Number System For SSC CGL Exam: As we know the number system is one of the most important topics in the SSC CGL Exam but many candidates don’t know the right approach to finding questions and related questions of the Number System for SSC CGL exam. Since it is a vast topic you must know how much you have to study for the SSC CGL exam. For that, you have to study smartly. The Number System for SSC CGL Exam consists of 2-3 questions at an easy to moderate level. In this blog, we have provided the basic concepts and questions related to the number system for SSC CGL exam along with the solution with explanations. You have to just download the PDF for free of number system basics and formulas and questions for SSC CGL exam.
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.
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. Here we have provided the number system questions for SSC CGL that have been asked in the previous years.
1. Find the sum of squares of the greatest value and smallest value of K in the number so that the number 45082K is divisible by 3.
(a) 68
(b) 64
(c) 100
(d) 50
2. Find the difference between squares of the greatest value and the smallest value of P if the number 5306P2 is divisible by 3.
(a) 60
(b) 68
(c) 36
(d) 6
3. What is the value of K such that the number 72K460K is divisible by 6?
(a) 4
(b) 9
(c) 7
(d) 8
4. Find the smallest value of a so that 42a48b (a > b) is divisible by 11.
(a) 4
(b) 5
(c) 0
(d) 9
5. Find the greatest value of b so that 30a68b (a > b) is divisible by 11.
(a) 4
(b) 9
(c) 3
(d) 6
6. Find the sum of all the possible values of (a + b) so that the number 4a067b is divisible by 11.
(a) 5
(b) 16
(c) 21
(d) 11
7. If the nine-digit number 7p5964q28 is completely divisible by 88, what is the value of (p² – q) for the largest value of q, where p and q are natural numbers?
(a) 72
(b) 9
(c) 0
(d) 81
8. If the 6-digit number 5x423y is divisible by 88, then what is the value of (5x – 8y)?
(a) 28
(b) 14
(c) 16
(d) 24
9. If the seven-digit number 94x29y6 is divisible by 72, then what is the value of (2x + 3y) for x ≠ y?
(a) 35
(b) 21
(c) 37
(d) 23
10. If the 8-digit number 888x53y4 is divisible by 72, then what is the value of (7x + 2y) for the maximum value of y?
(a) 19
(b) 15
(c) 23
(d) 27
Now it’s time to practice under time constraints and required to check how much you understand the number system. This can be done by attempting the number system test. Click on the link provided below to attempt the Number System Test for SSC CGL.
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 3rd and 7th terms of an arithmetic progression is 143 and 399 respectively. Find its 15th 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
ANSWER KEYS and SOLUTIONS:
1) – 2) | 2) – 3) | 3) – 3) | 4) – 1) | 5) – 3) | 6) – 4) |
7) – 2) | 8) – 1) | 9) – 4) | 10) – 3) |
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, 15th 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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Yes, the number system for SSC CGL Exam provides questions, download the PDF for free and start practicing.
A 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.
You can find the solutions in the PDF provided in this blog.
You can get the sample questions of the number system for SSC CGL in this blog.
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