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- Question 1 of 10
##### 1. Question

Directions:1-5)

**In the following questions, two equations numbered are given in variables a and b. You have to solve both the equations and find out the relationship between a and b. Then give answer accordingly-**I. 6a

^{2}+ 7a -3 = 0II. b (10b – 1) = 2

Correct6 a

^{2}+ 7a -3 = 0

6 a^{2}+ 9a – 2a – 3 = 0

So a = -1.5, 0.3

b (10b – 1) = 2

10b^{2}– b – 2 = 0

10b^{2}– 5b + 4b – 2 = 0

So b = 0.5, -0.4

-1.5… -0.4… 0.3… 0.5.Incorrect6 a

^{2}+ 7a -3 = 0

6 a^{2}+ 9a – 2a – 3 = 0

So a = -1.5, 0.3

b (10b – 1) = 2

10b^{2}– b – 2 = 0

10b^{2}– 5b + 4b – 2 = 0

So b = 0.5, -0.4

-1.5… -0.4… 0.3… 0.5. - Question 2 of 10
##### 2. Question

I. 4a

^{2}+ 3a – 27 = 0II. 15b

^{2}– 38b – 21 = 0Correct4a

^{2}+ 3a – 27 = 0

4a^{2}+ 12a – 9a – 27 = 0

So a =2.25, -3

15b^{2}– 38b – 21 = 0

15b^{2}– 45b + 7b – 21 = 0

So b = 3, – 0.46

-3… -0.46… 2.25…. 3.Incorrect4a

^{2}+ 3a – 27 = 0

4a^{2}+ 12a – 9a – 27 = 0

So a =2.25, -3

15b^{2}– 38b – 21 = 0

15b^{2}– 45b + 7b – 21 = 0

So b = 3, – 0.46

-3… -0.46… 2.25…. 3. - Question 3 of 10
##### 3. Question

I. 8a

^{2}+ 5a – 13 = 0II. 2b

^{2}+ 23b + 63 = 0Correct8a

^{2}+ 5a – 13 = 0

8a^{2}+ 13a – 8a – 13 = 0

So a = -1.625, 1

2b^{2}+ 23b + 63 = 0

2b^{2}+ 14b + 9b + 63 = 0

So b = -7, -4.5

-7…. -4.5 ….-1.625…. 1.Incorrect8a

^{2}+ 5a – 13 = 0

8a^{2}+ 13a – 8a – 13 = 0

So a = -1.625, 1

2b^{2}+ 23b + 63 = 0

2b^{2}+ 14b + 9b + 63 = 0

So b = -7, -4.5

-7…. -4.5 ….-1.625…. 1. - Question 4 of 10
##### 4. Question

I. 4a

^{2}+ 19a + 21 = 0II. 2b

^{2}– 25b – 27 = 0Correct4a

^{2}+ 19a + 21 = 0

4a^{2}+ 12a + 7a + 21 = 0,

So a = -3, – 1.75

2b^{2}– 25b – 27 = 0

2b^{2}– 27b + 2b – 27 = 0

So b = 13.5, -1

-3…. -1.75….. -1…..13.5.Incorrect4a

^{2}+ 19a + 21 = 0

4a^{2}+ 12a + 7a + 21 = 0,

So a = -3, – 1.75

2b^{2}– 25b – 27 = 0

2b^{2}– 27b + 2b – 27 = 0

So b = 13.5, -1

-3…. -1.75….. -1…..13.5. - Question 5 of 10
##### 5. Question

I. a

^{2}– 9a + 20 = 0II. b

^{2}– 11b + 30 = 0Correcta

^{2}– 9a + 20 = 0

a^{2}– 4a – 5a + 20 = 0

So a = 4, 5

b^{2}– 11b + 30 = 0

b^{2}– 5b -6b + 30 = 0

So b = 5, 6

4 …. 5….5……6.Incorrecta

^{2}– 9a + 20 = 0

a^{2}– 4a – 5a + 20 = 0

So a = 4, 5

b^{2}– 11b + 30 = 0

b^{2}– 5b -6b + 30 = 0

So b = 5, 6

4 …. 5….5……6. - Question 6 of 10
##### 6. Question

Directions:6-10)

