{"id":43071,"date":"2020-10-07T18:32:13","date_gmt":"2020-10-07T13:02:13","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43071"},"modified":"2020-10-07T18:32:13","modified_gmt":"2020-10-07T13:02:13","slug":"elementary-algebra-questions-for-rrb-ntpc","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/elementary-algebra-questions-for-rrb-ntpc\/","title":{"rendered":"Elementary Algebra Questions for RRB NTPC"},"content":{"rendered":"

Elementary Algebra Questions for RRB NTPC<\/strong><\/span><\/h2>\n

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

Download Elementary Algebra Questions for RRB NTPC<\/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>Find the value of x for which the expression $2 – 3x- 4x^{2}$ has the greatest value.<\/p>\n

a)\u00a0$-\\frac{41}{16}$<\/p>\n

b)\u00a0$\\frac{3}{8}$<\/p>\n

c)\u00a0$-\\frac{3}{8}$<\/p>\n

d)\u00a0$\\frac{41}{16}$<\/p>\n

Question 2:\u00a0<\/b>If 3x – 4 > 2 – x\/3 and 3x + 5 > 4x – 5; then \ufb01nd the value of x?<\/p>\n

a)\u00a07<\/p>\n

b)\u00a010<\/p>\n

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

d)\u00a01<\/p>\n

Question 3:\u00a0<\/b>If 2x – 3(x + 2) < 5 – 2x < – x + 2, then \ufb01nd the value of x.<\/p>\n

a)\u00a02<\/p>\n

b)\u00a00<\/p>\n

c)\u00a010<\/p>\n

d)\u00a012<\/p>\n

Question 4:\u00a0<\/b>If $3x^{2} = 10^{2} – 5^{2}$, find the value of x?<\/p>\n

a)\u00a07<\/p>\n

b)\u00a05<\/p>\n

c)\u00a09<\/p>\n

d)\u00a011<\/p>\n

Question 5:\u00a0<\/b>If 2x+ 4(x – 3) < -2 – x < 2x – 1, then find the value of x.<\/p>\n

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

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

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

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

Question 6:\u00a0<\/b>If $13x^2 = 17^2 – 9^2$, find the value of x?<\/p>\n

a)\u00a016<\/p>\n

b)\u00a012<\/p>\n

c)\u00a08<\/p>\n

d)\u00a04<\/p>\n

Question 7:\u00a0<\/b>If x +$\\frac{1}{x}$=c+$\\frac{1}{c}$ then find the value of x?<\/p>\n

a)\u00a0C,$\\frac{1}{c}$<\/p>\n

b)\u00a0C,$C^{2}$<\/p>\n

c)\u00a0C.2C<\/p>\n

d)\u00a00,1<\/p>\n

Question 8:\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

Question 9:\u00a0<\/b>If $a + b +c= 6$, $a^{2}+ b^{2}+ c^{2} = 14$, find the value of $bc + ca + ab$.<\/p>\n

a)\u00a022<\/p>\n

b)\u00a025<\/p>\n

c)\u00a020<\/p>\n

d)\u00a011<\/p>\n

Question 10:\u00a0<\/b>If $xy = -18$ and $x^{2} + y^{2} = 85$, then \ufb01nd the value of (x + y).<\/p>\n

a)\u00a08<\/p>\n

b)\u00a010<\/p>\n

c)\u00a09<\/p>\n

d)\u00a07<\/p>\n

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

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

Question 11:\u00a0<\/b>If a:b = 3:8, find the value of (5a – \u00ad3b)\/(2a + b).<\/p>\n

a)\u00a09\/14<\/p>\n

b)\u00a014\/9<\/p>\n

c)\u00a0-9\/14<\/p>\n

d)\u00a0-14\/9<\/p>\n

Question 12:\u00a0<\/b>If a:b = 5:8, find the value of (6a \u00ad- 5b)\/(a + 2b).<\/p>\n

a)\u00a010\/21<\/p>\n

b)\u00a0\u00ad21\/10<\/p>\n

c)\u00a0-21\/10<\/p>\n

d)\u00a0-10\/21<\/p>\n

Question 13:\u00a0<\/b>If $xy = -30$ and $x^{2} + y^{2} = 61$, then find the value of (x + y).<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a01<\/p>\n

d)\u00a04<\/p>\n

Question 14:\u00a0<\/b>If $7^{m+1}=2401$, then find the value of $2^{2m+2}$<\/p>\n

a)\u00a0224<\/p>\n

b)\u00a0256<\/p>\n

c)\u00a0264<\/p>\n

d)\u00a0286<\/p>\n

Question 15:\u00a0<\/b>$\u00a0If a + \\frac{1}{a} = 1, $ find the value of $\u00a0a^{3} + \\frac{1}{a^{3}} $<\/p>\n

a)\u00a02<\/p>\n

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

c)\u00a00<\/p>\n

d)\u00a01.5<\/p>\n\n

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

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

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

NOTE :- To find the min\/max value of an expression, we need to differentiate it, and put the first derivative equal to ‘0’ to find value of $x$<\/p>\n

