Hi students, Welcome to Amans Maths Blogs (AMBIPi). Are you preparing for SSC CGL Tier 1 and 2 and looking for SSC CGL 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
A positive integer when divided by 425 gives a remainder 45. When the same number is divided by 17, the remainder will be
Option A : 11
Option B : 8
Option C : 9
Option D : 10
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Option A : 11
The no. is of the form (425x + 45) First divisor (425) is multiple of second divisor (17).
Required remainder = Remainder obtained on dividing 45 by 17 = 11
SSC CGL Previous Year Math Questions: 2
A number x when divided by 289 leaves 18 as the remainder. The same number when divided by 17 leaves y as a remainder. The value of y is
Option A : 5
Option B : 2
Option C : 3
Option D : 1
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Option D : 1
Here, the first divisor (289) is a multiple of second divisor (17).
Required remainder = Remain-der obtained on dividing 18 by 17 = 1
SSC CGL Previous Year Math Paper: 3
216 –1 is divisible by
Option A : 11
Option B : 16
Option C : 17
Option D : 19
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Option C : 17
SSC CGL Mock Test Math Question: 4
Which one of the following will completely divide 571 + 572 +573 ?
Option A : 150
Option B : 160
Option C : 155
Option D : 30
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Option C : 155
SSC CGL Free Mock Test Math Question: 5
By which number should 0.022 be multiplied so that product becomes 66 ?
Option A : 3000
Option B : 3200
Option C : 4000
Option D : 3600
Show/Hide Answer Key
Option A : 3000
Let required number be x.
0.022 × x = 66
Þ x =66
0.022 = 3000
SSC CGL Advance Math Question: 6
(49)15 – 1 is exactly divisible by
Option A : 50
Option B : 51
Option C : 29
Option D : 8
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Option D : 8
SSC CGL Math Previous Year Question: 7
If a and b are two odd positive integers, by which of the follow-ing integers is (a4– b4) always divisible ?
Option A : 3
Option B : 6
Option C : 8
Option D : 12
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Option C : 8
SSC CGL Math Question Paper: 8
If m and n are positive integers and (m – n ) is an even number, then (m2– n2) will be always di-visible by
Option A : 4
Option B : 6
Option C : 8
Option D : 12
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Option A : 4
Let m = n = p and m – n = 2p m + n = 2p
(m – n) (m + n) = 4p2 Þ m2– n2 = 4
Advance Math Question Asked in SSC CGL: 9
If 5432*7 is divisible by 9, then the digit in place of * is :
Option A : 0
Option B : 1
Option C : 6
Option D : 9
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Option C : 6
A number is divisible by 9, if sum of its digits is divisible by 9. Let the number be x. Þ 5 + 4 + 3 + 2 + x + 7= 21 + x
x = 6
Previous Year Math Question Paper SSC CGL: 10
The least number, which must be added to 6709 to make it ex- actly divisible by 9, is
Option A : 5
Option B : 4
Option C : 7
Option D : 2
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Option A : 5
A number is divisible by 9 if the sum of its digits is divisible by 9. Here, 6 + 7 + 0 + 9 = 22 Now, 22 + 5 = 27, which is divis- ible by 9. Hence 5 must be add-ed to 6709.
SSC CGL Exam Math Number System Question: 11
The total number of integers be-tween 100 and 200, which are divisible by both 9 and 6, is :
Option A : 5
Option B : 6
Option C : 7
Option D : 8
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Option B : 6
A number is divisible by 9 and 6 both, if it is divisible by LCM of 9 and 6 i.e., 18. Hence, the numbers are 108, 126, 144, 162, 180, 198.
SSC CGL Previous Year Math Number System Questions: 12
How many 3-digit numbers, in all, are divisible by 6
Option A : 140
Option B : 150
Option C : 160
Option D : 170
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Option B : 150
First 3–digit number divisible by 6 = 102 Last such 3-digit number =996
996 = 102 + (n –1) 6
(n – 1)6 = 996 – 102 = 894
n – 1 =8946= 149
n = 150
SSC CGL Previous Year Math Number System Paper: 13
If ‘n’ be any natural number, then by which largest number (n3– n) is always divisible ?
Option A : 3
Option B : 6
Option C : 12
Option D : 18
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Option B : 6
SSC CGL Mock Test Math Number System Question: 14
If n is an integer, then (n3– n) is always divisible by :
Option A : 4
Option B : 5
Option C : 6
Option D : 7
Show/Hide Answer Key
Option B : 5
SSC CGL Free Mock Test Math Number System Question: 15
If a number is divisible by both 11 and 13, then it must be necessarily :
Option A : divisible by (11 + 13)
Option B : divisible by (13 – 11)
Option C : divisible by (11 × 13)
Option D : 429
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Option C : divisible by (11 × 13)
divisible by (11 × 13)
SSC CGL Advance Math Number System Question: 16
If * is a digit such that 5824* is divisible by 11, then * equals
Option A : 2
Option B : 3
Option C : 5
Option D : 6
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Option C: 5
SSC CGL Math Previous Year Number System Question: 17
A 4-digit number is formed by repeating a 2-digit number such as 2525, 3232, etc. Any number of this form is always exactly di-visible by
Option A : 7
Option B : 11
Option C : 13
Option D : Smallest 3-digit prime num-ber
Show/Hide Answer Key
Option D : Smallest 3-digit prime num-ber
Let the unit digit be x and ten’sdigit be y.
Number = 1000y + 100x + 10y + x = 1010y + 101x = 101(10y + x) Clearly, this number is divisible
by 101, which is the smallest three-digit prime number.
SSC CGL Math Number System Question Paper: 18
It is given that (232 + 1) is exactly divisible by a certain number. which one of the following is also definitely divisible by the same number ?
Option A : 296 + 1
Option B : 7 x 233
Option C : 216 – 1
Option D : 216 + 1
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Option A : 296 + 1
Advance Math Number System Question Asked in SSC CGL: 19
The greatest whole number, by which the expression n4 + 6n3 + 11n2 + 6n + 24 is divisible for every natural num- ber n, is
Option A : 6
Option B : 24
Option C : 12
Option D : 48
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Option D : 48
Previous Year Math Number System Question Paper SSC CGL: 20
How many numbers between 1000 and 5000 are exactly di- visible by 225 ?
Option A : 16
Option B : 18
Option C : 19
Option D : 12
Show/Hide Answer Key
Option B : 18
When we divide 1000 by 225, quotient = 4 When we divide 5000 by 225, quotient = 22
Required answer = 22 – 4 = 18
