Algebra For SBI PO : Set

Algebra For SBI PO : Set – 32

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

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

1) I. 3x² – 29x + 56 = 0

II. 3y² – 5y – 8 = 0

(a) x > y

(b) x ≥ y

(d) x ≤ y

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

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(b)

3x²– 29x + 56 = 0

or 3x²– 21x – 8x + 56 = 0

or 3x(x – 7) – 8(x – 7) = 0

or (3x – 8) (x – 7) = 0

x = , 7

II. 3y²– 5y – 8 = 0

or 3y²+ 3y – 8y – 8 = 0

or 3y(y + 1) – 8(y + 1) = 0

or (3y – 8) (y + 1) = 0

y = -1,

x≥ y

2) I. 5x² + 26x – 24 = 0

II. 5y² – 34y + 24 = 0

(a) x > y

(b) x ≥ y

(d) x ≤ y

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

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(d)

5x²+ 26x – 24 = 0

or 5x²+ 30x – 4x – 24 = 0

or 5x(x + 6) – 4(x + 6) = 0

or (5x – 4) (x + 6) = 0

x = 4/5, 6

II. 5y²– 30y – 4y + 24 = 0

or 5y(y – 6) – 4(y – 6) = 0

or (5y – 4) (y – 6) = 0 4

y =4/5 6

x≤y

3) I. x²– 7x = 0

II. 2y² + 5y + 3 = 0

(a) x > y

(b) x ≥ y

(d) x ≤ y

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

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(a)

x²– 7x = 0

or x (x – 7) = 0

x = 0, 7

II. 2y²+ 5y + 3 = 0

or 2y²+ 2y + 3y + 3 = 0

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

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

y = -1, -3/2

x > y

4) I. 7x – 4y = 40

II. 8x + 8y = 8

(a) x > y

(b) x ≥ y

(d) x ≤ y

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

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(a)

7x – 4y = 40 …(i)

and 8x + 8y = 8

or x+y= 1 …(ii)

Solving (i) and (ii), we have

x = 4, y = -3

x > y

5) I. 15x² – 41x + 14 = 0

II. 2y² – 13y + 20 = 0

(a) x > y

(b) x ≥ y

(d) x ≤ y

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

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(c)

15x²– 4x + 14 = 0

or 15x²– 6x – 35x + 14 = 0

or 3x(5x – 2) – 7(5x – 2) = 0

or (3x – 7)(5x – 2) = 0

x = 7/3, 2/5

II. 2y²– 13y + 20 = 0

or 2y²– 8y – 5y + 20 = 0

or 2y(y – 4) – 5(y – 4) = 0

or (2y – 5) (y – 4) = 0

y = 4, 5/2

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

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

6) I. x²-8√3x+45=0

II. y²-√2y-24=0

(a) if x > y

(b) if x ≥ y

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

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(e)

I.x²-8√3x + 45= 0

or x² – 5√3x + 3√3 (x – 5√3) = 0

or, (x + 3√3) (x – 5√3) = 0

X = 3√3, 5√3

II. y² – √2y – 24 = 0

Or y² – 4√2y + 3 √2y – 24 = 0

Or (y-4√2y) (y + 2√2)

y = -3 √2, 4√2

Hence relation cannot be established between x and y.

7) x-7√2x+24=0

II. y-5√2y+12=0

(a) if x > y

(b) if x ≥ y

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

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x – 7 √2x + 24 = 0

Or x – 4√2x – 3 √2x + 24 = 0

Or √x (√x – 4√2) – 3√2 (√x – 4√2) = 0

Or (√x – 3√2) (√x – 4√2) = 0

Now, if √x -3√2 = 0

then √x = 3√2

x = 9 × 2 = 18

If √x – 4√2 = 0

then √x = 4√2

x = 16 × 2 = 32

II. y – 5√2y + 12 = 0

y -3√2y – 2√2y + 12 = 0

Or √y (√y – 3√2) -2√2y + 12 = 0

Or (√y – 2√2) – (√y -3√2) = 0

If (√y -2√2) = 0

Then √y = 2√2

y = 4 2 = 18

If √y -3√2 = 0

Then, √y -√2

y = 9 × 2 = 18

x ≥ y

8. I. 12x² – 17x + 6 = 0

II. 20y² – 31y + 12 = 0

(a) if x > y

(b) if x ≥ y

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

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(d)

12x²– 17x + 6 = 0

or 12x²– 9x – 8x + 6 = 0

or 3x(4x – 3) – 2(4x – 3) = 0

or (3x – 2) (4x – 3) = 0

If 3x – 2 = 0

then 3x = 2

x= 2/3

If 4x – 3 = 0

then x = 3/4

II. 20y² – 31y + 12 = 0

or 20y² – 16y – 15y + 12 = 0

or 4y (5y – 4) – 3 (5y – 4) = 0

or (4y – 3) (5y – 4) = 0

y=3/4,4/5

Hence x ≤ y

9. I. 3x² – 8x + 4 = 0

II. 4y² – 15y + 9 = 0

(a) if x > y

(b) if x ≥ y

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

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(e)

3x²– 8x + 4 = 0

or 3x²– 6x – 2x + 4 = 0

or (3x – 2) (x – 2) = 0

x=2,2/3

II. 4y² – 15y + 9 = 0

or 4y² – 12y – 3y + 9 = 0

or 4y(y – 3) – 3(y – 3) = 0

or (4y – 3) (y – 3) = 0

y = 3/4, 3

Relation cannot be established between x and y.

10. I. x² -16x + 63 = 0

II. y² – 2y – 35 = 0

(a) if x > y

(b) if x ≥ y

(d) if x ≤ y

(e) if x = y or no relation can be established between x and y.

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(a)

I. x²– 16x + 63 = 0

or x² – 9x – 7x + 63 = 0

or x(x – 9) – 7(x – 9) = 0

or (x – 7) (x – 9) = 0

x = 7, 9

II. y² – 2y – 35 = 0

or y² – 17y + 5y – 35 = 0

or y(y – 7) + 5(y – 7) = 0

or (y + 5) (y – 7) = 0

y = -5, 7

Hence, x ≥ y

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Algebra For SBI PO : Set – 31