Algebra For SBI PO : Set

Algebra For SBI PO : Set – 31

D.1-5) : In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer if

(a) x > y

(b) x ≥ y

(d) x ≤ y

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

1) I. 15/√x-9/√x=(x)^(1/2)

y¹⁰ – (36)⁵ = 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)

15/√x- 9/√x = x^(1/2)

Or, (15-9)/√x = x^(1/2) = √x

or, x = 4

II. y¹⁰ – (36)⁵ = 0

or, y¹⁰ = (36)⁵

or y = 〖36]^(5/10)=〖36]^(1/2)

y = √36 = 6

x ≥ y

2) I. 5x + 2y = 96

II) 3(7x + 5y) = 489

(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)

5x + 2y = 96 … (i)

-21x + 15y = 489 … (ii)

Now, eqn (i) × 15 and eqn (ii) × 2

75x + 30y = 1440

42x + 30y = 978

(-) (-) (-)

————————

33x = 462

x = 14

Putting the value of x in equation (i), we get

5 × 14 + 2y = 96 or, 2y = 96 – 70 = 26

or, y = 26/2 = 13

x > y

3. I. (441)^(1/2)x-² 111=(15)²

II. √121 y²+(6)³=260

(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)

I. 〖(441) ^(1/2) x² – 111 = 15²

or 〖(21^)(2×1/2) x² = 225 + 111 = 336

or 21 x²= 336

x = 336/21 = 16

II. √121 y²+ 6³ = 260

or, 11y²+ 6³ = 260

or, 11y² = 260 -216 = 44

or, y²= 4 y = 2

x > y

4. I. 17x = (13)² + √196 + (e)² + 4x

II. 9y – 345 = 4y – 260

(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)

17x = 169+ 14 + 25 + 4x or, 13x = 208

x = 208/13 = 16

II. 9y – 4y = 345 – 260 = 85

or, 5y = 85

y = 17

x < y

5. I. 3x² – 13x + 14 = 0

II. y² – 7y + 12 = 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)

3x²– 13x + 14 = 0

or, 3x² – 7x – 6x + 14 = 0

or,3x(x – 2) -7(x – 2) = 0

or, (3x – 7) (x – 2) = 0

x = 7/3, 2

y² – 7y + 12 = 0 or, y² – 4y – 3y + 12 = 0

or, y(y – 4)

-3(y – 4) = 0

or,(y -3)(y – 4) = 0

y = 4, 3

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. 2x² – 15x + 28 = 0

II. 2y² + 3y-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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(b)

2x²– 8x – 7x + 28 = 0

or 2x(x – 4) – 7(x – 4) = 0

or (2x – 7) (x – 4) = 0

x = 4, 7/2

2y² + 10y – 7y – 35 = 0

or 2y(y + 5) – 7(y + 5) = 0

or (2y – 7) (y + 5) = 0

y = 7/2,-5

x ≥ y

7. I. 7x – 5y = 24

II. 4x + 3y = 43

(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)

28x – 20y = 96

28x + 21y = 301

(-) (-) (-)

———————-

-41y = -205

y = 5 and x = 7

x > y

8) I. x = ∛2744

II. y = √487

(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)

x = ∛2744 – 14

y = √487 = 22 √484 = 32

9) I. x² – 9x + 8 = 0

II) 2y² – 11y + 5 = 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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(e)

x²– x – 8x + 8 = 0

or x (x – 1) – 8(x – 1) = 0

or (x – 1) (x – 8) = 0

x = 1, 8

2y² – y – 10y + 5 = 0 or y (2y – 1) – 5 (2y – 1) = 0

or (y – 5) (2y – 1) = 0

y = 5, 1/2

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

10) I. 2x² + 3x + 1 = 0

II. 6y² + 17y + 12 = 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)

2x²+ 2x + x + 1 = 0

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

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

x = -1, 1/2 = 0.5

II. 6y²+ 9y + 8y + 12 = 0

or 3y (2y + 3) + 4 (2y + 3) = 0

or (3y + 4) (2y + 3) = 0

y = – 4/3, – 3/2 = – 1.33, -1.5

x >y

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Algebra For SBI PO : Set – 30