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

Directions: 1-5)

**In the following questions, two equations numbered I and II are given with variables a and b. Solve both the equations and find out the relationship between a and b. Then give answer accordingly****.**I. 3a

^{2}– a – 10 = 0

II. 3b^{2}– 11b + 6 = 0Correct3a

^{2}– a – 10 = 03a

^{2 }– 6a + 5a – 10 = 03a (a-2) + 5 (a-2) = 0

(3a+5) (a – 2) = 0

Gives a = -5/3, 2

3b^{2}– 11b + 6 = 03b

^{2}– 9b – 2b +6 =03b (b-3) – 2(b-3) =0

(3b -2) (b-3) =0

Gives b = 2/3, 3

Relationship cannot be determinedIncorrect3a

^{2}– a – 10 = 03a

^{2 }– 6a + 5a – 10 = 03a (a-2) + 5 (a-2) = 0

(3a+5) (a – 2) = 0

Gives a = -5/3, 2

3b^{2}– 11b + 6 = 03b

^{2}– 9b – 2b +6 =03b (b-3) – 2(b-3) =0

(3b -2) (b-3) =0

Gives b = 2/3, 3

Relationship cannot be determined - Question 2 of 10
##### 2. Question

I. 3a

^{2}– (3 – 2√2)a – 2√2 = 0

II. 3b^{2}– (1 + 3√3)b + √3 = 0Correct3a

^{2}– (3 – 2√2)a – 2√2 = 0

(3a^{2}– 3a) + (2√2a – 2√2) = 0

3a (a – 1) + 2√2 (a – 1) = 0

So a = 1, -2√2/3 (-0.9)

3b^{2}– (1 + 3√3)b + √3 = 0

(3b^{2}– b) – (3√3b – √3) = 0

b (3b – 1) – √3 (3b – 1) = 0

So, b = 1/3, √3 (1.7)Relationship cannot be determined

Incorrect3a

^{2}– (3 – 2√2)a – 2√2 = 0

(3a^{2}– 3a) + (2√2a – 2√2) = 0

3a (a – 1) + 2√2 (a – 1) = 0

So a = 1, -2√2/3 (-0.9)

3b^{2}– (1 + 3√3)b + √3 = 0

(3b^{2}– b) – (3√3b – √3) = 0

b (3b – 1) – √3 (3b – 1) = 0

So, b = 1/3, √3 (1.7)Relationship cannot be determined

- Question 3 of 10
##### 3. Question

I. a

^{2}+ (4 + √2)a + 4√2 = 0

II. 5b^{2}+ (2 + 5√2)b + 2√2 = 0Correcta

^{2}+ (4 + √2)a + 4√2 = 0

(a^{2}+ 4a) + (√2a + 4√2) = 0

a (a + 4) + √2 (a + 4) = 0

So a = -4, -√2 (-1.4)

5b^{2}+ (2 + 5√2)b + 2√2 = 0

(5b^{2}+ 2b) + (5√2b + 2√2) = 0

b (5b + 2) + √2 (5b + 2) = 0

So, b = -2/5 (-0.4), -√2 (-1.4)**a ≤ b**Incorrecta

^{2}+ (4 + √2)a + 4√2 = 0

(a^{2}+ 4a) + (√2a + 4√2) = 0

a (a + 4) + √2 (a + 4) = 0

So a = -4, -√2 (-1.4)

5b^{2}+ (2 + 5√2)b + 2√2 = 0

(5b^{2}+ 2b) + (5√2b + 2√2) = 0

b (5b + 2) + √2 (5b + 2) = 0

So, b = -2/5 (-0.4), -√2 (-1.4)**a ≤ b** - Question 4 of 10
##### 4. Question

I. 6a

^{2}– (3 + 4√3)a + 2√3 = 0,

II. 3b^{2}– (6 + 2√3)b + 4√3 = 0Correct6a

^{2}– (3 + 4√3)a + 2√3 = 0

(6a^{2}– 3a) – (4√3a – 2√3) = 0

3a (2a- 1) – 2√3 (2a – 1) = 0,

So a = 1/2, 2√3/3 (1.15)

3b^{2}– (6 + 2√3)b + 4√3 = 0

(3b^{2}– 6b) – (2√3b – 4√3) = 0

3b (b – 2) – 2√3 (b – 2) = 0

So, b = 2, 2√3/3**a ≤ b**Incorrect6a

^{2}– (3 + 4√3)a + 2√3 = 0

(6a^{2}– 3a) – (4√3a – 2√3) = 0

3a (2a- 1) – 2√3 (2a – 1) = 0,

So a = 1/2, 2√3/3 (1.15)

