Race IBPS Clerk 2015- Practice Aptitude Questions (Inequality) with Explanations Set-34

Race IBPS Clerk 2015- Practice Aptitude Questions (Inequality) with Explanations Set-34:
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 (Q.1-5): In each of these equations, two equations โ  and โก are given. You have to solve both the equations and give answer

1).โ .6x+3y=7xy            โก.3x+9y=11xy

2).โ .5x+1+52-x=53+1     โก.yยฒ-2y-3=0

3).โ .8xยฒ-22x-21=0       โก.(4/y)-3= 5/ (2y+3)

4).โ .4โ3xยฒ+5x-2โ3=0  โก.yยฒ+5y+6=0

5).โ .10x-1/x=3             โก.30yยฒ+11y+1=0

Directions (Q.6-8): In each question below two equations are provided On the basis of these, you have to findout the relationbetween p and q. Given answer:

6).โ .pยฒ+7p=3p+21          โก.4qยฒ+15q+9=0

7).โ . โp- (โ7/โp) =0       โก.q- (8/โq) =0

8).โ .2pยฒ-15p+22=0         โก.3qยฒ+7q-6=0

1). a)  2).e ) 3). e)  4). b)  5). e)  6). b)   7). a)   8). d)

Solutions:

1).Given 6x+3y=7xy   …(โฐ)
On dividing both sides by x y, we get 6/y+3/x=7     โฆ(โฑ)
Again, 3x+9y=11xy     โฆ(โฒ)
On dividing both sides in equation (โฒ) by x y, we get
Or, 3/y+9/x=11 โฆ(โณ)
Now, let 1/x=u and 1/y=v
Then, 6v+3u=7      โฆ(โด)
3v+9u=11 โฆ(โต)
Solving (โต)& (โต), we get
โดu=1 and v=2/3
โด u=1/x=1 so, v=1/y=2/3
Or, x=1                โดy=3/2=1.5
Hence x< y

2). โ . 5x+1+52-x=53+1
Or, 5x*5+52*5-x=126
Or, 5xร5+ (25/5x) =126
Let 5x= u (say)
Then, 5u2-126u+25=0
Or, 5u2-125u-u+25=0
Or, 5u (u-25)-1 (u-25) =0
Or, (u-25) (5u-1) =0
Or, u=25; u=1/5
Now, u=5x        or, 5x=52              โดx=2
Again, u=1/5=5-1
โด 5x=5-1
โด x=-1
โก.y2-2y-3=0  or, y2-3y+y-3=0  or, y(y-3) + (y-3) =0  or, (y+1) (y-3)=0
y=-1, 3
Hence relation canโt be established between x and y.

3).โ .8xยฒ-22x-21=0 or, 8xยฒ-28x+6x-21=0 or, 4x (2x-7) +3(2x-7) =0 or, (2x-7) (4x+3) =0
Or, 2x-7=0
Again, 4x+3=0 or, x=7/2   x=-3/4
โก.(4/y)-3=5/ (2y+3) or, (4-3y)/y=5/ (2y+3) or, (4-3y) (2y+3)=5y or, 6yยฒ+6y-12=0
Or, yยฒ+y-2=0 or, yยฒ+2y-y-2=0 or, y(y+2)-1(y+2) =0 or, y=1, -2
โด Relation cannot be established between x and y.

4).โ . 4โ3 xยฒ+5x-2โ3=0
Here 4โ3* (-2 โ3) =-24 and 8*-3=-24 and 8+ (-3) =5
โด 4โ3 xยฒ+5x-2 โ3=0 or, 4โ3 xยฒ+8x-3x-2โ3=0 or, 4x (โ3x+2) -โ3 (โ3x+2) =0
Or, (4x-โ3) (โ3x+2) =0
Or, โ3x+2=0 And x=(โ3/4) or, x=(-2/โ3)
II. yยฒ+5y+6 or, yยฒ+3y+2y+6=0 or, y+(y+3)=0 or, y=-2, -3  โด x>y
5). I.10x-1/x=3
Or, 10 xยฒ-3x-1=0 or, 10 xยฒ-5x+2x-1=0 or, 5x(2x-1)+1(2x-1)=0
Or, x=(1/2,-1/5)
II.30yยฒ+11y+1=0 or, 30 yยฒ+6y+5y+1=0 or, 6y(5y+1)+1(5y+1)=0
or , (5y+1)(6y+1)=0  or, (y=-1/5,-1/6)
โด No relation can be established between x and y.

6).โ .P2+7p=3p+21
or, p2+7p-3p-21=0
or, p(p+7)-3 (p+7) =0
or, (p-3) (p+7) =0
โด p=3, -7.
โก. 4q2+15q+9=0
or, 4q2+12q+3q+9=0
or, 4q (q+3) +3(q+3) =0
or, (4q+3) (q+3) =0
โดq=-3/4, -3.
Hence no relation can be established between p and q.

7).โ .โp- (โ7/โp) =0
or, p=โ7
โก.qรโq=8 or, q3/2=8
or, q=82/3 =641/3 =4
Hence p< q

8).โ .2pยฒ-15p+22=0
or, 2pยฒ-4p-11p+22=0
or, 2p (p-2)-11(p-2) =0
or, (2p-11) (p-2) =0
โด p=2, 11/2
โก.3q2+7q-6=0
or, 3q2+9q-2q-6=0
or, 3q (q+3)-2(q+3) =0
โด q=-3, +2/3
Hence, p> q