## Algebra For SBI PO : Set – 29

D.1-5) : In the following questions, two equations numbered I and H 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 relationship cannot be established

I. 4x + 3y = (1600)1/2

II) 6x – 5y = (484)1/2

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship cannot be established.

(a)

4x + 3y = 40 ……….(i) ×6

6x – 5y = 22 ………(ii) ×4

24x – 20y = 88

———————

38y = 152

y = 152/38  = 4

Putting the value of y in equation (i), we have 4x + 3 x 4 = 40

or, 4x = 40 12 = 28

x = 7

Hence, x > y.

2) I. 2x2-(4 ÷√13)x+2√13=0

II) (5y2-12)-(9y2-16)=0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship cannot be established.

(b)

2x2 – 4x – √13 x + 2  = 0…(i)

or,2x (x-2)  – √13 (x-2) = 0

or, (x-2) (2x – √13) = 0

x = 2, √13/2

10y2 – 18y – 5√13y +9√13 = 0…(ii)

or,2y (5y-9) – √13 (5y-9) = 0

or,(2y-√13) (5y – 9) = 0

y = 9/5, √13/2

10y2 – 18y – 5√13y + 9 √13 = 0 ……………(ii)

or 2y (5y – 9) – √13 (5y-9) = 0

y = 9/5, √13/2

Hence, x  y.

3. I. (6x2 + l7) – (3x2 + 20) = 0

II) (5y2 – 12) – (9y2 – 16) = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship cannot be established.

(e)

6x2  + 17 – 3x2  – 20 = 0 … (i) or,

3x2  = 3

x ± l

5y2– 12 -9y2+ 16 =0 …. (ii)

or, 4y2 = 4

y ± 1

Hence x = y.

4. I. (169)½ x+ √289=134

II) (361)1/2 y2 270 = 1269

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship cannot be established.

(b)

13x + 17 = 134 …. (i) .
x = 117/13 = 9

(361)1/2y2 – 270 = 1269

or, 19y2 = 1629 + 270 = 1539

y2 = 1539/19 = 81

y ± 9

Hence, x ≥ y

5) I. 821x2 – 757x2 = 256

II) √196 y3 12y3 =16

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

(e) if x = y or relationship cannot be established.

(d)

64x2 = 256 …. (i)

or, x2  = 4

x = ± 2

14y3 – 12y 3 = 16 …. (ii)

or, 2y3 = 16

y3 = 8

y = 2

Hence x ≥ y

D.6-10): 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.

6) I. 5x – 7y = -24

II) 13x + 3y = 86

(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)

15x – 21y = -72

91x + 21y = 602

——————–
106x = 530

x = 5,       y = 7

x < y

7) I. x2 – 13x + 40 = 0

II) y2 + 3y – 40 = 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 – 13x + 40 = 0

or x2 – 5x – 8x + 40 = 0

or x(x – 5) -8(x – 5) = 0

or (x – 5)(x – 8) = 0

x = 5, 8

II. y2 + 3y – 40 = 0

or y2 5y + 8y – 40 = 0

or y (y – 5) + 8 (y – 5) = 0

or (y – 5) (y + 8) = 0

y = 5, -8

Hence, x y

8) I. 8x2 – 26x+15 = 0

II) 2y2 – 17y + 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.

(d)

8x2 -26x + 15 = 0

or 8x2 – 20x – 6x + 15 = 0

or 4x (2x – 5) – 3 (2x – 5) = 0

or (4x – 3) (2x – 5) = 0

x =3/4 , 5

II. 2y2 -17y + 30 = 0

or 2y2 – 12y – 5y + 30 = 0

or 2y(y – 6) – 5(y – 6) = 0

or (2y – 5) (y 6) = 0

y = 5/2, 6

x ≤ y

9. I. x2 = 484

II) y2 – 45y + 506 = 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.

(d)

x2 = 484

x = 22

II. y2 – 45y + 506 = 0

or y2 – 22y – 23y + 506 = 0

or y (y – 22) – 23 (y – 22) = 0

or (y – 22) (y – 23) = 0

y = 22, 23

x ≤ y

10. I. 13x – 21=200 – 4x

II) y = ∛2197

(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. 13x -21 = 200- 4x

or 13x + 4x = 200 + 21

x = 221/17 = 13

II. y = ∛2197

y = 13

x = y