{"id":42827,"date":"2020-09-23T18:47:28","date_gmt":"2020-09-23T13:17:28","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42827"},"modified":"2020-09-23T18:47:28","modified_gmt":"2020-09-23T13:17:28","slug":"number-system-questions-for-ssc-cgl-tier-2-pdf-2","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/number-system-questions-for-ssc-cgl-tier-2-pdf-2\/","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 SSC CGL Tier 2 Number System 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

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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 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 2:\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 3:\u00a0<\/b>The unit digit in the sum of (124)372<\/sup> + (124)373<\/sup> is<\/p>\n

a)\u00a05<\/p>\n

b)\u00a04<\/p>\n

c)\u00a020<\/p>\n

d)\u00a00<\/p>\n

Question 4:\u00a0<\/b>The least number which must be added to 1728 to make it a perfect square is \u2026\u2026\u2026\u2026\u2026..<\/p>\n

a)\u00a036<\/p>\n

b)\u00a032<\/p>\n

c)\u00a038<\/p>\n

d)\u00a030<\/p>\n

Question 5:\u00a0<\/b>The square root of $(\\frac{\\sqrt{3} + \\sqrt{2}}{\\sqrt{3} – \\sqrt{2}})$<\/p>\n

a)\u00a0$\\sqrt{3} + \\sqrt{2}$<\/p>\n

b)\u00a0$\\sqrt{3} – \\sqrt{2}$<\/p>\n

c)\u00a0$\\sqrt{2} \\pm \\sqrt{3}$<\/p>\n

d)\u00a0$\\sqrt{2} – \\sqrt{3}$<\/p>\n

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

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Question 6:\u00a0<\/b>A positive integer when divided by 425 gives a remainder 45. When the same number is divided by 17, the remainder will be<\/p>\n

a)\u00a011<\/p>\n

b)\u00a08<\/p>\n

c)\u00a09<\/p>\n

d)\u00a010<\/p>\n

Question 7:\u00a0<\/b>The ten’s digit of a 2-digit number is greater than the units digit by 7. If we subtract 63 from the number, the new number obtained is a number formed by interchange of the digits. Find the number.<\/p>\n

a)\u00a081<\/p>\n

b)\u00a070<\/p>\n

c)\u00a092<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

Question 8:\u00a0<\/b>Common factor of $12a^4b^6, 18a^6c^2, 36a^2b^2$ is<\/p>\n

a)\u00a0$36a^2$<\/p>\n

b)\u00a0$108b^2$<\/p>\n

c)\u00a0$6a^2b^2$<\/p>\n

d)\u00a0$6a^2$<\/p>\n

Question 9:\u00a0<\/b>If $x^3 + 2x^2 – 5x + k$ is divisible by x + 1, then what is the value of k?<\/p>\n

a)\u00a0– 6<\/p>\n

b)\u00a0– 1<\/p>\n

c)\u00a00<\/p>\n

d)\u00a06<\/p>\n

Question 10:\u00a0<\/b>The sum of a fraction and 7 times its reciprocal is 11\/2. What is the fraction?<\/p>\n

a)\u00a07\/2<\/p>\n

b)\u00a02\/7<\/p>\n

c)\u00a03\/4<\/p>\n

d)\u00a04\/3<\/p>\n

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Question 11:\u00a0<\/b>What is the HCF of 6345 and 2160?<\/p>\n

a)\u00a045<\/p>\n

b)\u00a0135<\/p>\n

c)\u00a0270<\/p>\n

d)\u00a015<\/p>\n

Question 12:\u00a0<\/b>$(2^{51}+2^{52}+2^{53}+2^{54}+2^{55})$ is divisible by<\/p>\n

a)\u00a023<\/p>\n

b)\u00a058<\/p>\n

c)\u00a0124<\/p>\n

d)\u00a0127<\/p>\n

Question 13:\u00a0<\/b>Sum of the factors of $4b^2c^2 – (b^2 + c^2 – a^2)^2$ is<\/p>\n

