Number System Questions for SSC CGL PDF:
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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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