{"id":34673,"date":"2019-09-09T18:31:55","date_gmt":"2019-09-09T13:01:55","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=34673"},"modified":"2019-09-09T18:31:55","modified_gmt":"2019-09-09T13:01:55","slug":"algebra-questions-for-rrb-group-d-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/algebra-questions-for-rrb-group-d-set-2-pdf\/","title":{"rendered":"Algebra Questions for RRB Group – D Set – 2 PDF"},"content":{"rendered":"

Algebra Questions for RRB Group – D Set – 2 PDF<\/strong><\/span><\/h2>\n

Download Top-15 RRB Group-D Algebra Questions set-2 PDF.\u00a0RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.<\/p>\n

Download Algebra Questions for RRB Group – D Set – 2 PDF<\/a><\/p>\n\n

Download RRB Group-D Previous Papers PDF<\/a><\/p>\n

Take a RRB Group-D free mock test<\/a><\/p>\n

Question 1:\u00a0<\/b>If $\\frac{a}{1-a} + \\frac{b}{1-b} + \\frac{c}{1-c} = 1$ the the value of $\\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}$<\/p>\n

a)\u00a01<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a00<\/p>\n

Question 2:\u00a0<\/b>If $a^x = (x+y+z)^y$ , $a^y =(x+y+z)^z$ and $a^z = (x + y + z)^x$ , then the value of x + y + z (given a \u2260 0) is<\/p>\n

a)\u00a00<\/p>\n

b)\u00a0$a^3$<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0a<\/p>\n

Question 3:\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

Take a free mock test for RRB Group-D<\/a><\/p>\n

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Question 4:\u00a0<\/b>If $x = \\frac{\\sqrt{5}-2}{\\sqrt{5}+2}$, then $x^4 + x^{-4}$ is<\/p>\n

a)\u00a0a surd<\/p>\n

b)\u00a0a rational number but not an integer<\/p>\n

c)\u00a0an integer<\/p>\n

d)\u00a0an irrational number but not a surd<\/p>\n

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

a)\u00a00<\/p>\n

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

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

d)\u00a0$3$<\/p>\n

Question 6:\u00a0<\/b>If x and y are positive real numbers and xy = 8, then the minimum value of 2x + y is<\/p>\n

a)\u00a09<\/p>\n

b)\u00a017<\/p>\n

c)\u00a010<\/p>\n

d)\u00a08<\/p>\n

Question 7:\u00a0<\/b>If x = 3 + 2\u221a2 , then find the value of $x^2 + \\frac{1}{x^2}$<\/p>\n

a)\u00a036<\/p>\n

b)\u00a030<\/p>\n

c)\u00a032<\/p>\n

d)\u00a034<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

Daily Free RRB Online Test<\/a><\/p>\n

Question 8:\u00a0<\/b>if $\\frac{a}{b} + \\frac{b}{a} – 1$ = 0, then the value of $a^3 + b^3$ is<\/p>\n

a)\u00a03<\/p>\n

b)\u00a00<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0-1<\/p>\n

Question 9:\u00a0<\/b>If (a – b) = 3, (b – c) = 5 and (c – a) = 1, then the value of $\\frac{a^3 + b^3 + c^3 – 3abc}{a + b + c}$ is<\/p>\n

a)\u00a017.5<\/p>\n

b)\u00a020.5<\/p>\n

c)\u00a010.5<\/p>\n

d)\u00a015.5<\/p>\n

Question 10:\u00a0<\/b>$3 – \\frac{3+\\sqrt{5}}{4} – \\frac{1}{3 + \\sqrt{5}}$<\/p>\n

a)\u00a00<\/p>\n

b)\u00a03\/2<\/p>\n

c)\u00a0\u221a5\/2<\/p>\n

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

Question 11:\u00a0<\/b>If $x + \\frac{1}{x} = 12 $, the value of $x^2 + \\frac{1}{x^2}$ is<\/p>\n

a)\u00a0142<\/p>\n

b)\u00a0126<\/p>\n

c)\u00a0113<\/p>\n

d)\u00a0129<\/p>\n

RRB Group-D Important Questions (download PDF)<\/a><\/p>\n\n

Question 12:\u00a0<\/b>If $x^4 + \\frac{1}{x^4} = 23$, then the value of $(x-\\frac{1}{x})^2$ will be<\/p>\n

a)\u00a07<\/p>\n

b)\u00a0– 7<\/p>\n

c)\u00a0– 3<\/p>\n

d)\u00a03<\/p>\n

Question 13:\u00a0<\/b>If $ x = \\sqrt{\\frac{\\sqrt{5} + 1}{\\sqrt{5} – 1}}$, then the value of $5x^2 – 5x -1 $ is<\/p>\n

