Any competitive exam requires you to deal with maths and its applications. Not all can be Ramanujam, but with certain techniques and procedures, we can definitely be in a better position to deal with them and score well.
In fact, these maths tricks for competitive exams can be mastered not just for exam purposes but also to become efficient in solving problems. Remember anything learned never goes wasted.
So read on and practice the methods mentioned to become a whiz kid in Maths!
Finding the square root:
Calculating a square root takes uptime, which is precious during an exam. So you have to know a quicker way of doing it. Let us teach you how to work on a trick to make it happen.
Find the square root of 784.
Step 1: Check the digit at one’s place. Here it is 4.
Step 2 : Think about the square roots of all the digits from 1-10. Ask yourself, in which square root the digit 4 crops up at one’s place?
At 2 2 = 4 and 8 2 = 64.
Step 3 :
Now we look at the remaining number i.e. 7. To which number’s square root is 7 closer to ?
2 2 = 4 and 3 2 =9. We will opt for the smaller of the two numbers, i.e. 2.
Add the 2 to 2 and 8.
2 2 2 8
Add 1 to 2 , 1 + 2 = 3. Multiply 2 x 3 = 6.
Compare 7 to the product 6. Which is greater? 7 is. So 28 is our answer.
To check our answer : 28 x 28 = 784.
Practicing the technique a few times, it will increase your speed and you will have your answers in a jiffy!
Tip : It will help you if you can learn the square roots of numbers from 1-10
Tip : And if you observe, you will see the following pattern:
Unit’s place digit of both 12 and 92 is 1.
Unit’s place digit of both 22 and 82 is 4
Unit’s place digit of both 32 and 72 is 9
And unit’s place digit of both 42 and 62 is 6.
Find the cube root:
Similar to finding a technique to quickly calculate the square root, we have a trick for calculating cube root as well.
Find the cube of 39304.
Step 1 : Check the digit at one’s place: 4.
Step 2: Think of all the cube roots of numbers between 1-10, where 4 is in one’s place.
4 3 = 64. So we will put the number 4, whose cube root is 64, in the one’s place of the cube root of 39304.
∛39304 = ______ 4
Step 3: Now take the first two digits of the number, 39 and think between which two cube roots does it get placed.
3 3 = 27 and 4 3 = 64
Step 4: We will choose the smaller number which is 27and take the number it is the cube root value of i.e. 3.
We will place that 3 in front of the 4 we took for one’s place as the cube root of 39304
∛39304 = 34
We can check our answer: 34x 34x 34 = 39304
Tip: Again it will help you if you can remember the cube roots of digits from 1-10.
|13||1||Cube of 1 ends with 1|
|23||8||Cube of 2 ends with 8|
|33||27||Cube of 3 ends with 7|
|43||64||Cube of 4 ends with 4|
|53||125||Cube of 5 ends with 5|
|63||216||Cube of 6 ends with 6|
|73||343||Cube of 7 ends with 3|
|83||512||Cube of 8 ends with 2|
|93||729||Cube of 9 ends with 9|
|103||1000||Cube of 10 ends with 0|
Tip: To find the unit place of the cube root always remember the following points:
• If the last digit of a cube root is 8 then the unit digit will be 2.
• If the last digit of a cube root is 2 then the unit digit will be 8.
• If the last digit of a cube root is 3 then the unit digit will be 7.
• If the last digit of a cube root is 7 then the unit digit will be 3.
• If the last digit of a cube root is other than 2, 3, 7 and 8 then put the same number as the unit digit.
Not an easy topic to handle from Class 9 and even in competitive exams, they can be a challenge. The shortcut method can be used to calculate without any hiccups.
2x 2 — 11x –21 = 0
Step 1: This means, a+b= –11
And ab = 2 (–21) = –42
Step 2 : We know that –14 and 3 will give -11
And 14×3 = 42
We change the signs , -14 = 14, 3 = –3
Step 3 : Divide both numbers by co -efficient of x 2 = 2
14 /2=7 and –3/2= can’t be simplified
Ans= 7 and –3/2
A favourite for any person setting the question paper are number series questions. It is one problem that consumes a lot of effort as well during an exam.
The only way to do it is to determine the relationship between two consecutive numbers. This leads to the answer in the series or sequence. Let us see one such relationship.
5 ,11 ,24 ,51 ,106, 217, ________
Step 1: Check for the difference between the numbers. The difference between 11 and 5 is 6 ( 5×2+1) , the difference between 24 and 11 is 13 ( 11×2+2), the difference between 51 and 24 is 27 ( 24×2 + 3)
Step 2: Determine the pattern based on differences. Here is the pattern:
5 x 2 + 1 =11
11 x 2 + 2 = 24
24 x 2 + 3 = 51
51 x 2 + 4 = 106
106 x 2 + 5 = 217
217 x 2 + 6 = 440
So the next number in sequence is 440.