a)\u00a0a + b + c<\/p>\n

b)\u00a02(a + b + c)<\/p>\n

c)\u00a00<\/p>\n

d)\u00a01<\/p>\n

Question 14:\u00a0<\/b>If a number 657423547X46 is divisible by 11, then find the value of X.<\/p>\n

a)\u00a07<\/p>\n

b)\u00a09<\/p>\n

c)\u00a08<\/p>\n

d)\u00a06<\/p>\n

Question 15:\u00a0<\/b>$4^{11}+4^{12}+4^{13}+4^{14}\\ $is divisible by_________.<\/p>\n

a)\u00a07<\/p>\n

b)\u00a014<\/p>\n

c)\u00a017<\/p>\n

d)\u00a09<\/p>\n

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

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

1)\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

2)\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

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

Both of numbers have unit digit as 4 and it has a repeating cycle of 2 with unit digits as 4 and 6
\nso in first number power is 372 which is exactly divisible by 2 hence unit digit of first number will be 6.
\nand in second number power is 373 which exceeds one with the reapeating cycle of 2 hence its unit digit will be 4.<\/p>\n

now unit digit of the sum will be 6+4 = 10<\/p>\n

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

We know that, 41 * 41 = 1681<\/p>\n

and 42 * 42 = 1764<\/p>\n

So, the least number to be added in 1728 to make it a perfect square = 1764-1728<\/p>\n

= 36<\/p>\n

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

we need to calculate square root of $(\\frac{\\sqrt{3} + \\sqrt{2}}{\\sqrt{3} – \\sqrt{2}})$<\/p>\n

let $(\\frac{\\sqrt{3} + \\sqrt{2}}{\\sqrt{3} – \\sqrt{2}})$ be = y <\/span><\/p>\n

on rationalizing y , we get <\/span><\/p>\n

y = $(\\surd2 + \\surd3)^2$ <\/span><\/p>\n

hence square root of y => $\\surd(y)$ = $\\surd2 + \\surd3$<\/span><\/p>\n

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

Let the number be 45<\/p>\n

So, when 45 is divided by 425 => remainder = 45<\/p>\n

Now, when 45 is divided by 17<\/p>\n

=> Remainder = 45%17 = 11<\/p>\n

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

Let the unit’s digit of the number be $y$ and ten’s digit be $x$<\/p>\n

=> Number = $10x + y$<\/p>\n

According to ques, =>$x – y = 7$ ————–(i)<\/p>\n

According to question, => $10x + y – 63 = 10y + x$<\/p>\n

=> $9x – 9y = 63$<\/p>\n

=> $x – y = \\frac{63}{9} = 7$ ————–(ii)<\/p>\n

Equations (i) and (ii) are same and thus we have two variables and one equation<\/p>\n

