**Quadratic Equation Questions for RRB NTPC set-2 PDF**

Download RRB NTPC Quadratic Equation Questions set-2 PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Download Quadratic Equation Questions for RRB NTPC set-2 PDF

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**Question 1: **Sum of the roots of a quadratic equation Q1 is -1 and product is -156 then what is the quadratic equation having roots that are 2 less than the roots of the quadratic equation Q1 ?

a) $x^{2}+10x-150$=0

b) $x^{2}-5x-150$=0

c) $x^{2}+5x-200$=0

d) $x^{2}-5x-200$=0

e) $x^{2}+5x-150$=0

**Question 2: **In each question two equations numbered (I) and (II) are given. Student should solve both the equations and solve the question.

I. x$^{2}$ – 25x – 54 = 0

II. y$^{2}$ + 104 = 30y

a) x $\leq$ y

b) x = y or relation cannot be established between x and y

c) x $\geq$ y

d) x > y

e) x < y

**Question 3: **In the following questions two equations numbered I and II are given. Solve both the equations and choose the correct answer.

I. $X^{2} – 10X + 21 = 0$

II.$Y^{2} + 6Y – 27= 0$

a) X < Y

b) X $\geq$ Y

c) X $\leq$ Y

d) X > Y

e) X = Y or the relationship cannot be determined.

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**Question 4: **In the following questions two equations numbered I and II are given. Solve both the equations and choose the correct answer.

I. $7X^{2} – 2X – 9 = 0$

II.$Y^{2} – Y – 72= 0$

a) X < Y

b) X $\geq$ Y

c) X $\leq$ Y

d) X > Y

e) X = Y or the relationship cannot be determined.

**Question 5: **In the following questions two equations numbered I and II are given. Solve both the equations and choose the correct answer.

I. $2X^{2} – 4X – 16 = 0$

II.$Y^{2} + 8Y + 15 = 0$

a) X < Y

b) X $\geq$ Y

c) X $\leq$ Y

d) X > Y

e) X=Y or the relationship cannot be determined.

**Question 6: **I. $x^{2}+3x-28=0$

II. $y^{2} -y-20=0$

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

**Question 7: **$2x^2 + 3x + 1 = 0$

$4y^2 + 16y + 15 = 0$

a) $x > y$

b) $x < y$

c) $x \geq y$

d) $x \leq y$

e) $x = y$ or no relation

**Question 8: **Two quadratic equations are given below. Solve these equations and select the appropriate option.$x^{2} – 2x – 63 = 0$

$y^{2} – 8y + 15= 0$

a) If x is greater than or equal to y

b) If x is greater than y

c) If x is less than or equal to y

d) If x is less than y

e) If x = y or no relation can be established between x and y

**Question 9: **Two quadratic equations are given below. Solve these equations and select the appropriate option$x^{2} – 15x +26 = 0$

$y^{2} + 9y – 22= 0$

a) If x is greater than or equal to y

b) If x is greater than y

c) If x is less than or equal to y

d) If x is less than y

e) If x = y or no relation can be established between x and y

**Question 10: **Two quadratic equations are given below. Solve these equations and select the appropriate option.$x^{2} + 8x – 48 = 0$

$y^{2} + 9y – 10= 0$

a) If x is greater than or equal to y

b) If x is greater than y

c) If x is less than or equal to y

d) If x is less than y

e) If x = y or no relation can be established between x and y

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**Answers & Solutions:**

**1) Answer (E)**

As given sum of the roots=-1

Product of the roots=-156

So the quadratic equation Q1 is $x^{2}+x-156$=0

$x^{2}-12x+13x-156$=0

x(x-12)+13(x-12)=0

(x+13)(x-12)=0

x=-13 and x=12

So the required new roots are -15 and 10

Therefore required quadratic equation is (x+15)(x-10)=0

$x^{2}+5x-150$=0

**2) Answer (B)**

I. x$^{2}$ – 25x – 54 = 0

x$^{2}$ – 27x + 2x – 54 = 0

x(x – 27) + 2(x – 27) = 0

x = -2 or x = 27

II. y$^{2}$ + 104 = 30y

y$^{2}$ -30y + 104 = 0

y$^{2}$ – 26y – 4y + 104 = 0

y(y-26) – 4(y-26) = 0

y=26 or y=4

The relation between x and y cannot be established as one value of x is greater than and other value of x is less than y.

Hence, option B is the correct answer.

**3) Answer (B)**

On solving the equation I,

(X – 3)(X – 7) = 0,

X = 3, X = 7.

On solving the equation II,

(Y + 9)(Y – 3) = 0,

Y = -9, Y = 3.

One value of X=3 is equals to Y while the other value is higher than both the values of Y hence we can say that X $\geq$ Y.

**4) Answer (E)**

On solving the equation I,

$7X^{2} – 2X – 9 = 0$

$7X^{2} – 9X + 7X – 9 = 0$

(X + 1)(7X – 9) = 0,

X = -1, X = 9/7.

On solving the equation II,

(Y – 9)(Y + 8) = 0,

Y = 9, Y = -8.

Since X = -1 is greater than Y = -8 while at the same time X = -1 is smaller than Y = 9. Hence we can say that no relationship exists.

**5) Answer (D)**

On solving the equation I,

$2X^{2} – 4X – 16 = 0$

$X^{2} – 2X – 8 = 0$

(X – 4)(X + 2) = 0,

X = 4, X = -2.

On solving the equation II,

(Y + 5)(Y + 3) = 0,

Y = -5, Y = -3.

For both the roots X > Y. Hence answer d.

**6) Answer (E)**

I.$x^{2} + 3x – 28 = 0$

=> $x^2 + 7x – 4x – 28 = 0$

=> $x (x + 7) – 4 (x + 7) = 0$

=> $(x – 4) (x + 7) = 0$

=> $x = 4 , -7$

II.$y^{2} – y – 20 = 0$

=> $y^2 – 5y + 4y – 20 = 0$

=> $y (y – 5) + 4 (y – 5) = 0$

=> $(y + 4) (y – 5) = 0$

=> $y = -4 , 5$

$\therefore$ No relation can be established.

**7) Answer (A)**

$2x^2 + 3x + 1 = 0$

(2x + 1) (x + 1) = 0

x = -0.5 or -1

$4y^2 + 16y + 15 = 0$

$(2y + 3) (2y + 5) = 0$

y = -1.5 or -2.5

Hence, x > y.

**8) Answer (E)**

The first equation can be factorized as

(x + 7)(x- 9) = 0

So the roots of the first equation are x = -7 and x = 9

The second equation can be factorized as

(y- 5)(y-3) = 0

Thus, the roots of the second equation are y = 5 and y = 3

Thus, the possible combinations of x,y are

(-7,5), (-7,3), (9,5) and (9,3)

So we can see that no relation can be established between x and y.

**9) Answer (A)**

$x^{2} – 15x +26 = 0$

$(x-2)(x-13)=0$

X = 2 or 13

$y^{2} + 9y – 22= 0$

$(y+11)(y-2) = 0$

Y= 2 or -11

Therefore, x is greater than or equal to y. option A.

**10) Answer (E)**

The first equation can be factorized as

(x + 12)(x- 4) = 0

So the roots of the first equation are x = -12 and x = 4

The second equation can be factorized as

(y+10)(y-1) = 0

Thus, the roots of the second equation are y = 10 and y = 1

Thus, the possible combinations of x,y are

(-12,-10), (-12,1), (4,-10) and (4,1)

So we can see that no relation can be established between x and y.

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