{"id":42577,"date":"2020-09-09T20:51:05","date_gmt":"2020-09-09T15:21:05","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42577"},"modified":"2020-09-09T20:51:05","modified_gmt":"2020-09-09T15:21:05","slug":"algebra-questions-for-ssc-cgl-tier-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/algebra-questions-for-ssc-cgl-tier-2-pdf\/","title":{"rendered":"Algebra Questions for SSC CGL Tier 2 PDF\u00a0"},"content":{"rendered":"

Algebra Questions for SSC CGL Tier 2 PDF\u00a0<\/strong><\/span><\/h2>\n

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

Download Algebra Questions for SSC CGL Tier 2 PDF\u00a0<\/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

 <\/p>\n

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

Question 1:\u00a0<\/b>If x = \u221a5+ 2, then the value $\\frac{2x^2-3x-2}{3x^2-4x-3}$ is equal to<\/p>\n

a)\u00a00.185<\/p>\n

b)\u00a00.525<\/p>\n

c)\u00a00.625<\/p>\n

d)\u00a00.785<\/p>\n

Question 2:\u00a0<\/b>The value of $\\frac{(243)^\\frac{n}{5}\\times3^{2n+1}}{9^n\\times3^{n-1}}$ is<\/p>\n

a)\u00a03<\/p>\n

b)\u00a09<\/p>\n

c)\u00a06<\/p>\n

d)\u00a012<\/p>\n

Question 3:\u00a0<\/b>The value of 204 \u00d7 197 is<\/p>\n

a)\u00a040218<\/p>\n

b)\u00a040188<\/p>\n

c)\u00a040212<\/p>\n

d)\u00a039812<\/p>\n

Question 4:\u00a0<\/b>What number must be added to the expression $16a^{2} -12a$ to make it a perfect square ?<\/p>\n

a)\u00a0$\\frac{9}{4}$<\/p>\n

b)\u00a0$\\frac{11}{2}$<\/p>\n

c)\u00a0$\\frac{13}{2}$<\/p>\n

d)\u00a0$16$<\/p>\n

Question 5:\u00a0<\/b>The simplest form of the expression $\\frac{p^2-p}{2p^3+6p^2}+\\frac{p^2-1}{p^2+3p}+\\frac{p^2}{p+1}$<\/p>\n

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

b)\u00a0$\\frac{1}{2p^2}$<\/p>\n

c)\u00a0$p+3$<\/p>\n

d)\u00a0$\\frac{1}{p+3}$<\/p>\n

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

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Question 6:\u00a0<\/b>If $\\frac{5x}{2x^2+5x+1}=\\frac{1}{3}$ then the value of $(x+\\frac{1}{2x})$ is <\/span><\/p>\n

a)\u00a015<\/p>\n

b)\u00a010<\/p>\n

c)\u00a020<\/p>\n

d)\u00a05<\/p>\n

Question 7:\u00a0<\/b>If $x = \\frac{2\\sqrt{6}}{\\sqrt{3}+\\sqrt{2}}$, then the value of $\\frac{x+\\sqrt{2}}{x-\\sqrt{2}} + \\frac{x+\\sqrt{3}}{x-\\sqrt{3}}$ is<\/p>\n

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

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

c)\u00a0$\\sqrt{6}$<\/p>\n

d)\u00a0$2$<\/p>\n

Question 8:\u00a0<\/b>If $x^2 – 3x + 1= 0$ and x > 1, then the value of $(x – \\frac{1}{x})$<\/p>\n

a)\u00a0\u221a5 only<\/p>\n

b)\u00a01<\/p>\n

c)\u00a0\u00ad\u221a5 only<\/p>\n

d)\u00a0\u00b1\u221a5<\/p>\n

Question 9:\u00a0<\/b>If $(x-3)^2 + (y – 5)^2 + (z-4)^2 = 0$ then the value of $\\frac{x^2}{9} + \\frac{y^2}{25} + \\frac{z^2}{16}$<\/p>\n

a)\u00a012<\/p>\n

b)\u00a09<\/p>\n

c)\u00a03<\/p>\n

d)\u00a01<\/p>\n

Question 10:\u00a0<\/b>Coef\ufb01cient of x in (x + 9)(8 – 5x) is<\/p>\n

a)\u00a037<\/p>\n

b)\u00a0-53<\/p>\n

c)\u00a0-37<\/p>\n

d)\u00a053<\/p>\n

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

Question 11:\u00a0<\/b>Coefficient of x in (x + 8)(6 – 3x) is<\/p>\n

a)\u00a018<\/p>\n

b)\u00a030<\/p>\n

c)\u00a0-18<\/p>\n

d)\u00a0-30<\/p>\n

Question 12:\u00a0<\/b>If $3x^{2} = 10^{2} – 5^{2}$, find the value of x?<\/p>\n

a)\u00a07<\/p>\n

b)\u00a05<\/p>\n

c)\u00a09<\/p>\n

d)\u00a011<\/p>\n

Question 13:\u00a0<\/b>Simplify 437bxaz \/ 23ab.<\/p>\n

a)\u00a017xzb<\/p>\n

b)\u00a09xz<\/p>\n

c)\u00a019xz<\/p>\n

d)\u00a019ab<\/p>\n

Question 14:\u00a0<\/b>\u00ad\u00a0If 2(3x \u00ad-4) -\u00ad 2 < 4x -2 \u2265 2x \u00ad- 4\u037e then the value of x is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a05<\/p>\n

