{"id":27328,"date":"2019-04-10T11:26:03","date_gmt":"2019-04-10T05:56:03","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=27328"},"modified":"2019-04-18T12:26:57","modified_gmt":"2019-04-18T06:56:57","slug":"simplification-questions-for-rrb-je-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/simplification-questions-for-rrb-je-pdf\/","title":{"rendered":"Simplification Questions for RRB JE PDF"},"content":{"rendered":"

Simplification Questions for RRB JE PDF<\/strong><\/span><\/h1>\n

Download Top 20 RRB JE Simplification Questions and Answers PDF. RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam<\/p>\n

Download Simplification Questions for RRB JE PDF<\/a><\/p>\n\n

Download RRB JE Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>If the roots of the equation $2x^{2}-9x+7$=0 are p and q then the equation with roots as p+1 and q+1 is ?<\/p>\n

a)\u00a0$ 2x^{2}-13x+18$=0<\/p>\n

b)\u00a0$ 2x^{2}-11x+16$=0<\/p>\n

c)\u00a0$ 2x^{2}+11x-16$=0<\/p>\n

d)\u00a0$ 2x^{2}+13x-18$=0<\/p>\n

Question 2:\u00a0<\/b>If $b-\\frac{1}{b}$=7, then what is the value of $b+\\frac{1}{b}$<\/p>\n

a)\u00a0$\\sqrt{51}$<\/p>\n

b)\u00a0$\\sqrt{52}$<\/p>\n

c)\u00a0$\\sqrt{50}$<\/p>\n

d)\u00a0$\\sqrt{53}$<\/p>\n

Question 3:\u00a0<\/b>If $a+\\frac{1}{a}$=4, then what is the value of $a-\\frac{1}{a}$<\/p>\n

a)\u00a0$\\sqrt{11}$<\/p>\n

b)\u00a0$\\sqrt{12}$<\/p>\n

c)\u00a0$\\sqrt{10}$<\/p>\n

d)\u00a0$\\sqrt{13}$<\/p>\n

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Question 4:\u00a0<\/b>If $a^3+b^3+c^3 = 3abc$, then find the value of $(a+b)^2-c^2$.<\/p>\n

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

b)\u00a01<\/p>\n

c)\u00a00<\/p>\n

d)\u00a02<\/p>\n

Question 5:\u00a0<\/b>If $x^2 + y^2 + x^{-2} + y^{-2}$ = 4 then what is the value of $x^4 + y^4$ ?<\/p>\n

a)\u00a00<\/p>\n

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

c)\u00a02<\/p>\n

d)\u00a04<\/p>\n

Question 6:\u00a0<\/b>If $x^2 + 1 = 2x$, then find $x^3+\\dfrac{1}{x^3}$.<\/p>\n

a)\u00a04<\/p>\n

b)\u00a03<\/p>\n

c)\u00a02<\/p>\n

d)\u00a00<\/p>\n

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

Question 7:\u00a0<\/b>Find the value of x from the given equations.
\n7x + 3y + 5z = 27
\n6x + 6y + 10z = 14<\/p>\n

a)\u00a04<\/p>\n

b)\u00a011<\/p>\n

c)\u00a05<\/p>\n

d)\u00a09<\/p>\n

Question 8:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{7}}}}$.<\/p>\n

a)\u00a0$\\dfrac{38}{21}$<\/p>\n

b)\u00a0$\\dfrac{38}{23}$<\/p>\n

c)\u00a0$\\dfrac{36}{23}$<\/p>\n

d)\u00a0$\\dfrac{39}{20}$<\/p>\n

Question 9:\u00a0<\/b>If $x+\\dfrac{1}{x} = 3, x^6+\\dfrac{1}{x^6} = ?$<\/p>\n

a)\u00a0314<\/p>\n

b)\u00a0322<\/p>\n

c)\u00a0328<\/p>\n

d)\u00a0316<\/p>\n

Question 10:\u00a0<\/b>Find the value of x from the given equations.
\n7x + 3y + 5z = 27
\n6x + 6y + 10z = 14<\/p>\n

