# Maths Shortcut Tricks PDF For Competitive Exams

### Maths Shortcut Tricks PDF

#### 1. Square Root

Finding the square root of a number by estimating and multiplying can be a long procedure. Given below is a simpler method to find the square root of a number:

Example:Find the square root of 2116

To find the square root of 2116:

Step 1: See the digit in one’s place. In this case, it is 6. Now, check between 1-9, the square of what all numbers have “6” at one’s place. The answer is 42 = 16 and 62 = 36.

Step 2: Now, check the square of which number between 1 to 9 is closest to the first two digits of the given number. In this case, the sum of the number between 1 to 9 is closest to 21. The answer is 42 = 16 and 52 = 25.

So, one number among 44, 46, 54 and 56 is the square root of 2116.

Step 3: For the two numbers you got in step 2, multiply each of them with the next number in the number series. That is, 4×5 = 20 and 5×6 = 30. Since 20 is a closer number to 21, the answer has to be either 46 or 44. Multiply and check your answer.

Check yourself with the below-mentioned example:

Example: What is the square root of 1024?

Solution:

Step 1: 22 = 4 and 82 = 64

Step 2: 32 = 9

Step 3: 3×4 = 12. Since 12 is greater than 10, the square root will be 32.

#### 2. Cube Root

Follow the steps given below to find out the cube root of a number quickly.

Example: What is the cube root of 9261?

Step 1: Find the numbers between 1 to 9 whose cube is equal to the digit present at the one’s place; here, it is 1. So, we get 1×1×1 = 1.

Step 2: See the first digit of the number, in this case, 9. 9 lies between the cube of 2 (2×2×2 = 8) and (3×3×3 = 27). Since 8 is closest to 9, the cube root of 9261 is 21.

Note: To find the cube root of 5 digit number, use the first two digits instead of the first digit in step 2

Try one example by yourself to understand the trick even better:

Example: What is the cube root of 32768?

Step 1: 23 = 8

Step 2: 33 = 27 and 43 = 64

Since 27 is closer to 32, the cube root of 32768 will be 32.

Given below are two examples of quadratic equations solved with easy tricks to find the answer quickly:

Example: x² – 18x + 45 = 0

Step 1: Multiply the coefficient of x² and the constant in the equation. In this case, 1×45 = 45

Step 2: Multiply “-1” with the coefficient of x. In this case, -1× (-18) = 18

Step 3: Hence, the value of x will be 15 and 3 (3+15=18 & 3×15=45). Remember, for signs, if the answer obtained in both step 1 & 2 is positive, then both values of x will be positive. If even one is negative, then values of x will be negative.

Here, the value obtained in step 1 & 2 is positive hence the value of x will be positive. So, the answer is x = 15, 3

Example: x²-5x-6 = 0

Step 1: Multiply the coefficient of x² and the constant in the equation. In this case, 1×(-6) = (-6)

Step 2: Multiply “-1” with the coefficient of x. In this case, (-1)× (-5) = 5

Step 3: Hence, the value of x will be 6 and 1 (6-1=5 & 6×1=6). Remember, for signs, if the answer obtained in both step 1 & 2 is positive, then both values of x will be positive. If even one is negative, then one of the values of x will be negative.

Step 4: Here the answer in step 1 is negative. Thus, one value of x will be negative. If the answer in step 1 is negative, the smaller value of x will be negative. If the answer in step 2 is negative, the larger value will be negative.

So, x= 6, -1

#### 4. Compound Interest

Given below are a few formulas that may save you some time during the exam while solving the compound interest problems:

(a) If compound interest is x% for 1st interval of time and is y% for the second interval of time, Then,

Net Effective Rate of Interest after the 2 intervals = x + y + (xy/100)

Note: This is applicable if both the time intervals are equal)

(b) If a sum of money, say P, amounts to A1 in a certain duration of time, say T, at Compound Interest and the same sum of money amounts to A2 in “2T” time at Compound Interest,

Then,

P/A1 = A1/A2

(c) If a sum of money, say P, amounts to A1 in a certain time duration, say T, at compound interest and the same sum of money amounts to A2 after T+1 years at compound interest

Then,

Rate of Interest = {(A2-A1) / A1} × 100

For example: Raj pays compound interest at 16% per annum to Shyam, which is compounded quarterly. What is the effective rate of interest per annum paid by Raj?

Solution:

Annual interest rate = 16%

So, the interest is paid quarterly, which makes a 4 time installment. Therefore, the rate of interest per quarter = 16/4 = 4%

Using (a) x + y + (xy/100).

4 + 4 + {(4×4)/100} = 8 + 0.16 = 8.16% for two quarters

For four quarters, 8.16% + 8.16% = 16.32%

#### 5. Simple Interest

Take reference from the formulas given below and save some time while solving the questions in the final exam for the quantitative section:

(a) Difference between simple and compound interest for 2 years = {(PR)2/ (100)2}

(b) Difference between simple and compound interest for 3 years = {PR2 (300+R) / 1003}

For example: The difference between simple interest and compound interest for two years, on a certain sum of money at 4% per annum is Rs.800, when compounded annually. What is the sum of money on which the interest has been gained?

Solution:

Following (a) CI-SI = {(PR)2/ (100)2}

⇒800 = {(P×4)2/ (100)2}

⇒P = Rs. 707.11

#### 6. Time & Work

Given below is a simpler way to find out the time taken to complete a piece of work done by three people, when working together:

Example: Three labourers, Ajit, Sumit & Ramesh take 10, 8 and 20 days respectively to complete the same piece of work. How long will it take for all three of them if they work together?

Solution:

LCM of 10, 8 and 20 = 40

Efficiency of Ajit = 40/10 = 4

Efficiency of Sumit = 40/8 = 5

Efficiency of Ramesh = 40/20 = 2

Time Taken by all three together = {(LCM) / (Efficiency of all three)} = 40/11 days

So to calculate the time taken to complete the same work by 3 people = (Total Unit of Work) / Efficiency of all the works)

#### 7. Approximation

Simple multiplication is something that consumes maximum of our time while solving maths questions in competitive exams. Given below is a shortcut to multiply two numbers which may help you in questions related to approximation and simplification.

Example: Solve 32 × 34

Step 1: Multiply the first number (in this case, 32) with the digit at ten’s place in second number (in this case, 3)

We get, 32×3 = 96

Step 2: Add a “0” to the answer obtained in step 1. So the number now becomes “960”

Step 3: Multiply 32 with the one’s digit in the second number, we get, 32×4 = 128

Step 4: Add the result obtained in step 2 & step 3.

So answer is 960 + 128 =  1088