{"id":42713,"date":"2020-09-18T16:45:17","date_gmt":"2020-09-18T11:15:17","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42713"},"modified":"2020-09-18T16:45:17","modified_gmt":"2020-09-18T11:15:17","slug":"number-system-questions-for-ssc-cgl-tier-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/number-system-questions-for-ssc-cgl-tier-2-pdf\/","title":{"rendered":"Number System Questions for SSC CGL Tier – 2 PDF"},"content":{"rendered":"

Number System Questions for SSC CGL Tier – 2 PDF<\/strong><\/span><\/h2>\n

Download SSC CGL Tier 2 Number System Questions PDF. Top 15 questions based on asked questions in previous exam papers very important for the SSC exam.<\/p>\n

Download Number System Questions for SSC CGL Tier – 2 PDF <\/a><\/p>\n

Get 10 SSC CGL Tier-2 Mocks for Rs. 149<\/a><\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Download SSC CGL Tier-2 Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>A 4-digit number is formed by repeating a 2-digit number such as 1515, 3737, etc. Any number of this form is exactly divisible by<\/p>\n

a)\u00a07<\/p>\n

b)\u00a011<\/p>\n

c)\u00a013<\/p>\n

d)\u00a0101<\/p>\n

Question 2:\u00a0<\/b>A number, when divided by 136, leaves remainder 36. If the same number is divided by 17, the remainder will be<\/p>\n

a)\u00a09<\/p>\n

b)\u00a07<\/p>\n

c)\u00a03<\/p>\n

d)\u00a02<\/p>\n

Question 3:\u00a0<\/b>The sixth term of the sequence 2, 6, 11, 17 is<\/p>\n

a)\u00a024<\/p>\n

b)\u00a030<\/p>\n

c)\u00a032<\/p>\n

d)\u00a036<\/p>\n

Question 4:\u00a0<\/b>If a and b be positive integers such that $a^2-b^2 = 19$, then the value of a is<\/p>\n

a)\u00a019<\/p>\n

b)\u00a020<\/p>\n

c)\u00a09<\/p>\n

d)\u00a010<\/p>\n

Question 5:\u00a0<\/b>A number, when divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a07<\/p>\n

d)\u00a017<\/p>\n

Question 6:\u00a0<\/b>The ninth term of the sequence 0, 3, 8, 15, 24, 35, is<\/p>\n

a)\u00a063<\/p>\n

b)\u00a070<\/p>\n

c)\u00a080<\/p>\n

d)\u00a099<\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Download SSC CGL Tier-2 Previous Papers PDF<\/a><\/p>\n

Question 7:\u00a0<\/b>The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is<\/p>\n

a)\u00a01120<\/p>\n

b)\u00a04714<\/p>\n

c)\u00a05200<\/p>\n

d)\u00a05600<\/p>\n

Question 8:\u00a0<\/b>A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been<\/p>\n

a)\u00a0124<\/p>\n

b)\u00a0122<\/p>\n

c)\u00a0118<\/p>\n

d)\u00a0114<\/p>\n

Question 9:\u00a0<\/b>When ‘n’ is divided by 5 the remainder is 2. What is the remainder when n2<\/sup> is divided by 5?<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a01<\/p>\n

d)\u00a04<\/p>\n

Question 10:\u00a0<\/b>Which one of the following will completely divide 571<\/sup> + 572<\/sup> + 573<\/sup> ?<\/p>\n

a)\u00a0150<\/p>\n

b)\u00a0160<\/p>\n

c)\u00a0155<\/p>\n

d)\u00a030<\/p>\n

Question 11:\u00a0<\/b>The unit digit in the product $122^{173}$\u00a0is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a04<\/p>\n

c)\u00a06<\/p>\n

d)\u00a08<\/p>\n

Download SSC CGL General Science Notes PDF<\/a><\/p>\n

Question 12:\u00a0<\/b>I multiplied a natural number by 18 and another by 21 and added the products. Which one of the following could be the sum?<\/p>\n

a)\u00a02007<\/p>\n

b)\u00a02008<\/p>\n

c)\u00a02006<\/p>\n

d)\u00a02002<\/p>\n

Question 13:\u00a0<\/b>If’n’ be any natural number. then by which largest number $(n^3 – n)$ is always divisible ?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a06<\/p>\n

c)\u00a012<\/p>\n

d)\u00a018<\/p>\n

Question 14:\u00a0<\/b>Out of six consecutive natural numbers, if the sum of first three is 27, what is the sum of the other three ?<\/p>\n

a)\u00a036<\/p>\n

b)\u00a035<\/p>\n

c)\u00a025<\/p>\n

d)\u00a024<\/p>\n

Question 15:\u00a0<\/b>If p and q represent digits, what is the possible maximum value of q in the statement 5p9 + 327 + 2q8= 1114?<\/p>\n

a)\u00a09<\/p>\n

b)\u00a08<\/p>\n

c)\u00a07<\/p>\n

d)\u00a06<\/p>\n

18000+ Questions – Free SSC Study Material<\/a><\/p>\n

Free SSC Preparation (Videos Youtube)<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

let’s say digit is pqpq
\nor pq00 + pq
\nor pq*100 + pq
\nor pq (100 + 1)
\nor pq (101)
\nhence digit will always be divisible by 101<\/p>\n

