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

Directions:1-5)

**In the following questions, two equations I and II are given. Solve both the equations****and**g**ive Answer**I.(25 / p

^{2}) – (12 / p) + (9 / p^{2}) = (4 / p^{2})II.9.84 – 2.64 = 0.95 + q

^{2}CorrectI.(25/p

^{2}) + (9/p^{2 }) – (4/p^{2})^{ }= (12/p)(25 + 9 – 4) / p

^{2}= 12/p = 30/p^{2}= 12/p12p = 30

p = 30 / 12 = 5/2 = 2.5

II.9.84 -2.64 = 0.95 + q

^{2}7.2 – 0.95 = q

^{2}q = √(6.25) = ± (2.5)

Clearly p ≥ q

IncorrectI.(25/p

^{2}) + (9/p^{2 }) – (4/p^{2})^{ }= (12/p)(25 + 9 – 4) / p

^{2}= 12/p = 30/p^{2}= 12/p12p = 30

p = 30 / 12 = 5/2 = 2.5

II.9.84 -2.64 = 0.95 + q

^{2}7.2 – 0.95 = q

^{2}q = √(6.25) = ± (2.5)

Clearly p ≥ q

- Question 2 of 10
##### 2. Question

- √(900)p + √(1296) = 0
- (256)
^{1/4}q + (216)^{1/3}= 0

CorrectI.√(900)p + √(1296) =0

√(900)p = -√(1296)

30p = -36

p = -36 / 30 = -1.2

II.(256)

^{1/4}q = (216)^{1/3}(4

^{4})^{1/4}q = – (6^{3})^{1/3}à 4q = -6Q = -(6/4) = -1.5

Clearly, p > q

IncorrectI.√(900)p + √(1296) =0

√(900)p = -√(1296)

30p = -36

p = -36 / 30 = -1.2

II.(256)

^{1/4}q = (216)^{1/3}(4

^{4})^{1/4}q = – (6^{3})^{1/3}à 4q = -6Q = -(6/4) = -1.5

Clearly, p > q

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

I.[(3)

^{5}+ (7)^{3}]/ 3 = p^{3}II.7q

^{3}= – (15 × 2) + 17q^{3}^{ }CorrectI.[(3)

^{5}+ (7)^{3}] / 3 = p^{3}(243 + 343) / 3 = p

^{3}(586 / 3) = p

^{3}à p = 5.8II.7q

^{3}= -30 + 17q^{3}= 10q^{3}= 30q

^{3 }= 30/10 = 3 à q = 1.44Clearly, p > q

IncorrectI.[(3)

^{5}+ (7)^{3}] / 3 = p^{3}(243 + 343) / 3 = p

^{3}(586 / 3) = p

^{3}à p = 5.8II.7q

^{3}= -30 + 17q^{3}= 10q^{3}= 30q

^{3 }= 30/10 = 3 à q = 1.44Clearly, p > q

- Question 4 of 10
##### 4. Question

- (p
^{1/4}/ 16)^{2}= 144 / p^{3/2} - q
^{1/3 }× q^{2/3}× 3104 = 16 × q^{2}

CorrectI.(p

^{1/4}/ 16)^{2}= (144 / p^{3/2}) = (p^{1/2}/ 256) = (144 / p^{3/2})(p

^{1/2}) × (p^{3/2}) = 256 × 144p

^{2}= (256 × 144)p = √(256 × 144)

p = ± (16 × 12) = ±192

II.q

^{1/3}× q^{2/3}× 3104 = 16q^{2}q × 3104 = 16q

^{2}3104 = 16q

Q = 3104 / 16 = 194

Clearly, p < q

IncorrectI.(p

^{1/4}/ 16)^{2}= (144 / p^{3/2}) = (p^{1/2}/ 256) = (144 / p^{3/2})(p

^{1/2}) × (p^{3/2}) = 256 × 144p

^{2}= (256 × 144)p = √(256 × 144)

