{"id":44004,"date":"2021-01-08T14:10:12","date_gmt":"2021-01-08T08:40:12","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=44004"},"modified":"2021-01-08T14:10:12","modified_gmt":"2021-01-08T08:40:12","slug":"top-20-rrb-ntpc-algebra-questions","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/top-20-rrb-ntpc-algebra-questions\/","title":{"rendered":"Top-20 RRB NTPC Algebra Questions"},"content":{"rendered":"

RRB NTPC Algebra Questions:<\/strong><\/h1>\n

Download Top-20 Algebra questions for RRB NTPC\u00a0 exam. Most important Algebra questions based on asked questions in previous exam papers for RRB NTPC.<\/p>\n

Download Top-20 RRB NTPC Algebra Questions <\/a><\/p>\n

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

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

Question 1:\u00a0<\/b>The smallest positive integer n with 24 divisors considering 1 and n as divisors is<\/p>\n

a)\u00a0420<\/p>\n

b)\u00a0240<\/p>\n

c)\u00a0360<\/p>\n

d)\u00a0480<\/p>\n

Question 2:\u00a0<\/b>In an election a candidate gets 40% of votes polled and is defeated by the winning candidate by 298 votes. Find the total number of votes polled.<\/p>\n

a)\u00a01360<\/p>\n

b)\u00a01490<\/p>\n

c)\u00a01520<\/p>\n

d)\u00a01602<\/p>\n

Question 3:\u00a0<\/b>Two numbers are less than the third number by 30% and 37% respectively. By what percent is the second number less than the first number?<\/p>\n

a)\u00a015%<\/p>\n

b)\u00a010%<\/p>\n

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

d)\u00a020%<\/p>\n

Question 4:\u00a0<\/b>If 15 boys earn Rs.900 in 5 days, how much will 20 boys earn in 7 days ?<\/p>\n

a)\u00a0Rs. 1980<\/p>\n

b)\u00a0Rs. 1820<\/p>\n

c)\u00a0Rs. 1780<\/p>\n

d)\u00a0Rs. 1680<\/p>\n

Question 5:\u00a0<\/b>Srinivas is four times as old as his daughter. Five years ago, Srinivas was nine times as old as his daughter was at that time. His daughter’s present age is:<\/p>\n

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

b)\u00a08 years<\/p>\n

c)\u00a06 years<\/p>\n

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

Question 6:\u00a0<\/b>If $a$ is positive and $a^2 + \\frac{1}{a^2} = 7,$ then $a^3 + \\frac{1}{a^3} = ?$<\/p>\n

a)\u00a021<\/p>\n

b)\u00a0$3 \\sqrt7$<\/p>\n

c)\u00a018<\/p>\n

d)\u00a0$7 \\sqrt7$<\/p>\n

Question 7:\u00a0<\/b>Which of the following statement is sufficient to answer the question? Find the values of x, y, z from the given statements.
\nStatements:<\/strong>
\nI. x + y = 12 ; x + z = 4
\nII: x – y = 6<\/p>\n

a)\u00a0Only II is sufficient while I is not<\/p>\n

b)\u00a0Neither I nor II is sufficient<\/p>\n

c)\u00a0Both I and II are sufficient<\/p>\n

d)\u00a0Only I is sufficient while II is not<\/p>\n

Question 8:\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

Question 9:\u00a0<\/b>If $12x^2- ax + 7 = ax^2 + 9x + 3$ has only one (repeated) solution, then the positive integral solution of a is:<\/p>\n

a)\u00a02<\/p>\n

b)\u00a04<\/p>\n

c)\u00a03<\/p>\n

d)\u00a05<\/p>\n

Question 10:\u00a0<\/b>If $\\frac{x}{x^{2}-1}$=$\\frac{A}{x-1}+\\frac{B}{x+1}$ then find the values of A and B<\/p>\n

a)\u00a02,2<\/p>\n

b)\u00a02,-2<\/p>\n

c)\u00a0$\\frac{1}{2}$, $\\frac{-1}{2}$<\/p>\n

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

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Question 11:\u00a0<\/b>Find the value of $\\frac{1}{1\\times2}+\\frac{1}{2\\times3}+\\frac{1}{3\\times4}+\\frac{1}{4\\times5}+\\frac{1}{5\\times6}+….+\\frac{1}{9\\times10}$<\/p>\n