**In the following questions, two equations I and II are given. You have to solve both the equations and g****ive Answer as,**I. 4x

^{2}+ 17x + 18 = 0II. 4y

^{2}– 4y – 15 = 0CorrectI. 4x

^{2}+ 17x + 18 = 04x

^{2}+ 8x + 9x + 18 = 04x(x + 2) + 9(x + 2) = 0

(4x + 9) (x + 2) = 0

X = -9/4, -2 = -2.25, -2

II. 4y

^{2}– 4y – 15 = 04y

^{2}+ 6y – 10y – 15 = 02y (2y + 3) -5 (2y + 3) = 0

(2y – 5) (2y + 3) = 0

Y = 5/2, -3/2 = 2.5, -1.5

X < y

IncorrectI. 4x

^{2}+ 17x + 18 = 04x

^{2}+ 8x + 9x + 18 = 04x(x + 2) + 9(x + 2) = 0

(4x + 9) (x + 2) = 0

X = -9/4, -2 = -2.25, -2

II. 4y

^{2}– 4y – 15 = 04y

^{2}+ 6y – 10y – 15 = 02y (2y + 3) -5 (2y + 3) = 0

(2y – 5) (2y + 3) = 0

Y = 5/2, -3/2 = 2.5, -1.5

X < y

- Question 7 of 10
##### 7. Question

I. x

^{3}= 2197II. y

^{2}+ 26y + 169 = 0Correctx

^{3}= 2197X = 13

II. y

^{2}+ 26y + 169 = 0(y + 13) (y + 13) = 0

Y = -13, -13

X > y

Incorrectx

^{3}= 2197X = 13

II. y

^{2}+ 26y + 169 = 0(y + 13) (y + 13) = 0

Y = -13, -13

X > y

- Question 8 of 10
##### 8. Question

I. 3x = 5y – 5

II. x – 4y – 3 = 0

Correct3x = 5y – 5 —- > (1)

x – 4y – 3 = 0 –à (2)

By substituting (1) and (2), we get,

X = -5, y = -2

X < y

Incorrect3x = 5y – 5 —- > (1)

x – 4y – 3 = 0 –à (2)

By substituting (1) and (2), we get,

X = -5, y = -2

X < y

- Question 9 of 10
##### 9. Question

I. 3x

^{2}– 16x + 21 = 0II. 3y

^{2}– 18y + 27 = 0CorrectI. 3x

^{2}– 16x + 21 = 03x

^{2}– 9x – 7x + 21 = 03x(x – 3) -7(x – 3) = 0

(3x – 7) (x – 3) = 0

X = 7/3, 3 = 3, 2.33

II. 3y

^{2}– 18y + 27 = 03y

^{2}– 9y – 9y + 27 = 03y(y -3) -9(y – 3) = 0

(3y – 9) (y – 3) = 0

Y = 9/3, 3 = 3, 3

x ≤ y

IncorrectI. 3x

^{2}– 16x + 21 = 03x

^{2}– 9x – 7x + 21 = 03x(x – 3) -7(x – 3) = 0

(3x – 7) (x – 3) = 0

X = 7/3, 3 = 3, 2.33

II. 3y

^{2}– 18y + 27 = 03y

^{2}– 9y – 9y + 27 = 03y(y -3) -9(y – 3) = 0

(3y – 9) (y – 3) = 0

Y = 9/3, 3 = 3, 3

x ≤ y

- Question 10 of 10
##### 10. Question

I. x + √289 = √1089

II. y

^{1/2}= 64 ÷ yCorrectI. x + √289 = √1089

X + 17 = 33

X = 33 – 17 = 16

II. y

^{1/2}= 64 ÷ yY

^{1/2}* y = 64Y

^{3/2}= 64Y = (64)

^{2/3 }= (64^{1/3})^{2}Y = 4

^{2}= 16X = y

IncorrectI. x + √289 = √1089

X + 17 = 33

X = 33 – 17 = 16

II. y

^{1/2}= 64 ÷ yY

^{1/2}* y = 64Y

^{3/2}= 64Y = (64)

^{2/3 }= (64^{1/3})^{2}Y = 4

^{2}= 16X = y