**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.

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**Question 1: **What is the area between the lines 5x+4y=20 and x+y=15 in the first quadrant?

a) 117.50 sq units

b) 110 sq units

c) 105 sq units

d) 102.50 sq units

**Question 2: **If x and y are positive real numbers such that 15x + 12xy +20y = 96, find the minimum possible value of 6x + 8y ?

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

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**Question 4: **Find $x +y$ if $\frac{90}{3x – y} + \frac{46}{2x + 3y} = 7$ and $\frac{92}{2x + 3y} – \frac{36}{3x – y} = 2$.

**Question 5: **Solve the given system of equations in three variables:

x + 3y -z = 5

3x + y + 4z = 22

2x +6y -2z = 56

a) x = 35/34, y = 91/34, z = 69/17

b) x = -123/25, y = 147/25, z = 193/25

c) infinite solutions

d) no solution

**Question 6: **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?

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**Question 7: **If $(x-5)^{\frac{x^2 – 24x + 95}{x^2 – 13x + 36}} = 1$

What is the sum of all real values of ‘x’ which satisfy the given equation?

a) 21

b) 25

c) 29

d) 30

**Question 8: **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

a) 25, 43

b) 52, 43

c) 52, 67

d) None of the above

**Question 9: **Find the number of integral solutions of $\sqrt{x + 15 – 8\sqrt{x – 1}} + \sqrt{x + 24 – 10\sqrt{x – 1}} = 1$

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

a) 702

b) 970

c) 720

d) None of these

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**Question 11: **If the roots of $x^3 +3 x^2 + px – q = 0$ are in A.P, then which of the following is possible?

a) p = 5, q = -5

b) p = 4, q = 0

c) p = 1, q = -2

d) p = 2, q = 0

e) p = 0, q = -3

**Question 12: **If x =$\sqrt{2}-1$, find the value of $x^4+4x^3+6x^2+4x+8$.

a) 9

b) 10

c) 11

d) 12

**Question 13: **Solve the equation: $(7-x)^{4} + (9-x)^{4} = 16$

a) 9

b) 7

c) Both the above

d) None of the above

**Question 14: **Solve the equation: (x+2)(x+4)(x+6)(x+12)=-4$x^2$, $x^2$ = ?

a) $-10 \pm 2\sqrt {3}$

b) $-11 \pm 2\sqrt {3}$

c) $-12 \pm 2\sqrt {3}$

d) $-13 \pm 2\sqrt {3}$

**Question 15: **The maximum value of y = $ – x^2 + 6x- 8 $ is

a) 0

b) 1

c) 2

d) 3

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**Answers & Solutions:**

**1) Answer (D)**

Multiplying the second equation by 5 and subtracting from the first equation,

$5x+4y=20 $

$5x+5y=75$

Solving, $4y-5y = 20 – 75 \Rightarrow y=55$

Putting the value of $y$, $x = 15 – 55 = -40$

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.

For line 1 : $5x+4y=20$

Putting x=0, $y=\dfrac{20}{4}=5$

Putting y=0, $x=\dfrac{20}{5}=4$

Intercepts : (0,5) and (4,0)

For line 2 : $x+y=15$

Putting x=0, $y=15$ and putting y=0, $x=15$

Intercepts : (15,0) and (0,15)

Upon rough drawing, we get a graph something like this :

Point O is the origin while A,B and C,D are intercepts of line 2 and line 1 respectively.

The area of the between the two lines can be calculated by subtracting the area of $\triangle$DOC from the area of $\triangle$AOB

Thus, area of $\triangle$AOB $a_1 = \dfrac{1}{2} \times 15 \times 15$

$ \Rightarrow a_1 = \dfrac{225}{2} $ sq. units

Now, area of $\triangle$DOC $a_2 = \dfrac{1}{2} \times 5 \times 4$

$ \Rightarrow a_2 = 10$ sq units

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

**2) Answer: 24**

Given that: 15x + 12xy +20y = 96

$\Rightarrow$ 15x + 12xy +20y +25 = 96 +25

$\Rightarrow$ 5(3x + 5) + 4y(3x + 5) = 121

$\Rightarrow$ (3x + 5)(4y + 5) = 121

If thee product of two positive numbers (3x + 5) and (4y + 5) is constant, then the minimum value of sum occurs when the number are equal i.e. 3x + 5 = 4y + 5 = 11