a)\u00a00<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a05<\/p>\n

Question 14:\u00a0<\/b>The value of $\\frac{3\\sqrt{2}}{\\sqrt{3} + \\sqrt{6}} – \\frac{4\\sqrt{3}}{\\sqrt{6}+\\sqrt{2}} + \\frac{\\sqrt{6}}{\\sqrt{3}+\\sqrt{2}} $ is<\/p>\n

a)\u00a04<\/p>\n

b)\u00a00<\/p>\n

c)\u00a0\u221a2<\/p>\n

d)\u00a03\u221a6<\/p>\n

Question 15:\u00a0<\/b>If a + 1\/a+2 = 0, then the value of $(a+2)^{2}+\\frac{1}{(a+2)^{3}}$ is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a06<\/p>\n

c)\u00a04<\/p>\n

d)\u00a03<\/p>\n

General Science Notes for RRB Exams (PDF)<\/a><\/p>\n\n

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

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

Expression : $\\frac{a}{1-a} + \\frac{b}{1-b} + \\frac{c}{1-c} = 1$<\/p>\n

Let’s put each term equal to each other<\/p>\n

=> $3\\frac{a}{1 – a} = 1$<\/p>\n

=> $3a = 1 – a$<\/p>\n

=> $a = \\frac{1}{4} = b = c$<\/p>\n

To find : $\\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}$<\/p>\n

= $\\frac{1}{1 – \\frac{1}{4}} + \\frac{1}{1 – \\frac{1}{4}} + \\frac{1}{1 – \\frac{1}{4}}$<\/span><\/p>\n

= $3 \\times \\frac{4}{3} = 4$<\/span><\/p>\n

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

Expressions : $a^x = (x+y+z)^y$<\/p>\n

$a^y =(x+y+z)^z$<\/p>\n

$a^z = (x + y + z)^x$<\/p>\n

Multiplying above equations, we get :<\/p>\n

=> $a^x \\times a^y \\times a^z = (x + y + z)^x \\times (x + y + z)^y \\times (x + y + z)^z$<\/p>\n

=> $(a)^{x + y + z} = (x + y + z)^{x + y + z}$<\/span><\/span><\/span><\/p>\n

Since the power on both sides is same, thus :<\/p>\n

=> $x + y + z = a$<\/p>\n

3)\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

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

Given\u00a0:\u00a0$x = \\frac{\\sqrt{5}-2}{\\sqrt{5}+2}$<\/p>\n

=>\u00a0$x = \\frac{\\sqrt{5}-2}{\\sqrt{5}+2}\\times(\\frac{\\sqrt5-2}{\\sqrt5-2})$<\/p>\n

=> $x=\\frac{(\\sqrt5-2)^2}{5-4}$<\/p>\n

=> $x=9-4\\sqrt5$ ————(i)<\/p>\n

Similarly, $\\frac{1}{x}=9+4\\sqrt5$ ————-(ii)<\/p>\n

To find\u00a0:\u00a0$x^4 + x^{-4}=x^4+\\frac{1}{x^4}$<\/p>\n

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

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

Substituting values from equations (i) and (ii), we get :<\/p>\n

= $[(9-4\\sqrt5+9+4\\sqrt5)^2-2]^2-2$<\/p>\n

= $[324-2]^2-2$<\/p>\n

= $103684-2=103682$, which is an integer.<\/p>\n

=> Ans – (C)<\/p>\n

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

$(\\sqrt{5}+\\sqrt{3})(\\frac{3\\sqrt{3}}{\\sqrt{5}+\\sqrt{2}} – \\frac{\\sqrt{5}}{\\sqrt{3}+\\sqrt{2}})$<\/p>\n

=$(\\sqrt{5}+\\sqrt{3})(\\frac{3\\sqrt{3}}{\\sqrt{5}+\\sqrt{2}} * \\frac{\\sqrt{5}-\\sqrt{2}}{\\sqrt{5}-\\sqrt{2}}- {\\frac{\\sqrt{5}}{\\sqrt{3}+\\sqrt{2} }x \\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}})$<\/p>\n

=\u00a0$(\\sqrt{5}+\\sqrt{3})(\\sqrt{15} – \\sqrt{6} – \\sqrt{15} + \\sqrt{10})$ = $(\\sqrt{5}+\\sqrt{3})(\\sqrt{10}-\\sqrt{6})$ = $2\\sqrt{2}$<\/p>\n