2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Number will be (136n + 36) where n is quotient
\nhence when it is divided by 17 remainder for $\\frac{136n +36}{17}$ will be 2 as 136 is divisible by 17 and 36=34+2<\/p>\n

3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

As we can see there difference of the sequence is in arithmatic progression of difference 1
\nSecond term = first term + 4
\nThird term = second term + 5
\nFourth term = third term + 6
\nnth term = (n-1)th term + (4 + (n-1)*1)
\nAdding above equation we get
\nnth term = first term + (4+5+6….+(3+(n-1)))
\nhence 6th term will be 32.<\/p>\n

4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$a^2-b^2 = 19$
\n19 is a prime number
\n(a + b)(a – b) = 19 x 1
\nSo a + b = 19
\na – b = 1
\nAdding both the equations we get 2a = 20
\na = 10<\/p>\n

5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the given number be x<\/em>
\nLet a be the quotient when x<\/em>\u00a0is divided by 114
\nSo $\\frac{x}{114}$ = a$\\frac{21}{114}$
\nso x<\/em>\u00a0= 114a + 21
\nwhen x<\/em>\u00a0is divided by 19 it can be written as
\n$\\frac{x}{19} = \\frac{114a + 21}{19}$
\n114 is divisible by 19 and 21 leaves a remainder of 2.<\/p>\n

6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Given:
\nSequence of numbers 0,3,8,15,24,35
\nTo know:
\n9th term of the sequence
\nAs we can see difference of sequence is in arthmatic progression of 3,5,7,9,11,13 and so on.
\n6th term of sequence is 35
\nand adding common difference for it (that is 13)\u00a0, hence 7th term will be 48
\naccordingly 9th term will be 80<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given: Numbers- First = 5834
\nSecond = x (Suppose)
\nAnd number (5834 – x) is divisible by each of 20,28,32,35
\nLet’s say it is y
\nHence 5834 – x = y
\nor x = 5834 – y
\nNow for x to be greatest y should be least
\nhence y should be least common multiple of 20,28,32,35
\ny = 1120
\nnow x = 5834 – 1120
\nx = 4714<\/p>\n

 <\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let’s say number is N
\nSo according to student result is $112= \\frac{N+12}{6} $
\nor N = 660
\nCorrect answer will be = $\\frac{660}{6} +12 = 110+12 = 122$<\/p>\n

9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

n = 5k+2 (where k is quotient )
\nso $n^2 = 25k^2 + 4 + 20k$
\nNow when $n^2$ will divided by 5 , remainder will be 4.<\/p>\n

10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Among all options only option C has unit digit 5, and in given equation unit digit will also be 5.
\nSo only 155 can divide the given equation completely.<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

As we know a number with unit digit 2 have repeating cycle of 2,4,8,6 after every fourth power
\nas power is 173 or (172+1) where till 172 , 43rd cycle will get complete and next unit digit will be 2.<\/p>\n

12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

let’s say one number is n and another number is p
\nso acc. to question sum will be 18n+21p
\nand this number will be divisible by 3 so answer will be (A)<\/p>\n

13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$(n^{3} – n) $ can be written as n(n-1)(n+1)
\nfor n to be any natural number, $n^{3} – n$ is a product 3 consecutive numbers starting from 1.
\nhence for any value of a min. product of 6 will be there hence it is always be divisible by 6.<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

let’s say 6 consecutive numbers are (a-d), a, (a+d), (a+2d), (a+3d), (a+4d)
\nwhere d is the common difference i.e. 1 (given) and a is second term
\nsummation of first three terms will be 3a = 27
\nhence second term a = 9
\nnow sequence is 8,9,10,11,12,13,
\nso sum of last three terms 36<\/p>\n

15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

We can break the sum according to 500 + 10p +9 + 327 + 200 + 10q + 8 = 1114
\nor 1027 + 17 + 10(p+q) = 1114
\nor (p+q) = 7
\nso for q to be maximum p will be zero then q = 7<\/p>\n

Free SSC Online Coaching<\/a><\/p>\n

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We hope this Number System Questions pdf for SSC CGL Tier 2 exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Number System Questions for SSC CGL Tier – 2 PDF Download SSC CGL Tier 2 Number System Questions PDF. Top 15 questions based on asked questions in previous exam papers very important for the SSC exam. Take SSC CGL Tier-2 Mock Tests Download SSC CGL Tier-2 Previous Papers PDF Question 1:\u00a0A 4-digit number is formed […]<\/p>\n","protected":false},"author":41,"featured_media":42715,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,9,504,378,1441],"tags":[22,462,4111,461],"class_list":{"0":"post-42713","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-stenographer","13":"tag-number-systems","14":"tag-ssc-cgl","15":"tag-ssc-cgl-tier-2","16":"tag-ssc-exams"},"better_featured_image":{"id":42715,"alt_text":"Number System Questions for SSC CGL Tier \u2013 2 PDF","caption":"Number System Questions for SSC CGL Tier \u2013 2 PDF","description":"Number System Questions 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