Algebra For SBI PO : Set – 24

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 the relationship between x and y cannot be established.

1) I. x2 + 12x + 36 = 0

II) y2 + 15y + 56 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(a)

x2 + 12x + 36 = 0

or, (x + 6)2 = 0

or, x + 6 = 0

or, x = – 6

II. y2 + 15y + 56 = 0

or, y2 + 7y + 8y + 56 = 0

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

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

y = -7, -8

x > y

2) I. x2 = 35

II) y2 + 13y + 42 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(a)

x2 = 35

x = ± √35

II. y2 + 13y + 42 = 0

or, y2 + 6y + 7y + 42 = 0

or, y(y + 6) + 7(y + 6) = 0

or, (y + 6) (y + 7) = 0

y = -6, – 7

x > y

3) I. 2x2 – 3x – 35 = 0

II) y2 – 7y + 6 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(e)

2x2 – 3x – 35 = 0

or, 2x2 – 10x + 7x – 35 = 0

or, 2x(x – 5) + 7(x – 5) = 0

or, (2x + 7) (x – 5) = 0

x = – 7/2, 5

II. y2 – 7y + 6 = 0

or, y2 – y – 6y + 6 = 0

or, y(y – 1) – 6(y – 1)

or, (y – 1)(y – 6) = 0

y = 1, 6

No relation can be established between x and y.

4) I. 6x2 – 29x + 35 = 0

II) 2y2 – 19y + 35 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(d)

6x2 – 29x + 35 = 0

or, 6×2- 15x – 14x + 35 = 0

or, 3x(2x – 5) -7(2x – 5) = 0

or, (3x – 7) (2x – 5) = 0

∴x= 7/3, 5/2

2y2 – 19y + 35 = 0

or, 2y2 – 14y – 5y + 35 = 0

or, 2y(y – 7) -5 (y – 7) = 0

or, (2y – 5)(y – 7) = 0

∴y= 5/2, 7

5) I. 12x2 – 47x + 40 = 0

II) 4y2 + 3y – 10 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(b)

12x2 – 47x + 40 = 0

or, 12x2 – 32x – 15x + 40 = 0

or, 4x(3x – 8) -5(3x – 8) = 0

or, (4x – 5) (3x – 8) = 0

x = 5/4, 8/3

4y2 + 3y – 10 = 0

or, 4y2 + 8y – 5y – 10 = 0

or, 4y(y + 2) -5(y + 2) = 0

or, (4y – 5) (y + 2) = 0

y = 5/4, -2

 

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 (a) if x > y       n ‘x’ and ‘y’.

6) I. x2 + 3x – 28 = 0

II) y2 – 11y + 28 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(d)

x2 + 7x – 4x – 28 = 0

or, x(x + 7) – 4 (x + 7) = 0

or, (x – 4)(x + 7) = 0

x = 4, – 7

II. y2 – 11y + 28 = 0

or, y2 – 7y – 4y + 28 = 0

or, y (y – 7) -4(y – 7) = 0

or, (y – 4) (y – 7) = 0

y = 4, 7

7) I. 6x2 – 17x + 12 = 0

II) 6y2 – 7y + 2 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(a)

6x2 – 17x + 12 = 0

or, 6x2 – 9x – 8x + 12 = 0

or, 3x (2x – 3) – 4 (2x – 3) = 0

or, (3x – 4) (2x – 3) = 0

x = 4/3, 3/2

6y2 – 3y – 4y + 2 = 0

or, 3y (2y – 1) – 2 (2y – 1) =0

or, (3y – 2) (2y – 1) = 0

y = 2/3, 1/2

8) I. x=√256/√576

3y2 + y-2 = 0

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(b)

x = √256/√576

∴x= 16/24= 2/3

3y2 + y – 2 = 0

or, 3y2 + 3y – 2y – 2 = 0

or, 3y (y + 1) – 2(y + 1) = 0

or, (3y – 2) (y + 1) = 0

y = 2/3, -1

9) I. x2 = 64

II) y2 = 9y

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(e)

x2 = 64

x = ± 8

II. y2 = 9y

or, y2 – 9y = 0

or, y (y – 9) = 0

y = 0, 9

no relationship can be established between x and y.

10) I. x2 + 6x – 7 = 0

II) 41y + 17 = 140

(a) if x > y

(b) if x ≥ y

(c)if x < y

(d) if x ≤ y

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

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(c)

x2 + 6x – 7 = 0

or, x2 + 7x – x – 7 = 0

or, x(x + 7) -1 (x + 7)= 0

or, (x – 1) (x + 7) = 0

x = 1, -7

II. 41y + 17 = 140

or, 41y = 140 – 17 = 123
y = 123/41=3

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