Short Tricks on Number System in Quant Section

By Ashwini Shivhare|Updated : November 4th, 2022

The number system is an important topic for upcoming SSC & Railways Exams. Here, we are going to help you with Basic Concepts & Short Tricks on Number System in Quant Section. We will be providing you with details of the topic to make the Quant Section and calculation easier for you all to understand. We hope you all will like the post.

Introduction

(1)Natural Numbers: Numbers starting from 1, 2, 3 and so on and so forth are counted as Natural numbers. They are 1, 2, 3, 4...

Exceptions: Zero, negative and decimal numbers are not counted in this list.

(2)Whole numbers: Zero and all other natural numbers are known as natural numbers. They are 0, 1, 2, 3, 4...

(3)Integers: They are the numbers which include all the whole numbers and their negatives. They are ...-4, -3, -2, -1, 0, 1, 2, 3, 4....

(4)Rational Numbers: All the numbers which are terminating, repeating and can be written in the form p/q, where p and q are integers and q should not be equal to 0 are termed as rational numbers.

Example: 0.12121212....

(5)Irrational Numbers: All the numbers which are non-terminating, non-repeating and cannot be written in the form p/q, where p and q are integers and q should not be equal to 0 are termed as irrational numbers.

Example: pie, e 

(6)Real numbers: All the numbers existing on the number line are real numbers. The group is made up of all rational and irrational numbers. 

(7)Imaginary Numbers: Imaginary numbers are the numbers formed by the product of real numbers and imaginary unit 'i'.

This imaginary unit is defined as the following:

i2= -1, multiplication of this 'i' is calculated according to the above value. Example: 8i

(8)Complex Number: The numbers formed by the combination of real numbers and imaginary numbers are called the complex number. Every complex number is written in the following form:

A+iB, where A is the real part of the number and B is the imaginary part. 

(9)Prime numbers: All the numbers having only two divisors, 1 and the number itself is called a prime number. Hence, a prime number can be written as the product of the number itself and 1.

Example: 2, 3, 5, 7 etc. 

(10)Composite Numbers: All the numbers which are not prime are called composite numbers. This number has factors other than one and itself.

Example: 4, 10, 99, 105, 1782 etc.

(11)Even & Odd Numbers: All the numbers divided by 2 are even numbers. Whereas the ones not divisible by 2 are odd numbers.

Example: 4, 6, 64, 100, 10004 etc are all even numbers.

3, 7, 11, 91, 99, 1003 are all odd numbers. 

(12)Relative Prime Numbers/Co-prime Numbers: Numbers which do not have any common factor other than 1 are called co-prime numbers.

Example: 5 and 17 are co-primes.

(13)Perfect Numbers: All the numbers are called perfect numbers if the sum of all the factors of that number, excluding the number itself and including 1, equalizes the to the number itself then the number is termed as a perfect number. 

Example:6 is a perfect number. As the factors of 6= 2 and 3. 

As per the rule of perfect numbers, sum= 2+3+1 = 6. Hence, 6 is a perfect number.

Some important properties of Numbers:

  1. The number 1 is neither prime nor composite.
  2. The only number which is even is 2.
  3. All the prime numbers greater than 3 can be written in the form of (6k+1) or (6k-1) where k is an integer.
  4. Square of every natural number can be written in the form 3n or (3n+1) and 4n or (4n+1).
  5. The tens digit of every perfect square is even unless the square is ending in 6 in which case the tens digit is odd.
  6. The product of n consecutive natural numbers is always divisible by n!, where n!= 1X2X3X4X….Xn (known as factorial n).

To test whether a given number is a prime number or not

If you want to test whether any number is a prime number or not, take an integer larger than the approximate square root of that number. Let it be ‘x’. test the divisibility of the given number by every prime number less than ‘x’. if it not divisible by any of them then it is prime number; otherwise it is a composite number (other than prime).

Example:  Is 349 a prime number?

Solution:

The square root of 349 is approximately 19. The prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17.

Clearly, 349 is not divisible by any of them. Therefore, 349 is a prime number.

 

Rules of Simplification

(i) In simplifying an expression, first of all vinculum or bar must be removed. For example: we have known that – 8 – 10 = -18

But,image004 = - (-2) = 2

(ii) After removing the bar, the brackets must be removed, strictly in the order (), {} and [].

(iii) After removing the brackets, we must use the following operations strictly in the order given below. (a) of (b) division (c) multiplication (d) addition and (e) subtraction.

Note: The rule is also known as the rule of ‘VBODMAS’ where V, B, O, D, M, A and S stand for Vinculum, Brackets, Of, Division, Multiplication, Addition and Subtraction respectively.

Example: Simplify  image005

Solution: 

image006

Ascending or Descending Order in Rational Numbers

Rule 1: When the numerator and the denominator of the fractions increase by a constant value, the last fraction is the biggest. 

