I- 12 p 2-7 p -1II- 6 q 2-7 q -2-0A- If p - qB- If p - qC- If p-q D- If p - qE- If p - q

The correct option is D If p lt; qEquation I 12p^2 7p = 1 ⇒ 12p^2 7p + 1 = 0 ⇒ 12p^2 4p 3p + 1 =0 ⇒ 4p (3p 1) 1(3p 1)=0 ⇒ (3p 1)(4p 1)=0 ∴ p = 1/3 ...

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Quantitative Aptitude

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I. 12 p 2-7 p...

Question

I. $12 p^{2} - 7 p = - 1$
II. $6 q^{2} - 7 q + 2 = 0$

p < q

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p > q

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p \leq q

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p \geq q

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p = q

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Solution

The correct option is A If $p < q$
Equation I
$12 p^{2} - 7 p = - 1$
$\Rightarrow 12 p^{2} - 7 p + 1 = 0$
$\Rightarrow 12 p^{2} - 4 p - 3 p + 1 = 0$
$\Rightarrow 4 p (3 p - 1) - 1 (3 p - 1) = 0$
$\Rightarrow (3 p - 1) (4 p - 1) = 0$
$∴ p = \frac{1}{3}$ or $p = \frac{1}{4}$

Equation II
$6 q^{2} - 7 q + 2 = 0$
$\Rightarrow 6 q^{2} - 4 q - 3 q + 2 = 0$
$\Rightarrow 2 q (3 q - 2) + (3 q - 2) = 0$
$\Rightarrow (3 q - 2) (2 q - 1) = 0$
$∴ q = \frac{2}{3}$ or $\frac{1}{2}$
Hence, $p < q$

Suggest Corrections

I. 12p2−7p=−1 II. 6q2−7q+2=0

The correct option is A If p<qEquation I 12p2−7p=−1 ⇒12p2−7p+1=0 ⇒12p2−4p−3p+1=0 ⇒4p(3p−1)−1(3p−1)=0 ⇒(3p−1)(4p−1)=0 ∴p=13 or p=14 Equation II 6q2−7q+2=0 ⇒6q2−4q−3q+2=0 ⇒2q(3q−2)+(3q−2)=0 ⇒(3q−2)(2q−1)=0 ∴q=23 or 12 Hence, p<q

I. $12 p^{2} - 7 p = - 1$
II. $6 q^{2} - 7 q + 2 = 0$