{"id":29124,"date":"2019-05-16T18:55:51","date_gmt":"2019-05-16T13:25:51","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=29124"},"modified":"2019-05-16T18:55:51","modified_gmt":"2019-05-16T13:25:51","slug":"simplification-questions-for-rrb-group-d-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/simplification-questions-for-rrb-group-d-set-2-pdf\/","title":{"rendered":"Simplification Questions for RRB Group-D Set-2 PDF"},"content":{"rendered":"

Simplification Questions for RRB Group-D Set-2 PDF<\/strong><\/span><\/h2>\n

Download Top 15 RRB Group-D Simplification Set-2 Questions and Answers PDF. RRB Group-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.<\/p>\n

Download Simplification Questions for RRB Group-D Set-2 PDF<\/a><\/p>\n\n

Download RRB Group-D Previous Papers PDF<\/a><\/p>\n

Take a RRB Group-D free mock test<\/a><\/p>\n

Question 1:\u00a0<\/b>What is the area between the lines 5x+4y=20 and x+y=15 in the first quadrant?<\/p>\n

a)\u00a0117.50 sq units<\/p>\n

b)\u00a0110 sq units<\/p>\n

c)\u00a0105 sq units<\/p>\n

d)\u00a0102.50 sq units<\/p>\n

Question 2:\u00a0<\/b>If x and y are positive real numbers such that 15x + 12xy +20y = 96, find the minimum possible value of 6x + 8y ?<\/p>\n

Question 3:\u00a0<\/b>How man Y different pairs(a,b) of positive integers are there such that $a\\geq b$ and $\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{9}$?<\/p>\n

Take a free mock test for RRB Group-D<\/a><\/p>\n

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Question 4:\u00a0<\/b>Find $x +y$ if $\\frac{90}{3x – y} + \\frac{46}{2x + 3y} = 7$ and $\\frac{92}{2x + 3y} – \\frac{36}{3x – y} = 2$.<\/p>\n

Question 5:\u00a0<\/b>Solve the given system of equations in three variables:
\nx + 3y -z = 5
\n3x + y + 4z = 22
\n2x +6y -2z = 56<\/p>\n

a)\u00a0x = 35\/34, y = 91\/34, z = 69\/17<\/p>\n

b)\u00a0x = -123\/25, y = 147\/25, z = 193\/25<\/p>\n

c)\u00a0infinite solutions<\/p>\n

d)\u00a0no solution<\/p>\n

Question 6:\u00a0<\/b>Roots of the quadratic equation $px^2 + p^2(x + 1) – 4p = 0$ are reciprocal to each other. How many values of $p$ are possible?<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

Daily Free RRB Online Test<\/a><\/p>\n

Question 7:\u00a0<\/b>If $(x-5)^{\\frac{x^2 – 24x + 95}{x^2 – 13x + 36}} = 1$
\nWhat is the sum of all real values of \u2018x\u2019 which satisfy the given equation?<\/p>\n

a)\u00a021<\/p>\n

b)\u00a025<\/p>\n

c)\u00a029<\/p>\n

d)\u00a030<\/p>\n

Question 8:\u00a0<\/b>If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ then the closest approximate values of a and b are<\/p>\n

a)\u00a025, 43<\/p>\n

b)\u00a052, 43<\/p>\n

c)\u00a052, 67<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 9:\u00a0<\/b>Find the number of integral solutions of $\\sqrt{x + 15 – 8\\sqrt{x – 1}} + \\sqrt{x + 24 – 10\\sqrt{x – 1}} = 1$<\/p>\n

Question 10:\u00a0<\/b>If $x^4 – 79x^2 + 1 = 0$, find the value of $x^3 + \\frac{1}{x^3}$ if x>0.<\/p>\n

a)\u00a0702<\/p>\n

b)\u00a0970<\/p>\n

c)\u00a0720<\/p>\n

d)\u00a0None of these<\/p>\n

RRB Group-D Important Questions (download PDF)<\/a><\/p>\n\n

Question 11:\u00a0<\/b>If the roots of $x^3 +3 x^2 + px – q = 0$ are in A.P, then which of the following is possible?<\/p>\n

a)\u00a0p = 5, q = -5<\/p>\n

b)\u00a0p = 4, q = 0<\/p>\n

c)\u00a0p = 1, q = -2<\/p>\n

d)\u00a0 p = 2, q = 0<\/p>\n

e)\u00a0p = 0, q = -3<\/p>\n

Question 12:\u00a0<\/b>If x =$\\sqrt{2}-1$, find the value of $x^4+4x^3+6x^2+4x+8$.<\/p>\n

