{"id":26072,"date":"2019-03-14T18:26:17","date_gmt":"2019-03-14T12:56:17","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26072"},"modified":"2019-03-14T18:26:17","modified_gmt":"2019-03-14T12:56:17","slug":"rrb-ntpc-algebra-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/rrb-ntpc-algebra-questions-pdf\/","title":{"rendered":"RRB NTPC Algebra Questions PDF"},"content":{"rendered":"

RRB NTPC Algebra Questions PDF<\/span><\/h1>\n

Download RRB NTPC Alghebra Questions and Answers PDF. Top 25 RRB NTPC\u00a0 Algebra questions based on asked questions in previous exam papers very important for the Railway NTPC exam<\/p>\n

Download RRB NTPC Algebra Questions PDF<\/a><\/p>\n

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

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Question 1:\u00a0<\/b>A two digit number when reverse decreases its value by 63 and the sum of the digits of the number is 11.Fin the value of the number.<\/p>\n

a)\u00a056<\/p>\n

b)\u00a047<\/p>\n

c)\u00a038<\/p>\n

d)\u00a029<\/p>\n

Question 2:\u00a0<\/b>A two digit number when reverse decreases its value by 27 and the sum of the digits of the number is 7.Fin the value of the number.<\/p>\n

a)\u00a034<\/p>\n

b)\u00a016<\/p>\n

c)\u00a025<\/p>\n

d)\u00a043<\/p>\n

Question 3:\u00a0<\/b>If $a+b=0$ then what is the value of $a^3+b^3$?<\/p>\n

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

b)\u00a00<\/p>\n

c)\u00a01<\/p>\n

d)\u00a0Can\u2019t be determined<\/p>\n

Question 4:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{7}{3}}}}$.<\/p>\n

a)\u00a0$\\frac{37}{16}$<\/p>\n

b)\u00a0$\\frac{25}{14}$<\/p>\n

c)\u00a0$\\frac{26}{17}$<\/p>\n

d)\u00a0$\\frac{36}{23}$<\/p>\n

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Question 5:\u00a0<\/b>A person can buy 54 gifts with the money he has and if the cost of the gift is decreased by 4 rupees then he can buy 60 gifts and he is left with 18 rupees. What is the total money he has ?<\/p>\n

a)\u00a01900<\/p>\n

b)\u00a02035<\/p>\n

c)\u00a01998<\/p>\n

d)\u00a01961<\/p>\n

Question 6:\u00a0<\/b>A number is increased by 12 and divided by 19 to get the result as 9 and if the number is decreased by 40 and divided by 17 then the remainder obtained is ?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a00<\/p>\n

c)\u00a01<\/p>\n

d)\u00a02<\/p>\n

Question 7:\u00a0<\/b>If x+y+z = 20 and xy+yz+xz = 10, find the value of $x^3 + y^3 + z^3 -3xyz$.<\/p>\n

a)\u00a06800<\/p>\n

b)\u00a07525<\/p>\n

c)\u00a07400<\/p>\n

d)\u00a06400<\/p>\n

Question 8:\u00a0<\/b>If $x+\\dfrac{1}{x} = 3$, then find the value of $x^3+\\dfrac{1}{x^3}$.<\/p>\n

a)\u00a016<\/p>\n

b)\u00a014<\/p>\n

c)\u00a018<\/p>\n

d)\u00a022<\/p>\n

Question 9:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{5}{4}}}}$.<\/p>\n

a)\u00a0$\\frac{37}{11}$<\/p>\n

b)\u00a0$\\frac{35}{22}$<\/p>\n

c)\u00a0$\\frac{25}{44}$<\/p>\n

d)\u00a0$\\frac{35}{11}$<\/p>\n

Question 10:\u00a0<\/b>If 7\/2(2x\/3 – 1\/2) + 11\/2 = 2x\/3, then what is the value of 1\/x?<\/p>\n

a)\u00a0-9\/4<\/p>\n

b)\u00a0-4\/9<\/p>\n

c)\u00a04\/9<\/p>\n

d)\u00a09\/4<\/p>\n

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Question 11:\u00a0<\/b>If $a+\\frac{1}{a} = 5$, then find the value of $a^2+\\frac{1}{a^2}$.<\/p>\n

a)\u00a018<\/p>\n

b)\u00a023<\/p>\n

c)\u00a027<\/p>\n

d)\u00a014<\/p>\n

Question 12:\u00a0<\/b>Find the value of $1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{3}}}}$.<\/p>\n

