# Quantitative Aptitude Tricks and Shortcuts for Bank Exams

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Quantitative aptitude is a very important subject and comes in every banking exam. You can score well in this subject if you are well versed with the basic concepts but if you want to score high in this subject you must be aware of various Quantitative Aptitude Tricks. The quantitative aptitude section includes Data Interpretation, Arithmetic problems, Number series, Quadratic quantity comparison, Simplification/Approximation, etc.

Some topics like Simplification/Approximation, Number Series, and Quadratic Equations Comparison can be solved very effectively and, in less time, if you know how to do a fast calculation and how to apply tricks. These topics contain approximately more than 50% weightage in the quantitative aptitude section of bank (IBPS/SBI and others) exams.

In this article, we will discuss some of the Quantitative Aptitude Tricks that will help to increase your calculation speed so that you can score well in this section.

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## 1. Quantitative Aptitude Tricks for Simplification/Approximation:

Simplification and approximation questions generally contain a weightage of 5 – 10 marks in the exam. Simplification means replacing a complex mathematical expression with a simpler one to get the final answer. In the case of approximation you, need not find the exact answer. All you have to do is to convert a complex mathematical expression into a similar (not exactly equal) to get to the final answer.

#### Trick 1: Do memorize multiplication tables of integers up to 30, squares of integers up to 50, and cubes of integers up to 30.

It will help you to calculate a complex mathematical sum in a short time and, save time you can use in the remaining questions.

#### Trick 2: Conversion of percentages into fractions. Have a look at the conversion table below:

Example: What is 87.5% of 64?

Solution:

= 87.5% of 64
= 7/8 * 64
= 56

## 2. Quantitative Aptitude Tricks for Number Series:

Number Series questions generally contain a weightage of 5 – 6 marks in the exam. There is some special format of number series that is frequently asked in the bank prelims exam. Different type of number series is given below:

A series in which the next term is obtained by adding/subtracting a constant to its previous term or the difference between two consecutive terms forms a pattern.

Example 1: 5, 18, 31, 44, 57, ___a___

Solution:

The next term is obtained by adding 13 to the previous term.

5 + 13 = 18
18 + 13 = 31
31 + 13 = 44
44 + 13 = 57
57 + 13 = 70 = a

Example 2: 13, 22, 33, 46, 61, __b__

Solution:

The difference between two consecutive terms forms an increasing arithmetic series.

22 – 13 = 9
33 – 22 = 11
46 – 33 = 13
61 – 46 = 15
78 – 61 = 17
b = 78

Difference forms an increasing arithmetic sequence.

#### Type 2: Multiplication/Division Series:

A series in which the next term is obtained by multiplying/dividing a constant to its previous term or ratio of two consecutive terms forms a pattern.

Example 1: 3, 12, 48, 192, 768, __c__

Solution:

The difference between two consecutive terms forms a geometric series.

3 * 4 = 12
12 * 4 = 48
48 * 4 = 192
192 * 4 = 768
768 * 4 = 3072 = c

Example 2: 5, 10, 40, 240, 1920, __d__

Solution:

The difference between two consecutive terms forms a geometric series.

d = 19200

Ratios form an increasing arithmetic sequence.

#### Type 3: Perfect square/cube series:

This is the most common pattern followed. In this either different between terms are perfect square/cube or terms itself are perfect square/cube.

Example 1: 27, 125, 343, 729, 1331, __a__

Solution:

Individual terms are perfect cubes.

27 = 33
125 = 53
343 = 73
729 = 93
1331 = 113
a = 2197 = 133

Example 2: 19, 23, 32, 48, 73, __b__

Solution:

The difference between terms forms a perfect square series:

19 + 22 = 23
23 + 32 = 32
32 + 42 = 48
48 + 52 = 73
73 + 62 = 109 = b

#### Type 4: (n2 + 1), (n2 – 1), (n2 + n) and (n2 – n) series.

Example 1: 26, 37, 50, 65, 82, __a__

Solution:

52 + 1 = 26
62 + 1 = 37
72 + 1 = 50
82 + 1 = 65
92 + 1 = 82
102 + 1 = 101 = a

#### Type 5: (n3 + 1), (n3 – 1), (n3 + n) and (n3 – n) series.

Example 1: 24, 60, 120, 210, 336, __b__

Solution:

33 – 3 = 24
43 – 4 = 60
53 – 5 = 120
63 – 6 = 210
73 – 7 = 336
83 – 8 = 504 = b

## 3. Quantitative Aptitude Tricks for Quadratic Equations:

The quadratic quantity comparison questions generally contain a weightage of 5 – 6 marks in the exam. It is one of the trickiest topics of the section and plays a vital role in the bank prelims exams.

The equation of the form ax2 + bx + c = 0 is a general form of a quadratic equation. Where, ‘a’, ‘b’, and ‘c’ are constants and ‘x’ is a variable.

To compare two quadratic equations, we need to find their roots first and after that compare roots of the both equations.

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Steps to find the roots of a quadratic equation:

1. Multiply ‘a’ and ‘c’.
2. Split ‘b’ such that their sum is equal to ‘b’ and their product is equal to the product of ‘a’ and ‘c’.
3. Change sign.
4. Divide by ‘a’.

Example 1: 6x2 + 2x – 20 = 0

Solution:
6x2 + 2x – 20 = 0
Compare with: ax2 + bx + c = 0
a = 6, b = 2, c = -20

Multiply ‘a’ and ‘c’ => 6 * -20 = -120
Split ‘b’ => 12 and -10
Change sign => -12 and 10
Divide by ‘a’ => -12/6 and 10/6

Roots are -2 and 5/3

But to find roots of a quadratic equation is not an easy task so, we can compare the roots of two quadratic by looking at their signs only.

Now let us learn quantitative aptitude tricks to find the sign of roots of any quadratic equation:

a) When ‘c’ is positive, then a sign of both roots will be the same which is the opposite of the sign of ‘b’.
b) When ‘c’ is negative, then the sign of both roots will be different irrespective of the sign of ‘b’.

Sign of ‘c’ Sign of ‘b’ Sign of the first root Sign of the second root
+ +
+ + +
+ +
+ +

I: 2x2 – 13x + 15 = 0
II: 2y2 + 7y + 3 = 0

Solution:
The signs of the roots of the first equation are both positive.
The signs of the roots of the second equation are both negative.
Hence, x > y.

I: 6x2 + 23x + 7 = 0
II: 4y2 – 21y + 5 = 0

Solution:

The signs of the roots of the first equation are both negative.
The signs of the roots of the second equation are both positive.
Hence, x < y.

I: 2x2 + 7x – 15 = 0
II: 3y2 – 32y – 11 = 0

Solution:

The sign of one root of the first equation is positive and the other is negative.
The sign of one root of the second equation is positive and the other is negative.
Hence, the relation between x and y cannot be determined.