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

**Directions (1- 5): In the following questions, two equations I and II are given. You have to solve both the equations and g****ive answer**I. 4p² + 8p + 3 = 0

II. 4q² – 29q + 45 = 0Correct4p² + 8p + 3 = 0

4p^{2}+2p+6p+3 =0

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

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

p = -0.5, -1.5

4q² – 29q + 45 = 0

4q^{2}-20q-9q+45 =0

4q(q-5) -9 (q-5) =0

(4q-9) (q-5) =0

q = 2.25, 5

P < qIncorrect4p² + 8p + 3 = 0

4p^{2}+2p+6p+3 =0

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

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

p = -0.5, -1.5

4q² – 29q + 45 = 0

4q^{2}-20q-9q+45 =0

4q(q-5) -9 (q-5) =0

(4q-9) (q-5) =0

q = 2.25, 5

P < q - Question 2 of 10
##### 2. Question

I. 2p² – 23p + 21 = 0

II. q² + 42q + 272 = 0Correct2p² – 23p + 21 = 0

2p^{2}-21p-2p +21 =0

2p (p-1) -21 (p-1) =0

(2p-21)(p-1) =0

p = 10.5, 1

q² + 42q + 272 = 0

q^{2}+8q+34q+272 =0

q(q+8)+34 (q+8) =0

(q+8)(q+34)=0

q = -8, -34Incorrect2p² – 23p + 21 = 0

2p^{2}-21p-2p +21 =0

2p (p-1) -21 (p-1) =0

(2p-21)(p-1) =0

p = 10.5, 1

q² + 42q + 272 = 0

q^{2}+8q+34q+272 =0

q(q+8)+34 (q+8) =0

(q+8)(q+34)=0

q = -8, -34 - Question 3 of 10
##### 3. Question

**I.5p² – 26p + 21 = 0**

**II. 2q² – 17q + 21 = 0**Correct5p² – 26p + 21 = 0

5p^{2}-5p-21p+21=0

(5p-21) (p-1) =0

p = 4.2, 0.2

2q² – 17q + 21 = 0

2q^{2}-14q-3q+21 =0

2q (q-7) -3(q-21) =0

(2q-3) (q-7) =0

q = 7, 1.5

Relationship cannot be determinedIncorrect5p² – 26p + 21 = 0

5p^{2}-5p-21p+21=0

(5p-21) (p-1) =0

p = 4.2, 0.2

2q² – 17q + 21 = 0

2q^{2}-14q-3q+21 =0

2q (q-7) -3(q-21) =0

(2q-3) (q-7) =0

q = 7, 1.5

Relationship cannot be determined - Question 4 of 10
##### 4. Question

I. p² – 21p + 104 = 0

II. q² – 33q + 260 = 0Correctp² – 21p + 104 = 0

P^{2}-13p-8p+104 =0

(p-8) (p-13) =0

p = 13, 8

q² – 33q + 260 = 0

q^{2}– 13q- 20q + 260 =0

(q-13) (q-20) =0

q = 13, 20

P ≤ qIncorrectp² – 21p + 104 = 0

P^{2}-13p-8p+104 =0

(p-8) (p-13) =0

p = 13, 8

q² – 33q + 260 = 0

q^{2}– 13q- 20q + 260 =0

(q-13) (q-20) =0

q = 13, 20

P ≤ q - Question 5 of 10
##### 5. Question

I. p² – 31p + 240 = 0

II. q² – 28q + 195 = 0Correctp² – 31p + 240 = 0

p^{2}– 15p- 16p + 240 =0

(p-15) (p-16) =0

p = 15, 16

q² – 28q + 195 = 0

q^{2}– 13q – 15q +195 =0

(q-13) (q-15) =0

q = 13, 15

P ≥ qIncorrectp² – 31p + 240 = 0

p^{2}– 15p- 16p + 240 =0

(p-15) (p-16) =0

p = 15, 16

q² – 28q + 195 = 0

q^{2}– 13q – 15q +195 =0

(q-13) (q-15) =0

q = 13, 15

P ≥ q - Question 6 of 10
##### 6. Question

Directions (6-10):

