{"id":42571,"date":"2020-09-09T20:43:36","date_gmt":"2020-09-09T15:13:36","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42571"},"modified":"2020-09-09T20:43:36","modified_gmt":"2020-09-09T15:13:36","slug":"simplification-questions-for-railway-exams-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/simplification-questions-for-railway-exams-pdf\/","title":{"rendered":"Simplification Questions for Railway Exams PDF"},"content":{"rendered":"

Simplification Questions for Railway Exams PDF<\/strong><\/span><\/h1>\n

Download Simplification Questions for Railway Exams PDF. Top 15 questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

Download Simplification Questions for Railway Exams PDF<\/a><\/p>\n

Take a free mock test for RRB NTPC<\/a><\/p>\n

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

Question 1:\u00a0<\/b>Simplify : $5 + 4 \\times 5^2 + 4 \\times 5^3 + 4 \\times 5^4 + 4 \\times 5^5$<\/p>\n

a)\u00a0$5^8$<\/p>\n

b)\u00a0$5^4$<\/p>\n

c)\u00a0$5^6$<\/p>\n

d)\u00a0$5^2$<\/p>\n

Question 2:\u00a0<\/b>simplify $\\sqrt{25+10\\sqrt{6}}+\u00a0\\sqrt{25-10\\sqrt{6}}$?<\/p>\n

a)\u00a0$2\\sqrt{15}$<\/p>\n

b)\u00a0$2\\sqrt{5}$<\/p>\n

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

d)\u00a0$\\sqrt{55}$<\/p>\n

Question 3:\u00a0<\/b>Simplify: $6 \\div [2(1 + 2)]$<\/p>\n

a)\u00a03<\/p>\n

b)\u00a09<\/p>\n

c)\u00a01<\/p>\n

d)\u00a02<\/p>\n

Question 4:\u00a0<\/b>$(8-(28-53))\\div(-4\\times5-(-9))=?$<\/p>\n

a)\u00a011<\/p>\n

b)\u00a0-3<\/p>\n

c)\u00a03<\/p>\n

d)\u00a0-11<\/p>\n

Question 5:\u00a0<\/b>$45-[38-(80\\div4-(8-12\\div3)\\div4)]=?$<\/p>\n

a)\u00a025<\/p>\n

b)\u00a027<\/p>\n

c)\u00a026<\/p>\n

d)\u00a028<\/p>\n

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RRB NTPC Previous Papers [Download PDF]<\/a><\/p>\n

Question 6:\u00a0<\/b>The value of $ 14 \\div\u00a0 $ { ( 5 of 2 – 3 ) } $ \\times 4 ( 7 – 2 )\u00a0 $ is :<\/p>\n

a)\u00a044<\/p>\n

b)\u00a040<\/p>\n

c)\u00a0$\u00a0\\frac{1}{10} $<\/p>\n

d)\u00a0$\u00a0\\frac{14}{19} $<\/p>\n

Question 7:\u00a0<\/b>$\u00a04 + \\frac{1}{6} \\times $ [ { $\u00a0– 12 \\times ( 24 – 13 – 3) $ } $\u00a0\\div ( 20 – 4) $ ] = ?<\/p>\n

a)\u00a04<\/p>\n

b)\u00a06<\/p>\n

c)\u00a05<\/p>\n

d)\u00a03<\/p>\n

Question 8:\u00a0<\/b>$384 \\div 2^5 \\times 3 + 8 = ?$<\/p>\n

a)\u00a012<\/p>\n

b)\u00a0132<\/p>\n

c)\u00a03<\/p>\n

d)\u00a044<\/p>\n

Question 9:\u00a0<\/b>$\\frac{3}{4} + \\left\\{\\frac{3}{4} + \\frac{3}{4} \\div (\\frac{3}{4} + \\frac{3}{4})\\right\\} = ?$<\/p>\n

a)\u00a0$\\frac{3}{4}$<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a0$2\\frac{3}{4}$<\/p>\n

