{"id":36568,"date":"2019-10-24T12:19:09","date_gmt":"2019-10-24T06:49:09","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=36568"},"modified":"2019-10-24T12:19:09","modified_gmt":"2019-10-24T06:49:09","slug":"algebra-questions-for-rrb-group-d-set-3-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/algebra-questions-for-rrb-group-d-set-3-pdf\/","title":{"rendered":"Algebra Questions for RRB Group-D Set-3 PDF"},"content":{"rendered":"

Algebra Questions for RRB Group-D Set-3 PDF<\/strong><\/span><\/h1>\n

Download Top-10 RRB Group-D Algebra Questions set – 3 PDF.\u00a0RRB GROUP-D 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-3 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 a^3 – b^3 = 56 and a – b = 2, then the value of (a^2 + b^2 + ab) is :<\/p>\n

a)\u00a0\u00ad10<\/p>\n

b)\u00a0\u00ad12<\/p>\n

c)\u00a028<\/p>\n

d)\u00a018<\/p>\n

Question 2:\u00a0<\/b>If x = 2 + \u221a3 , then the value, $\\sqrt{x} + \\frac{1}{\\sqrt{x}}$<\/p>\n

a)\u00a0\u221a3<\/p>\n

b)\u00a0\u221a6<\/p>\n

c)\u00a02 \u221a6<\/p>\n

d)\u00a06<\/p>\n

Question 3:\u00a0<\/b>If $2\\sqrt{x} = \\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}} + \\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$<\/p>\n

a)\u00a06<\/p>\n

b)\u00a030<\/p>\n

c)\u00a0\u221a15<\/p>\n

d)\u00a016<\/p>\n

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Question 4:\u00a0<\/b>If a + 1\/a = 1, then the value of a^2 + 1\/a^2 is :<\/p>\n

a)\u00a0-2<\/p>\n

b)\u00a02<\/p>\n

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

d)\u00a04<\/p>\n

Question 5:\u00a0<\/b>If ax + by = 6, bx – ay = 2 and x^2 + y^2 = 4, then the value of (a^2 + b^2) would be:<\/p>\n

a)\u00a010<\/p>\n

b)\u00a02<\/p>\n

c)\u00a04<\/p>\n

d)\u00a05<\/p>\n

Question 6:\u00a0<\/b>If $ a =\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}$ and $b=\\frac{\\sqrt{3}+\\sqrt{2}}{\\sqrt{3}-\\sqrt{2}}$ then $\\frac{a^2}{b}+\\frac{b^2}{a}$<\/p>\n

a)\u00a0900<\/p>\n

b)\u00a0970<\/p>\n

c)\u00a01030<\/p>\n

d)\u00a0930<\/p>\n

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

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Question 7:\u00a0<\/b>If (5x^2 – 3y^2): xy = 11:2, and x,y are positive, then the value of x\/y is:<\/p>\n

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

b)\u00a05\/2<\/p>\n

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

d)\u00a05\/3<\/p>\n

Question 8:\u00a0<\/b>If $x=\\frac{1}{2+\\sqrt{3}}$ and $y=\\frac{1}{2-\\sqrt{3}}$ then the value of $\\frac{1}{x+1} + \\frac{1}{y+1}$ is<\/p>\n

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

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

c)\u00a0$1$<\/p>\n

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

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Question 9:\u00a0<\/b>The value of $\\sqrt{-\\sqrt{3}+\\sqrt{3+8\\sqrt{7+4\\sqrt{3}}}}$ is<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a01<\/p>\n

Question 10:\u00a0<\/b>If $\\sqrt{4x-9}+\\sqrt{4x+9}=5+\\sqrt{7}$, then the value of $x$ is<\/p>\n

a)\u00a04<\/p>\n

b)\u00a05<\/p>\n

c)\u00a07<\/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

it is given that $a^3 – b^3$ = 56<\/p>\n

we know $a^3 – b^3 = (a-b)(a^2 + b^2 + ab)$<\/p>\n

a- b = 2 (Given)<\/p>\n

hence<\/p>\n

$(a^2 + b^2 + ab)$ = $\\frac{56}{2}$ = 28<\/p>\n

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

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

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

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

$(\\sqrt{x} + \\frac{1}{\\sqrt{x}})^{2}$ = 4 + 2 = 6<\/p>\n

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

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

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

it is given that<\/p>\n

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

here , $\\frac{\\sqrt5+\\sqrt3}{\\sqrt5-\\sqrt3}$ = $\\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}$ x $\\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$ = $\\frac{(\\sqrt5 + \\sqrt3)^2}{2}$<\/p>\n

 <\/p>\n

similarly , $\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$ = $\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$ x $\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}$ = $\\frac{(\\sqrt5 – \\sqrt3)^2}{2}$<\/p>\n

$\\frac{(\\sqrt5 + \\sqrt3)^2}{2}$ + $\\frac{(\\sqrt5 – \\sqrt3)^2}{2}$ = 2$\\sqrt(x)$<\/p>\n

