Algebra For SBI PO : Set – 08

D.1-5) Each of the following questions contains two equation I & II. Solve those equation and find the value of x and y.

1)

I. 6x2+11x+4=0

II. 4y2-7y-2=0

a) If x > y

b) If x < y

c) If x ≥ y

d) If x ≤ y

e) If x = y or can’t be determined

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b)

X=-1.3, -0.5       y=-0.25, 2

2)

I. x2-463=321

II. Y2-421=308

a) If x > y

b) If x < y

c) If x ≥ y

d) If x ≤ y

e) If x = y or can’t be determined

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e)

X=28, -28   y=27, -27

3)

I. 9x2-45x+56=0

II. 4y2-17x+18=0

a) If x > y

b) If x < y

c) If x ≥ y

d) If x ≤ y

e) If x = y or can’t be determined

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a)

X=2.33, 2.6 y=2.25, 2

4)

I. 4x2+16x+15=0

II. 2y2+3y+1=0

a) If x > y

b) If x < y

c) If x ≥ y

d) If x ≤ y

e) If x = y or can’t be determined

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b)

X=-1.5, -2.5                   y=-0.5, -1

5)

I. 2x2-7x+3=0

II. 2y2-7y+6=0

a) If x > y

b) If x < y

c) If x ≥ y

d) If x ≤ y

e) If x = y or can’t be determined

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e)

X=0.5, 3      y=1.5, 2

D.6-10) In each of these questions, two equations are given. You have to solve these equations and find out the values of x and y and then select the correct option based on the relationship between x and y.

6)

I. 12x + 21y = 627

II. 12x ­-14y = ­38

a) x < y

b) x > y

c) x ≤ y

d) x ≥ y

e) x = y

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e)

Multiply equation I by 3

12x + 21y = 627

12x ­-14y = ­38 (equation 2)

Subtracting equation 2 from equation 1

35y = 665

Y = 19

4x + 7 X 19 = 209

4x = 76 X = 19

Hence, X = Y

7)

I. 17x2 +48x -9 = 0

II.  13y2 – 32y +12= 0

a) x < y

b) x > y

c) x ≤ y

d) x ≥ y

e) x = y

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a)

Consider statement I: 17x2 +48x -9 = 0

17x2 + 51x -3x -9 = 0

17x(x+3) ­-3(x+3) = 0

(x+3)(17x-­3) = 0

X = -3 or 3/17

Consider statement II: 13y2 – 32y +12= 0

13y 2­- 26y – 6y + 12 = 0

13y(y­-2) ­-6(y-­2) = 0

(y­-2)(13y-­6) = 0

Y = 2 or 6/13

Hence, X < Y

8)

II. 12y2 -22y +8 =0

a) x < y

b) x > y

c) x ≤ y

d) x ≥ y

e) x = y

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e)

Consider statement I: 8x^2 +6x +6 =5

8x2 +6x +1 = 0

8x2 + 4x + 2x + 1 =0

4x(2x+1) + 1(2x+1) = 0

(2x+1)(4x+1) = 0

X = (­-1/2) or (-­1/4)

Consider statement II: 12y2 -22y +8 =0

12y 2-6y -16y + 8 = 0

6y(2y­-1) ­ -8(2y-­1) = 0

(2y­-1)(6y­-8) = 0

Y = 1/2 or 4/3

Hence, X < Y

 9)

I. 18x2 +18x +4 =0

II. 12y2 + 29y +14 =0

a) x < y

b) x > y

c) x ≤ y

d) x ≥ y

e) x = y

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d)

Consider statement I: 18x2 +18x +4 =0

18x2 + 6x + 12x + 4 = 0

6x(3x+1) + 4(3x+1) = 0

(3x+1)(6x+4) = 0

X = (­-1/3) or (-­2/3)

Consider statement II: 12y2 + 29y +14 =0

12y2 + 8y + 21y + 14 = 0

4y(3y+2) + 7(3y+2) = 0

(3y+2)(4y+7) = 0

Y = (­-2/3) or (­-7/4)

Hence, X ≥ Y

10)

I. 16x2+20x +6 =0

II. 10y2 + 38y +24 =0

a) x < y

b) x > y

c) x ≤ y

d) x ≥ y

e) x = y

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b)

Consider statement I: 16x2+20x +6 =0

16x2 +8x + 12x + 6 = 0

8x(2x+1) + 6(2x+1) = 0

(8x+6)(2x+1) = 0

X = -(­3/4) or -(­1/2)

Consider statement II: 10y2 + 38y +24 =0

10y2 + 30y + 8y + 24 = 0

10y(y+3) + 8(y+3) = 0

(y+3)(10y+8) = 0

Y = ­-3 or -(­4/5)

Hence, X is always greater than Y