**In the following questions, two equations numbered are given in variables p and q. You have to solve both the equations and find out the relationship between p and q. Then give answer accordingly****.**I. 3p

^{2}– 17p + 10 = 0

II. 3q^{2}+ 4q – 4 = 0Correct3p

^{2}– 17p + 10 = 0

3p^{2}– 15p – 2p + 10 = 0

Gives p = 2/3, 5

3q^{2}+ 4q – 4 = 0

3q^{2}+ 6q – 2q – 4 = 0

Gives q = -2, 2/3

-2…. 2/3…… 5

p ≥ qIncorrect3p

^{2}– 17p + 10 = 0

3p^{2}– 15p – 2p + 10 = 0

Gives p = 2/3, 5

3q^{2}+ 4q – 4 = 0

3q^{2}+ 6q – 2q – 4 = 0

Gives q = -2, 2/3

-2…. 2/3…… 5

p ≥ q - Question 7 of 10
##### 7. Question

I. 3p

^{2}– 14p + 8 = 0

II. 3q^{2}– 20q + 12 = 0Correct3p

^{2}– 14p +8 = 0

3p^{2}– 12p – 2p + 8 = 0

Gives p = 2/3, 4

3q^{2}– 20q + 12 = 0

3q^{2}– 18q – 2q + 12 = 0

Gives q = 6, 2/3

2/3….. 4….. 6

Relation cannot be determinedIncorrect3p

^{2}– 14p +8 = 0

3p^{2}– 12p – 2p + 8 = 0

Gives p = 2/3, 4

3q^{2}– 20q + 12 = 0

3q^{2}– 18q – 2q + 12 = 0

Gives q = 6, 2/3

2/3….. 4….. 6

Relation cannot be determined - Question 8 of 10
##### 8. Question

I. 3p

^{2}– 19p + 28 = 0

II. 4q^{2}– 5q – 6 = 0Correct3p

^{2}– 19p + 28 = 0

3p^{2}– 12p – 7p + 28 = 0

Gives p = 7/3, 4

4q^{2}– 5q – 6 = 0

4q^{2}– 8q + 3q – 6 = 0

Gives q = -3/4, 2

-3/4…… 2…… 7/3…. 4

p > qIncorrect3p

^{2}– 19p + 28 = 0

3p^{2}– 12p – 7p + 28 = 0

Gives p = 7/3, 4

4q^{2}– 5q – 6 = 0

4q^{2}– 8q + 3q – 6 = 0

Gives q = -3/4, 2

-3/4…… 2…… 7/3…. 4

p > q - Question 9 of 10
##### 9. Question

I. 6p

^{2}+ 23p + 21 = 0

II. 3q^{2}– 14q – 5 = 0Correct6p

^{2}+ 23p + 21 = 0

6p^{2}+ 9p + 14p + 21 = 0

Gives p = -7/3, -3/2

3q^{2}– 14q – 5 = 0

3q^{2}– 15q + q – 5 = 0

Gives q = -1/3, 5

-7/3…… -3/2….. -1/3…… 5

p < qIncorrect6p

^{2}+ 23p + 21 = 0

6p^{2}+ 9p + 14p + 21 = 0

Gives p = -7/3, -3/2

3q^{2}– 14q – 5 = 0

3q^{2}– 15q + q – 5 = 0

Gives q = -1/3, 5

-7/3…… -3/2….. -1/3…… 5

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

I. 2p

^{2}– 7p + 3 = 0

II. 2q^{2}+ 11q + 12 = 0Correct2p

^{2}– 7p + 3 = 0

2p^{2}– 6p – p + 3 = 0

Gives p = 3, 1/2

2q^{2}+ 11q + 12 = 0

2q^{2}+ 8q + 3q + 12 = 0

Gives q = -3/2, -4

-4……. -3/2…… 1/2….. 3

p > qIncorrect2p

^{2}– 7p + 3 = 0

2p^{2}– 6p – p + 3 = 0

Gives p = 3, 1/2

2q^{2}+ 11q + 12 = 0

2q^{2}+ 8q + 3q + 12 = 0

Gives q = -3/2, -4

-4……. -3/2…… 1/2….. 3

p > q