SBI Clerk Mains 2016- Practice Aptitude Questions (Quadratic Equation)

SBI Clerkย Mainsย Exam 2016- Practice Aptitude Questions (Quadratic Equation) Set-85:

Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

Directions (Q. 1-10): In the following questions, two Equations I and II are given. You have to solve both the equation.
a)ย ย If x> y
b)ย ย If x โฅ y
c)ย ย If x< y
d)ย ย If x โค y
e)ย ย If x = y or the relationship cannot be established
ย
1).I. 5x2 โ 18x + 9 = 0
II. 20y2 โ 13y + 2 = 0
ย
2).I. x3 โ 878 = 453
II. y2 โ 82 = 39
ย
3).I. 3/โx + 4/โx = โx
II. y3 โ (7)7/2 /โy = 0
ย
4).I. 9x โ 15.45 = 54.55 + 4x
II. โ(y + 155) โ โ36 = โ49
ย
5).I. x2 + 11x + 30 = 0
II. y2 + 7y + 12 = 0
ย
6).I. x2 โ 19x + 84 = 0
II. y2 โ 25y + 156 = 0
ย
7).I. x3 โ 468 = 1729
II. y2 โ 1733 + 1564 = 0
ย
8).I. 9/โx + 19/โx = โx
II. y5 โ (2 ร 14)11/2 /โy = 0
ย
9).I. โ(784)x + 1234 = 1486
II. โ(1089)y + 2081 = 2345
ย
10).I. 12/โx – 23/โx = 5โx
II. โy/12 – 5โy/12 = 1/โy
1).a)ย ย  2).b)ย ย  3).e)ย ย  4).e)ย ย  5).c)ย ย  6).d)ย ย  7).b)ย ย  8).e)ย ย  9).a)ย ย  10).a)ย
ย
Solution:
ย
1).I. 5x2 โ 18x + 9 = 0
5x2โ 15x โ 3x + 9 = 0
5x (x -3) โ 3(x – 3) = 0
(x – 3) (5x โ 3) = 0
x = 3 or 3/5
II. 20y2 โ 13y + 2 = 0
20y2โ 8y โ 5y + 2 = 0
4y(5y – 2) -1(5y – 2) = 0
(4y – 1) (5y – 2) = 0
ย y = 1/4 or 2/5
Clearly x> y
ย
2).I. x3 โ 878 = 453
x =3โ1331 = 11ร x = 11
II. y2 โ 82 = 39
y2= 82 + 39 = 121
y = ยฑ11
x โฅ y
ย
3).I. 3/โx + 4/โx = โx
3 +4 = x
x = 7
II. y3 โ (7)7/2 /โy = 0
y3+1/2โ (7)7/2 = 0
y7/2= 77/2ร y = 7
Clearly, x = y
ย
4).I. 9x โ 15.45 = 54.55 + 4x
9x – 4x = 70
5x = 70ร  x = 14
II. โ(y + 155) โ โ36 = โ49
โ(y + 155) = 6 + 7ร โ(y + 155) = 13
y + 155 = 169
y = 169 โ 155 = 14
Clearly, x = y
ย
5).I. x2 + 11x + 30 = 0
x2+ 6x + 5x + 30 = 0
x(x +6) + 5(x +ย 6) = 0
(x + 5) (x +ย 6) = 0
x = -5 (or) -6
II. y2 + 7y + 12 = 0
y2+ 4y + 3y + 12 = 0
y(y+4) + 3(y+4) = 0
(y+3) (y+4) = 0
y = -3 (or) -4
x<y
ย
6).I. x2 โ 19x + 84 = 0
x2โ 12x โ 7x + 84 = 0
x(x – 12) -7(x – 12) = 0
x = 7 (or) 12
II. y2 โ 25y + 156 = 0
y2โ 12y โ 13y + 156 = 0
y(y – 12) โ 13(y – 12) = 0
(y – 12) (y – 13) = 0
Y = 12 (or) 13
Clearly, x โค y
ย
7).I. x3 โ 468 = 1729
x3= 1729 + 468 = 2197
x =3โ2197 = 13
II. y2 โ 1733 + 1564 = 0
y2โ 169 = 0
y = ยฑ13
x โฅ y
ย
8).I. 9/โx + 19/โx = โx
(9 + 19) / โx = โx
28 = โx รโxร x = 28
II. y5 โ (2 ร 14)11/2 /โy = 0
[y5โy โ (28)11/2] / โy = 0ร y11/2 โ (28)11/2 = 0
y = 28
Clearly, x = y
ย
9).I. โ(784)x + 1234 = 1486
โ(784)x = 1486 โ 1234ร โ(784)x = 252
x = 252 / 28ร x = 9
II. โ(1089)y + 2081 = 2345
โ(1089)y = 2345 โ 2081ร โ(1089)y = 264
y = 264 / 33ร y = 8
x> y
ย
10).I. 12/โx – 23/โx = 5โx
(12 – 23) / โx = 5xร ย -11
x = -11/5
II. โy/12 – 5โy/12 = 1/โy
(โy-5โy) / 12 = 1/โyร -4โy รโy = 12
-4y = 12ร  y = -3
Clearly, x>y