8 = 2$\\sqrt(x)$<\/p>\n

x = 16<\/p>\n

 <\/p>\n

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

it is given that a + 1\/a = 1<\/p>\n

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

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

= $1^2$ – 2 = -1<\/span><\/span><\/p>\n

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

it is given that<\/p>\n

ax + by = 6……….(1)<\/p>\n

bx – ay = 2……..(2)<\/p>\n

and x^2 + y^2 = 4<\/p>\n

now multiply 1 and 2nd equation by a and b respectively<\/p>\n

we get<\/p>\n

$a^2$x + $ab$y = 6a<\/p>\n

$b^2$x – aby = 2b<\/p>\n

adding above equations we get,<\/p>\n

$a^2 + b^2$ x = 6a +2b<\/p>\n

x = $\\frac{6a+2b}{a^2 + b^2}$<\/p>\n

Similarly ,<\/p>\n

we get y = $\\frac{6a-2b}{a^2 + b^2}$<\/p>\n

putting above values in x^2 + y^2 = 4<\/span><\/p>\n

we get , a = 1 and b = 3 <\/span><\/span><\/p>\n

hence $1^2 + 3^2$ = 10 <\/span><\/span><\/p>\n

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

Give\u00a0:\u00a0$ a =\\frac{\\sqrt{3}-\\sqrt{2}}{\\sqrt{3}+\\sqrt{2}}$<\/p>\n

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

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

=> $a=5-2\\sqrt6$<\/p>\n

Squaring both sides, we get\u00a0: $a^2=49-20\\sqrt6$<\/p>\n

Similarly, $b=5+2\\sqrt6$ and $b^2=49+20\\sqrt6$<\/p>\n

To find\u00a0:\u00a0$\\frac{a^2}{b}+\\frac{b^2}{a}$<\/p>\n

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

= $\\frac{[(5-2\\sqrt6)+(5+2\\sqrt6)][(49-20\\sqrt6)+(49+20\\sqrt6)-(5-2\\sqrt6)(5+2\\sqrt6)]}{(5-2\\sqrt6)(5+2\\sqrt6)}$<\/p>\n

= $\\frac{10[49+49-(25-24)]}{25-24}$<\/p>\n

= $10\\times97=970$<\/p>\n

=> Ans – (B)<\/p>\n

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

it is given that (5x^2 – 3y^2): xy = 11:2<\/p>\n

let divide the numerator and denominator of left hand side of the given equation by y^2 <\/span><\/p>\n

we will get , <\/span><\/p>\n

$\\frac{5( \\frac{x}{y})^2 – 3}{\\frac{x}{y}}$ = $\\frac{11}{2}$<\/span><\/p>\n

let $\\frac{x}{y}$ = k<\/span><\/p>\n

So, $\\frac{5k^2-3}{k}$ = $\\frac{11}{2}$ <\/span><\/p>\n

Solving this we get , k = $\\frac{3}{2}$ <\/span><\/p>\n

hence $\\frac{x}{y}$ = $\\frac{3}{2}$<\/span><\/p>\n

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

$x=\\frac{1}{2+\\sqrt{3}}$ , on rationalizing x = 2 – $\\surd3$<\/p>\n

$y=\\frac{1}{2-\\sqrt{3}}$ , on rationalizing y = 2 + $\\surd3$
\n<\/span><\/p>\n

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

using above values of x and y <\/span><\/span><\/p>\n

$\\frac{1}{x+1} + \\frac{1}{y+1}$ = $\\frac{1}{2 + \\surd3 +1} + \\frac{1}{2 – \\surd3 +1}$
\n<\/span><\/span><\/span><\/p>\n

now on rationalizing , <\/span><\/span><\/span><\/span><\/p>\n

$\\frac{1}{x+1} + \\frac{1}{y+1}$ = $\\frac{3 – \\surd3 + 3 + \\surd3}{6}$ = 1<\/p>\n

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

we need to find value of $\\sqrt{-\\sqrt{3}+\\sqrt{3+8\\sqrt{7+4\\sqrt{3}}}}$<\/p>\n

$\\sqrt{7 + \\sqrt3}$ = $\\sqrt{2^2 + (\\sqrt3)^2 + 2 \\times 2 \\times \\sqrt3}$ = $2 + \\sqrt3$<\/p>\n

$\\sqrt{3 + 4^2 + 2 \\times 4 \\times \\sqrt3}$ = $4 + \\sqrt3$<\/p>\n

$\\sqrt{-\\surd3 + 4 + \\surd3}$ = 2\\<\/p>\n

hence $\\sqrt{-\\sqrt{3}+\\sqrt{3+8\\sqrt{7+4\\sqrt{3}}}}$ = 2<\/p>\n

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

given that<\/p>\n

$\\sqrt{4x-9}+\\sqrt{4x+9}=5+\\sqrt{7}$<\/p>\n

$\\sqrt{4x-9} – 5 = \\sqrt{7} – \\sqrt{4x+9}$<\/p>\n

on squaring both sides<\/p>\n

4x – 9 + 25 + 10$\\sqrt{4x+9}$ = 7 + 4x + 9 – 2$\\sqrt{28x+63}$<\/p>\n

10$\\sqrt{4x-9}$\u00a0 = 2$\\sqrt{28x+63}$<\/p>\n

on squaring both sides again<\/p>\n

400x – 900 = 112x + 252<\/p>\n

288x = 1152<\/p>\n

x = 4<\/p>\n\n

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

We hope this Algebra Questions set – 3 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-3 PDF Download Top-10 RRB Group-D Algebra Questions set – 3 PDF.\u00a0RRB GROUP-D 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 a^3 – b^3 = 56 and […]<\/p>\n","protected":false},"author":41,"featured_media":36571,"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],"tags":[1637,1976,2795,2052],"class_list":{"0":"post-36568","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":"tag-rrb-exam","12":"tag-rrb-group-d-2019","13":"tag-rrb-group-d-mocks","14":"tag-rrb-maths"},"better_featured_image":{"id":36571,"alt_text":"Algebra Questions for RRB Group-D Set-3 PDF","caption":"Algebra Questions for RRB Group-D Set-3 PDF","description":"Algebra Questions for RRB Group-D Set-3 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