{"id":43201,"date":"2020-10-16T17:39:39","date_gmt":"2020-10-16T12:09:39","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=43201"},"modified":"2020-10-16T17:39:39","modified_gmt":"2020-10-16T12:09:39","slug":"algebra-questions-for-rrb-ntpc-set-4-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/algebra-questions-for-rrb-ntpc-set-4-pdf\/","title":{"rendered":"Algebra Questions for RRB NTPC Set-4 PDF"},"content":{"rendered":"

Algebra Questions for RRB NTPC Set-4 PDF<\/strong><\/span><\/h2>\n

Download RRB NTPC Top-10 Algebra Questions Set-4 PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n

Download Algebra Questions for RRB NTPC Set-4 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>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>If 3 (a^{2} + b^{2} + c^{2} ) = (a + b + c)^{2} , then the relation between a, b and c is<\/p>\n

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

b)\u00a0a = b \u2260 c<\/p>\n

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

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

Question 3:\u00a0<\/b>If $(x-3)^2+(y-5)^2+(z-4)^2=0$, then the value of $\\frac{x^2}{9}+\\frac{y^2}{25}+\\frac{z^2}{16}$ is<\/p>\n

a)\u00a012<\/p>\n

b)\u00a09<\/p>\n

c)\u00a03<\/p>\n

d)\u00a01<\/p>\n

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

a)\u00a064<\/p>\n

b)\u00a0194<\/p>\n

c)\u00a081<\/p>\n

d)\u00a0124<\/p>\n

Question 5:\u00a0<\/b>If $\\frac{5x}{2x^2+5x+1}=\\frac{1}{3}$ then the value of $(x+\\frac{1}{2x})$ is <\/span><\/p>\n

a)\u00a015<\/p>\n

b)\u00a010<\/p>\n

c)\u00a020<\/p>\n

d)\u00a05<\/p>\n

FREE RRB NTPC MOCK TEST<\/a><\/p>\n

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

Question 6:\u00a0<\/b>If $2\\sqrt{x} = \\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}} + \\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$<\/p>\n

a)\u00a06<\/p>\n

b)\u00a030<\/p>\n

c)\u00a0\u221a15<\/p>\n

d)\u00a016<\/p>\n

Question 7:\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 8:\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 9:\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 10:\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

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

Download General Science Notes PDF<\/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(A)<\/strong><\/p>\n

Given , 3 (a^{2} + b^{2} + c^{2} ) = (a + b + c)^{2}
\n3 (a^{2} + b^{2} + c^{2} )=3 (a^{2} + b^{2} + c^{2}) +2[ab+bc+ca]
\n2 (a^{2} + b^{2} + c^{2} )= +2[ab+bc+ca]
\n(a^{2} + b^{2} + c^{2} )= [ab+bc+ca]
\n(a^{2} -ab) + (b^{2} -bc) + (c^{2} -ca) = 0
\na(a-b) + b(b-c) + c(c-a) = 0
\nThis is only possible when a=b=c .<\/p>\n

\u00a0<\/del><\/sup><\/sub><\/u><\/i><\/strong><\/p>\n

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

$(x-3)^2+(y-5)^2+(z-4)^2=0$
\nSince square values are always positive or equal to zero, x must be 3, y must be 5 and 4 must be 4.
\nSubstituting these values in $\\frac{x^2}{9}+\\frac{y^2}{25}+\\frac{z^2}{16}$, we get the value as 1+1+1 = 3.
\nOption C is the right answer.<\/p>\n

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

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

squaring both sides<\/p>\n

$x^{2}+\\frac{1}{x^2} +2*x*\\frac{1}{x}$ = 16<\/p>\n

$x^{2}+\\frac{1}{x^2}$ = 16-2 = 14<\/p>\n

again squaring both sides<\/p>\n

$x^{4}+\\frac{1}{x^4} +2*x^{2}*\\frac{1}{x^2}$= 196 <\/span><\/p>\n

$x^{4}+\\frac{1}{x^4}$ = 196 – 2 = 194<\/span><\/p>\n

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

Expression : $\\frac{5x}{2x^2+5x+1}=\\frac{1}{3}$<\/p>\n

=> $2x^2 + 5x + 1 = 15x$<\/p>\n

=> $2x^2 + 1 = 10x$<\/p>\n

To find : $(x+\\frac{1}{2x})$<\/p>\n

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

= $\\frac{10x}{2x}$<\/span><\/p>\n

= 5<\/span><\/p>\n\n

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

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

it is given that<\/p>\n

$2\\sqrt{x} = \\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}} – \\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$<\/p>\n

here , $\\frac{\\sqrt5+\\sqrt3}{\\sqrt5-\\sqrt3}$ = $\\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}$ x $\\frac{\\sqrt{5}+\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$ = $\\frac{(\\sqrt5 + \\sqrt3)^2}{2}$<\/p>\n

similarly , $\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$ = $\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}+\\sqrt{3}}$ x $\\frac{\\sqrt{5}-\\sqrt{3}}{\\sqrt{5}-\\sqrt{3}}$ = $\\frac{(\\sqrt5 – \\sqrt3)^2}{2}$<\/p>\n

$\\frac{(\\sqrt5 + \\sqrt3)^2}{2}$ + $\\frac{(\\sqrt5 – \\sqrt3)^2}{2}$ = 2$\\sqrt(x)$<\/p>\n

8 = 2$\\sqrt(x)$<\/p>\n

x = 16<\/p>\n

7)\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

8)\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

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

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

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

We hope this Top-10 Algebra Questions Set-4 PDF for RRB NTPC exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Algebra Questions for RRB NTPC Set-4 PDF Download RRB NTPC Top-10 Algebra Questions Set-4 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:\u00a0The smallest positive integer n with 24 divisors considering 1 and n as divisors is a)\u00a0420 b)\u00a0240 […]<\/p>\n","protected":false},"author":42,"featured_media":43204,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,31,3171,1701,1679,1605,1603,1673],"tags":[4269,4117,492,1635,1593],"class_list":{"0":"post-43201","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-railways-en","11":"category-rpf-si-and-constable","12":"category-rrb-group-d","13":"category-rrb-je","14":"category-rrb-ntpc","15":"category-rrb-para-medical-categories","16":"tag-algebra-questiona","17":"tag-railway-exams","18":"tag-rrb-group-d","19":"tag-rrb-je","20":"tag-rrb-ntpc"},"better_featured_image":{"id":43204,"alt_text":"Algebra Questions for RRB NTPC Set-4 PDF","caption":"Algebra Questions for RRB NTPC Set-4 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