Algebra For SBI PO : Set – 26

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, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

1) I. 9x2 = 1

II) 4y2+ 11y – 3 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(c)

I. 9x2 = 1

x2 =1/9

x = ± 1/3

II. 4y2 + 11y – 3 = 0

or, 4y2 + 12y – y – 3 = 0

or, 4y(y + 3) – 1(y + 3) = 0

∴y= 1/4, -3

Hence, there is no relation between x and y.

2) I. 3x2 + 5x – 2 = 0

II) 2y2– 7y + 5 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(a)

3x2+ 5x – 2 = 0

or, 3x2 + 6x – x – 2 = 0

or, 3x(x + 2) – 1(x + 2) = 0

or, (3x – 1) (x + 2) = 0

∴x= -2,1/3

2y2 – 2y – 5y + 5 = 0

or, 2y(y – 1) – 5(y – 1) = 0

y = 1,5/2

Hence, x < y

3) I. 6x2 + 13x + 5 = 0

II) 3y2+ 11y + 10 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(e)

6x2 + 13x + 5 = 0 or 6x2 + 3x + 10x + 5 = 0

or, 3x(2x + 1) + 5(2x + 1) = 0

or, (3x + 5) (2x + 1) = 0

x =-5/3,1/2

3y2+ 11y + 10 = 0

or, 3y2 + 6y + 5y + 10 = 0

or, 3y(y + 2) + 5(y + 2) = 0

or, (3y + 5) (y + 2) = 0

y = -5/3,2 

4) I. 7x – 4y = 29

II) 5x + 3y – 50 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(d)

I . 7x – 4y = 29

II. 5x + 3y = 50

(I) × 3 + (II) × 4

21x – 12y = 87

20x + 12y = 200

x = 7

Putting the value of x in (I), we get y = 5

Hence, x > y

5) I. x2 – 5 = 0

II) 4y2– 24y + 35 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(a)

x2 = 5

x = ±√5 ≈ ± 2.236

4y2 – 24y + 35 = 0 or, 4y2 – 14y – 10y + 35 = 0 or,

2y (2y – 7) – 5 (2y – 7) = 0

or, (2y – 5) (2y – 7) = 0

y=5/2,7/2= 2.5,

3.5

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. 35x2– 53x + 20 = 0

II) 56y2-97y + 42 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(c)

35x2 – 28x – 25x + 20 = 0

or 7x(5x – 4) – 5(5x – 4) =0

or (7x – 5) (5x – 4) = 0

∴x=5/7,4/5

56y2– 48y – 49y + 42 = 0

or 8y (7y – 6) – 7 (7y – 6) = 0

or (8y – 7) (7y – 6) = 0

∴y=7/8,6/7

x < y

7) I. x = ∛4913

II) 13y + 3x = 246

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(a)

x = ∛4913

x = 17

13y = 246 – 3x or 13y = 246 – 51 = 195

y = 15

y

8) I. x2 – 5x – 14 = 0

II) y2+ 7y + 10 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(b)

x2– 7x + 2x – 14 = 0

or x(x – 7) + 2(x – 7) = 0

(x + 2) (x – 7) = 0

x = -2, 7

y2+ 5y + 2y + 10 = 0

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

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

y = -2, -5

x≥y

9) I. x2 – 3481 = 0

II) 3y2= ∛216000

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(e)

x2 = 3481

x = ± 59

3y2∛216000 3y260

y = ± 20

No relation

10) I. 5x2 + 2x – 3 = 0

II) 2y2+ 7y + 6 = 0

(a) if x < y

(b) if x ≤ y

(c) if x = y, or no relation can be established between x and y

(d) if x > y

(e) if x ≥ y

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(a)

5x2+ 5x – 3x – 3 = 0

or 5x (x + 1) – 3(x + 1) = 0

or (5x – 3) (x + 1) = 0

x = 3/5, -1

2y2 + 4x + 3y + 6 = 0

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

y = –3/2, -2

ie x > y