a)\u00a04<\/p>\n

b)\u00a011<\/p>\n

c)\u00a05<\/p>\n

d)\u00a09<\/p>\n

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Question 11:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{7}}}}$.<\/p>\n

a)\u00a0$\\dfrac{38}{21}$<\/p>\n

b)\u00a0$\\dfrac{38}{23}$<\/p>\n

c)\u00a0$\\dfrac{36}{23}$<\/p>\n

d)\u00a0$\\dfrac{39}{20}$<\/p>\n

Question 12:\u00a0<\/b>If $x+\\dfrac{1}{x} = 3, x^6+\\dfrac{1}{x^6} = ?$<\/p>\n

a)\u00a0314<\/p>\n

b)\u00a0322<\/p>\n

c)\u00a0328<\/p>\n

d)\u00a0316<\/p>\n

Question 13:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{4}}}}$.<\/p>\n

a)\u00a0$\\dfrac{21}{15}$<\/p>\n

b)\u00a0$\\dfrac{23}{14}$<\/p>\n

c)\u00a0$\\dfrac{24}{13}$<\/p>\n

d)\u00a0$\\dfrac{23}{11}$<\/p>\n

Question 14:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{2}{7}}}}$.<\/p>\n

a)\u00a0$\\frac{37}{25}$<\/p>\n

b)\u00a0$\\frac{41}{22}$<\/p>\n

c)\u00a0$\\frac{41}{25}$<\/p>\n

d)\u00a0$\\frac{35}{19}$<\/p>\n

Question 15:\u00a0<\/b>If $x^{4}+(\\frac{1}{x})^{4}$=34 then what is the value of $x-1\/x$<\/p>\n

a)\u00a02<\/p>\n

b)\u00a06<\/p>\n

c)\u00a08<\/p>\n

d)\u00a04<\/p>\n

Daily Free Online Tests for RRB Exams<\/a><\/p>\n

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Question 16:\u00a0<\/b>If $x^{4}+(\\frac{1}{x})^{4}$=727 then what is the value of $x^{3}-(\\frac{1}{x})^{3}$<\/p>\n

a)\u00a0140<\/p>\n

b)\u00a0138<\/p>\n

c)\u00a0134<\/p>\n

d)\u00a0136<\/p>\n

Question 17:\u00a0<\/b>If $x^{4}+(\\frac{1}{x})^{4}$=119 then what is the value of $x^{3}-(\\frac{1}{x})^{3}$<\/p>\n

a)\u00a036<\/p>\n

b)\u00a024<\/p>\n

c)\u00a042<\/p>\n

d)\u00a030<\/p>\n

Question 18:\u00a0<\/b>If $p + \\frac{1}{p} = 3$, then find the value of $p^6 + \\large\\frac{1}{p^6}$.<\/p>\n

a)\u00a0322<\/p>\n

b)\u00a0348<\/p>\n

c)\u00a0329<\/p>\n

d)\u00a0342<\/p>\n

Question 19:\u00a0<\/b>If $px^3 + qx^2 – 19x – 30 = 0$ is completely divided by $x^2 + 5x + 6$, find the value of $(p, q)$.<\/p>\n

a)\u00a0(1,0)<\/p>\n

b)\u00a0(2,1)<\/p>\n

c)\u00a0(1,0)<\/p>\n

d)\u00a0(4,1)<\/p>\n

Question 20:\u00a0<\/b>If $l^{3} + m^{3}$ = -218 and $lm = -35$ , then what is the value of $l + m$?<\/p>\n

a)\u00a0-6<\/p>\n

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

c)\u00a0-4<\/p>\n

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

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

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

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

Given $2x^{2}-9x+7=0$
\n$ 2x^{2}-2x-7x+7$=0
\n$2x(x-1)-7(x-1)$=0
\n$(2x-7)(x-1)$=0
\np=3.5 and q=1
\nRoots of the new equation are p+1 an q+1 i.e 4.5 and 2
\n$(x-4.5)(x-2)$=0
\n$ 2x^{2}-13x+18$=0<\/p>\n

