Practice Quantitative Aptitude Questions For Upcoming Exams (Quadratic Equation)

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

Dear Readers, Important Practice Aptitude Questions for Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can 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)   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.2x2+ 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.9x2– 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
Answers:
1) e  2) e  3) e  4) c   5) e  6) a  7) e   8) a    9) b  10) e
Solution:      
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
Answer: e)
 
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
Answer: e)
 
3). I.2x2+ 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
Answer: e)
 
4). I. 56x2– 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
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
Answer: c)
 
 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.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
Answer: e)
 
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, 12y2+ 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
Answer: a)
 
7). 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
 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
Answer: e)
 
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
Answer: a)
 
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
Il.y2+ 11y – 12= 0
or, y2+ 12y – y -12 = 0
or, y(y + 12) – 1(y + 12) = 0 or, (y – 1) (y + 12) = 0
y = 1, -12
Therefore, x ≥ y
Answer: b)
 
10). I  8x2+ 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.6y2– 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.

 

Answer: e)
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