Practice Quantitative Aptitude Questions For Upcoming Exams (Quadratic Equation)

Practice Quantitative Aptitude Questions For Upcoming Exams (Quadratic Equation)ย  Set-100:

Dear Readers, Important Practice Aptitude Questions forย Upcoming Exams was given here with Solutions. Aspirants those who areย preparingย for the Bank Examination and other Competitive Examination canย use this.

Directions (Q. 1-5):In these questions, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. Give answer
a)ย ย ย if x>y
b)ย ย ย if xโค y
c)ย ย ย if x<y
d)ย ย ย if xโฅy
e)ย ย ย if x = y or relationship between x and y can’t be established.
1). I.5x2โ 34x + 45 = 0
II.4y2โ 19y + 21 = 0
ย
2).ย  I.7x2 – 89x + 60 = 0
ย II.6y2 – 86y + 168 = 0
ย
3). I.2x2+ 33x + 108 โ0
II.5y2+ 56y + 99 = 0
ย
4). I.56x2– 81x + 28=0
II.4y2– 37y + 63 =0
ย
5). I.3x2– 29x + 70 =0
II.3y2– 23y + 40 = 0
ย
Directions (Q. 6-10):In each question two ego numbered I and II are given. You have to solve both equations and mark the appropriate answer.
a)ย ย ย if x>y
b)ย ย ย if xโฅy
c)ย ย ย if x<y
d)ย ย ย if xโคy
e)ย ย ย if x = y or no relation can be established between x and y.
6). I.9x2– 41x+46=0
II.12y2 + 43y+ 38 = 0
ย
7). I.6x2 + 13x – 169 =0
II.y2 + 8y – 65 = 0
ย
8). I.3x + 5y = 4
II.6x – 7y= 25
ย
9). I.x2 – 5x + 4 = 0
II.y2 + 11y โ 12 = 0
ย
10). I.8x2 + 50x + 57 = 0
II.6y2โ y โ 57 = 0
1) eย  2) eย ย 3) eย ย 4) cย ย  5) eย ย 6) aย ย 7) eย ย  8) aย ย ย  9) bย ย 10) e
Solution:ย ย ย ย ย ย
1) . I. 5x2 โ 34x + 45 = 0ย
orย  5x – 25x โ 9x + 45 = 0
or, 5x (x – 5) –ย 9 (x – 5)
or (x – 5) (5x – 9) = 0
therefore, x = 5, x = 9/5
II. 4y2 – 19y + 21 = 0
or,ย 4y2 – 12y – 7y + 21 =0
or, 4y(y – 3) – 7(y – 3) = 0
ย  or, (y – 3) (4y – 7) = 0
therefore, y = 3, y = 7/4
Hence no relationship can be established
ย
2).ย  I.7x2 –ย  89x + 60 = 0
or, 7x2 – 84x – 5x + 60 = 0
ย  or, 7x(x – 12) – 5(x – 12) = 0
or, (x – 12)ย (7x โ 5) = 0
therefore, x = 12, 5/7
II.6y2 – 86y + 168 = 0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย
or, 6y2 – 72x — 14y + 4 + 168 = 0
or, 6y(y – 12) – 14(y – 12) = 0
therefore, y = 12, 14/6
Hence no relationship can be established
ย
3).ย I.2x2+ 33x + 108 โ0ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย
or, 2x2 + 24x + 9x + 108 = 0
or, 2x(x + 12) + 9(x + 12) =0
or, (x + 12) (2x + 9) = 0
therefore, x = -12, -9/2
ย II. 5y2+ 56y + 99 = 0
or, 5y2 + 45y + 11y + 99 = 0
or, 5y(y + 9) + 11(y + 9) = 0 or,(y + 9) (5y + 11) = 0
therefore, y = -9, -11/5
Hence relations canโt be established
ย
4).ย I. 56x2– 81x + 28=0
or, 56x2 – 32x โ 49x + 28 = 0
or, 8x(7x – 4) – 7(7xย  – 4) = 0 or, (7x – 4) (8x – 7) = 0
x = 4/7, 7/8
4y2– 37y + 63 = 0
or, 4y2โ 28y โ 9y + 63 = 0
ย or, 4y(y – 7) – 9(y – 7) = 0
or, (y 7) (4y – 9)ย = 0
y = 7, 9/4
Hence x< y
ย
ย 5). I. 3×2 – 29x + 70 = 0
or, 3x2 – 15x – 14x + 70 = 0
or, 3x(x – 5) – 14(x – 5) = 0
or, (x – 5) (3x 14)=0
x = 5, 14/3
ย II.3y2– 23y + 40=0
or, 3y2 – 15y – 8y + 40 = 0
or, 3y(y – 5) – 8(y – 5) = 0
y = 5, 8/3
Hence no relationship can he established
ย
6). I 9×2 – 41x + 46 = 0
or, 9x2 – 18x – 23x + 46 = 0
or, 9x(x โ 2) – 23(x –ย 2) = 0
or, (9x – 23) (x – 2) =0
x = 23/9, 2
ย II. 12y2–ย 43y + 38 =0ย ย ย
or, 12y2+ 24y + 19y + 38 = 0
or, 12y(y + 2) + 19(y + 2) = 0
or, (12y + 19) (y+ 2) = 0
y = 19/12, -2
ย Therefore x> y
ย
7).ย I.6x2+ I3x – 169 = 0
or, 6x2 – 26x + 39x – 169 =0
or, 2x(3x – 13) + 13(3x – 13) = 0
or, (2x + 13) (3x – 13) = 0
x = -13/2, 13/3
ย II. y2 + 8y – 65 = 0
or, y2 + 13y – 5y – 65 = 0
or, y(y + 13) – 5(y + 13) = 0
or, (y – 5) (y + 13) = 0.
y = 5, -13
ย Therefore relation can’t be established between x and y
ย
8). I. 3x + 5y = 4
ย II.ย ย  6x โ 7y = 25
Solving equation (I) and (ii), we get
ย ย ย ย ย ย  6x + 10y = 8
ย ย ย ย ย ย  6x โย  7yย  = 25
ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  17y = -17
y = -1 and x =3,ย  Therefore x> y
ย
9). I. x2 – 5x + 4 = 0ย ย ย ย ย ย
or, x2 โ 4x โ x + 4 = 0
or, x(x – 4) โ 1(x -4) = 0
or, (x -1) (x – 4) = 0
x = 1, 4
Il.y2+ 11y – 12= 0
or, y2+ 12y โ y -12 = 0
or, y(y + 12) โ 1(y + 12) = 0 or, (y – 1) (y + 12) = 0
y = 1, -12
Therefore, x โฅ y
ย
10). Iย  8x2+ 50x + 57 = 0
or, 8x2 + 12x + 38x +57=0
or, 4x(2x + 3) + 19(2x + 3) = 0
or, (4x + 19) (2x + 3) = 0
ย ย  x = -19/4, -3/2
ย II.6y2โ y – 57=0ย
or, 6y2 + 18y – 19y – 57 = 0
or, 6y(y + 3) – 19(y + 3) = 0
or, (6y – 19) (y + 3) = 0
y= 19/6, -3
Therefore relation can’t be established.