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

b)\u00a0$\\frac{9}{10}$<\/p>\n

c)\u00a0$\\frac{5}{11}$<\/p>\n

d)\u00a0$\\frac{2}{5}$<\/p>\n

Question 12:\u00a0<\/b>simplify $\\sqrt{0.01+\\sqrt{0.0225}}$<\/p>\n

a)\u00a016<\/p>\n

b)\u00a00.4<\/p>\n

c)\u00a04<\/p>\n

d)\u00a00.04<\/p>\n

Question 13:\u00a0<\/b>Simplify $ \\frac{121}{3\\frac{2}{3}}+\\frac{92}{7\\frac{1}{3}}$ ?<\/p>\n

a)\u00a0$43\\frac{11}{19}$<\/p>\n

b)\u00a0$41\\frac{12}{13}$<\/p>\n

c)\u00a0$40\\frac{13}{11} $<\/p>\n

d)\u00a0$45\\frac{6}{11} $<\/p>\n

Question 14:\u00a0<\/b>If $x^{3}$ = n and the units digit of ‘n’ is a prime number, what are the possible choices for ‘x’ among
\nthe numbers from 1 to 9.<\/p>\n

a)\u00a03, 5, 7, 8<\/p>\n

b)\u00a03, 5 ,7<\/p>\n

c)\u00a02, 3, 4, 5<\/p>\n

d)\u00a01, 5, 3<\/p>\n

Question 15:\u00a0<\/b>simplify$\\sqrt{10+\\sqrt{25}+\\sqrt{108}+\\sqrt{154}+\\sqrt{225}}$?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a08<\/p>\n

c)\u00a04<\/p>\n

d)\u00a06<\/p>\n

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Question 16:\u00a0<\/b>If $\\frac{a}{b}+\\frac{b}{a}=1 $then find $a^{3}+b^{3}$<\/p>\n

a)\u00a02<\/p>\n

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

c)\u00a00<\/p>\n

d)\u00a01<\/p>\n

Question 17:\u00a0<\/b>Evaluate $1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+……\u00a0$?<\/p>\n

a)\u00a0$\\frac{1}{50}$<\/p>\n

b)\u00a03<\/p>\n

c)\u00a0$\\frac{1}{22}$<\/p>\n

d)\u00a02<\/p>\n

Question 18:\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

Question 19:\u00a0<\/b>If A + B = C, D – C = A and E – B = C,then what does D + F stand for?<\/p>\n

a)\u00a0C<\/p>\n

b)\u00a0F<\/p>\n

c)\u00a0J<\/p>\n

d)\u00a0Q<\/p>\n

Question 20:\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

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

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

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

For any given number, that can be represented as $ A^{x} \\times B ^ {y} $, etc<\/p>\n

The number of factors is denoted by (x+1) x ( y+1), etc<\/p>\n

360 = $ 2 ^ {3} \\times 3 ^ {2} \\times 5^ {1}$<\/p>\n

So the number of factors = (3 +1) x (2+ 1) (1+ 1) = 4x3x2 = 24<\/p>\n

For 240, it is\u00a0$ 2 ^ {4} \\times 3 ^ {1} \\times 5^ {1}$<\/p>\n

Number of factors = 5 x 2 x 2 = 20 only<\/p>\n

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

There is an assumption in the question that there are only two candidates participating in the election. One candidate got 40% votes and the other candidate got 60% votes. The difference is 20% votes which are 298. If 298 votes are 20%, 100% is how much.<\/p>\n

= 298\/20% = 1490<\/p>\n

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

Let the third number be x. So, the first number is .7x<\/p>\n

The second number is .63x<\/p>\n

So, the second number is less than the first number by .7 ie 10% of the first number.<\/p>\n