Question 10:\u00a0<\/b>The simplest form of $\\frac{391}{667}$ is<\/p>\n

a)\u00a0$\\frac{23}{31}$<\/p>\n

b)\u00a0$\\frac{19}{23}$<\/p>\n

c)\u00a0$\\frac{15}{19}$<\/p>\n

d)\u00a0$\\frac{17}{29}$<\/p>\n

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Question 11:\u00a0<\/b>$\\frac{9x^{2}-24xy+16y^{2}}{3x-4y}$ = ?<\/p>\n

a)\u00a03x+4y<\/p>\n

b)\u00a03x-4y<\/p>\n

c)\u00a04x+3y<\/p>\n

d)\u00a04x-3y<\/p>\n

Question 12:\u00a0<\/b>Evaluate :- $\\frac{16 \\times2^{n+1 }-4\\times2^{n}}{16\\times 2^{n+2}-2\\times2^{n+2}}$<\/p>\n

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

b)\u00a0$0$<\/p>\n

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

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

Question 13:\u00a0<\/b>$ (1-\\frac{1}{2}) (1-\\frac{1}{3}) (1-\\frac{1}{4}) …..(1-\\frac{1}{40}) = ?$<\/p>\n

a)\u00a0$ \\frac{1}{40} $<\/p>\n

b)\u00a0$ \\frac{1}{20} $<\/p>\n

c)\u00a0Countless<\/p>\n

d)\u00a0Zero<\/p>\n

Question 14:\u00a0<\/b>Evaluate: $\\left[\\frac{\\sqrt{3} + 1}{\\sqrt{3} – 1}\\right]^2\u00a0 + \\left[\\frac{\\sqrt{3} – 1}{\\sqrt{3} + 1}\\right]^2$<\/p>\n

a)\u00a016<\/p>\n

b)\u00a012<\/p>\n

c)\u00a014<\/p>\n

d)\u00a024<\/p>\n

Question 15:\u00a0<\/b>Simplify: $\\frac{1\\frac{1}{4} \\div 1\\frac{1}{2}}{\\frac{1}{15} + 1- \\frac{9}{10}}$<\/p>\n

a)\u00a02<\/p>\n

b)\u00a05<\/p>\n

c)\u00a03<\/p>\n

d)\u00a04<\/p>\n

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

FREE RRB NTPC YOUTUBE VIDEOS<\/a><\/p>\n

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

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

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

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

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

$(8-(28-53)) = 8-28+53 = 33$
\n$(-4\\times5-(-9)) = -20+9 = -11$
\nTherefore,\u00a0$(8-(28-53))\\div(-4\\times5-(-9))=\\dfrac{33}{-11} = -3$<\/p>\n

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

$45-[38-(80\\div4-(8-12\\div3)\\div4)] = 45 – (38 – (20 – \\dfrac{8-4}{4}))$
\n$= 45 – (38 -(20-\\dfrac{4}{4})) = 45-(38-(20-1))$
\n$= 45-(38-19) = 45-19 = 26$<\/p>\n

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

$ 14 \\div\u00a0 $ { ( 5 of 2 – 3 ) } $ \\times 4 ( 7 – 2 )\u00a0 $
\n=$ 14 \\div\u00a0 $\u00a0 \u00a07\u00a0 $ \\times 4 ( 5)\u00a0 $
\n=2$ \\times 4 ( 5)\u00a0 $
\n=2$ \\times 20\u00a0 $
\n=40<\/p>\n

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

$\u00a04 + \\frac{1}{6} \\times $ [ { $\u00a0– 12 \\times ( 24 – 13 – 3) $ } $\u00a0\\div ( 20 – 4) $ ]
\n=$\u00a04 + \\frac{1}{6} \\times $ [ { $\u00a0– 12 \\times 8 $ } $\u00a0\\div 16 $ ]
\n=$\u00a04 + \\frac{1}{6} \\times\u00a0 -6$
\n=4-1
\n=3<\/p>\n

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

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

$\\frac{3}{4} + \\left\\{\\frac{3}{4} + \\frac{3}{4} \\div (\\frac{3}{4} + \\frac{3}{4})\\right\\}$<\/p>\n

First solve small bracket as per BODMAS rule.<\/p>\n

$\\frac{3}{4} + \\left\\{\\frac{3}{4} + \\frac{3}{4} \\div (\\frac{6}{4} )\\right\\}$<\/p>\n

$\\frac{3}{4} + \\left\\{\\frac{3}{4} + \\frac{1}{2} \\right\\}$<\/p>\n

Now solve curly\u00a0bracket as per BODMAS rule.<\/p>\n

$\\frac{3}{4} + \\frac{5}{4} $<\/p>\n

$\\frac{8}{4} $<\/p>\n

2<\/p>\n

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

$\\frac{391}{667}$ = $\\frac{17 \\times 23}{29 \\times 23}$ = $\\frac{17}{29}$<\/p>\n

