Hi students, Welcome to **Amans Maths Blogs (AMBIPi)**. Are you preparing for **SSC CGL Tier 1 and 2** and looking for * SSC CGL Exam Free Mock Test Math Number System Question with Solutions AMBIPi*? In this article, you will get Previous Year Mathematics Questions asked in SSC CGL Tier 1 and Tier 2, which helps you in the preparation of government job exams of SSC CGL.

### SSC CGL Exam Math Number System Question with Solutions

**SSC CGL Exam Math Question: 1**

The least number, which is to be added to the greatest number of 4 digits so that the sum may be divisible by 345, is

**Option A** : 50

**Option B** : 6

**Option C** : 60

**Option D** : 5

**Show/Hide Answer Key**

**Option B : 6**

The largest 4-digit number= 9999

Required number = 345 – 339= 6

**SSC CGL Previous Year Math Questions: 2**

The difference of a number consisting of two digits from the number formed by interchanging the digits is always divisible by

**Option A** : 10

**Option B** : 9

**Option C** : 11

**Option D** : 6

**Show/Hide Answer Key**

**Option B : 9**

Let the number be 10x + y where y < x.

Number obtained by interchanging the digits = 10y + x

Difference = 10x + y – 10y – x= 9x – 9y = 9 (x – y)

Hence, the difference is always exactly divisible by 9.

**SSC CGL Previous Year Math Paper: 3**

Which one of the numbers is di-visible by 25 ?

**Option A** : 303310

**Option B** : 373355

**Option C** : 303375

**Option D** : 22040

**Show/Hide Answer Key**

**Option C : 303375**

**SSC CGL Mock Test Math Question: 4**

The least number which must be added to the greatest number of 4 digits in order that the sum may be exactly divisible by 307 is

**Option A** : 132

**Option B** : 32

**Option C** : 43

**Option D** : 75

**Show/Hide Answer Key**

**Option A : 132**

307 × 32 = 9824

307 × 33 = 10131

Required number= 10131 – 9999 = 132

**SSC CGL Free Mock Test Math Question: 5**

If a = 4011 and b = 3989 then value of ab = ?

**Option A** : 15999879

**Option B** : 15899879

**Option C** : 15989979

**Option D** : 15998879

**Show/Hide Answer Key**

**Option A : 15999879**

a = 4011, b = 3989

ab = 4011 × 3989= (4000 + 11) (4000 – 11)= (4000)^{2} – (11)^{2} = 16000000 – 121 = 15999879

**SSC CGL Advance Math Question: 6**

For any integral value of n, 3^{2n}+ 9n + 5 when divided by 3 will leave the remainder

**Option A** : 1

**Option B** : 2

**Option C** : 0

**Option D** : 5

**Show/Hide Answer Key**

**Option B : 2**

**SSC CGL Math Previous Year Question: 7**

In a farm there are cows and hens. If heads are counted they are 180, if legs are counted they are 420. The number of cows in the farm is

**Option A** : 130

**Option B** : 150

**Option C** : 50

**Option D** : 30

**Show/Hide Answer Key**

**Option D : 30**

A cow and a hen each has a head.

If the total number of cows be x, then

Number of hens = 180 – x A cow has four legs and a hence

has two legs.

(180 – x) × 2 + 4x = 420

Þ 360 – 2x + 4x = 420

Þ 2x = 420 – 360 = 60 Þ x =602= 30

**SSC CGL Math Question Paper: 8**

The number which can be writ- ten in the form of n (n + 1) (n + 2), where n is a natural number, is

**Option A** : 7

**Option B** : 3

**Option C** : 5

**Option D** : 6

**Show/Hide Answer Key**

**Option D : 6**

On putting n = 1 n(n +1) (n + 2) = 1 × 2 × 3 = 6

**Advance Math Question Asked in SSC CGL: 9**

A number when divided by 2736 leaves the remainder 75. If the same number is divided by 24, then the remainder is

**Option A** : 12

**Option B** : 3

**Option C** : 0

**Option D** : 23

**Show/Hide Answer Key**

**Option D : 23**

2736 ÷ 24 = 114

Hence, first divisor (2736) is a

multiple of second divisor (24).