p = ± (16 × 12) = ±192

II.q

^{1/3}× q^{2/3}× 3104 = 16q^{2}q × 3104 = 16q

^{2}3104 = 16q

Q = 3104 / 16 = 194

Clearly, p < q

- (p
- Question 5 of 10
##### 5. Question

I.3p

^{2}– 19p +28 = 0II.5q

^{2}– 18q + 16 = 0CorrectI.3p

^{2}– 9p + 28 = 03p

^{2}– 12p – 7p + 28 = 03p (p – 4) – 7 (p – 4) = 0

(p – 4) (3p – 7) = 0

p = 4, 7/3

II.5q

^{2}– 18q + 16 = 05q

^{2}– 10q – 8q + 16 = 05q (q – 2) – 8 (q – 2) = 0

(q – 2) (5q – 8) = 0

Q = 2, 8/5

Clearly, p > q

IncorrectI.3p

^{2}– 9p + 28 = 03p

^{2}– 12p – 7p + 28 = 03p (p – 4) – 7 (p – 4) = 0

(p – 4) (3p – 7) = 0

p = 4, 7/3

II.5q

^{2}– 18q + 16 = 05q

^{2}– 10q – 8q + 16 = 05q (q – 2) – 8 (q – 2) = 0

(q – 2) (5q – 8) = 0

Q = 2, 8/5

Clearly, p > q

- Question 6 of 10
##### 6. Question

Durections:6-10)

**In each question, two equations are given. Find p and q and give the answer:**a)(p³ – 12p

^{2}– p + 12)/(p+1) = 0b)(q³ + 5q² – 2q – 24)/(q-2) = 0

Correcta)(p³ – 12p

^{2}– p + 12)/(p+1) = 0[p

^{2}(p-12) -1(p-12)]/(p+1) = 0[(p-1)(p+1)(p-12)]/(p+1) = 0

p² – 13p + 12 = 0

Solving, p = 12, 1

b)(q³ + 5q² – 2q – 24)/(q-2) = 0

(q³ + 7q² – 2q² -14q + 12q – 24)/(q-2) = 0

(q-2)(q² + 7q + 12)/(q-2) = 0

Solving, q = -4, -3

Thus, p>q

Incorrecta)(p³ – 12p

^{2}– p + 12)/(p+1) = 0[p

^{2}(p-12) -1(p-12)]/(p+1) = 0[(p-1)(p+1)(p-12)]/(p+1) = 0

p² – 13p + 12 = 0

Solving, p = 12, 1

b)(q³ + 5q² – 2q – 24)/(q-2) = 0

(q³ + 7q² – 2q² -14q + 12q – 24)/(q-2) = 0

(q-2)(q² + 7q + 12)/(q-2) = 0

Solving, q = -4, -3

Thus, p>q

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

a)q = 2p + 1

b)2q = 3p – 1

CorrectSolving the two using substitution method,

p = -3

q = -5

Thus, p>q

IncorrectSolving the two using substitution method,

p = -3

q = -5

Thus, p>q

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

a)9p² – 29p + 22 = 0

b)q² – 7q + 12 = 0

Correcta)9p² – 18p – 11p + 22 = 0

9p(p-2) – 11(p-2) = 0

(9p-11)(p-2) = 0

p = 11/9, 2.

b)q² – 4q – 3q + 12 = 0

q(q-4) – 3(q-4) = 0

(q-4)(q-3) = 0

q = 3, 4

Thus, p<q

Incorrecta)9p² – 18p – 11p + 22 = 0

9p(p-2) – 11(p-2) = 0

(9p-11)(p-2) = 0

p = 11/9, 2.

b)q² – 4q – 3q + 12 = 0

q(q-4) – 3(q-4) = 0

(q-4)(q-3) = 0

q = 3, 4

Thus, p<q

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

a)3p² – 4p – 32 = 0

b)12q² – 109q + 247 = 0

Correcta)3p² – 12p + 8p – 32 = 0

3p(p-4) + 8(p-4) = 0

(3p+8)(p-4) = 0

p = -8/3, 4

b)12q² – 52q – 57q + 247 = 0

4q (3q – 13) – 19(3q – 13) = 0

(4q-19)(3q-13) = 0

q = 19/4, 13/3

Thus, p<q

Incorrecta)3p² – 12p + 8p – 32 = 0

3p(p-4) + 8(p-4) = 0

(3p+8)(p-4) = 0

p = -8/3, 4

b)12q² – 52q – 57q + 247 = 0

4q (3q – 13) – 19(3q – 13) = 0

(4q-19)(3q-13) = 0

q = 19/4, 13/3

Thus, p<q

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

a)4p + 7q = 42

b)3p – 11q = -1

CorrectSol. Solving the two equations using the substitution method,

p = 7

q = 2

Thus, p>q

IncorrectSol. Solving the two equations using the substitution method,

p = 7

q = 2

Thus, p>q