## Algebra For SBI PO : Set – 34

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. x2 – 2x -15 = 0

II. y2 + 5y + 6 = 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)

x2 – 2x – 15 = 0

or,x2 – 5x + 3x – 15 = 0

or, x(x – 5) + 3(x – 5) = 0

or,(x – 5) (x + 3) = 0

x = 5, -3

II. y 2 + 5y + 6 = 0

or, y 2 + 3y + 2y + 6 = 0

or, y(y + 3) + 2(y + 3) = 0

or,(y + 3)(y + 2) = 0

y = -3, -2

No relation

2) I. x2 – x – 12 = 0

II. y2 – 3y + 2 = 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)

I . x2  – x – 12 = 0

or, x2  – 4x + 3x – 12 = 0

or, x(x – 4) + 3(x – 4) = 0

or, (x – 4) (x + 3) = 0

x = 4, -3

II . y 2 -3y + 2 = 0

or, y 2 – 2y – y + 2 = 0

or, y(y – 2) – 1 (y – 2) = 0

or, (y – 2)(y – 1) = 0

y = 2, 1

Hence, no relation can be established

3) I. x – √169= 0

II. y2 – 169 = 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 – √169 = 0

Or x = √169

x = 13

II.  y2 – 169 = 0

or, y2 = 169

y = ±13

Hence, x ≥ y

4) I. x2 – 32 = 112

II. y – √256 = 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)

x2 – 32 = 112

or, x2 = 112 + 32 = 144

or, x = 144

x = ±12

II. y – 256 = 0

or, y = 256

y = 16

Hence, x < y

5) I. x2 – 25 = 0

II. y2 – 9y + 20 = 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)

x2 – 25 = 0

or, x2 = 25

or, x = 25

x = ±5

II. y2 – 9y + 20 = 0

or, y2 – 5y – 4y + 20 = 0

or, y(y – 5) – 4(y – 5) = 0

or, (y – 5) (y – 4) = 0

y = 5, 4

Hence, no relation can be established.

D.6-10): In the following questions, three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately or by any other method and give answer

(a) if x = y > z

(b) if x < y = z

(c) if x < y > z

(d) if x = y = z or if none of the above relationship can be established.

(e) if x ≤ y < z

6) I. 3x + 5y = 69

II. 9x + 4y = 108

III. x + z = 12

(a) if x = y > z

(b) if x < y = z

(c) if x < y > z

(d) if x = y = z or if none of the above relationship can be established.

(e) if x ≤ y < z

(c)

3x + 5y = 69 … (i)

9x + 4y = 108 … (ii)

x + z = 12 … (iii)

Now, from (i) and (ii), we have

3x + 5y = 69 … (i) × 4

9x + 4y = 108 … (ii) × 5

12x + 20y = 276

45x + 20y = 540

– 33x = – 264

On subtracting, we get

or, 33x = 264

x=264/3=8

Putting the value of x in equation (i), we get 3 × 8

+ 5y = 69

or, 5y = 69 – 24 = 45

∴y=45/5=9

Again, putting the value of x in equation (iii),

we get

x + z = 12

or, z = 12 – 8 = 4

Hence, x < y > z

7) I. y=√((729)(1/3) ×(6541) (1/4) )

II. 2x + 5z = 54

III. 6x+ 4z = 74

(a) if x = y > z

(b) if x < y = z

(c) if x < y > z

(d) if x = y = z or if none of the above relationship can be established.

(e) if x ≤ y < z

y = √(9(3×1/3) ×9(4×1/4) )= √(9 ×9 )……..(I)

2x + 5z = 54 .. (ii)

III. 6x + 4z = 74

or, 3x + 2z = 37 … (iii)

From equation (ii) × 2 – (iii) × 5, we get

4x + 10z = 108

15x + 10z = 185

– 11x = – 77

or, 11x = 77

x = 7

Putting the value of x in equation (ii), we get

2 × 7 + 5z = 54

or, 5z = 40

z = 8

Hence, x < y > z

8) I. 2x + 3y + 4z = 66

II. 2x + y + 3z = 42

III. 3x + 2y + 4z = 63

(a) if x = y > z

(b) if x < y = z

(c) if x < y > z

(d) if x = y = z or if none of the above relationship can be established.

(e) if x ≤ y < z

(b) 2x + 3y + 4z = 66 … (i)

2x + y + 3z = 42 … (ii)

3x + 2y + 4z = 63 … (iii)

From (iii) and (i),

x – y = – 3 …(iv)

From equation (i) × 3 – equation (ii) × 4

6x + 9y + 12z = 198

8x + 4y + 12z = 168

2x + 5y = 30 … (v)

Solving equation (iv) and (v), we get

x = 5, y = 8

Now, on putting the value of x and y in equation

(i),

10 + 24 + 4z = 66

or, 4z = 32

Hence, x < y = z

9) I. (x + z)3 = 1728

II. 2x + 3y = 35

III. x – z = 2

(a) if x = y > z

(b) if x < y = z

(c) if x < y > z

(d) if x = y = z or if none of the above relationship can be established.

(e) if x ≤ y < z

(a)

(x + z)3 = 1728 = 123

or, x + z = 12 …(i)

2x + 3y = 35 … (ii)

III. x – z = 2 …(iii)

Now, equation (i) and (ii),

x = 7, z = 5

Putting the value x in question (ii) we have,

2 × 7 + 3y = 35

or, 3y = 35 – 14 = 21

or, y = 25/5=5

Hence, x = y > z

10) I. 4x + 5y = 3

II. x + z = 8

III. 7x + 3y = 36

(a) if x = y > z

(b) if x < y = z

(c) if x < y > z

(d) if x = y = z or if none of the above relationship can be established.

(e) if x ≤ y < z

(b)

4x + 5y = 37 … (i)

x + z = 8 … (ii)

7x + 3y = 36 … (iii)

From equation (i) and (iii),

4x + 5y = 37 … (i) × 3

7x + 3y = 36 … (ii) × 5

or, 12x + 15y = 111

35x + 15y = 180

-23x = – 69

x = 3

Putting the value of x in equation (i)

4 × 3 + 5y = 37

or, y = 25/5 =5

Now, putting the value of x in equation (ii)

z = 5. Hence, x < y = z