## Algebra For SBI PO : Set – 33

D.1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give 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.

1) I. 63x-94√x+35

II. 32y-52√y+21=0

(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.

(e)

63x – 94+ 35 = 0

Or, 63x, – 49 – 45 + 35 = 0

Or, (9 – 7) (7 – 5) = 0

x = 49/81, 25/45

32y – 52√y + 21 = 0

Or 32y – 28√y – 24√y + 21 = 0

Or,  (4√y – 3). (8√y -7) = 0

y = 9/16, 49/64

Therefore relation can ’t be established between x
and y.

2. I. x2-7√3x- 35 √15=5√5 x

II. y2-5√5 y+30=0

(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.

(a)

x2 -7√3x–35 √15 =5√5x

Or x2 -5√5x–7√3x – 35√15 = 0

Or, (x -7√3) (x- 5√5) = 0

= 7√3, 5√5

y2 – 5√5y + 30 = 0

Or, y2 – 5√5y + 30 = 0

Or, (y -3√5), (y-2√5) = 0

y = 3√5, 2√5

3) I. 14x2+ 11x – 15 = 0

II. 20y2– 31y + 12 = 0

(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.

(c)

14x2 + 11x – 15 = 0

or (7x – 5) (2x + 3) = 0

x =5/7, – 3/2

II. 20y2 – 31y + 12 = 0

or (4y – 3), (5y – 4) = 0

y = 3/4, 4/5

x < y

4. I. √25 x+ √16 y=41

II. √16 x- √25 y=40

(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.

a)

5x + 4y = 41 … (i)

4x + 5y = 40 … (ii)

On solving both equations, we have x = 5 and y = 4

x > y

5. I. √x-  (18)(15/2)/x2 =

II. √y=(19)(9/2)/y=0

(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.

(c)

√x- 〖(18] (15/2)/y = 0

or x (5/2) = (18)(5/2)

x = 183

II √y- 〖(19)](9/2)/y = 0

Or y(3/2) = 〖19 ](9/2)

Y = 193

x < y

D.6-10): In each of these questions, two equations (I) and (II) are given. Solve both the equations and give 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. 63x – 194√x+143=0

II. 99 y-255 √y+150=0

(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’.

(e)

63x – 194√x + 143 = 0

Or 63x – 117 √x – 77√x + 143 = 0

Or (7√x – 13) (9 √x – 11) = 0

x = 169/49, 121/81

99y – 225√y + 150 = 0

Or (11√y -10) (9 √y – 15) = 0

y = 100/121, 225/81

Therefore relation cannot be established between x and y

7) I. 16x2 – 40x – 39 = 0

II. 12y2 – 113y + 255 = 0

(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’.

(b)

16x2 – 40x – 3 = 0

or 16x2 – 52x + 12x – 39 = 0

or (4x-13) (4x+3)

x = 13/4, -3/4

12y2 – 113y + 255 = 0

Or 12y2 – 45y – 68y + 255 = 0

Or (4y-15) (3y-17) = 0

y = 15/4, 17/3

Therefore y > x

8) I. x-7√3 x +36=0

II. y-12√2y+70=0

(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’.

(b)

x – 7 √3x + 36 = 0

Or x – 7 √3x. √x + 36 = 0

Or x – 3√3.√x-4√3.√x + 36 = 0

Or (√x-3√3) (√x-4√3)

x = 27, 28

y – 5 √2y -7√2y + 70 = 0

Y – 5 √2.√y -7√2y + 70 = 0

Or (√y-5√2) (√y-7√2) = 0

x = 50, 28

9) I. x2-7√7x+84=0

II. y2-5√5y+30=0

(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’.

(a)

x2 -7√7x + 84 = 0

Or (x-4√7)(x – 3√7) = 0

y = 4√7, 3√7

y2 – 5√5y + 30 = 0

Or (y2-2√5) (y-3√5) = 0

Y = 2√5, 3√5

x > y

10) I. 10x – 6y = 13

II. 45x + 24y = 56

(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’.