Tips and Tricks to score 40+ in SSC CGL Maths Tier I 2020
Preparing for the upcoming SSC CGL Exam 2020? Here are a few Tips, Tricks & Practices to score 40+ in SSC CGL Maths or Quantitative Aptitude examination. Go through the post to get expert guidance for clearing all stages of the SSC CGL Exam.
- It is common that Quantitative Aptitude is the trickiest section and students tend to find it difficult.
- To ease the candidate’s worries about clearing the Maths section, here are tips and tricks to score more marks in the SSC CGL Maths section.
- All the sections in the SSC CGL Exam are qualifying. It is mandatory for candidates to score above the cut off in order to get shortlisted for the next stage of the exam.
- If you are looking to get the best result, then joining the best SSC CGL Online Coaching will help in clearing all stages of CGL.
Direct Link to Download SSC CGL Tier 1 Admit Card 2020
Tips and Tricks to score 40+ in SSC CGL Maths Tier I 2020
Tips and Tricks to score 40+ in SSC CGL Maths:- Mathematics or quantitative aptitude is one of the most important sections with regards to exam preparation for SSC CGL. One must give extra emphasis on SSC CGL Maths Preparation in order to score well in the examination.
- In SSC CGL Tier I exam, Quantitative Aptitude/ Maths carry 25 questions for 50 marks.
- To learn tips, tricks & best practices for SSC CGL Maths Preparation for Tier I exam, read our post on Quantitative Aptitude Preparation Tips for SSC CGL
- Strategizing your exam preparation for SSC CGL is the foremost thing one must do in order to pave a way for successful exam preparation, read our article on SSC CGL Preparation Tips to learn more.
Must learn Tips and Tricks to score 40+ in SSC CGL Maths 2020
- Do not try to jump directly to the test series. The very first thing you need to do is to work on your fundamentals. Try to cover every topic from every chapter and solve as many questions as possible to smoothen your basics.
- Once you have learned the basics, solve some good problems ( Level 1 of Quantum CAT ) which will check your conceptual skill which you learned during initial preparation.
- Take up the SSC CGL Mock Test Series regularly to analyze your preparation levels.
- Do not waste your time in doing advanced level questions of Quantum CAT or from any other book if you have got less time. It’s better that you utilize that time in preparing GA and English.
- Do not try to solve problems from different books every time you start solving questions. Try to keep your resources limited and solve problems from the same book you learned basics from.
Level of Difficulty – Tips and Tricks to score 40+ in SSC CGL Maths
Now let’s break the Maths/ Quantitative Aptitude syllabus into smaller segments in order to have a clear vision of what needs to be given more emphasis during Quantitative Aptitude/ Maths Preparation for SSC CGL Tier I examination.
Basically, questions in SSC CGL Maths is prepared from various difficulty levels. One must start with the easiest sections in order to prepare efficiently.
Easy Topics |
No of Questions (Tier I) | Moderate Topics | No of Questions (Tier I) | Difficult Topics | No of Questions (Tier I) |
Simplification | 1-2 Qs | Speed Time and Distance | 1-2 Qs | Time and Work | 1-2 Qs |
Interest | 1-2 Qs | Profit and Loss | 2 Qs | Mixture Problems | 1 Qs |
Percentage | 1 Qs | Number Series | 1 Qs | Algebra | 2-3 Qs |
Average | 1 Qs | Mensuration | 2-3 Qs | Geometry | 3-4 Qs |
Ratio & Proportion | 1-2 Qs | Data Interpretation | 2-4 Qs | Trigonometry | 2-3 Qs |
Problems on Ages | 1 Qs | Number System | 1 Qs | – | – |
Level of Difficulty
As mentioned in the table above, we have categorized SSC CGL Maths Topics as per their level of difficulty. Well, the trick is still the same. Start with the basics, perfect yourself in those basic areas and keep on revising until you become proficient.
- Easy Topics will cover 8- 10 questions which almost everyone will be able to solve and it can easily fetch you a little less than the target of 40 marks in SSC CGL Maths.
- Moderate Topics will cover another 8-10 questions but this section is a bit lengthy in terms of the length of the solution. so, remain calm while solving one of those and you will boss it soon.
- Difficult Topics will cover the rest. To say, you can expect 6-8 Questions and this will be the icebreaker because if you can solve at least half of them, you can score well. Try to be proficient in this section and it will help you immensely.
If you follow these steps and understand the nature of the exam well, you can definitely score 40 plus marks in SSC CGL Maths.
SSC CGL Maths Short Tricks and Tips
Number System Quick Maths Formulas
- 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
HCF and LCM Quick Maths Formulas
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.
Co-primes
Two numbers are said to be co-prime 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 the 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.
Simplification Quick Maths Formulas
‘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.
Square roots & Cube Roots Quick Maths Formulas
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
Problems on Ages Quick Maths Formulas
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 Quick Maths Formulas
- 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 Quick Maths Formulas
- 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 the quantity is taken by the first, y% of the remaining is taken by the second and z% of the remaining is taken by the 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 the quantity is added. Again y% of the 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 and Loss Quick Maths Formulas
- 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).
Time and Work Quick Maths Formulas
- 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
Pipe & Cisterns Quick Maths Formulas
- 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 them 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 is taken by the leak to empty the tank = xy / (y-x) hours.
Time and Distance Quick Maths Formulas
- 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)
Problem on Train Quick Maths Formulas
- When x and y trains are moving in the opposite direction, then their relative speed = Speed of x + Speed of y
- When x and y trains are moving in the 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)
Boats and Streams Quick Maths Formulas
- 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 the 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)
Allegation Quick Maths Formulas
Simple Interest Quick Maths Formulas
Compound Interest Quick Maths Formulas
Mensuration Quick Maths Formulas
- 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}.
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- 90-100 Hours of Online Live Classes. Each class of 2 hours.
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- Live Classes & Videos can be played on the Web and App both
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