Then, we need to differentiate it again and put value of $x$, if second derivative is less than zero, then $x$ is maxima otherwise $x$ is minima.<\/p>\n

\n

Expression : $y$ = $2 – 3x- 4x^{2}$<\/p>\n

=> $\\frac{dy}{dx} = -3 – 8x$<\/p>\n

Now, putting it equal to 0, we get :<\/p>\n

=> $y’ = -3 -8x = 0$<\/p>\n

=> $x = \\frac{-3}{8}$<\/p>\n

Differentiating it again :<\/p>\n

=> $\\frac{d^2y}{dx^2} = -8$<\/p>\n

Since, it is less than ‘0’ ,=> $x = \\frac{-3}{8}$ is maximum value.<\/p>\n

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

Expression 1\u00a0:\u00a03x – 4 > 2 – x\/3<\/p>\n

=> $3x + \\frac{x}{3}$\u00a0> $2 + 4$<\/p>\n

=> $\\frac{10x}{3}$ > $6$<\/p>\n

=> $x$ > $\\frac{18}{10} = 1.8$ ——–(i)<\/p>\n

Expression 2\u00a0:\u00a03x + 5 > 4x – 5<\/p>\n

=> $4x – 3x$ < $5 + 5$<\/p>\n

=> $x$ < $10$ ———(ii)<\/p>\n

Combining inequalities (i) and (ii), we get\u00a0: $1.8$ < $x$ < $10$<\/p>\n

Thus, the only value that $x$ can take among the options = 7<\/p>\n

=> Ans – (A)<\/p>\n

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

Expression\u00a01\u00a0:\u00a0$5 – 2x < -x + 2$<\/p>\n

=> $2x – x$\u00a0> $5 – 2$<\/p>\n

=> $x$\u00a0> $3$ ———-(i)<\/p>\n

Expression 2\u00a0:\u00a0$2x – 3(x + 2) < 5 – 2x$<\/p>\n

=> $-x – 6$ < $5 – 2x$<\/p>\n

=> $2x – x$ < $5 + 6$<\/p>\n

=> $x$ < $11$ ——(ii)<\/p>\n

Combining inequalities (i) and (ii), we get\u00a0: $3$ < $x$ < $11$<\/p>\n

Thus, only value that $x$ can take among the options = 10<\/strong><\/p>\n

=> Ans – (C)<\/p>\n

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

Expression\u00a0:\u00a0$3x^{2} = 10^{2} – 5^{2}$<\/p>\n

=> $3x^2 = 100 – 25$<\/p>\n

=> $3x^2 = 75$<\/p>\n

=> $x^2 = \\frac{75}{3} = 25$<\/p>\n

=> $x = \\sqrt{25} = 5$<\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a01\u00a0:\u00a0$-2 – x < 2x – 1$<\/p>\n

=> $2x + x$\u00a0> $1 – 2$<\/p>\n

=> $x$\u00a0> $\\frac{-1}{3}$ ———-(i)<\/p>\n

Expression 2\u00a0:\u00a0$2x + 4(x – 3) < -2 – x$<\/p>\n

=> $6x – 12$ < $-2 – x$<\/p>\n

=> $6x + x$ < $12 – 2$<\/p>\n

=> $x$ < $\\frac{10}{7}$ ——(ii)<\/p>\n

Combining inequalities (i) and (ii), we get\u00a0: $\\frac{-1}{3}$ < $x$ < $\\frac{10}{7}$<\/p>\n

