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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