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

Amount of money earned by a boy in a day = $\\frac{900}{15*5}$ = Rs. 12<\/p>\n

Hence, amount of money earned\u00a0by 20 boys in 7 days = 20*7*12 = Rs. 1680<\/p>\n

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

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

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

Addition 2 in both sides of equation.<\/p>\n

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

$a^2 + \\frac{1}{a^2} + 2 = 9$\u00a0 \u00a0 Eq.(1)<\/p>\n

Eq.(1) is making the formula of $(a+\\frac{1}{a})^{2}$.<\/p>\n

After removing the square got $(a+\\frac{1}{a}) =\u00a0\\pm 3$<\/p>\n

In question, it is mentioned that value of a<\/strong> is positive.<\/p>\n

So\u00a0$(a+\\frac{1}{a}) = 3$\u00a0 \u00a0 Eq.(2)<\/p>\n

In Eq.(2) apply formula $(a+\\frac{1}{a})^{3}$.<\/p>\n

So\u00a0$(a+\\frac{1}{a})^{3} =\u00a0a^{3} + (\\frac{1}{a})^{3} + 3 \\times a \\times (\\frac{1}{a}) [a + \\frac{1}{a}]$<\/p>\n

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

Put\u00a0Eq.(2) in\u00a0Eq.(3).<\/p>\n

$(3)^{3} = a^{3} + (\\frac{1}{a})^{3} + 3\\times3$<\/p>\n

$27 = a^{3} + (\\frac{1}{a})^{3} + 9$<\/p>\n

$27-9 = a^{3} + (\\frac{1}{a})^{3}$<\/p>\n

$18 = a^{3} + (\\frac{1}{a})^{3}$<\/p>\n

$a^{3} + (\\frac{1}{a})^{3} = 18$<\/p>\n

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

8)\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

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

Given,\u00a0$12x^2- ax + 7 = ax^2 + 9x + 3$
\n$(a-12)x^2 + (a+9)x-4 = 0$
\nIf $ax^2+bx+c=0$ has equal roots, then $b^2 = 4ac$
\n$(a+9)^2 = 4(a-12)(-4)$
\n$a^2+81+18a = 192-16a$
\n$a^2+34a-111 = 0$
\nOn solving above equation, we get a = 3 and a = -37.
\nHere, The positive integral solution will be 3.<\/p>\n

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

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

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

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

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

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

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

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

$1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+……\u00a0$<\/p>\n

Use GP formula as<\/p>\n

$\\frac{(1-r^n)}{(1-r)}$<\/p>\n

here $n = \\infty$<\/p>\n

therefore, $\\left(\\frac{1 – \\left(\\frac{1}{2}\\right)^{\\infty}}{1 – \\left(\\frac{1}{2}\\right)}\\right)$<\/p>\n

$\\frac{1}{\\left(\\frac{1}{2}\\right)} = 2$<\/p>\n

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

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

20)\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

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 <\/p>\n","protected":false},"excerpt":{"rendered":"

RRB NTPC Algebra Questions: Download Top-20 Algebra questions for RRB NTPC\u00a0 exam. Most important Algebra questions based on asked questions in previous exam papers for RRB NTPC. Take a free RRB NTPC mock test Download RRB NTPC Previous Papers PDF Question 1:\u00a0The smallest positive integer n with 24 divisors considering 1 and n as divisors […]<\/p>\n","protected":false},"author":53,"featured_media":44010,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[31,1603],"tags":[4279,4101,3407],"class_list":{"0":"post-44004","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-railways","8":"category-rrb-ntpc","9":"tag-analogy-questions","10":"tag-rrb-ntpc-2020","11":"tag-rrb-ntpc-exam"},"better_featured_image":{"id":44010,"alt_text":"top 20 rrb ntpc algebra questions","caption":"top 20 rrb ntpc algebra questions","description":"top 20 rrb ntpc algebra 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