Example: Which of the following fractions is the greatest?

image007

Solution:

We see that the numerators as well as denominators of the above fraction increase by 1, so the last fraction, i.e. image008  is the greatest fraction.

Rule 2: The fraction whose numerator after cross-multiplication given the greater value is greater.

Example: Which is greater : image009

Solution:

Students generally solve these questions by changing the fractions into decimal values or by equating the denominators. But, we suggest you a better method for getting the answer more quickly. 

Step 1: Cross –multiply the two given fractions.

image010

We have, 5 × 14 = 70 and 8 ×9 =72

Step II. As 72 is grater than 70 and the numerator involved with the greater value is 9, the fraction image011 is the greater of the two.

Example: Which is greater: image012

Solution:

Step I: 4 ×23 > 15 ×6

Step II: As the greater value has the numerator 4 involved with it, image013  is greater.

You can see how quickly this method works. After good practice, you won’t need to calculate before answering the question.

The arrangement of fractions into the ascending or descending order becomes easier now. Choose two fractions at a time. See which one is grater. This way you may get a quick arrangement of fractions.

Note:  Sometimes, when the values are smaller (i.e., less than 10), the conventional method, i.e., changing the values into decimals or equating the denominators after getting LCM, will prove more convenient for some of you.

Example: Arrange the following in ascending order.

image014

Solution: Method I

The LCM of 7,5,9,2,5, is 630.

Now, to equate the denominators, we divide the LCM by the denominators and multiply the quotient by the respectively numerators.

Like for image015 , 630 ÷ 7 = 90, so, multiply 3 by 90.

Thus, the fractions change to image016

The fraction which has a larger numerator is naturally larger. So,

image017

Method II:

Change the fractions into decimals like

image015= 0.428, image018= 0.8, image019= 0.777, image020 = 0.5, image021 = 0.6

Clearly,

image022

Method III:

Rule of CM (cross-multiplication)

Step I: Take the first two fractions. Find the greater one by the rule of CM.

image023

3 × 5< 7×4

image024 image025

Step II: Take the third fraction. Apply CM with the third fraction and the larger value obtained is step I.

image026

4 × 9 > 5 × 7

image027

Now we see that image019 can lie after image015 or between image018 and image015.

Therefore, we apply CM with image015 and image019 see that image028.

image029

Step III: Take the next fraction. Apply CM with image015 and image020 and see that image030. Next, we apply CM with image019 and image020 and see that image032.

Therefore,

Capture

Step IV: With similar applications, we get the final result as:

image033

Note: This rule has some disadvantages also. But if you act fast, it gives faster results. Don’t reject this method at once. This can prove to be the better method for you.

 

Formulas

1. Sum of all the first n natural numbers = image001

For example:  1+ 2 +3 +…..+105= image002

2. Sum of first n odd numbers =image003

For example: 1+3+5+7=image004=16(as there are four odd numbers)

3. Sum of first n even numbers = n (n+1)

For example : 2+4+6+8+….+100 (or 50th even number) = 50×(50+1)= 2550

4. Sum of squares of first n natural numbers = image005

For example: image006

image007

5. Sum of cubes of first n naturals numbers =image008

For example :

image009

Example:

(1) What is the total of all the even numbers from 1 to 400?

Solution:

From 1 to 400, there are 400 numbers. So, there are 400/2= 200 even numbers.

Hence, sum = 200(200+1) = 40200     (From Rule III)

(2) What is the total of all the even numbers from 1 to 361?

Solution:

From 1 to 361, there are 361, there are 361 numbers; so there areimage010 even numbers. Thus, sum = 180(180+1)=32580

(3) What is the total of all the odd numbers from 1 to 180?

Solution:

Therefore are 180/2 = 90 odd numbers between the given range. So, the sum =image011

(4) What is the total of all the odd numbers from 1 to 51?

Solution

There are image012odd numbers between the given range. So, the sum =image013

(5) Find the of all the odd numbers from 20 to 101.

Solution:

The required sum = Sum of all the odd numbers from 1 to 101.

Sum of all the odd numbers from 1 to 20

= Sum of first 51 odd numbers – Sum of first 10 odd numbers

=image014

Miscellaneous

1. In a division sum, we have four quantities – Dividend, Divisor, Quotient and Remainder. These are connected by the relation.

Dividend = (Divisor × Quotient) + Remainder

2. When the division is exact, the remainder is zero (0). In this case, the above relation becomes

Dividend = Divisor × Quotient

Example: 1: The quotient arising from the divisor of 24446 by a certain number is 79 and the remainder is 35; what is the divisor?

Solution:

Divisor × Quotient = Dividend -  Remainder

79×Divisor = 24446 -35 =24411

Divisor = 24411 ÷ 79 = 309.