Rule of 72
It is one of the most amazing rules in this subject that can be easily understood and applied in questions where a sum of money has to be doubled in a specific period of time at a given rate of interest.
Formulas to remember:
Number of years invested = 72/ Annual Investment Rate
Investment Rate = 72/ Number of years Invested
Investment rate x Number of years invested = 72
If you invested Rs.1000 /- in a business, then how much time will it take to double your investment, if the rate of interest is 5%?
Step 1: According to rule of 72,
Time duration in which the amount will be doubled at 5% interest rate = 72/5 = 14.4 years
Step 2 :To re check the answer
Investment rate x Number of years invested = 72
5 x 14.4 =72
Time and Work
A favourite question for maths instructors is the time work question. These problems usually relate to how much time would a work be done in, by a given number of people.
If Satish completes his work in 8 days and Mahesh completes the same work in 10 days, how many days do they require to finish the work together?
Step 1 : Calculate the LCM of 8 and 10 = 40
Step 2: Determine the efficiency of both individuals
Efficiency of Satish = 40/ 8= 5
Efficiency of Mahesh= 40/ 10= 4
Step 3: Time required by them together = LCM/ Total Efficiency = 40/ 9 = 5 days
A number of questions demand multiplication operation in the exams and we all dread losing time in it. Practice the following method a few times to reduce the time spent on multiplication.
Multiply: 334 x 217
Follow the steps below:
Step 4 : 3 3 4
X 2 1 7 Multiply 3×2=6, 3×1= 3 . Add 6+3+3= 12 . Carry forward 1
3 3 4
X 2 1 7
_______ Multiply 3×2= 6. Add 6 + 1= 7
7 2 4 7 8
Calculate Cost Price with Profit / Loss Percentage given
How to calculate cost price using selling price and profit percent?
We know that cost price = selling price – profit
Formula to calculate cost price if selling price and profit percentage are given:
CP = ( SP * 100 ) / ( 100 + percentage profit).
Formula to calculate cost price if selling price and loss percentage are given:
CP = ( SP * 100 ) / ( 100 – percentage loss ).
A book was sold for Rs.575 thereby gaining 15%. Find the cost price of the book.
Given selling price = Rs. 575
Gain% = 15%
We know, cost price = selling price × 100/100 + gain%
Therefore, cost price = 575 × 100/100 + 15
= 575 × 100/115
Therefore, the cost price of the book is $500.
Simple Interest and Compound Interest
Different formulas have been created to calculate the various aspects regarding Simple Interest and Compound Interest. Go through them to get a better hang of the calculations to be done.
If a sum becomes 2 times at a certain rate of interest in 5 years, then the time taken for the same amount to be 8 times at the same rate of interest will be:
(n–1) / 2x t , where n= number of times the amount will be , t= time
8–1 / 2×5 = 7 / 2×5 = 17.5 years
If the sum of money becomes 6 times in 20 years as simple Interest, what is the rate of interest?
R = 100 (n-1) / T %
= 100 ( 6-1)/ 20
=100 x 5/ 20
= 25 %
Rs. 2500 was borrowed for 3 years. What will be the compound interest if the rate of interest for first year is 3% per annum, second year is 4% per annum and for third year is 5% per annum respectively?
A = P ( 1 + r 1 / 100) ( 1 + r 2 /100) ( 1 + r 3 /100)
= 2500 x 103 /100 x 26/25 x 21/ 20
C.I. = 2811.90– 2500 = Rs. 311.90
The difference of S.I and C.I on an amount of Rs. 30000 for 2 years is Rs. 147. What is the rate of Interest?
C.I. = S.I. = P ( r /100) 2
147 = 30000 ( r/100 ) 2
49/ 10000 = (r/100) 2
7/100= r/ 100
How to ace in the competitive exams?
- Thorough study of basics: the shortcuts and techniques provided above can only work if you have mastered the principles of the subject. Otherwise, all your efforts to try and understand the tricks will go in vain.
- Don’t stop practicing: devise a schedule where you solve maths problems every day. Knowledge of shortcuts can’t fetch results if they aren’t worked with every day. Even the Iron man had to work on his suit constantly to make him invincible!
- Absorb the formulas: maths is all about formulas and their systematic application. A formula can find its use in various types of problems. So get them at your fingertips.
- Understand the question: read it a couple of times and think over what it asks for. What information is it providing? Based on it attempt it and work out a solution.
- Join Edulyte: an innovative learning platform, Edulyte provides world-class guidance in the field of mathematics. With its state of the art learning tools, highly qualified faculty, and extensive resources, learners are exposed to much more than the knowledge in the books. Enjoy the benefits of online and offline learning together in one place.
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