Thus, only value that $x$ can take among the options = 1<\/b><\/p>\n

=> Ans – (B)<\/p>\n

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

Expression\u00a0:\u00a0$13x^2 = 17^2 – 9^2$<\/p>\n

=> $13 x^2 = (17 – 9) (17 + 9)$<\/p>\n

=> $13 x^2 = 8 \\times 26$<\/p>\n

=> $x^2 = 8 \\times 2 = 16$<\/p>\n

=> $x = \\sqrt{16} = 4$<\/p>\n

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

Solving the equation , you will get ,<\/p>\n

x = $frac{(c+1)^2 }{2}\u00a0 ,\u00a0\u00a0frac{(c-1)^2 }{2}$<\/p>\n

8)\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

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

It is given that : $a + b + c = 6$ and $a^{2}+ b^{2}+ c^{2} = 14$<\/p>\n

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

=> $6^2 = 14 + 2(ab + bc + ca)$<\/p>\n

=> $ab + bc + ca = 22\/2 = 11$<\/p>\n

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

Given\u00a0: $(x^2 + y^2) = 85$ and $xy = -18$<\/p>\n

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

=> $(x + y)^2 = 85 + (2 \\times -18)$<\/p>\n

=> $(x + y)^2 = 85 – 36 = 49$<\/p>\n

=> $(x + y) = \\sqrt{49} = 7$<\/p>\n

=> Ans – (D)<\/p>\n

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

It is given that $a$ : $b$ = 3\u00a0: 8<\/p>\n

Let $a = 3$ and $b = 8$<\/p>\n

To find\u00a0: $\\frac{5a – 3b}{2a + b}$<\/p>\n

= $\\frac{(5 \\times 3) – (3 \\times 8)}{(2 \\times 3) + (8)}$<\/p>\n

= $\\frac{(15 – 24)}{(6 + 8)} = \\frac{-9}{14}$<\/p>\n

=> Ans – (C)<\/p>\n

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

It is given that a\u00a0: b = 5\u00a0: 8<\/p>\n

Let $a = 5$ and $b = 8$<\/p>\n

To find\u00a0: $\\frac{6a – 5b}{a + 2b}$<\/p>\n

= $\\frac{(6 \\times 5) – (5 \\times 8)}{(5) + (2 \\times 8)}$<\/p>\n

= $\\frac{30 – 40}{5 + 16} = \\frac{-10}{21}$<\/p>\n

=> Ans – (D)<\/p>\n

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

Given\u00a0: $(x^2 + y^2) = 61$ and $xy = -30$<\/p>\n

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

=> $(x + y)^2 = 61 + (2 \\times -30)$<\/p>\n

=> $(x + y) = \\sqrt{61 – 60} = 1$<\/p>\n

=> Ans – (C)<\/p>\n

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

Given\u00a0:\u00a0$7^{m+1}=2401$<\/p>\n

=>\u00a0$7^{m+1}=7^4$<\/p>\n

=> $m+1=4$<\/p>\n

=> $m=4-1=3$<\/p>\n

$\\therefore$\u00a0$2^{2m+2}=2^{2\\times3+2}$<\/p>\n

= $2^8=256$<\/p>\n

=> Ans – (B)<\/p>\n

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

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

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

$a^{3} +\u00a0\\frac{1}{a^{3}} + 3(a +\u00a0\\frac{1}{a}) = 1$<\/p>\n

$ a^{3} + \\frac{1}{a^{3}} = -2$ (\u00a0as we know $a + \\frac{1}{a} = 1$)<\/p>\n

Hence, option B is the correct answer.<\/p>\n

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We hope this Top-15 Elementary Algebra Questions pdf for RRB NTPC exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Elementary Algebra Questions for RRB NTPC Download RRB NTPC Top-15 Elementary Algebra Questions PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Download RRB NTPC Previous Papers PDF Question 1:\u00a0Find the value of x for which the expression $2 – 3x- 4x^{2}$ has the greatest value. a)\u00a0$-\\frac{41}{16}$ […]<\/p>\n","protected":false},"author":42,"featured_media":43074,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,31,1701,1679,1897,1605,1603,1673],"tags":[2308,4241,4117,492,1635,1593],"class_list":{"0":"post-43071","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-rpf-si-and-constable","11":"category-rrb-group-d","12":"category-rrb-group-d-2","13":"category-rrb-je","14":"category-rrb-ntpc","15":"category-rrb-para-medical-categories","16":"tag-algebra","17":"tag-elementary-algebra","18":"tag-railway-exams","19":"tag-rrb-group-d","20":"tag-rrb-je","21":"tag-rrb-ntpc"},"better_featured_image":{"id":43074,"alt_text":"Elementary Algebra Questions for RRB NTPC","caption":"Elementary Algebra Questions for 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