a)\u00a09<\/p>\n

b)\u00a010<\/p>\n

c)\u00a011<\/p>\n

d)\u00a012<\/p>\n

Question 13:\u00a0<\/b>Solve the equation: $(7-x)^{4} + (9-x)^{4} = 16$<\/p>\n

a)\u00a09<\/p>\n

b)\u00a07<\/p>\n

c)\u00a0Both the above<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 14:\u00a0<\/b>Solve the equation: (x+2)(x+4)(x+6)(x+12)=-4$x^2$, $x^2$ = ?<\/p>\n

a)\u00a0$-10 \\pm 2\\sqrt {3}$<\/p>\n

b)\u00a0$-11 \\pm 2\\sqrt {3}$<\/p>\n

c)\u00a0$-12 \\pm 2\\sqrt {3}$<\/p>\n

d)\u00a0$-13 \\pm 2\\sqrt {3}$<\/p>\n

Question 15:\u00a0<\/b>The maximum value of y = $ – x^2 + 6x- 8 $ is<\/p>\n

a)\u00a00<\/p>\n

b)\u00a01<\/p>\n

c)\u00a02<\/p>\n

d)\u00a03<\/p>\n

General Science Notes for RRB Exams (PDF)<\/a><\/p>\n\n

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

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

Multiplying the second equation by 5 and subtracting from the first equation,<\/p>\n

$5x+4y=20 $<\/p>\n

$5x+5y=75$<\/p>\n

Solving, $4y-5y = 20 – 75 \\Rightarrow y=55$<\/p>\n

Putting the value of $y$, $x = 15 – 55 = -40$<\/p>\n

Since this point is outside the first quadrant, we need to find the intercepts of the lines on the $x$ and $y$ axis to get an idea of the shape of the area.<\/p>\n

For line 1 : $5x+4y=20$<\/p>\n

Putting x=0, $y=\\dfrac{20}{4}=5$<\/p>\n

Putting y=0, $x=\\dfrac{20}{5}=4$<\/p>\n

Intercepts : (0,5) and (4,0)<\/p>\n

For line 2 : $x+y=15$<\/p>\n

Putting x=0, $y=15$ and putting y=0, $x=15$<\/p>\n

Intercepts : (15,0) and (0,15)<\/p>\n

Upon rough drawing, we get a graph something like this :<\/p>\n

<\/figure>\n

Point O is the origin while A,B and C,D are intercepts of line 2 and line 1 respectively.<\/p>\n

The area of the between the two lines can be calculated by subtracting the area of $\\triangle$DOC from the area of $\\triangle$AOB<\/p>\n

Thus, area of $\\triangle$AOB $a_1 = \\dfrac{1}{2} \\times 15 \\times 15$<\/p>\n

$ \\Rightarrow a_1 = \\dfrac{225}{2} $ sq. units<\/p>\n

Now,\u00a0area of $\\triangle$DOC $a_2 = \\dfrac{1}{2} \\times 5 \\times 4$<\/p>\n

$ \\Rightarrow a_2 = 10$ sq units<\/p>\n

Therefore, area between the lines = $\\dfrac{225}{2} – 10 = \\dfrac{205}{2} = 102.50$ sq. units<\/p>\n

2)\u00a0Answer:\u00a024<\/b><\/p>\n

Given that: 15x + 12xy +20y = 96<\/p>\n

$\\Rightarrow$ 15x + 12xy +20y +25 = 96 +25<\/p>\n

$\\Rightarrow$ 5(3x + 5) + 4y(3x + 5)\u00a0 = 121<\/p>\n

$\\Rightarrow$ (3x + 5)(4y + 5)\u00a0 = 121<\/p>\n

If thee product of two positive numbers (3x + 5)\u00a0 and (4y + 5) is constant, then the minimum value of sum occurs when the number are equal i.e.\u00a0\u00a03x + 5 =\u00a04y + 5 = 11<\/p>\n

$\\therefore$ The minimum value of\u00a0\u00a0(3x + 5) + (4y + 5) = 11 + 11<\/p>\n

Hence, the minimum value of\u00a0 3x + 4y = 22 – 10 = 12, hence minimum value of 6x + 8y = 2*12 = 24 (Answer)<\/p>\n

3)\u00a0Answer:\u00a03<\/b><\/p>\n

$\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{9}$
\n=> $ab = 9(a + b)$
\n=> $ab – 9(a+b) = 0$
\n=> $ ab – 9(a+b) + 81 = 81$
\n=> $(a – 9)(b – 9) = 81, a > b$
\nHence we have the following cases,
\n$ a – 9 = 81, b – 9 = 1$ => $(a,b) = (90,10)$
\n$ a – 9 = 27, b – 9 = 3$ => $(a,b) = (36,12)$
\n$ a – 9 = 9, b – 9 = 9$ => $(a,b) = (18,18)$
\nHence there are three possible positive integral values of (a,b)<\/p>\n