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

$\\frac{9x^{2}-24xy+16y^{2}}{3x-4y}$<\/p>\n

=$\\frac{(3x-4y)^{2}}{3x-4y}=3x-4y$<\/p>\n

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

$\\frac{16 \\times2^{n+1 }-4\\times2^{n}}{16\\times 2^{n+2}-2\\times2^{n+2}}$<\/p>\n

=$\\frac{2^{n}(32-4)}{2^{n+2}(16-2)}$<\/p>\n

=$\\frac{28}{4(14)}$=$\\frac{1}{2}$<\/p>\n

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

1 – 1\/2 = 1\/2<\/p>\n

1 – 1\/3 = 2\/3<\/p>\n

1 – 1\/4 = 3\/4<\/p>\n

The denominator of the first term gets cancelled by the numerator of the second term and so on…<\/p>\n

So, the final value = 1\/40<\/p>\n

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

Using the identities<\/p>\n

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

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

$ (a + b)(a – b) = a^2 – b^2 $<\/p>\n

Rationalizing the denominator,<\/p>\n

$\\left[\\frac{\\sqrt{3} + 1}{\\sqrt{3} – 1}\\right] = \\frac{\\sqrt{3} + 1}{\\sqrt{3} -1} \\times\u00a0\\frac{\\sqrt{3} + 1}{\\sqrt{3}\u00a0+1} $<\/p>\n

Solving the equation using identities we get<\/p>\n

$ \\frac{\\sqrt{3} + 1}{\\sqrt{3} -1} \\times\u00a0\\frac{\\sqrt{3} + 1}{\\sqrt{3}\u00a0+1} = \\frac{4 + 2\\sqrt3}{2} $<\/p>\n

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

$\\left[\\frac{\\sqrt{3} + 1}{\\sqrt{3} – 1}\\right]^2 =\u00a0\u00a0(2 + \\sqrt 3)^2$<\/p>\n

= $ 7 + 4\\sqrt3 $<\/p>\n

Rationalizing the denominator,<\/p>\n

$\\left[\\frac{\\sqrt{3} – 1}{\\sqrt{3} + 1}\\right] = \\frac{\\sqrt{3} – 1}{\\sqrt{3} +1} \\times\u00a0\\frac{\\sqrt{3} -1}{\\sqrt{3} -1} $<\/p>\n

Solving the equation using identities we get<\/p>\n

$ \\frac{\\sqrt{3} -1}{\\sqrt{3} +1} \\times\u00a0\\frac{\\sqrt{3} – 1}{\\sqrt{3} -1} = \\frac{4 – 2\\sqrt3}{2} $<\/p>\n

= $ 2 – \\sqrt 3 $<\/p>\n

$\\left[\\frac{\\sqrt{3} -1}{\\sqrt{3} + 1}\\right]^2 =\u00a0\u00a0(2 – \\sqrt 3)^2$<\/p>\n

= $ 7 – 4\\sqrt3 $<\/p>\n

Thus,<\/p>\n

$\\left[\\frac{\\sqrt{3} + 1}{\\sqrt{3} – 1}\\right]^2 + \\left[\\frac{\\sqrt{3} – 1}{\\sqrt{3} + 1}\\right]^2 = 7 + 4\\sqrt 3 + 7 – 4\\sqrt 3 $<\/p>\n

= 14<\/p>\n

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

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

We hope this\u00a0 Simplification Questions for Railway Exams PDF for RRB NTPC exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Simplification Questions for Railway Exams PDF Download Simplification Questions for Railway Exams PDF. Top 15 questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Download RRB NTPC Previous Papers PDF Question 1:\u00a0Simplify : $5 + 4 \\times 5^2 + 4 \\times 5^3 + 4 \\times 5^4 + 4 […]<\/p>\n","protected":false},"author":41,"featured_media":42583,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,31,1679,1603],"tags":[489,1593],"class_list":{"0":"post-42571","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-railways","10":"category-rrb-group-d","11":"category-rrb-ntpc","12":"tag-railway-exam","13":"tag-rrb-ntpc"},"better_featured_image":{"id":42583,"alt_text":"Simplification Questions for Railway Exams PDF","caption":"Simplification Questions for Railway Exams PDF","description":"Simplification Questions for Railway Exams 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