Number System For SSC CGL Exam, Learn Concept & Questions

Number System For SSC CGL Exam

Number System For SSC CGL Exam: The SSC CGL consists of four sections but here we will talk about the Number System for SSC CGL Exam which comes under the SSC CGL Quantitative Aptitude section. Most of the candidates are already familiar with this topic and the good part is that this is a very small and easy topic. The Number System for SSC CGL Exam consists of 2-3 questions in easy to moderate level. In this blog, we have provided the basic concepts and some questions related to the number system for SSC CGL exam along with the solution with explanations. Candidates are advised to practice these questions and boost their performance.

Concept of Number System

Number System is a mathematical concept, it is the representation of numbers or writing the numbers by using some set of digits or other symbols.

A number is a numerical value or mathematical value that is used for counting or measuring objects. Examples are

• Rational Numbers (Q):- 1/2, 2/3, 7/5, etc
• Irrational Numbers:- Square root of 3, pi
• Natural Numbers (N):- Numbers starting from 1. Example:- 1,2,3,4…….
• Whole Numbers (W):- Numbers starting from 0 and so on. Example:- 0,1,2,3,4………..
• Integers (Z):- Numbers including negative numbers, zero, and positive numbers. Example:- -3,-2,-1,0,1,2……
• Real Numbers (R): Real numbers include both rational and irrational numbers.

Questions on Number System

Now it’s time to practice the questions on different types of questions on the Number System. Practicing questions is very important to boost preparation and score the highest in the exam. Let’s start

Number System Questions in the English Language

Question 1: If (50/3)% of a number is added to itself then the number obtained is 4956 then find the original number.

A) 8428

B) 4248

C) 4824

D) 4284

Question 2: The 3^rd and 7^th terms of an arithmetic progression is 143 and 399 respectively. Find its 15^th term.

A) 749

B) 865

C) 911

D) 857

Question 3: When a number is divided by 13 then it leaves a remainder 8 and when the quotient thus obtained, is divided by 5, it leaves a remainder 3. What will be the remainder if the number is divided by 65?

A) 28

B) 37

C) 47

D) 42

Question 4: By interchanging the digits of a two-digit number we get a number which is four times the original number minus 24. If the unit’s digit of the original number exceeds its ten’s digit by 7, then find the number.

A) 29

B) 18

C) 36

D) 58

Question 5: When a number is divided by 49 the remainder is 24. Find the remainder when the same number is divided by 7.

A) 6

B) 4

C) 3

D) 5

Question 6: (4²⁵ + 4²⁶ + 4²⁷ + 4²⁸) is divisible by which of the following?

A) 21

B) 19

C) 11

D) 17

Question 7: Find the remainder when (10⁵⁶ + 5) is divided by 9

A) 3

B) 6

C) 2

D) 4

Question 8: If six-digit number 5x2y6z is divisible by 7, 11 and 13, then the value of 2x + 3y – 4z is:

A) 19

B) 11

C) 21

D) 23

Question 9: How many natural numbers less than 500 is divisible by 4 or 6 but not 24?

A) 187

B) 240

C) 120

D) 167

Question 10: Find the value of (2² + 22²) – (1² + 11²).

A) 512

B) 424

C) 366

D) 242

Number System Questions in the Hindi Language

प्रश्न 1: यदि किसी संख्या में संख्या का (50/3)% जोड़ा जाए तो प्राप्त संख्या 4956 है, तो वास्तविक संख्या ज्ञात करें ?

A) 8428

B) 4248

C) 4824

D) 4284

प्रश्न 2: अंकगणितीय प्रगति का तीसरा और सातवां पद क्रमशः 143 और 399 है। इसका पन्द्रहवाँ पद ज्ञात करें?

A) 749

B) 865

C) 911

D) 857

प्रश्न 3: जब एक संख्या को 13 से विभाजित किया जाता है तो वह शेषफल 8 छोड़ती है और जब इस तरह प्राप्त भागफल को 5 से विभाजित किया जाता है, तो यह शेषफल 3 छोड़ता है।जब संख्या को 65 से विभाजित किया जाता है तो शेषफल कितना प्राप्त होगा?

A) 28

B) 37

C) 47

D) 42

प्रश्न 4: दो-अंकीय संख्या के अंको को आपस में बदल देने से हमें वह संख्या प्राप्त होती है जो वास्तविक संख्या से चार गुना से 24 कम होती है।यदि वास्तविक संख्या की इकाई अंक उसके दहाई अंकों के 7 से अधिक है, तो संख्या ज्ञात करें?

