# Quadratic Equation Questions for RRB NTPC set-2 PDF

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## 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.

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.

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

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

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.

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.

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.

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.

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.

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

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.

$x^{2} – 15x +26 = 0$
$(x-2)(x-13)=0$
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.