Example: 2: A number when divided by 12 leaves a remainder 7. What remainder will be obtained by dividing the same number by 7?

Solution:

We see that in the above example, the first divisor 12 is not a multiple of the second divisor 7. Now, we take the two numbers 139 and 151, which when divided by 12, leave 7 as the remainder. But when we divide the above two numbers by 7, we get the respective remainder as 6 and 4. Thus, we conclude that the question is wrong.

Important Questions on Number System for SSC with Explanation

1.What is the value of 143 + 163 + 183 + ... + 303?

A. 134576
B. 120212
C. 115624
D. 111672

Answer ||| D

Solution ||| Sum of cube of first n natural numbers:
byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
byjusexamprep
= 111672

2.Ifbyjusexamprep, then byjusexamprep is equals to:

A. 84
B. 116
C. 98
D. 126

Answer ||| D

Solution ||| Required value
byjusexamprep
byjusexamprep
byjusexamprep

3.A number being divided by 54 gives remainder of 53. If the number is divided by 18, then remainder is?

A. 1
B. 3
C. 17
D. 13

Answer ||| C

Solution ||| Let a number be N
If N is divided by 54, then remainder is 53.
Or N= (54Q + 53)
Now if N is divided by 18
N/18 = (54Q)/18+ 17/18 = 0 + 17
So the remainder  is 17

4.(12 + 22 + 32 + ………… + 102) is equal to how much?

A. 380
B. 385
C. 390
D. 392

Answer ||| B

Solution ||| 12+22+32+…………….+n2 = [n*(n+1)*(2n+1)]/6
where, n=10 for the given given series
Hence, required answer =10(10+1)(20+1)/6 = 385

5.Which is the smallest fraction among 3/4, 2/3, 4/7 and 5/3?

A. 2/3
B. 3/4
C. 4/7
D. 5/3

Answer ||| C

Solution |||

First we will make all the denominators equal

So,

byjusexamprep

byjusexamprep

The smallest fraction is 48/84 i.e. 4/7

Therefore, option C is the right answer.

 

6.What is the largest two-digit number which when divided by 6 and 5 leaves remainder 1 in each case?

A. 61
B. 93
C. 91
D. 97

Answer ||| C

Solution |||

LCM (6, 5) = 30

The required number will be given by = 30K + 1 

Here, 30K + 1 < 100 …….( as we require largest two digit number )

Putting K = 3,

⇒ 30×3 + 1 = 91 

Hence the number is 91.

 

7.In three consecutive numbers the difference between highest and lowest number is equal to 4th root of middle number. What is the highest number?

A. 15
B. 17
C. 16
D. 21

Answer ||| B

Solution ||| Let number be xx + 1 and x + 2
(x + 2 – x) = (x+1)(1/4)
16 = x + 1
∴ x = 15
Highest number = 15 + 2 = 17

8.Let byjusexamprepbe the greatest number which when divides 6475, 4984 and 4132, the remainder in each case is the same. What is the sum of digits ofbyjusexamprep?

A. 4
B. 7
C. 5
D. 6

Answer ||| D

Solution |||

byjusexamprepbe the greatest number which when divides 6475, 4984 and 4132, the remainder in each case is the same.

So, 6475 = byjusexamprep , where byjusexamprep…….(1)

4984 = byjusexamprep , where byjusexamprep…….(2)

4132 = byjusexamprep , where byjusexamprep…..(3)

Subtract (2) from (1)

 1491 = byjusexamprep……….(4)

Subtract (3) from (1)

 2343 = byjusexamprep……….(5)

Now, 1491 and 2343 both are divided by x and x is the greatest number.

So, Required value of x = HCF (1491,2343)

Now, byjusexamprep

byjusexamprep

Hence, HCF (1491,2343) = byjusexamprep

Sum of digits of 213 = 2+1+3 = 6

 

9.How many three digits numbers are there in which all the digits are odd?

A. 100
B. 125
C. 500
D. 250

Answer ||| B

Solution ||| Short Trick:
There are 5 odd and 5 even numbers. So there are 5 possible values of each place. And there are 3 places. So total number of 3 digit numbers in which all the digits are odd = 5*5*5 = 125
Basic Explanation:
First odd number in which all the digits are odd is 111, and last one is 999
Let us see all such numbers starting from 1.
They are 111, 113, 115,117,119,131,133, 135, 137, 139, 151, 153, 155, 157, 159, 171, 173, 175, 177, 179, 191, 193, 195, 197, 199
So there are 25 odd numbers starting with 1 in which all digits are odd. So same for 3, 5, 7 and 9
Hence total odd number in which all the digits are odd are 25× 5=125

10.What is the value of x so that the seven-digit number 55350X2 is divisible by 72?

A. 7
B. 8
C. 1
D. 3

Answer ||| A

Solution |||

Factors of 72= 8×9 so no. which is divisible by 72 must be divisible by 8 & 9 also.