 <\/p>\n

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

As $x$ & $y$ are positive real numbers, we can use AM $\\geq$ GM<\/p>\n

Let two numbers be $2x$ and $y$<\/p>\n

=> $\\frac{2x + y}{2} \\geq \\sqrt{2xy}$<\/p>\n

=> $2x + y \\geq 2\\sqrt{16}$<\/p>\n

=> $2x + y \\geq 8$<\/p>\n

=> Minimum value of $(2x + y)$ is 8<\/p>\n

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

Expression : $x = 3 + 2\\sqrt{2}$<\/p>\n

=> $\\frac{1}{x} = \\frac{1}{3 + 2\\sqrt{2}}$<\/p>\n

=> $\\frac{1}{x} = \\frac{1}{3 + 2\\sqrt{2}} \\times \\frac{3 – 2\\sqrt{2}}{3 – 2\\sqrt{2}}$<\/p>\n

=> $\\frac{1}{x} = 3 – 2\\sqrt{2}$<\/p>\n

$\\therefore$ $x + \\frac{1}{x} = 3 + 2\\sqrt{2} + 3 – 2\\sqrt{2} = 6$<\/p>\n

Squaring both sides, we get :<\/p>\n

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

=> $x^2 + \\frac{1}{x^2} + 2 = 36$<\/p>\n

$\\therefore x^2 + \\frac{1}{x^2} = 34$<\/p>\n

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

Expression : $\\frac{a}{b} + \\frac{b}{a} – 1 = 0$<\/p>\n

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

=> $a^2 + b^2 = ab$ ———-Eqn(1)<\/p>\n

To find : $a^3 + b^3$<\/p>\n

= $(a + b) (a^2 + b^2 – ab)$<\/p>\n

Using eqn(1), we get :<\/p>\n

= $(a + b) (ab – ab)$<\/p>\n

= 0<\/p>\n

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

it is given that\u00a0(a – b) = 3, (b – c) = 5 and (c – a) = 1<\/p>\n

we need to find the value of\u00a0$\\frac{a^3 + b^3 + c^3 – 3abc}{a + b + c}$<\/p>\n

as we know that ${a^3 + b^3 + c^3}$ = $(a+b+c)({a^2 + b^2 + c^2 – ab – bc – ac})$ ……..(5)<\/p>\n

and ${(a-b)^2 = a^2 + b^2 – 2ab}$…….(1)<\/p>\n

${(b-c)^2 = b^2 + c^2 – 2cb}$………..(2)<\/p>\n

${(c-a)^2 = a^2 + c^2 – 2ac}$……….(3)<\/p>\n

adding 1 , 2 and 3<\/p>\n

17.5 =\u00a0 $({a^2 + b^2 + c^2 – ab – bc – ac}$ …..(4)<\/p>\n

Now using 4 and 5 statement<\/p>\n

$\\frac{a^3 + b^3 + c^3 – 3abc}{a + b + c}$\u00a0 = 17.5<\/p>\n

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

Expression\u00a0:\u00a0$3 – \\frac{3+\\sqrt{5}}{4} – \\frac{1}{3 + \\sqrt{5}}$<\/p>\n

=\u00a0$3 – \\frac{3+\\sqrt{5}}{4} – [\\frac{1}{3 + \\sqrt{5}}\\times(\\frac{3-\\sqrt5}{3-\\sqrt5})]$<\/p>\n

= $\\frac{12-3-\\sqrt5}{4}-(\\frac{3-\\sqrt5}{9-5})$<\/p>\n

= $\\frac{9-\\sqrt5}{4}+ (\\frac{-3+\\sqrt5}{4})$<\/p>\n

= $\\frac{9-3-\\sqrt5+\\sqrt5}{4}=\\frac{6}{4}=\\frac{3}{2}$<\/p>\n

=> Ans – (B)<\/p>\n

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

it is given that $x + \\frac{1}{x}$ = 12<\/p>\n

we need to find value of $x^2 + \\frac{1}{x^2}$<\/p>\n

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

= $12^2 – 2$ = 142<\/p>\n

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

it is given that $x^4 + \\frac{1}{x^4} = 23$<\/p>\n

$x^2 + \\frac{1}{x^2}$ = $\\surd(25)$ = 5<\/p>\n

we need to calculate $(x-\\frac{1}{x})^2$ = $x^2 + \\frac{1}{x^2} – 2$ <\/span><\/p>\n

= 5 – 2 = 3<\/span><\/p>\n

\u00a0<\/span><\/p>\n

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

Given\u00a0:\u00a0$ x = \\sqrt{\\frac{\\sqrt{5} + 1}{\\sqrt{5} – 1}}$<\/p>\n

=>\u00a0$ x = \\sqrt{\\frac{\\sqrt{5} + 1}{\\sqrt{5} – 1}\\times (\\frac{\\sqrt5+1}{\\sqrt5+1})}$<\/p>\n