A) 29

B) 18

C) 36

D) 58

प्रश्न 5: जब एक संख्या को 49 से विभाजित किया जाता है तो शेषफल 24 होता है।उसी संख्या को 7 से विभाजित करने पर शेषफल ज्ञात करें।

A) 6

B) 4

C) 3

D) 5

प्रश्न 6: (4²⁵ + 4²⁶ + 4²⁷ + 4²⁸) निम्नलिखित में से किससे विभाजित होता है?

A) 21

B) 19

C) 11

D) 17

प्रश्न 7: (10⁵⁶ + 5) को 9 से विभाजित करने पर प्राप्त शेषफल ज्ञात करें|

A) 3

B) 6

C) 2

D) 4

प्रश्न 8: यदि छह अंकीय संख्या 5x2y6z 7, 11 और 13 से विभाज्य है, तो 2x + 3y – 4z का मान है:

A) 19

B) 11

C) 21

D) 23

प्रश्न 9: 500 से कम ऐसी कितनी प्राकृतिक संख्या है जो 4 या 6 से विभाजित है पर 24 से नहीं?

A) 187

B) 240

C) 120

D) 167

प्रश्न 10: (2² + 22²) – (1² + 11²) का मान ज्ञात करें|

A) 512

B) 424

C) 366

D) 242

ANSWER KEYS and SOLUTIONS:

Number System Solutions with Explanation

Solution 1: 2)

Let the original number be 6x

Since, (50/3)% = 1/6

New number = 6x + 6x × (1/6) = 7x

According to question,

7x = 4956

x = 708

Original number = 6x = 6 × 708 = 4248

Hence, option b.

Solution 2: 3)

Let the first term and common difference of the series be ‘a’ and ‘d’ respectively

According to the question,

{a + (7 – 1)d} – {a + (3 – 1)d} = 399 – 143

Or, 4d = 256

Or, d = 64

Therefore, a = 143 – 128 = 15

Therefore, 15^th term of the series = a + (15 – 1)d = 911

Hence, option c.

Solution 3: 3)

Let the number be ‘x’ and the quotient be ‘y’

So, x = 13y + 8………………..(1)

And, y = 5z + 3

Putting value of ‘y’ in (1) we get

x = 13 × (5y + 3) + 8

x = 65y + 39 + 8

x = 65y + 47

So, when x is divided by 65 we will get remainder 47

Hence, option c.

Solution 4: 1)

Let the original number be ‘10x + y’

According to question;

y – x = 7……………………………………….(1)

10y + x = 4 × (10x + y) – 24

10y + x = 40x + 4y – 24

39x – 6y = 24…………………………………(2)

From equation (1) and (2), we get

39x – 6(x + 7) = 24

39x – 6x – 42 = 24

33x = 66

x = 2

y = 9

So, the desired number is 29.

Hence, option a.

Solution 5: 3)

According to the question,

Number = 49x + 24

Required remainder = (49x + 24)/7 = 7x + (7 × 3) + 3

Hence, option c.

Solution 6: 4)

(4²⁵ + 4²⁶ + 4²⁷ + 4²⁸) = 4²⁵(1 + 4 + 4² + 4³) = 4²⁵ × 85 = 4²⁴ × 340

Therefore, it is divisible by 17 only

Hence, option d.

Solution 7: 2)

(10⁵⁶ + 5) can be written as (9 + 1)⁵⁶ + 5

Therefore, remainder = 1⁵⁶ + 5 = 6

Hence, option b.

Solution 8: 1)

LCM of 7, 11 and 13 = 1001

So, a six digit number will be divisible by 1001 if it is in the form of abcabc

So, comparing first three and last three digits of 5x2y6z, we get x = 6, y = 5 an z = 2

So, 2x + 3y – 4z = 2 × 6 + 3 × 5 – 4 × 2 = 19

Hence, option a.

Solution 9: 4)

Numbers less than 500 which are divisible by 4 = Quotient of (499/4) = 124

Number less than 500 which are divisible by 6 = Quotient of (499/6) = 83

Numbers less than 500 which are divisible by 24 = Quotient of (499/24) = 20

Therefore, required number of natural numbers = 124 + 83 – (2 × 20) = 167

Hence, option d.

Solution 10: 3)

Given, (2² + 22²) – (1² + 11²) = 4(1 + 11²) – (1 + 11²) = (1 + 11²)(4 – 1) = 122 × 3 = 366

Hence, option c.

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