Algebra For SBI PO : Set – 30

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 p > q

(b) if p ≥ q

(c) if p < q

(d) if p ≤ q

(e) if p = q or there is no relation between ‘p’ and ‘q’.

1) I. (p + q)2 = 3136

II. q + 2513 = 2569

(a) if p > q

(b) if p ≥ q

(c) if p < q

(d) if p ≤ q

(e) if p = q or there is no relation between ‘p’ and ‘q’.

(c)

1) I. (p + q)2 = 3136

p + q =  56

II. q + 2513 = 2569

or, q = 2569 – 2513 = 56

Putting the value of q in (I) we have,

p = 0. -112

p < q

2. I. 4p2– 16p +15 = 0

II . 2q2 + 5q – 7 = 0

(a) if p > q

(b) if p ≥ q

(c) if p < q

(d) if p ≤ q

(e) if p = q or there is no relation between ‘p’ and ‘q’.

(a)

4p2 – 16p + 15 = 0

or, 4p2 – 10p – 6p + 15 = 0

or, 2p (2p – 5) – 3 (2p – 5) = 0 or, (2p – 3) (2p – 5) = 0 p =

p =3/2, 5/2

II. 2q2 + 5q – 7 = 0

or, 2q2 + 7q – 2q – 7 = 0

or, q (2q + 7) – 1(2q + 7) = 0

or, (q – 1) (2q + 7) = 0

q = 1, -7/2

3. I. p2 = 49

II . q2+15q + 56 = 0

(a) if p > q

(b) if p ≥ q

(c) if p < q

(d) if p ≤ q

(e) if p = q or there is no relation between ‘p’ and ‘q’.

(b)

I. p2 = 49

P = ±7

II. q2 + 15q + 56 = 0

or, q2 + 8q + 7q + 56 = 0

or, q(q + 8) + 7(q + 8) = 0

or, (q + 7) (q + 8) = 0

q = -7, -8

p ≥ q

4) I. 2p2 + 5p – 12 = 0

II. 2q2 – q – 1 = 0

(a) if p > q

(b) if p ≥ q

(c) if p < q

(d) if p ≤ q

(e) if p = q or there is no relation between ‘p’ and ‘q’.

(e)

I. 2p2 + 5p – 12 = 0

or, 2p2 + 8p – 3p – 12 = 0

or, 2p (p + 4) – 3(p + 4) = 0

or, (2p – 3) (p + 4) = 0

p = 3/2, -4

II. 2q2 – q – 1 = 0

or, 2q2 – 2q + q – 1 = 0

or, 2q (q – 1) + 1(q – 1) = 0

or, (2q + 1) (q – 1) = 0

q = 1, – 1/2

No reaction between ‘p ’and ‘q ’.

5) I. p2– 12p + 35 = 0

II. q2– 25 = 0

(a) if p > q

(b) if p ≥ q

(c) if p < q

(d) if p ≤ q

(e) if p = q or there is no relation between ‘p’ and ‘q’.

(b)

p2 – 12p + 35 = 0 or p2 – 5p – 7p + 35 = 0

or p (p – 5) – 7(p – 5) = 0

or (p – 7) (p – 5) = 0

p = 5, 7

II. q2 – 25 = 0

or, q2 = 25

q = ± 5

p ≥ q

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 there is no relation between ‘x’ and ‘y’.

6) I. 3x2 + 7x + 2 = 0

II. 2y2 + 9y + 10 = 0

(a) if x > y

(b) if x < y

(c) if x ≥ y

(d) if x ≤ y

(e) if x = y or there is no relation between ‘x’ and ‘y’.

(c)

3x2 + 6x + x + 2 = 0

or 3x(x + 2) + 1(x + 2) = 0

or (x + 2) (3x + 1) = 0 x = –2,     1/3

II. 2y2 + 4y + 5y + 10 = 0

or 2y (y + 2) + 5 (y + 2) = 0

or (2y + 5) (y + 2) = 0

y = –2,- 5/2

x ≥ y

7) I. x2 + x – 2 = 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 there is no relation between ‘x’ and ‘y’.

(d)

x2 + 2x – x – 2 = 0 or x(x + 2)– 1(x + 2) = 0

or (x – 1)

(x + 2) = 0 x = 1, –2

II. y 2 – y – 2y + 2 = 0

or y (y – 1) – 2 (y – 1) = 0 or

(y – 1) (y – 2) y = 1, 2

x ≤ y

8) I. 20x2 – 51x + 27 = 0

II. 15y2 – 16y + 4 = 0

(a) if x > y

(b) if x < y

(c) if x ≥ y

(d) if x ≤ y

(e) if x = y or there is no relation between ‘x’ and ‘y’.

(a)

20x2 – 15x – 36x + 27 = 0

or 5x(4x – 3) – 9(4x – 3) = 0

or (5x – 9) (4x – 3) = 0

x = 9/53/4

II. 5y2 – 10y – 6y + 4 = 0

or 5y (3y – 2) – 2(3y – 2) = 0

or (5y – 2) (3y – 2)

y= 2/5, 2/3

x > y

9. I. 7x2 + 16x – 15 = 0

II. y2 – 6y – 7 = 0

(a) if x > y

(b) if x < y

(c) if x ≥ y

(d) if x ≤ y

(e) if x = y or there is no relation between ‘x’ and ‘y’.

(e)

7x2 + 21x – 5x – 15 = 0

or 7x(x + 3) – 5(x + 3) = 0

or (x + 3) (7x – 5) = 0

x = -3,

II. y2 – 7y + y – 7 = 0

or y (y – 7) + 1 (y – 7) = 0

or (y + 1) (y – 7)

y =– 1, 7

Therefore, no relation between ‘x ’and ‘y ’.

10) I. x2 = 729

II. y2 + 58y + 840 = 0

(a) if x > y

(b) if x < y

(c) if x ≥ y

(d) if x ≤ y

(e) if x = y or there is no relation between ‘x’ and ‘y’.

(a)

I. x2 = 729

x = ±27

II. y2 + 58y + 840 = 0

or y2 + 28y + 30y + 840 = 0

or y (y + 28) + 30 (y + 28) = 0

or (y + 30) (y + 28) = 0

y = –30, –28

x > y