## Algebra For SBI PO : Set – 12

D.1-6) In the following questions, two equations numbered I and II are given. You have to solve both the equations and find the Solutions.

1)

I. 5X – 7Y = 3;

II. 6X – 8Y = 4

a) X>Y

b) X<Y

c) X≤Y

d) X≥Y

e) X=Y or Can’t be determined

a) X>Y

X = 2;

Y = 1

2)

I. 2X + 3Y = -5;

II. 5X + 2Y = 4

a) X>Y

b) X<Y

c) X≤Y

d) X≥Y

e) X=Y or Can’t be determined

a) X>Y

X = 2;          Y = -3

3)

I. X + Y = 8;

II. 2X + 3Y = 20

a) X>Y

b) X<Y

c) X≤Y

d) X≥Y

e) X=Y or Can’t be determined

e) X=Y or Can’t be determined

X = 4;          Y = 4

4)

I. 8x2 + 6x = 5

II. 12y2 – 22y + 8 = 0

a) If x > y

b) If x ≥ y

c) If x < y

d) If x ≤ y

e) If x = y

(d)

I. 8x2 + 6x -5 = 0

8 × 5 = – 40

10 × (- 4) = – 40

10 – 4 = 6

II. 12y2 – 22y + 8 = 0

12 × 8 = 96

-16 × – 6 = 96

– 16 + (-6) = -22

x ≤ y

5)

I. 18x2 + 18x + 4 = 0

II. 12y2 + 29y + 14 = 0

a) If x > y

b) If x ≥ y

c) If x < y

d) If x ≤ y

e) If x = y

(b)

I. 18x2 + 18x + 4 = 0

18 × 4 = 72

12 × 6 = 72

12 × 6 = 18

II. 12y2 + 29y + 14 = 0

12 × 14 = 168

21 × 8 = 168

21 + 8 = 29

x ≥ y

6)

I. x2 – 16 = 0

II. y2 – 9y + 20 = 0

a) If x > y

b) If x ≥ y

c) If x < y

d) If x ≤ y

e) If x = y

(d)

I. x2 = 16 => x = ±4

X = (4, -4)

II. y2 – 9x + 20 = 0

1 × 20 = 20

(-5) × (-4) = -9

Y = (5, 4)

Hence, x ≤ y

7) If x2 + y2 – 4x – 4y + 8 = 0, then the value of x-y is

a) 4

b) -4

c) 0

d) 8

e) 2

C)

8) X=2+√3 then find the value of (X6+X4+X2+1)/X3

a) 52

b) 68

c) 80

d) 56

e) 46

(d)

9) What number should be subtracted from x3+4x2−7x+12, if it is to be perfectly divisible by x+3?

a) 42

b) 39

c) 13

d) 23

e) None of these

(a)

In this case, x + 3 divides x3 + 4x2 – 7x + 12 – K perfectly.

We know that the remainder is 0, when the value of x is substituted by – 3.

(-3)3 + 4 (-3)2 – 7 (-3) + 12 – K = 0

-27 + 36 + 21 + 12 = K

K = 42

10) If a = 2 , b = -3 then the value of 27 a3 – 54a2b+ 36 ab2 – 8b3 is

a) 1562

b) 1616

c) 1676

d) 1728

e) 1626