c)\u00a0\u00ad4<\/p>\n

d)\u00a0-5<\/p>\n

Question 15:\u00a0<\/b>If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is<\/p>\n

a)\u00a010<\/p>\n

b)\u00a06<\/p>\n

c)\u00a04<\/p>\n

d)\u00a02<\/p>\n

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 <\/p>\n

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

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

Given , x = (\u221a5+ 2)
\n$\\frac{2x^2-3x-2}{3x^2-4x-3}$ = $\\frac{2(\\sqrt{5}+ 2)^2-3(\\sqrt{5}+ 2)-2}{3(\\sqrt{5}+ 2)^2-4(\\sqrt{5}+ 2)-3}$
\n= $\\frac{21.18}{33.88}$
\n= 0.625<\/p>\n

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

=$\\frac{(3)^\\frac{5n}{5}\\times3^{2n+1}}{3^2n\\times3^{n-1}}$<\/p>\n

=$\\frac{(3)^{3n+1}}{3^{3n-1}}$<\/p>\n

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

=$(3)^{2}$ = 9 (B)<\/p>\n

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

$200\\times197=(200+4)(200-3)=(200)^{2}+4(200)-3(200)-12=40000+200-12=40188$<\/p>\n

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

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

Using the rule, $a^2 – 2ab + b^2 = (a-b)^2$<\/p>\n

=> $16a^2 – 12a = (4a)^2 – 2*4a*\\frac{3}{2}$<\/p>\n

= $(4a)^2 – 2*4a*\\frac{3}{2} + (\\frac{3}{2})^2 – \\frac{9}{4}$<\/p>\n

= $(4a – \\frac{3}{2})^2$<\/p>\n

=> $\\frac{9}{4}$ should be added to make the above expression a perfect square.<\/p>\n

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

$\\frac{p^2-p}{2p^3+6p^2}$ = $\\frac{p(p-1)}{2p^2(p+3)}$ = $\\frac{(p-1)}{2p(p+3)}$<\/p>\n

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

$\\frac{p^2}{p+1}$ = $\\frac{p^2}{p+1}$<\/p>\n

$\\frac{(p-1)}{2p(p+3)}$ + $\\frac{(p-1)(p+1)}{p(p+3)}$ + $\\frac{p^2}{p+1}$ = $\\frac{1}{2p^2}$<\/span><\/span>
\n<\/span><\/p>\n

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

Expression : $\\frac{5x}{2x^2+5x+1}=\\frac{1}{3}$<\/p>\n

=> $2x^2 + 5x + 1 = 15x$<\/p>\n

=> $2x^2 + 1 = 10x$<\/p>\n

To find : $(x+\\frac{1}{2x})$<\/p>\n

= $\\frac{2x^2 + 1}{2x}$<\/span><\/p>\n

= $\\frac{10x}{2x}$<\/span><\/p>\n

= 5<\/span><\/p>\n

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

Given\u00a0:\u00a0$x = \\frac{2\\sqrt{6}}{\\sqrt{3}+\\sqrt{2}}$<\/p>\n

=>\u00a0$x = \\frac{2\\sqrt{6}}{\\sqrt{3}+\\sqrt{2}}\\times(\\frac{\\sqrt3-\\sqrt2}{\\sqrt3-\\sqrt2})$<\/p>\n

=> $x=\\frac{2\\sqrt6(\\sqrt3-\\sqrt2)}{3-2}$<\/p>\n

=> $x=2\\sqrt{18}-2\\sqrt{12}$<\/p>\n

=> $x=6\\sqrt2-4\\sqrt3$ —————(i)<\/p>\n

To find\u00a0:\u00a0$\\frac{x+\\sqrt{2}}{x-\\sqrt{2}} + \\frac{x+\\sqrt{3}}{x-\\sqrt{3}}$<\/p>\n

=\u00a0$\\frac{6\\sqrt2-4\\sqrt3+\\sqrt{2}}{6\\sqrt2-4\\sqrt3-\\sqrt{2}} + \\frac{6\\sqrt2-4\\sqrt3+\\sqrt{3}}{6\\sqrt2-4\\sqrt3-\\sqrt{3}}$ \u00a0 \u00a0 [Using (i)]<\/p>\n

=\u00a0$\\frac{7\\sqrt{2}-4\\sqrt3}{5\\sqrt{2}-4\\sqrt3} + \\frac{6\\sqrt2-3\\sqrt{3}}{6\\sqrt2-5\\sqrt{3}}$<\/p>\n

