Number System Questions for SSC CGL PDF

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Number System Questions for SSC CGL PDF
Number System Questions for SSC CGL PDF

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

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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

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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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