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

$b-\\frac{1}{b}$=7
\nSquaring on both sides we have
\n$b^{2}-2(b)(1\/b)+(\\frac{1}{b})^{2}$=49
\n$b^{2}+(\\frac{1}{b})^{2}$=49+2 = 51
\nAdding 2 on both sides
\n$b^{2}+2+(\\frac{1}{b})^{2}$=51+2
\n$(b+\\frac{1}{b})^{2}$=53
\n$b+\\frac{1}{b}$=$\\sqrt{53}$<\/p>\n

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

$a+\\frac{1}{a}$=4
\nSquaring on both sides we have
\n$a^{2}+2(a)(1\/a)+(\\frac{1}{a})^{2}$=16
\n$a^{2}+(\\frac{1}{a})^{2}$=14
\nSubtracting 2 on both sides
\n$a^{2}-2+(\\frac{1}{a})^{2}$=14-2
\n$(a-\\frac{1}{a})^{2}$=12
\n$a-\\frac{1}{a}$=$\\sqrt{12}$<\/p>\n

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

If $a^3+b^3+c^3 = 3abc$, then $a+b+c = 0$.
\n$(a+b)^2-c^2 = (a+b+c)(a+b-c) = 0$<\/p>\n

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

$x^2 + y^2 + x^{-2} + y^{-2} = 4$
\n\u21d2 $(x^2 + \\frac{1}{x^2} – 2) + (y^2 + \\frac{1}{y^2} – 2) = 0$
\n\u21d2 ${(x – \\frac{1}{x})}^2 + {(y – \\frac{1}{y})}^2 = 0$
\nSum of two squares is zero when both the numbers are zero.
\n\u21d2 $x – \\frac{1}{x}$ = 0 and $y – \\frac{1}{y} = 0$
\n\u21d2 $x = 1$or $-1$ and $y = 1$ or $-1$
\nIn all the cases, $x^4 + y^4 = 2$
\nHence, option C is the correct answer.<\/p>\n

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

Given, $x^2 + 1 = 2x$
\nDividing with x on both sides
\n\u21d2 $x+\\dfrac{1}{x} = 2$
\nCubing on both sides
\n\u21d2 $(x+\\dfrac{1}{x})^3 = 2^3$
\n\u21d2 $x^3+\\dfrac{1}{x^3}+3\\times x\\times \\dfrac{1}{x}(x+\\dfrac{1}{x}) = 8$<\/p>\n

\u21d2 $x^3+\\dfrac{1}{x^3}+3\\times2 = 8$
\n\u21d2 $x^3+\\dfrac{1}{x^3} = 8-6 = 2$<\/p>\n

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

7x + 3y + 5z = 27 \u2192 (1)
\n6x + 6y + 10z = 14 \u2192 (2)
\nMultiplying equation (1) by 2 and subtracting (2) from (1) we get
\n8x = 40
\n\u21d2 x = 5
\nHence, option C is the correct answer.<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{7}}}} = 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{8}{7}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{7}{8}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{15}{8}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{8}{15}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{23}{15}}$<\/p>\n

$= 1+\\dfrac{15}{23}$<\/p>\n

$= \\dfrac{38}{23}$<\/p>\n

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

Given, $x+\\dfrac{1}{x} = 3$
\nSquaring on both sides
\n$(x+\\dfrac{1}{x})^2 = 3^2$
\n$x^2+\\dfrac{1}{x^2}+2\\times x \\times \\dfrac{1}{x} = 9$<\/p>\n