3b^{2}– (6 + 2√3)b + 4√3 = 0

(3b^{2}– 6b) – (2√3b – 4√3) = 0

3b (b – 2) – 2√3 (b – 2) = 0

So, b = 2, 2√3/3**a ≤ b** - Question 5 of 10
##### 5. Question

I. 8a

^{2}+ (4 + 2√2)a + √2 = 0

II. b^{2}– (3 + √3)b + 3√3 = 0Correct8a

^{2}+ (4 + 2√2)a + √2 = 0

(8a^{2}+ 4a) + (2√2a + √2) = 0

4a (2a + 1) + √2 (2a + 1) = 0

So a = -1/2 (-0.5), -√2/4 (-0.35)

b^{2}– (3 + √3)b + 3√3 = 0

(b^{2}– 3b) – (√3b – 3√3) = 0

b (b – 3) – √3 (b – 3) = 0

So b = 3, √3 (1.73)**a < b**Incorrect8a

^{2}+ (4 + 2√2)a + √2 = 0

(8a^{2}+ 4a) + (2√2a + √2) = 0

4a (2a + 1) + √2 (2a + 1) = 0

So a = -1/2 (-0.5), -√2/4 (-0.35)

b^{2}– (3 + √3)b + 3√3 = 0

(b^{2}– 3b) – (√3b – 3√3) = 0

b (b – 3) – √3 (b – 3) = 0

So b = 3, √3 (1.73)**a < b** - Question 6 of 10
##### 6. Question

Directions:6-10)

**In each of the following questions, two equations are given. You have to solve these questions and find out the values of p and q.**

I.16p² + 20p + 6 = 0

II.10q² + 38q + 24 = 0CorrectI. 16p² + 20p + 6 = 0

or, 8p² + 10p + 3 = 0

or, (4p + 3 ) ( 2p + 1) = 0

Therefore, p = -3/4 = -0.75 or p = – 1/2 = – 0.5II. 10q² + 38q + 24 = 0

or, 5q² + 19q + 12 = 0

or, ( q + 3) (5q + 4 ) = 0

Therefore, q = – 3 or q = – 4/5 = -0.8

Hence p > qIncorrectI. 16p² + 20p + 6 = 0

or, 8p² + 10p + 3 = 0

or, (4p + 3 ) ( 2p + 1) = 0

Therefore, p = -3/4 = -0.75 or p = – 1/2 = – 0.5II. 10q² + 38q + 24 = 0

or, 5q² + 19q + 12 = 0

or, ( q + 3) (5q + 4 ) = 0

Therefore, q = – 3 or q = – 4/5 = -0.8

Hence p > q - Question 7 of 10
##### 7. Question

I.18p² + 18p + 4 = 0

II.12q² + 29q + 14 = 0Correct- 18p² + 18p + 4 = 0

or, 9p² + 9p + 2 = 0

or, ( 3p + 2 ) ( 3p + 1) = 0

Therefore, p = – 2/3 = -0.67 or p = – 1/3 = -0.33 - 12q² + 29q + 14 = 0

or, (3q + 2) ( 4q + 7 ) = 0

Therefore, q = – 2/3 = -0.67 or q = -7/4 = -1.75

Hence p ≥ q

Incorrect- 18p² + 18p + 4 = 0

or, 9p² + 9p + 2 = 0

or, ( 3p + 2 ) ( 3p + 1) = 0

Therefore, p = – 2/3 = -0.67 or p = – 1/3 = -0.33 - 12q² + 29q + 14 = 0

or, (3q + 2) ( 4q + 7 ) = 0

Therefore, q = – 2/3 = -0.67 or q = -7/4 = -1.75

Hence p ≥ q

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

I.8p² + 6p = 5

II.12q² – 22q + 8 = 0CorrectI. 8p² + 6p – 5 = 0

or, (4p + 5) (2p – 1) = 0

Therefore, p = – 5/4 = -1.25 or p = 1/2 = 0.5II. 12q² – 22q + 8 = 0

or, 6q² -11q + 4 = 0

or, (2q -1) (3q – 4) = 0

Therefore, q = 1/2 = 0.5 or q = 4/3 = 1.33

Hence p ≤ qIncorrectI. 8p² + 6p – 5 = 0

or, (4p + 5) (2p – 1) = 0

Therefore, p = – 5/4 = -1.25 or p = 1/2 = 0.5II. 12q² – 22q + 8 = 0

or, 6q² -11q + 4 = 0

or, (2q -1) (3q – 4) = 0

Therefore, q = 1/2 = 0.5 or q = 4/3 = 1.33

Hence p ≤ q - Question 9 of 10
##### 9. Question

I.17p² + 48p – 9 = 0

II.13q² = 32q – 12CorrectI. 17p² + 48p – 9 = 0

or, (p + 3) (17p – 3) = 0

Therefore, p = -3 or p = 3/17 = 0.18II. 13q² – 32q + 12 = 0

or, (q -2) (13q – 6) = 0

Therefore, q = 2 or q = 6/13 = 0.46

Hence p < qIncorrectI. 17p² + 48p – 9 = 0

or, (p + 3) (17p – 3) = 0

Therefore, p = -3 or p = 3/17 = 0.18II. 13q² – 32q + 12 = 0

or, (q -2) (13q – 6) = 0

Therefore, q = 2 or q = 6/13 = 0.46

Hence p < q - Question 10 of 10
##### 10. Question

I.821p² – 757p² = 256

II.√196 q³ – 12q³ = 16Correct64p² = 256

or, p² = 4

or, p = ±2

AND

14q³ – 12q³ = 16

or, 2q³ = 16

or, q³ = 8

or, q = 2

Hence p ≤ qIncorrect64p² = 256

or, p² = 4

or, p = ±2

AND

14q³ – 12q³ = 16

or, 2q³ = 16

or, q³ = 8

or, q = 2

Hence p ≤ q