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Home / Quantitative Aptitude QUiz: Quadratic Equations For Upcoming Exams

# Quantitative Aptitude QUiz: Quadratic Equations For Upcoming Exams

Dear Aspirants,

One of the most expected questions for IBPS Clerk Mains & South India Bank Clerk examination.  Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this.

Directions (Q1-Q5): 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)if x>y

b)if x≤ y

c)if x<y

d)if x≥y

e)if x = y or relationship between x and y can’t be established.

1). I.5x2 — 34x + 45 = 0

II. 4y2— 19y + 21 = 0

2).  I.7x2 – 89x + 60 = 0

II.6y2 – 86y + 168 = 0

3). I.2x+ 33x + 108 —0

II. 5y2+ 56y + 99 = 0

4). I.56x2– 81x + 28=0

II.4y2– 37y + 63 =0

5). I. 3x2– 29x + 70 =0

II.3y2 – 23y + 40 = 0

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)if x >y

b)if x≥y

c)if x<y

d)if x≤y

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

6). I. 9x– 41x+46=0

II. 12y2+ 43y+ 38 = 0

7). I. 6x2 + 13x – 169 =0

II. y2+ 8y – 65 = 0

8). I. 3x + 5y = 4

II. 6x – 7y= 25

9). I. x2 – 5x + 4 = 0

II. y2+ 11y – 12 = 0

10). I. 8x2 + 50x + 57 = 0

II. 6y2– y – 57 = 0

1) . I. 5x2 — 34x + 45 = 0

or  5x – 25x – 9x + 45 = 0

or, 5x (x – 5) –  9 (x – 5)

or (x – 5) (5x – 9) = 0

therefore, x = 5, x = 9/5

II. 4y2– 19y + 21 = 0

or,  4y2 – 12y – 7y + 21 =0

or, 4y(y – 3) – 7(y – 3) = 0

or, (y – 3) (4y – 7) = 0

therefore, y = 3, y = 7/4

Hence no relationship can be established

2).  I.7x2 –  89x + 60 = 0

or, 7x2 – 84x – 5x + 60 = 0

or, 7x(x – 12) – 5(x – 12) = 0

or, (x – 12)  (7x – 5) = 0

therefore, x = 12, 5/7

II.6y2 – 86y + 168 = 0

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

3).  I.2x+ 33x + 108 —0

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

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 can’t be established

4).  I. 56x– 81x + 28=0

or, 56x2 – 32x — 49x + 28 = 0

or, 8x(7x – 4) – 7(7x  – 4) = 0 or, (7x – 4) (8x – 7) = 0

x = 4/7, 7/8

II. 4y2 – 37y + 63 = 0

or, 4y2 – 28y – 9y + 63 = 0

or, 4y(y – 7) – 9(y – 7) = 0

or, (y 7) (4y – 9)  = 0

y = 7, 9/4

Hence x < y

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

II. 3y– 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

6). I 9×2 – 41x + 46 = 0

or, 9x2 – 18x – 23x + 46 = 0

or, 9x(x — 2) – 23(x –  2) = 0

or, (9x – 23) (x – 2) =0

x = 23/9, 2

II. 12y2–  43y + 38 =0

or, 12y+ 24y + 19y + 38 = 0

or, 12y(y + 2) + 19(y + 2) = 0

or, (12y + 19) (y+ 2) = 0

y = 19/12, -2

Therefore x > y

7).  I.6x+ 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

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

Therefore relation can’t be established between x and y

8). I. 3x + 5y = 4

II. 6x – 7y = 25

Solving equation (I) and (ii), we get

6x + 10y = 8

6x –  7y  = 25

17y = -17

y = -1 and x =3,  Therefore x > y

9). I. x2 – 5x + 4 = 0

or, x2 – 4x – x + 4 = 0

or, x(x – 4) – 1(x -4) = 0

or, (x -1) (x – 4) = 0

x = 1, 4

II. y+ 11y – 12= 0

or, y+ 12y – y -12 = 0

or, y(y + 12) – 1(y + 12) = 0 or, (y – 1) (y + 12) = 0

y = 1, -12

Therefore, x ≥ y

10). I  8x+ 50x + 57 = 0

or, 8x2 + 12x + 38x +57=0

or, 4x(2x + 3) + 19(2x + 3) = 0

or, (4x + 19) (2x + 3) = 0

x = -19/4, -3/2

II.6y– y – 57=0

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.