a)\u00a0$\\dfrac{18}{11}$<\/p>\n

b)\u00a0$\\dfrac{16}{11}$<\/p>\n

c)\u00a0$\\dfrac{27}{11}$<\/p>\n

d)\u00a0$\\dfrac{12}{11}$<\/p>\n

Question 13:\u00a0<\/b>Find the value of $x$ (from the given options), if $5 = 40-12x+x^2$ ?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a04<\/p>\n

c)\u00a05<\/p>\n

d)\u00a06<\/p>\n

Question 14:\u00a0<\/b>Find $(a+b)(a^2-ab+b^2+3ab)$ = ?<\/p>\n

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

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

c)\u00a0$(a+b)^2$<\/p>\n

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

Question 15:\u00a0<\/b>Find the value of $x$ (from the given options), if $5x^2-13x-6 = 0$ ?<\/p>\n

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

b)\u00a03\/5<\/p>\n

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

d)\u00a0-2\/5<\/p>\n

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Question 16:\u00a0<\/b>If $x+y+z = 9$, $xy+yz+zx = 18$, then find $x^2+y^2+z^2$<\/p>\n

a)\u00a081<\/p>\n

b)\u00a045<\/p>\n

c)\u00a056<\/p>\n

d)\u00a063<\/p>\n

Question 17:\u00a0<\/b>The product of $(a+b+c) \\times (a+2b-c)$ is ?<\/p>\n

a)\u00a0$a^2+2b^2-c^2+3ab+bc$<\/p>\n

b)\u00a0$a^2+2b^2-c^2+3ab+2bc$<\/p>\n

c)\u00a0$a^2+2b^2-c^2+2ab+bc$<\/p>\n

d)\u00a0$a^2+2b^2-c^2+3ab+3bc$<\/p>\n

Question 18:\u00a0<\/b>If $b+1\/b = 6$, then find $b^2+1\/b^2$ ?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a027<\/p>\n

c)\u00a083<\/p>\n

d)\u00a034<\/p>\n

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Question 19:\u00a0<\/b>FInd $x^4+1\/x^4$, if $x+1\/x = 2$ ?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a02<\/p>\n

Question 20:\u00a0<\/b>There are some sheep and hens. If there are a total of 36 legs and there are 14 animals and birds in total, how many sheep are there ?<\/p>\n

a)\u00a04<\/p>\n

b)\u00a010<\/p>\n

c)\u00a08<\/p>\n

d)\u00a012<\/p>\n

Question 21:\u00a0<\/b>Sum of 4 times of a fraction and its reciprocal is 4, find that fraction ?<\/p>\n

a)\u00a02\/3<\/p>\n

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

c)\u00a03\/4<\/p>\n

d)\u00a01\/2<\/p>\n

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Question 22:\u00a0<\/b>Find $ab+bc+ca$, if $a^2+b^2+c^2 = 24$ & $a+b+c=4\\sqrt3$ ?<\/p>\n

a)\u00a012<\/p>\n

b)\u00a013<\/p>\n

c)\u00a014<\/p>\n

d)\u00a015<\/p>\n

Question 23:\u00a0<\/b>[4x-1\/4]3 = x + 4\/3, then find x ?<\/p>\n

a)\u00a0123\/25<\/p>\n

b)\u00a0132\/25<\/p>\n

c)\u00a025\/132<\/p>\n

d)\u00a025\/123<\/p>\n

Question 24:\u00a0<\/b>Find which of $a$ & $b$ is smaller, $(a+b)^3 = 27$ & $(a-b)^2 = 1$ ?<\/p>\n

a)\u00a0a<\/p>\n

b)\u00a0b<\/p>\n

c)\u00a0a=b<\/p>\n

d)\u00a0cannot be determined<\/p>\n

Question 25:\u00a0<\/b>3[2x-4\/3] + 4[2-3x\/2] = 4, then find x ?<\/p>\n

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

b)\u00a00<\/p>\n

c)\u00a0x can be any number<\/p>\n

d)\u00a01<\/p>\n

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

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

let the number be xy
\nGiven x+y=11
\n10y+x-(10x+y)=63
\n9(y-x)=63
\ny-x=7
\n2y=18
\ny=9
\nx=2
\nRequired number is 29<\/p>\n

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

let the number be xy
\nGiven x+y=7
\n10y+x-(10x+y)=27
\n9(y-x)=27
\ny-x=3
\n2y=10
\ny=5
\nx=2
\nRequired number is 25<\/p>\n