$x^2+\\dfrac{1}{x^2}+2 = 9$<\/p>\n

$x^2+\\dfrac{1}{x^2} = 7$
\nCubing above equation on both sides
\n$(x^2+\\dfrac{1}{x^2})^3 = 7^3$<\/p>\n

\u21d2<\/p>\n

$x^6+\\dfrac{1}{x^6}+3\\times x^2\\times \\dfrac{1}{x^2}((x^2+\\dfrac{1}{x^2}}) = 343$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6}+3\\times7 = 343$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6}+21 = 343$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6} = 343-21 = 322$<\/p>\n

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

7x + 3y + 5z = 27 \u2192 (1)
\n6x + 6y + 10z = 14 \u2192 (2)
\nMultiplying equation (1) by 2 and subtracting (2) from (1) we get
\n8x = 40
\n\u21d2 x = 5
\nHence, option C is the correct answer.<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{7}}}} = 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{8}{7}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{7}{8}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{15}{8}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{8}{15}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{23}{15}}$<\/p>\n

$= 1+\\dfrac{15}{23}$<\/p>\n

$= \\dfrac{38}{23}$<\/p>\n

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

Given, $x+\\dfrac{1}{x} = 3$
\nSquaring on both sides
\n$(x+\\dfrac{1}{x})^2 = 3^2$
\n$x^2+\\dfrac{1}{x^2}+2\\times x \\times \\dfrac{1}{x} = 9$<\/p>\n

$x^2+\\dfrac{1}{x^2}+2 = 9$<\/p>\n

$x^2+\\dfrac{1}{x^2} = 7$
\nCubing above equation on both sides
\n$(x^2+\\dfrac{1}{x^2})^3 = 7^3$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6}+3\\times x^2\\times \\dfrac{1}{x^2}((x^2+\\dfrac{1}{x^2}}) = 343$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6}+3\\times7 = 343$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6}+21 = 343$<\/p>\n

\u21d2 $x^6+\\dfrac{1}{x^6} = 343-21 = 322$<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{4}}}} = 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{5}{4}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{4}{5}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{9}{5}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{5}{9}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{14}{9}}$<\/p>\n

$= 1+\\dfrac{9}{14}$<\/p>\n

$= \\dfrac{23}{14}$<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{2}{7}}}} = 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{9}{7}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{7}{9}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{16}{9}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{9}{16}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{25}{16}}$<\/p>\n

$= 1+\\dfrac{16}{25}$<\/p>\n

$= \\dfrac{41}{25}$<\/p>\n

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

$x^{4}+(\\frac{1}{x})^{4}$=34
\nAdd 2 on both sides
\n$x^{4}+2+(\\frac{1}{x})^{4}$=34+2
\n($x^{2}+\\frac{1}{x^{2}})^{2}$=36
\n$x^{2}+(\\frac{1}{x})^{2}$=6
\nSubtracting 2 on both sides
\n$x^{2}+(\\frac{1}{x})^{2}-2$=4<\/p>\n

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

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

$x^{4}+(\\frac{1}{x})^{4}$=727
\nAdd 2 on both sides
\n$x^{4}+2+(\\frac{1}{x})^{4}$=727+2
\n($x^{2}+\\frac{1}{x^{2}})^{2}$=729
\n$x^{2}+(\\frac{1}{x})^{2}$=27
\nSubtracting 2 on both sides
\n$x^{2}+(\\frac{1}{x})^{2}-2$=27-2
\nx-1\/x=5
\nCubing on both sides we have
\n$x^{3}-(\\frac{1}{x})^{3}-15$=125
\n$x^{3}-(\\frac{1}{x})^{3}$=125+15=140<\/p>\n

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

$x^{4}+(\\frac{1}{x})^{4}$=119
\nAdd 2 on both sides
\n$x^{4}+2+(\\frac{1}{x})^{4}$=119+2<\/p>\n