$\therefore$ The minimum value of (3x + 5) + (4y + 5) = 11 + 11

Hence, the minimum value of 3x + 4y = 22 – 10 = 12, hence minimum value of 6x + 8y = 2*12 = 24 (Answer)

**3) Answer: 3**

$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$

=> $ab = 9(a + b)$

=> $ab – 9(a+b) = 0$

=> $ ab – 9(a+b) + 81 = 81$

=> $(a – 9)(b – 9) = 81, a > b$

Hence we have the following cases,

$ a – 9 = 81, b – 9 = 1$ => $(a,b) = (90,10)$

$ a – 9 = 27, b – 9 = 3$ => $(a,b) = (36,12)$

$ a – 9 = 9, b – 9 = 9$ => $(a,b) = (18,18)$

Hence there are three possible positive integral values of (a,b)

**4) Answer: 10**

Let

$\frac{1}{3x – y} = a$ and $\frac{1}{2x + 3y} = b$

Therefore,

$92b – 36a = 2$ and

$46b + 90a = 7$

Multiplying ii by 2 and subtracting i from it,

$216a = 12$

$a = \frac{1}{18}$

Substituting in ii,

$b = \frac{1}{23}$

Therefore,

$3x – y = 18$ and $2x + 3y = 23$

Multiplying the first equation by 3 and adding both the equations.

$11x = 77$, $ x = 7$

Hence, $y = 3$

$x + y = 10$

**5) Answer (D)**

The coefficient matrix of the given system of equations = M =

$\begin{bmatrix}

1 &3 &-1 \\

3&1 &4 \\

2&6 &-2

\end{bmatrix}$

|M| = D = 1(1x-2-6×4)-3(3x-2-6x-1)+2(3×4-1x-1) = 0

The x numerator matrix of the given system of equations = M$_x$ =

$\begin{bmatrix}

5 & 3 &-1 \\

22&1 &4 \\

56&6 &-2 \end{bmatrix}$

|M$_x$| = D$_x$ = 5(1x-2-6×4)-22(3x-2-6x-1)+56(3×4-1x-1) = 598

Since, at-least one of the numerator matrix is non-zero, the given set of equations have no solution

**6) Answer: 1**

Since the roots are reciprocal to each other , product of roots must be equal to 1.

Writing the given equation as $px^2 + p^2x + p^2-4p$ = 0

So, $\frac{p^2 – 4p}{p}$ = 1

or, $p^2 – 5p$ = 0

Solving this equation, we get roots as $p$ = 0 and $p$ = 5

But, $p$ cannot be 0 because then the given equation would not be quadratic. Hence, only one value of $p$ is possible

**7) Answer (B)**

We have $(x-5)^{\frac{x^2 – 24x + 95}{x^2 – 13x + 36}} = 1$

$ x^2 – 24x + 95 = (x-5)(x-19) $

$ x^2 – 13x + 36 = (x-4)(x-9) $

So, $(x-5)^{ \frac{(x-5)(x-19)}{(x-4)(x-9)} } = 1$

Case 1 –

Consider following possibility,

x – 5 = 1 i.e. x = 6

So $\frac{(x-5)(x-19)}{(x-4)(x-9)} = \frac{1*-13}{2*-3}$

So x=6 satisfies

Case 2 –

Consider, x-5 = -1

x = 4

So $\frac{(x-5)(x-19)}{(x-4)(x-9)} = \frac{-1*-15}{0*-3} $

So, x=4 does not satisfy

Case 3 –

Consider $\frac{(x-5)(x-19)}{(x-4)(x-9)} = 0 $

$ (x-5)(x-19) = 0$

x = 5 or x = 19

For x = 5, the base becomes 0. So not possible.

For x = 19, the equation will be satisfied.

Thus, 2 values of x, i.e. 6 and 19 satisfy the given equation.