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

Given, $a+b=0$
\n$a^3+b^3$ can be written as
\n$a^3+b^3=(a+b)(a^2-ab+b^2)$
\n\u21d2 $a^3+b^3=0*( a^2-ab+b^2)$
\n\u21d2 $a^3+b^3=0$<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{7}{3}}}} = 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{10}{3}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{3}{10}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{13}{10}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{10}{13}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{23}{13}}$<\/p>\n

$= 1+\\dfrac{13}{23}$<\/p>\n

$= \\dfrac{36}{23}$<\/p>\n

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

let the cost of gift be x
\n54x=((x-4)*60)+18
\n54x=60x-240+18
\n6x=222
\nx=37
\nIf x=37 total money with him is 54*37=Rs 1998<\/p>\n

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

let the number be x
\n(x+12)\/19=9
\nx=171-12
\nx=159
\nAnd then (159-40)=119
\n119\/17 we get 0 as the remainder.<\/p>\n

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

We know that $x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2 +z^2 -xy -yz -zx)$
\nNow, $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
\n=> $x^2 + y^2 +z^2 -xy -yz -zx = (x+y+z)^2 – 3(xy+yz+zx)$
\n=> $x^3 + y^3 + z^3 -3xyz = (x+y+z)[ (x+y+z)^2 – 3(xy+yz+zx)] = 20 [400 – 30] = 20 \\times 370 = 7400$<\/p>\n

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

Given $x+\\dfrac{1}{x} = 3$
\nCubing on both sides<\/p>\n

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

$x^3+\\dfrac{1}{x^3}+3\\times x \\times \\dfrac{1}{x}(x+\\dfrac{1}{x}) = 27$<\/p>\n

\u21d2 $x^3+\\dfrac{1}{x^3}+3\\times3 = 27$<\/p>\n

\u21d2 $x^3+\\dfrac{1}{x^3}+9 = 27$<\/p>\n

\u21d2 $x^3+\\dfrac{1}{x^3} = 18$<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{5}{4}}}} = 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{9}{4}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{4}{9}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{13}{9}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{9}{13}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{22}{13}}$<\/p>\n

$= 1+\\dfrac{13}{22}$<\/p>\n

$= \\dfrac{35}{22}$<\/p>\n

\"RRB<\/a><\/p>\n

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

7\/2(2x\/3 – 1\/2) + 11\/2 = 2x\/3
\n7x\/3 – 7\/4 + 11\/2 = 2x\/3
\n7x\/3 – 2x\/3 = 7\/4 – 11\/2
\n5x\/3 = -15\/4
\nx = -9\/4
\n1\/x = -4\/9
\nSo the answer is option B.<\/p>\n

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

Given, $a+\\frac{1}{a} = 5$<\/p>\n

Squaring on both sides
\n$(a+\\frac{1}{a})^2 = 5^2$<\/p>\n

$a^2+\\frac{1}{a^2}+2\\times a\\times\\frac{1}{a} = 25$<\/p>\n

$a^2+\\frac{1}{a^2}+2 = 25$<\/p>\n

$a^2+\\frac{1}{a^2} = 23$<\/p>\n

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

$1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{3}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{4}{3}}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{1+\\frac{3}{4}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{1}{\\dfrac{7}{4}}}$<\/p>\n