Number can be = 92 , 81 and thus the solution cannot be determined.<\/p>\n

=> Ans – (D)<\/p>\n

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

Factors of\u00a0:<\/p>\n

$12a^4b^6$ = $(2 \\times 6) \\times (a^2 \\times a^2) \\times b^6$<\/p>\n

$18a^6c^2$ = $(3 \\times 6) \\times (a^2 \\times a^4) \\times c^2$<\/p>\n

$36a^2b^2$ = $(6 \\times 6) \\times (a^2) \\times b^2$<\/p>\n

The common factor in the 3 terms = $6a^2$<\/p>\n

=> Ans – (D)<\/p>\n

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

$f(x)=x^3 + 2x^2 – 5x + k$ is divisible by $(x+1)$<\/p>\n

=> $x=-1$ is a factor of\u00a0$f(x)$<\/p>\n

=> $f(-1)=0$<\/p>\n

=> $(-1)^3+2(-1)^2-5(-1)+k=0$<\/p>\n

=> $-1+2+5+k=0$<\/p>\n

=> $k=-6$<\/p>\n

=> Ans – (A)<\/p>\n

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

Let the fraction be\u00a0$x$<\/p>\n

According to ques,<\/p>\n

=> $x+\\frac{7}{x}=\\frac{11}{2}$<\/p>\n

=> $\\frac{x^2+7}{x}=\\frac{11}{2}$<\/p>\n

=> $2x^2+14=11x$<\/p>\n

=> $2x^2-11x+14=0$<\/p>\n

=> $2x^2-4x-7x+14=0$<\/p>\n

=> $2x(x-2)-7(x-2)=0$<\/p>\n

=> $(x-2)(2x-7)=0$<\/p>\n

=> $x=2,\\frac{7}{2}$<\/p>\n

=> Ans – (A)<\/p>\n

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

6345 = 3*3*3*5*47<\/p>\n

2160 = 3*3*3*5*16<\/p>\n

HCF = product of common prime factors =\u00a0\u00a03*3*3*5 = 135<\/p>\n

So the answer is option B.<\/p>\n

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

Expression\u00a0:\u00a0$(2^{51}+2^{52}+2^{53}+2^{54}+2^{55})$<\/p>\n

= $2^{51}(1+2+2^2+2^3+2^4)$<\/p>\n

= $2^{51}(1+2+4+8+16)$<\/p>\n

= $2^{51} \\times 31$<\/p>\n

Thus, the above expression is divisible by $31k$ and the only option that is divisible by 31 = 124<\/p>\n

=> Ans – (C)<\/p>\n

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

Expression\u00a0:\u00a0$4b^2c^2 – (b^2 + c^2 – a^2)^2$<\/p>\n

= $(2bc)^2-(b^2+c^2-a^2)^2$<\/p>\n

Using, $x^2-y^2=(x-y)(x+y)$, where $x=2bc$ and\u00a0$y=b^2+c^2-a^2$<\/p>\n

= $(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)$<\/p>\n

= $[a^2-(-2bc+b^2+c^2)][(2bc+b^2+c^2)-a^2]$<\/p>\n

= $[a^2-(b-c)^2][(b+c)^2-a^2]$<\/p>\n

= $[(a-b+c)(a+b-c)][(b+c-a)(b+c+a)]$<\/p>\n

Thus, sum of factors = $(a-b+c)+(a+b-c)+(b+c-a)+(b+c+a)$<\/p>\n

= $2a+2b+2c=2(a+b+c)$<\/p>\n

=> Ans – (B)<\/p>\n

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

For a number $657423547X46$ to be divisible by 11, the difference between the sum of numbers at even position and odd position should be either ‘0’ or ’11’.<\/p>\n

Sum of digits at odd position (starting from right) = $6+X+4+3+4+5=(22+X)$<\/p>\n

Even positions = $4+7+5+2+7+6=31$<\/p>\n

=> $22+X-31=0$<\/p>\n

=> $X=9$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression :\u00a0$4^{11}+4^{12}+4^{13}+4^{14}\\ $<\/p>\n

= $4^{11}(1+4+4^2+4^3)$<\/p>\n

= $4^{11}\\times(1+4+16+64)$<\/p>\n

= $4^{11}\\times(85)$<\/p>\n

$\\because$ $85$ is divisible by 17, hence the above expression is also divisible by\u00a017<\/strong><\/p>\n

=> Ans – (C)<\/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 SSC CGL Tier 2 Number System 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 […]<\/p>\n","protected":false},"author":42,"featured_media":42834,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1741,1441],"tags":[2452,3415,462,4111,358,500,461,1711,1847],"class_list":{"0":"post-42827","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-mts","14":"category-stenographer","15":"tag-number-system","16":"tag-quant-questions","17":"tag-ssc-cgl","18":"tag-ssc-cgl-tier-2","19":"tag-ssc-chsl","20":"tag-ssc-cpo","21":"tag-ssc-exams","22":"tag-ssc-je","23":"tag-ssc-mts"},"better_featured_image":{"id":42834,"alt_text":"Number 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