= $\\frac{(84-35\\sqrt6-24\\sqrt6+60)+(60-15\\sqrt6-24\\sqrt6+36)}{60-25\\sqrt6-24\\sqrt6+60}$<\/p>\n

= $\\frac{240-98\\sqrt6}{120-49\\sqrt6}$<\/p>\n

= $\\frac{2(120-49\\sqrt6)}{120-49\\sqrt6}=2$<\/p>\n

=> Ans – (D)<\/p>\n

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

Expression : $x^2 – 3x + 1= 0$<\/p>\n

=> $x^2 + 1 = 3x$<\/p>\n

Dividing by $(x)$ on both sides<\/p>\n

=> $x + \\frac{1}{x} = 3$<\/p>\n

$\\because (x – \\frac{1}{x})^2 = (x + \\frac{1}{x})^2 – 4$<\/p>\n

=> $x – \\frac{1}{x} = \\sqrt{9 – 4}$<\/p>\n

=> $(x – \\frac{1}{x}) = \\pm\\sqrt{5}$<\/p>\n

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

Expression : $(x-3)^2 + (y – 5)^2 + (z-4)^2 = 0$<\/p>\n

Since, all the terms are positive, the only way the sum can be ‘0’ is, if each term is equal to 0.<\/p>\n

=> $(x – 3)^2 = 0$<\/p>\n

=> $x = 3$<\/p>\n

Similarly, $y = 5$ & $z = 4$<\/p>\n

To find : $\\frac{x^2}{9} + \\frac{y^2}{25} + \\frac{z^2}{16}$<\/p>\n

= $\\frac{3^2}{9} + \\frac{5^2}{25} + \\frac{4^2}{16}$<\/span><\/p>\n

= 1 + 1 + 1 = 3<\/span><\/p>\n

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

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg :\u00a0In $ax^2$, coefficient is $a$<\/p>\n

Expression\u00a0: $(x + 9)(8 – 5x)$<\/p>\n

= $8x – 5x^2 + 72 – 45x$<\/p>\n

= $-5x^2 – 37x + 72$<\/p>\n

$\\therefore$ Coefficient of $x$ = -37<\/p>\n

=> Ans – (C)<\/p>\n

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

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg :\u00a0In $ax^2$, coefficient is $a$<\/p>\n

Expression\u00a0: $(x + 8)(6 – 3x)$<\/p>\n

= $6x – 3x^2 + 48 – 24x$<\/p>\n

= $-3x^2 – 18x + 48$<\/p>\n

$\\therefore$ Coefficient of $x$ = -18<\/p>\n

=> Ans – (C)<\/p>\n

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

Expression\u00a0:\u00a0$3x^{2} = 10^{2} – 5^{2}$<\/p>\n

=> $3x^2 = 100 – 25$<\/p>\n

=> $3x^2 = 75$<\/p>\n

=> $x^2 = \\frac{75}{3} = 25$<\/p>\n

=> $x = \\sqrt{25} = 5$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0: $\\frac{437 b x a z}{23 a b}$<\/p>\n

= $\\frac{437}{23} \\times \\frac{abxz}{ab}$<\/p>\n

= $19 x z$<\/p>\n

=> Ans – (C)<\/p>\n

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

Expression 1\u00a0:\u00a02(3x \u00ad-4) -\u00ad 2 < 4x -2<\/p>\n

=> $6x-8-2$ < $4x-2$<\/p>\n

=> $6x-4x$ < $-2+10$<\/p>\n

=> $2x$ < $8$<\/p>\n

=> $x$ < $4$ ————(i)<\/p>\n

Expression 2\u00a0:\u00a04x -2 \u2265 2x \u00ad- 4<\/p>\n

=> $4x-2x \\geq -4+2$<\/p>\n

=> $2x \\geq -2$<\/p>\n

=> $x \\geq -1$ ————(ii)<\/p>\n

Combining inequalities (i) and (ii), we get : $-1 \\leq x$ < $4$<\/p>\n

The only value that $x$ can take among the given options = 2<\/p>\n

=> Ans – (A)<\/p>\n

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

For 3*5 put a=3 and b=5 in given equation
\nand for 5*3 put a=5 and b=3 in equation
\nnow add both values<\/p>\n

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

DOWNLOAD APP FOR SSC FREE MOCKS<\/a><\/p>\n

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

Algebra Questions for SSC CGL Tier 2 PDF\u00a0 Download SSC CGL Tier 2 Algebra Questions PDF. Top 15 SSC CGL Tier 2 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:\u00a0If x = […]<\/p>\n","protected":false},"author":42,"featured_media":42580,"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":[2308,4093,462,4111,461],"class_list":{"0":"post-42577","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-algebra","16":"tag-algebra-questions","17":"tag-ssc-cgl","18":"tag-ssc-cgl-tier-2","19":"tag-ssc-exams"},"better_featured_image":{"id":42580,"alt_text":"Algebra Questions for SSC CGL Tier 2 PDF\u00a0","caption":"Algebra Questions for SSC CGL 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