$= 1+\\dfrac{1}{1+\\dfrac{4}{7}}$<\/p>\n

$= 1+\\dfrac{1}{\\dfrac{11}{7}}$<\/p>\n

$= 1+\\dfrac{7}{11} = \\dfrac{18}{11}$<\/p>\n

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

$5 = 40-12x+x^2$<\/p>\n

$x^2-12x+35 = 0$<\/p>\n

$x^2-7x-5x+35 = 0$<\/p>\n

$x(x-7)-5(x-7) = 0$<\/p>\n

$(x-7)(x-5) = 0$<\/p>\n

$x = 7$ or $x = 5$<\/p>\n

So the answer is option C.<\/p>\n

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

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

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

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

$5x^2-13x-6 = 0$<\/p>\n

$5x^2-15x+2x-6 = 0$<\/p>\n

$5x(x-3)+2(x-3) = 0$<\/p>\n

$(x-3)(5x+2) = 0$<\/p>\n

$ x = 3 (or) x = -2\/5$<\/p>\n

So the answer is option D.<\/p>\n

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

$(x+y+z)^2$ = $x^2+y^2+z^2+2(xy+yz+zx)$<\/p>\n

$(9)^2$ = $x^2+y^2+z^2+2(18)$<\/p>\n

$81$ = $x^2+y^2+z^2+2(18)$<\/p>\n

$x^2+y^2+z^2 = 81-36 = 45$<\/p>\n

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

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

$(a+b+c)(a+2b-c)$<\/p>\n

=$a^2+2ab-ac+ab+2b^2-bc+ac+2bc-c^2$<\/p>\n

=$a^2+2b^2-c^2+3ab+bc$<\/p>\n

So the answer is option A.<\/p>\n

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

$b+1\/b = 6$<\/p>\n

Squaring on both sides<\/p>\n

$(b+1\/b)^2 = 36$<\/p>\n

$b^2+1\/b^2 +2 = 36$<\/p>\n

$b^2+1\/b^2 = 36-2 = 34$<\/p>\n

So the answer is option D.<\/p>\n

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

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

squaring on both sides<\/p>\n

$(x+1\/x)^2 = 4$<\/p>\n

$x^2+1\/x^2+2 = 4$<\/p>\n

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

squaring on both sides<\/p>\n

$(x^2+1\/x^2)^2 = 4$<\/p>\n

$x^4+1\/x^4+2 = 4$<\/p>\n

$x^4+1\/x^4 = 2$<\/p>\n

So the answer is option D.<\/p>\n

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

A sheep has 4 legs and hen has 2 legs<\/p>\n

4x+2y = 36 ===> 2x+y = 18—-(1)<\/p>\n

x+y = 14 —-(2)<\/p>\n

(1)-(2) ==> 2x+y-x-y = 18-14 = 4<\/p>\n

So the answer is option A.<\/p>\n

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

Let the fraction be 1\/x<\/p>\n

$4(1\/x)+x = 4$<\/p>\n

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

$x^2-4x+4 = 0$<\/p>\n

$(x-2)^2 = 0$<\/p>\n

$x-2 = 0$<\/p>\n

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

Hence the fraction is 1\/x = 1\/2<\/p>\n

So the answer is option D.<\/p>\n

\"RRB<\/a><\/p>\n

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

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

$(4\\sqrt3)^2 = 24 + 2(ab+bc+ca)$<\/p>\n

$48 = 24 + 2(ab+bc+ca)$<\/p>\n

$24 = 2(ab+bc+ca)$<\/p>\n

$(ab+bc+ca) = \\frac{24}{2} = 12$<\/p>\n

So the answer is option A.<\/p>\n

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

[4x-1\/4]3 = x + 4\/3<\/p>\n

12x – 3\/4 = x + 4\/3<\/p>\n

11x =4\/3 + 3\/4<\/p>\n

11x = 25\/12<\/p>\n

x = 25\/132<\/p>\n

So the answer is option C.<\/p>\n

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

$(a+b)^3 = 27 \\rightarrow a+b = 3$<\/p>\n

$(a-b)^2 = 1 \\rightarrow a-b = \\pm 1$<\/p>\n

If $a-b = -1$, & $a+b = 3$, then $a = 1$ & $b = 2$<\/p>\n

If $a-b = +1$, & $a+b = 3$, then $a = 2$ & $b = 1$<\/p>\n

So we can’t find exact values of $a$ & $b$<\/p>\n

So the answer is option D.<\/p>\n

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

3[2x-4\/3] + 4[2-3x\/2] = 4<\/p>\n

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

0 = 0<\/p>\n

here we have 6x and -6x, So for any value of x, 6x-6x = 0<\/p>\n

So the answer is option C.<\/p>\n

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

RRB NTPC Algebra Questions PDF Download RRB NTPC Alghebra Questions and Answers PDF. Top 25 RRB NTPC\u00a0 Algebra questions based on asked questions in previous exam papers very important for the Railway NTPC exam Question 1:\u00a0A two digit number when reverse decreases its value by 63 and the sum of the digits of the number […]<\/p>\n","protected":false},"author":41,"featured_media":26075,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603,1601],"tags":[489,491,42,1619],"class_list":{"0":"post-26072","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-ntpc","11":"category-rrb-ntpc-je","12":"tag-railway-exam","13":"tag-rrb","14":"tag-rrb-mock-test","15":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":26075,"alt_text":"RRB NTPC Algebra Questions PDF","caption":"RRB NTPC Algebra 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