4)\u00a0Answer:\u00a010<\/b><\/p>\n

Let
\n$\\frac{1}{3x – y} = a$ and $\\frac{1}{2x + 3y} = b$<\/p>\n

Therefore,<\/p>\n

$92b – 36a = 2$ and<\/p>\n

$46b + 90a = 7$<\/p>\n

Multiplying ii by 2 and subtracting i from it,<\/p>\n

$216a = 12$
\n$a = \\frac{1}{18}$
\nSubstituting in ii,
\n$b = \\frac{1}{23}$<\/p>\n

Therefore,
\n$3x – y = 18$ and $2x + 3y = 23$<\/p>\n

Multiplying the first equation by 3 and adding both the equations.<\/p>\n

$11x = 77$, $ x = 7$<\/p>\n

Hence, $y = 3$<\/p>\n

$x + y = 10$<\/p>\n

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

The coefficient matrix of the given system of equations = M =
\n$\\begin{bmatrix}
\n1 &3 &-1 \\\\
\n3&1 &4 \\\\
\n2&6 &-2
\n\\end{bmatrix}$
\n|M| = D = 1(1x-2-6×4)-3(3x-2-6x-1)+2(3×4-1x-1) = 0<\/p>\n

The x numerator matrix of the given system of equations = M$_x$ =
\n$\\begin{bmatrix}
\n5 & 3 &-1 \\\\
\n22&1 &4 \\\\
\n56&6 &-2 \\end{bmatrix}$<\/p>\n

|M$_x$| = D$_x$ = 5(1x-2-6×4)-22(3x-2-6x-1)+56(3×4-1x-1) = 598
\nSince, at-least one of the numerator matrix is non-zero, the given set of equations have no solution<\/p>\n

6)\u00a0Answer:\u00a01<\/b><\/p>\n

Since the roots are reciprocal to each other , product of roots must be equal to 1.
\nWriting the given equation as \u00a0$px^2 + p^2x + p^2-4p$ = 0<\/p>\n

So, $\\frac{p^2 – 4p}{p}$ = 1<\/p>\n

or, $p^2 – 5p$ = 0<\/p>\n

Solving this equation, we get roots as $p$ = 0 and $p$ = 5
\nBut, $p$ cannot be 0 because then the given equation would not be quadratic. Hence, only one value of $p$ is possible<\/p>\n

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

We have $(x-5)^{\\frac{x^2 – 24x + 95}{x^2 – 13x + 36}} = 1$
\n$ x^2 – 24x + 95 = (x-5)(x-19) $
\n$ x^2 – 13x + 36 = (x-4)(x-9) $
\nSo, $(x-5)^{ \\frac{(x-5)(x-19)}{(x-4)(x-9)} } = 1$
\nCase 1 –
\nConsider following possibility,
\nx – 5 = 1 i.e. x = 6
\nSo $\\frac{(x-5)(x-19)}{(x-4)(x-9)} = \\frac{1*-13}{2*-3}$
\nSo x=6 satisfies
\nCase 2 –
\nConsider, x-5 = -1
\nx = 4
\nSo $\\frac{(x-5)(x-19)}{(x-4)(x-9)} = \\frac{-1*-15}{0*-3} $
\nSo, x=4 does not satisfy
\nCase 3 –
\nConsider $\\frac{(x-5)(x-19)}{(x-4)(x-9)} = 0 $
\n$ (x-5)(x-19) = 0$
\nx = 5 or x = 19
\nFor x = 5, the base becomes 0. So not possible.<\/p>\n

For x = 19, the equation will be satisfied.
\nThus, 2 values of x, i.e. 6 and 19 satisfy the given equation.
\nSum = 6 + 19 = 25
\nHence, option B is the right answer.<\/p>\n

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

If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$
\nThen x = -5 and x = 2 will give $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ = 0
\nSubstituting x = -5 we get,
\n$3(-5)^{4}+2(-5)^{3}-a(-5)^{2}+b(-5)-a+b-4 = 0$
\nSolving we get,
\n$26a+4b = 1621$…….(i)
\nSubstituting x = 2 we get,
\n$3(2)^{4}+2(2)^{3}-a(2)^{2}+b(2)-a+b-4 =0$
\n=> $5a-3b = 60$……..(ii)
\nSolving i and ii we get
\na and b $\\approx 52, 67$
\nHence, option C is the correct answer.<\/p>\n