Sum = 6 + 19 = 25

Hence, option B is the right answer.

**8) Answer (C)**

If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$

Then x = -5 and x = 2 will give $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ = 0

Substituting x = -5 we get,

$3(-5)^{4}+2(-5)^{3}-a(-5)^{2}+b(-5)-a+b-4 = 0$

Solving we get,

$26a+4b = 1621$…….(i)

Substituting x = 2 we get,

$3(2)^{4}+2(2)^{3}-a(2)^{2}+b(2)-a+b-4 =0$

=> $5a-3b = 60$……..(ii)

Solving i and ii we get

a and b $\approx 52, 67$

Hence, option C is the correct answer.

**9) Answer: 10**

Let $\sqrt{x – 1} = t$

$x = t^2 + 1$

Substituting the value of t in the given equation,

$\sqrt{t^2 – 8t + 16} + \sqrt{t^2 – 10t + 25} = 1$

$\sqrt{(t – 4)^2} + \sqrt{(t – 5)^2} = 1$

|t – 4| +|t – 5| = 1

The following equation is true for $4 \leq t \leq 5$

The corresponding values of x are $ 17 \leq x \leq 26$

Hence, 10 integral values exist.

**10) Answer (A)**

$x^4 – 79x^2 + 1 = 0$

Dividing both sides by $x^2$

=> $x^2 -79 + \frac{1}{x^2} = 0$

=> $x^2 + \frac{1}{x^2} = 79$

Adding 2 on both sides.

$x^2 + \frac{1}{x^2} + 2 \frac{1}{x^2}x^2 = 81$

=> $(x+\frac{1}{x}) = \pm 9$

But since x>0, $(x+\frac{1}{x}) = 9$

Now, $x^3 + \frac{1}{x^3} = (x+\frac{1}{x})(x^2 + \frac{1}{x^2} – 1)$

=> $x^3 + \frac{1}{x^3} = 9 (79 – 1) = 702$

**11) Answer (D)**

Let the roots of the given equation be a -d, a and a + d

Sum of roots of the equation = a – d + a + a + d = 3a = -3

=> a = -1

Sum of product of roots taken two at a time = (a – d)a + a( a + d) + (a – d)(a – d) = $3a^2 – d^2$ = p

$3 – p = d^2$

=> 3 – p $\geq$ 0, Thus, p $\leq$ 3

Product of roots of the equation = $(a – d)a(a + d) = a(a^2 – d^2) = q$

$1 – d^2 = -q$

=> q $\geq$ -1

From the options only option D satisfies the given conditions.

**12) Answer (C)**

x =$\sqrt{2}-1$

x+1 =$\sqrt{2}$

$(x+1)^4 = 4$

$x^4+4x^3+6x^2+4x+1$ = 4

$x^4+4x^3+6x^2+4x+8$ = 4+7 = 11

**13) Answer (C)**

Let z = (7-x + 9 – x )/2 = 8-x

So, the equation becomes $ (z-1)^4 + (z+1)^4 = 16$

$z^4+6z^2-7=0$

= > $ (z^2+7)(z^2-1)=0$

Taking only real values, z = -1 or 1

X = 9 or 7

**14) Answer (C)**

Since (-2)(-12) = (-4)(-6), we can re-arrange the equation as (x+2)(x+12)(x+4)(x+6)

= > ($x^2+14x+24)(x^2+10x+24$)

Since x= 0 is not a root, dividing by $x^2$ on both sides,

We get (x + 14 + 24/x)(x+10+24/x) = -4

Putting x + 24/x = y, we get

(y + 14)(y + 10) = -4

$ y^2 + 24y+144=0$

Y = -12

X+24/x = -12

$x^2 +12x+24=0$

x = $\frac{ -12 \pm \sqrt {144-96}}{2}$

=$ -12 \pm \sqrt {12}$

=$-12 \pm 2\sqrt {3}$

**15) Answer (B)**

$ – x^2 + 6x- 8 $ can be rewritten as – ($ x^2 – 6x + 8 $)

= – (x-2)(x-4)

The 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

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