Required remainder = Remainder obtained on

dividing 75 by 24 = 3

**Previous Year Math Question Paper SSC CGL: 10**

The sum of four numbers is 48. When 5 and 1 are added to the first two; and 3 and 7 are sub- tracted from the 3rd and 4th, the numbers will be equal. The numbers are

**Option A** : 9, 7, 15, 17

**Option B** : 4, 12, 12, 20

**Option C** : 5, 11, 13, 19

**Option D** : 16, 10, 14, 18

**Show/Hide Answer Key**

**Option D : 16, 10, 14, 18**

**SSC CGL Exam Math Number System Question: 11**

The least number that should be added to 2055, so that the sum is exactly divisible by 27 is

**Option A** : 28

**Option B** : 24

**Option C** : 27

**Option D** : 31

**Show/Hide Answer Key**

**Option B : 24**

Required number = 27 – 3 = 24

**SSC CGL Previous Year Math Number System Questions: 12**

A number when divided by 361 gives a remainder 47. If the same number is divided by 19, the re- mainder obtained is

**Option A** : 3

**Option B** : 8

**Option C** : 9

**Option D** : 1

**Show/Hide Answer Key**

**Option C : 9**

Here, the first divisor (361) is a multiple of second divisor (19).

Required remainder = Remainder obtained on dividing 47 by 19 = 9

**SSC CGL Previous Year Math Number System Paper: 13**

Which one of the following is the minimum value of the sum of two integers whose product is 24?

**Option A** : 25

**Option B** : 11

**Option C** : 8

**Option D** : 10

**Show/Hide Answer Key**

**Option D : 10 **

xy = 24

(x, y)= (1 × 24), (2 ×12), (3 × 8), (4 × 6)

Minimum value of (x + y)= 4 + 6 = 10.

**SSC CGL Mock Test Math Number System Question: 14**

If the sum of the digits of a three digit number is subtracted from that number, then it will always be divisible by

**Option A** : 3 only

**Option B** : 9 only

**Option C** : Both 3 and 9 only

**Option D** : All of 3, 6 and 9

**Show/Hide Answer Key**

**Option C : Both 3 and 9 only**

Let the 3–digit number be

100x + 10y + z.

Sum of the digits = x + y + z

According to the question,

Difference= 100x + 10y + z – (x + y + z)= 99x + 9y= 9 (11x + y)

Clearly, it is a multiple of 3 and9.

**SSC CGL Free Mock Test Math Number System Question: 15**

The greater of the two numbers whose product is 900 and sum exceeds their difference by 30 is

**Option A** : 60

**Option B** : 75

**Option C** : 90

**Option D** : 100

**Show/Hide Answer Key**

**Option A : 60**

**SSC CGL Advance Math Number System Question: 16**

In a division sum, the divisor ‘d’ is 10 times the quotient ‘q’ and 5 times the remainder ‘r’. If r = 46, the dividend will be

**Option A** : 5042

**Option B** : 5328

**Option C** : 5336

**Option D** : 4276

**Show/Hide Answer Key**

**Option C : 5336**

According to the question,

Divisor (d) = 5r = 5 × 46 = 230

Again, Divisor (d) = 10 × Quotient (q)

Þ 230 = q × 10

Þ q =23010 = 23

\ Dividend = Divisor × Quotient+ Remainder= 230 × 23 + 46= 5290 + 46 = 5336

**SSC CGL Math Previous Year Number System Question: 17**

If the sum of a number and its reciprocal be 2, then the number is

**Option A** : 0

**Option B** : 1

**Option C** : -1

**Option D** : 2

**Show/Hide Answer Key**

**Option B : 1**

**SSC CGL Math Number System Question Paper: 18**

When a number is divided by 56, the remainder will be 29. If the same number is divided by 8, then the remainder will be

**Option A** : 6

**Option B** : 7

**Option C** : 5

**Option D** : 3

**Show/Hide Answer Key**

**Option C : 5**

First divisor (56) is a mul-tiple of second divisor (8).Required remainder= Remainder obtained after dividing 29 by 8 = 5

**Advance Math Number System Question Asked in SSC CGL: 19**

A positive number when de-creased by 4, is equal to 21 times the reciprocal of this number. The number is :

**Option A** : 3

**Option B** : 7

**Option C** : 5

**Option D** : 9

**Show/Hide Answer Key**

**Option B : 7**

**Previous Year Math Number System Question Paper SSC CGL: 20**

When n is divided by 4, the remainder is 3. The remainder when 2n is divided by 4 is

**Option A** : 1

**Option B** : 2

**Option C** : 3

**Option D** : 6

**Show/Hide Answer Key**

**Option D : 2**

Let quotient be 1.n = 4 × 1 + 3 = 7

2n = 2 × 7 = 14,On dividing 14 by 4, remainder= 2