$(x^{2}+\\frac{1}{x^{2}})^{2}$=121<\/p>\n

$x^{2}+(\\frac{1}{x})^{2}$=11
\nSubtracting 2 on both sides
\n$x^{2}+(\\frac{1}{x})^{2}-2$=11-2
\nx-1\/x=3
\nCubing on both sides we have
\n$x^{3}-(\\frac{1}{x})^{3}-9$=27<\/p>\n

$x^{3}-(\\frac{1}{x})^{3}$=36<\/p>\n

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

Given,
\n$p + \\large\\frac{1}{p}$ $=$ $3$<\/p>\n

Squaring both sides,<\/p>\n

$p^2$ $+$ $\\large\\frac{1}{p^2}$ $+$ $2 \\times p \\times \\large\\frac{1}{p}$ $=$ $9$<\/p>\n

$p^2 + \\large\\frac{1}{p^2}$ $=$ $7$<\/p>\n

Cubing on both sides,<\/p>\n

$p^6 + \\large\\frac{1}{p^6}$ $+ 3\u00a0\\times p^2 \\times\u00a0\\large\\frac{1}{p^2}(p^2 + \\large\\frac{1}{p^2}) $ $= 343$<\/p>\n

$p^6 + \\large\\frac{1}{p^6}$ $+ 3 (p^2 + \\large\\frac{1}{p^2})$ $=343$<\/p>\n

$p^6 + \\large\\frac{1}{p^6}$ $+ 3 (7)$= $343$<\/p>\n

$p^6 + \\large\\frac{1}{p^6}$\u00a0 $=$ $343-21$<\/p>\n

$p^6 + \\large\\frac{1}{p^6}$\u00a0 $=$ $322$<\/p>\n

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

Given, $px^3 + qx^2 – 19x – 30 = 0$ is divisible by $x^2 + 5x + 6$<\/p>\n

$x^2 + 5x + 6$ can be factorized as $(x + 2)(x + 3)$<\/p>\n

x = -2 and x = -3 should satisfy\u00a0$px^3 + qx^2 – 19x – 30 = 0$.<\/p>\n

Substituting the values,<\/p>\n

When, $x = -2$<\/p>\n

$-8p + 4q + 8 = 0$ or $2p – q – 2 = 0$<\/p>\n

When, $x = -3$<\/p>\n

$-27p + 9q + 27 = 0$ or $3p – q – 3 = 0$<\/p>\n

Subtracting both, $p = 1$<\/p>\n

Substituting the value of $p = 1$, we get $q = 0$<\/p>\n

Hence,\u00a0$(p, q)$=\u00a0$(1, 0)$.<\/p>\n

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

Given, $l^{3} + m^{3}$ = -218 and $lm = -35$<\/p>\n

$(l+m)^{3} = l^{3}+ m^{3} + 3lm(l+m)$<\/p>\n

$(l+m)^{3} = -218 + 3(-35)(l+m)$<\/p>\n

$(l+m)^{3} = -218 -105(l+m)$<\/p>\n

we need to solve 3rd degree equation<\/p>\n

so,without solving, by verification, we get $l+m = -2$,<\/p>\n

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

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

We hope this Simplification Questions for RRB JE Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Simplification Questions for RRB JE PDF Download Top 20 RRB JE Simplification Questions and Answers PDF. RRB JE Maths questions based on asked questions in previous exam papers very important for the Railway JE exam Download RRB JE Previous Papers PDF Question 1:\u00a0If the roots of the equation $2x^{2}-9x+7$=0 are p and q then the […]<\/p>\n","protected":false},"author":41,"featured_media":27331,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1605],"tags":[489,491,1635,1647],"class_list":{"0":"post-27328","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-je","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-je","14":"tag-rrb-mocks"},"better_featured_image":{"id":27331,"alt_text":"Simplification Questions for RRB JE PDF","caption":"Simplification Questions for RRB JE 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