9)\u00a0Answer:\u00a010<\/b><\/p>\n

Let $\\sqrt{x – 1} = t$
\n$x = t^2 + 1$
\nSubstituting the value of t in the given equation,
\n$\\sqrt{t^2 – 8t + 16} + \\sqrt{t^2 – 10t + 25} = 1$
\n$\\sqrt{(t – 4)^2} + \\sqrt{(t – 5)^2} = 1$
\n|t – 4| +|t – 5| = 1<\/p>\n

The following equation is true for $4 \\leq t \\leq 5$<\/p>\n

The corresponding values of x are $ 17 \\leq x \\leq 26$<\/p>\n

Hence, 10 integral values exist.<\/p>\n

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

$x^4 – 79x^2 + 1 = 0$
\nDividing both sides by $x^2$
\n=> $x^2 -79 + \\frac{1}{x^2} = 0$
\n=> $x^2 + \\frac{1}{x^2} = 79$
\nAdding 2 on both sides.
\n$x^2 + \\frac{1}{x^2} + 2 \\frac{1}{x^2}x^2 = 81$
\n=> $(x+\\frac{1}{x}) = \\pm 9$
\nBut since x>0, $(x+\\frac{1}{x}) = 9$
\nNow, $x^3 + \\frac{1}{x^3} = (x+\\frac{1}{x})(x^2 + \\frac{1}{x^2} – 1)$
\n=> $x^3 + \\frac{1}{x^3} = 9 (79 – 1) = 702$<\/p>\n

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

Let the roots of the given equation be a -d, a and a + d
\nSum of roots of the equation = a – d + a + a + d = 3a = -3
\n=> a = -1
\nSum of product of roots taken two at a time = (a – d)a + a( a + d) + (a – d)(a – d) = $3a^2 – d^2$ = p
\n$3 – p = d^2$
\n=> 3 – p $\\geq$ 0, Thus, p $\\leq$ 3
\nProduct of roots of the equation = $(a – d)a(a + d) = a(a^2 – d^2) = q$
\n$1 – d^2 = -q$
\n=> q $\\geq$ -1
\nFrom the options only option D satisfies the given conditions.<\/p>\n

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

x =$\\sqrt{2}-1$
\nx+1 =$\\sqrt{2}$
\n$(x+1)^4 = 4$
\n$x^4+4x^3+6x^2+4x+1$ = 4
\n$x^4+4x^3+6x^2+4x+8$ = 4+7 = 11<\/p>\n

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

Let z = (7-x + 9 – x )\/2 = 8-x
\nSo, the equation becomes $ (z-1)^4 + (z+1)^4 = 16$
\n$z^4+6z^2-7=0$
\n= > $ (z^2+7)(z^2-1)=0$
\nTaking only real values, z = -1 or 1
\nX = 9 or 7<\/p>\n

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

Since (-2)(-12) = (-4)(-6), we can re-arrange the equation as (x+2)(x+12)(x+4)(x+6)
\n= > ($x^2+14x+24)(x^2+10x+24$)
\nSince x= 0 is not a root, dividing by $x^2$ on both sides,
\nWe get (x + 14 + 24\/x)(x+10+24\/x) = -4
\nPutting x + 24\/x = y, we get
\n(y + 14)(y + 10) = -4
\n$ y^2 + 24y+144=0$
\nY = -12
\nX+24\/x = -12
\n$x^2 +12x+24=0$
\nx = $\\frac{ -12 \\pm \\sqrt {144-96}}{2}$
\n=$ -12 \\pm \\sqrt {12}$
\n=$-12 \\pm 2\\sqrt {3}$<\/p>\n

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

$ – x^2 + 6x- 8 $ can be rewritten as – ($ x^2 – 6x + 8 $)
\n= – (x-2)(x-4)
\nThe roots of the equation are 2,4. The midpoint of the roots is x = 3. That is where the maximum of the function occurs. Substituting x = 3 in the function, we get the maximum point as -9 + 18 – 8 = 1<\/p>\n\n

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We hope this Simplification Set-2 Questions for RRB Group-D Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Simplification Questions for RRB Group-D Set-2 PDF Download Top 15 RRB Group-D Simplification Set-2 Questions and Answers PDF. RRB Group-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam. Download RRB Group-D Previous Papers PDF Take a RRB Group-D free mock test Question 1:\u00a0What is the area […]<\/p>\n","protected":false},"author":41,"featured_media":29128,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1679],"tags":[489,491,1819,1647],"class_list":{"0":"post-29124","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-group-d","11":"tag-railway-exam","12":"tag-rrb","13":"tag-rrb-group-d-exam","14":"tag-rrb-mocks"},"better_featured_image":{"id":29128,"alt_text":"Simplification Questions for RRB Group-D Set-2 PDF","caption":"Simplification Questions for RRB Group-D Set-2 PDF","description":"Simplification Questions for RRB Group-D Set-2 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