=> $x=\\sqrt{\\frac{(\\sqrt5+1)^2}{5-1}}$<\/p>\n

=> $x=\\frac{\\sqrt5+1}{2}$ ————–(i)<\/p>\n

Squaring both sides, we get\u00a0: $x^2=\\frac{6+2\\sqrt5}{4}$ ————–(ii)<\/p>\n

To find\u00a0:\u00a0$5x^2 – 5x -1 $<\/p>\n

= $5(x^2-x)-1$<\/p>\n

Substituting values from equations (i) and (ii), we get\u00a0:<\/p>\n

= $5[(\\frac{6+2\\sqrt5}{4})-(\\frac{\\sqrt5+1}{2})]-1$<\/p>\n

=\u00a0$5\\times(\\frac{6+2\\sqrt5-2\\sqrt5-2}{4})-1$<\/p>\n

= $5\\times1-1=4$<\/p>\n

=> Ans – (C)<\/p>\n

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

Expression\u00a0:\u00a0$\\frac{3\\sqrt{2}}{\\sqrt{3} + \\sqrt{6}} – \\frac{4\\sqrt{3}}{\\sqrt{6}+\\sqrt{2}} + \\frac{\\sqrt{6}}{\\sqrt{3}+\\sqrt{2}} $<\/p>\n

= $\\frac{3\\sqrt2(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)-4\\sqrt3(\\sqrt3+\\sqrt6)(\\sqrt3+\\sqrt2)+\\sqrt6(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)}{(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)}$<\/p>\n

= $\\frac{1}{(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)}\\times$ $[3\\sqrt2(\\sqrt{18}+\\sqrt{12}+\\sqrt6+\\sqrt4)-4\\sqrt3(\\sqrt9+\\sqrt6+\\sqrt{18}+\\sqrt{12}) + \\sqrt6(\\sqrt{18}+ \\sqrt6+\\sqrt{36}+\\sqrt{12})]$<\/p>\n

= $\\frac{1}{(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)}\\times$\u00a0$[3\\sqrt2(3\\sqrt2+2\\sqrt3+\\sqrt6+2)-4\\sqrt3(3+\\sqrt6+3\\sqrt2+2\\sqrt3)+ \\sqrt6(3\\sqrt2+\\sqrt6+6+2\\sqrt3)]$<\/p>\n

= $\\frac{1}{(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)}\\times$\u00a0$[(18+6\\sqrt6+6\\sqrt3+6\\sqrt2)+(-12\\sqrt3-12\\sqrt2-12\\sqrt6-24)+ (6\\sqrt3+6+6\\sqrt6+6\\sqrt2)]$<\/p>\n

= $\\frac{1}{(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)}\\times$\u00a0$[(18+6-24)+(6\\sqrt2-12\\sqrt2+6\\sqrt2)+(6\\sqrt3-12\\sqrt3+6\\sqrt3)+(6\\sqrt6-12\\sqrt6+6\\sqrt6)]$<\/p>\n

= $\\frac{1}{(\\sqrt3+\\sqrt6)(\\sqrt6+\\sqrt2)(\\sqrt3+\\sqrt2)}\\times0=0$<\/p>\n

=> Ans – (B)<\/p>\n

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

it is given that $a + \\frac{1}{a}$ = -2<\/p>\n

it is possible only when a = -1<\/p>\n

hence a + 2 = 1<\/p>\n

and so<\/p>\n

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

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

We hope this Algebra Questions set-2\u00a0 PDF for RRB Group-D Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Algebra Questions for RRB Group – D Set – 2 PDF Download Top-15 RRB Group-D Algebra Questions set-2 PDF.\u00a0RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam. Download RRB Group-D Previous Papers PDF Take a RRB Group-D free mock test Question 1:\u00a0If $\\frac{a}{1-a} + \\frac{b}{1-b} […]<\/p>\n","protected":false},"author":41,"featured_media":34678,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1679,1601],"tags":[1637,1819,2052,1647],"class_list":{"0":"post-34673","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-group-d","11":"category-rrb-ntpc-je","12":"tag-rrb-exam","13":"tag-rrb-group-d-exam","14":"tag-rrb-maths","15":"tag-rrb-mocks"},"better_featured_image":{"id":34678,"alt_text":"Algebra Questions for RRB Group - D Set - 2 PDF","caption":"Algebra Questions for RRB Group - D Set - 2 PDF","description":"Algebra Questions for RRB Group - D Set - 2 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