Race IBPS Clerk 2015- Practice Aptitude Questions (Inequality) with Solutions Set-38

Dear Readers, Important Practice Aptitude Questions for Upcoming IBPS Clerk V Exam was given here with solutions. Aspirants those who are preparing for the examination can use this.
Directions ( 1-5 ): In the given questions, two equations numbered Ⅰ and Ⅱ are given. Solve both the equations and mark the appropriate answer.
 
1).Ⅰ. 6x+5y=25          Ⅱ.8x-7y=6
2).Ⅰ.12x²+17x-40=0  Ⅱ.12y²-5y-25=0
3).Ⅰ.5x²+4√10x+8=0 Ⅱ.y²-22y+121=0
4).Ⅰ.x²+x-30=0          Ⅱ.y²-12y+35=0
5).Ⅰ.x²-20x+99=0      Ⅱ.2x²-23y+56=0

 

Directions (Q.6-10): In the given questions two equations numbered Ⅰ and Ⅱ are given. You have to solve both the equations and mark the appropriate answer.
 
 
 
 
 
 
 
 
6).Ⅰ. 3x²+6x-9=0         Ⅱ.2y²-7y+3=0
7).Ⅰ.3x2+14x +15=0   Ⅱ.4y2+16y+15=0
8).Ⅰ.2x2+17x +36=0    Ⅱ.2y2+11y+14=0
9).Ⅰ.4x2-7x-2=0           Ⅱ.6y2+11x+4=0
10).Ⅰ.5x2+16x +12=0   Ⅱ.3y²+32y+45=0
Answers:
1). a) 2). e) 3). c) 4). d) 5).a ) 6).c ) 7). c) 8). b) 9).a ) 10). c)
 
Solutions:
1).Ⅰ. 6x+5y=25 …(ⅰ)
Ⅱ. 8x-7y=6 …(ⅱ)
Solving (ⅰ) and (ⅱ), we get
          6x+5y=25 …(ⅰ)×7
         8x-7y =6  …(ⅱ)×5
         42x+35y=175
      40x-35y=30    
82x=205
∴ x=205/80=2.5
∴ y= (25-15)/5=2
Hence x>y
Answer: (a)
2).Ⅰ. 12x²+32x-15x-40
Or, (3x+8) (4x-5) =0
Or, x=-8/3, 5/4
Ⅱ. 12y²+15y-20y-25=0
Or, (3y-5) (4y+5) =0
Or, y=5/3, -5/4
Relationship between x and y can’t be determined.
Answer: (e)
3).Ⅰ. 5x² + 2 × √5x  × √8 + 8=0
Or, (√5x+√8)²=0
Or, x=-√8/√5
Ⅱ. y²-22y+121=0
Or, (y-11)² =0
Or, y=11
Hence x
Answer: (c)
4).Ⅰ. x²+6x-5x-30=0
Or, (x-5) (x+6) =0
Or, x=5, -6
Ⅱ. y² -7y-5y+35=0
Or, (y-5) (y-7) =0
Or, y=5, 7
Hence x<= y
Answer: (d)
5).Ⅰ. x²-11x-9x+99=0
Or, (x-9) (x-11) =0
∴ x=9, 11
Ⅱ. 2y² -16y-7y+56=0
Or, (y-8) (2y-7) =0
∴ y=8, 7/2
Hence x>y
Answer: (a)
6).Ⅰ. 3x²+6x-9=0
Or, 3x²+9x-3x-9=0
Or, 3x(x+3) -3(x+3) =0
Or, (3x-3) (x+3) =0
∴ x=+1, -3
Ⅱ. 2y²-7y+3=0
Or, 2y²-6y-y+3=0
Or, 2y(y-3) -1(y-3) =0
Or, (2y-1) (y-3) =0
∴y=1/2, 3
Answer: (c)
7). Ⅰ. 3x²+14x+15=0
Or, 3x²+9x+5x+15=0
Or, 3x(x+3) +5(x+3) =0
Or, (3x+5) (x+3) =0
∴ x=-5/3, -3
Ⅱ. 4y²+16y+15=0
Or, 4y²+10y+6y+15=0
Or, 2y (2y+5) +3y (2y+15) =0
Or, (2y+3) (2y+5) =0
∴y=-3/2, -5/2
Hence relation can’t be established
Answer: (c)
8). Ⅰ. 2x²+17x+36=0
Or, 2x²+9x+8x+36=0
Or, 2x(x+4) +9(x+4) =0
Or, (2x+9) (x+4) =0
∴ x=-9/2=-4.5, -4
Ⅱ. 2y²+11y+14=0
Or, 2y²+7y+4y+14=0
Or, 2y(y+2) +7(y+2) =0
Or, (2y+7) (y+2) =0
∴y=-7/2, -2
Hence x
Answer: (b)
9). Ⅰ. 4x²-7x-2=0
Or, 4x²-8x+x-2=0
Or, 4x(x-2) +1(x-2) =0
Or, (4x+1) (x-2) =0
∴ x=-1/4, 2
Ⅱ.6y²+11y+4=0
Or, 6y²+8y+3y+4=0
Or, 3y (2y+1) +4(2y+1) =0
Or, (3y+4) (2y+1) =0
∴y=-4/3, -1/2
Hence x>y
Answer: (a)
10).Ⅰ. 5x²+16x+12=0
Or, 5x²+10x+6x+12=0
Or, 5x(x+2) +6(x+2) =0
Or, (5x+6) (x+2) =0
∴ x=-6/5, -2
Ⅱ. 3y²+32y+45=0
Or, 3y²+27y+5y+45=0
Or, 3y(y+9) +5(y+9) =0
∴y=-9, -5/3
Hence no relation can be established.
Answer: (c)

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