volume of cuboid = (l*b*h)

Whole Surface of cuboid = 2 * (lb + bh + lh) sq. units

Diagonal of Cuboid = Root of (l^{2} + b^{2} + h^{2})

Volume of a cube = a^{3}

Whole Surface Area of cube = (6*a^{2})

Diagonal of Cube = Root of (3) * a

Volume of Cylinder = (22/7) * r^{2} * h

Curved Surface area of Cylinder = 2*(22/7)*r*h

Total Surface Area of Cylinder = [2*(22/7)*r*h] + {2*(22/7)*r^{2})

Volume of Sphere = (4/3) * (22/7) * r^{3}

Surface Area of Sphere = 4 * (22/7) * r^{2}

Volume of hemisphere = (2/3) * (22/7) * r^{3}

Curved Surface area of hemisphere = 2 * (22/7) * r^{2}

Whole Surface Area of hemisphere = 3 * (22/7) * r^{2}.

** **

** **

**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.

**Thank You !!**

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