# Number System Questions for SSC CGL PDF:

Download Number system questions for SSC CGL Tier-1 and Tier-2 PDF. Maths number system questions for competitive exams like SSC CGL. Number System Questions with answers explained with detailed solutions and explanations using tricks.

Number System Questions for SSC CGL PDF

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Question 1:

The least number which when divided by 25, 40 and 60 leaves the remainder 7 in each case is

a) 609

b) 593

c) 1207

d) 607

**Question 2:**

The digit in the unit place in the square root of 66049 is

a) 3

b) 7

c) 8

d) 2

**Question 3:**

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

a) 10

b) 9

c) 11

d) 6

**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

a) 132

b) 32

c) 43

d) 75

**Question 5:**

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

a) 15999879

b) 15899879

c) 15989979

d) 15998879

**Solutions 1 to 5**

1) Answer (d)

The least number which when divided by 25, 40 and 60 leaves the remainder 7

= L.C.M. (25,40,60) + 7

= 600 + 7 = 607

**2) Answer (b)**

Square root of 66049 = 257

Thus, unit’s digit = 7

SSC CGL Questions and Answers PDF

**3) Answer (b)**

let the digits of the no. be X and Y

number = 10x+y

reverse of no. = 10y+ x

now (10x+y) – (10y+x) = 9x-9y = 9(x-y)

as shown above that 9 will always be the factor, irrespective of the difference and the number

so it can be concluded that the resulting number will always be divisible by 9

**4) Answer (a)**

Greatest 4 digit number = 9999

When 9999 is divisible by 307, remainder obtained = 132

Least number which must be added = 132

To verify, 9999+132 = 10131, which is divisible by 307.

**5) Answer (a)**

If a = 4011 and b = 3989

ab = 4011 * 3989 = (4000 + 11) (4000 – 11)

= 4000^2 -11^2

= 16000000 – 121

= 15999879

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