We know if last three digit of a no. is divisible by 8 then the no. is also divisible by 8.

So here we get two possible value of x 3 or 7.

Now from the divisibility test by 9 we know that if the sum of digit of a no. is divisible by 9 then the no. is also divisible by 9.

By above constraint 3 is eliminated because by taking value of x as 3 the sum becomes 23 which is not divisible by 9.

So, the only possible value of x is 7.

 

 

11.If byjusexamprep = a + b√2, then what is the value of (3a + 4b)?

A. byjusexamprep
B. 98
C. byjusexamprep
D. 97

Answer ||| C

Solution |||

Given:

byjusexamprep = a + b√2

⇒ byjusexamprep = byjusexamprep

⇒ byjusexamprep = byjusexamprep

⇒ byjusexamprep = byjusexamprep

⇒ byjusexamprepbyjusexamprep

⇒ byjusexamprepbyjusexamprep

On comparing both sides:

a = byjusexamprep & b = byjusexamprep

Now, required (3a + 4b)

byjusexamprep = 98byjusexamprep

 

12.Which of the following is TRUE?
I. byjusexamprep
II. byjusexamprep
III. byjusexamprep
IV. byjusexamprep

A. Only I
B. Only II
C. Only III
D. Only IV

Answer ||| C

Solution ||| byjusexamprep
byjusexamprep
byjusexamprep
Hence by comparing
byjusexamprep.

13.If (333 + 333 + 333) (233 + 233) = 6x, then what is the value of x?

A. 34
B. 35
C. 33
D. 33.5

Answer ||| A

Solution ||| (333 + 333 + 333) (233 + 233) = 6x
3× 333 × 2× 233 = 6x
6× 633 = 6x
x = 34

14.Which of the following is a recurring decimal number?

A. byjusexamprep
B. byjusexamprep
C. byjusexamprep
D. byjusexamprep

Answer ||| D

Solution |||

Recurring numbers are those numbers which keep on repeating the same value after decimal point.

byjusexamprep = byjusexamprep

byjusexamprep

byjusexamprep

byjusexamprep

 

15.Check which of the given Numbers is a Prime Number.

A. 733
B. 843
C. 943
D. 633

Answer ||| A

Solution |||

To check the Prime No. out of the given four, we will go by options as –

(a). 733 -When we take the square root of 733, it is approximate 27, so we will consider the prime Numbers till 28.

Now, we divide 733 by all the Prime Numbers below 28 viz. 2, 3,5,7,11,13,17,19,23

Since 733 is not divisible by any of the Prime Numbers below 28. So it is a prime Number.

 

16.Let ab, a ≠ b, is a 2-digit prime number such that ba is also a prime number. The sum of all such numbers is:

A. 407
B. 418
C. 396
D. 374

Answer ||| B

Solution |||

Such numbers are 13, 17, 31, 37, 71, 73, 79 and 97.

Sum = 13 + 17 + 31 + 37 + 71 + 73 + 79 + 97 = 418

 

17.The value of byjusexamprep is:

A. byjusexamprep
B. byjusexamprep
C. byjusexamprep
D. byjusexamprep

Answer ||| B

Solution |||

Given, byjusexamprep

byjusexamprep

byjusexamprep

byjusexamprepbyjusexamprep

byjusexamprep = byjusexamprep

 

 

18.If byjusexamprep, then the value of a + b is equal to:

A. 16
B. 15
C. 18
D. 24

Answer ||| C

Solution |||

Given, byjusexamprep

byjusexamprepbyjusexamprepbyjusexamprep

byjusexamprepbyjusexamprepbyjusexamprep

 

byjusexamprepbyjusexamprepbyjusexamprep

byjusexamprep7√3 – 11 = a√3 – b

By comparing,

a = 7 and b = 11

Now, required = a + b = 7 + 11 = 18

 

 

19.Let x = (633)24 – (277)38 + (266)54. What is the units digit of x?

A. 6
B. 4
C. 7
D. 8

Answer ||| D

Solution |||

We will check cyclicity of 3 and 7, as 6 raised to any power will have unit digit as 6. Hence,

byjusexamprep

byjusexamprep

byjusexamprep = -2
Since Unit digit can't be negative;
So, 10 + (-2) = 8 will be the unit digit.

20.Let x be the greatest number which when divides 955, 1027, 1075, the remainder in each case is the same. Which of the following is NOT a factor of x?

A. 4
B. 16
C. 8
D. 6

Answer ||| B

Solution |||

Here, x must be the HCF of difference between these numbers.

Difference between numbers = 1027 – 955 = 72 & 1075 – 1027 = 48

HCF of 72 & 48 = 24

By going through options, we can find that 16 is not a factor of 24 but all other options are.

Hence, option B is the correct option.

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