In the following question, one or two equ… | Answer | TuitionWave

Subject

Quantitative Aptitude

Topic

Algebra

In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘p’ and ‘q’ and mark correct answer. 1. $\frac{{135}}{{\sqrt p }} - 22\sqrt p = 23\sqrt p$ 2. $\frac{{2 \times 324 \div \sqrt q }}{{108}} - 2\sqrt q = 0$
Reading

ক)

p > q

খ)

p ≥ q

গ)

p < q

ঘ)

p ≤ q

ঙ)

p = q or the relation cannot be determined

Explanation

We will solve both the equations separately. Equation I:

\frac{{135}}{{\sqrt p }} - 22\sqrt p = 23\sqrt p

Multiplying both sides by √p, we get ⇒ 135 – 22p = 23p ⇒ 23p + 22p = 135 ⇒ 45p = 135 ⇒ p = 135/45 ⇒ p = 3 Equation II:

\begin{array}{l} \frac{{2 \times 324 \div \sqrt q }}{{108}} - 2\sqrt q = 0\\ \Rightarrow \frac{{2 \times 324}}{{108 \times \sqrt q }} = 2\sqrt q \end{array}

Multiplying both sides by √q, we get

\Rightarrow q = \;\frac{{2 \times 324}}{{108 \times 2}}

⇒ q = 3 Comparing the values of p and q, we notice that p = q

\frac{a}{3} = \frac{b}{4} = \frac{c}{7}

then what would be the value of

\frac{{a + b + c}}{a}

?Multiplication and difference of two numbers are 93 and 28 respectively. Find the difference of reciprocals of the numbers.if

{P^2} + {1 \over {{P^2}}} = 7

, then find the value of

{P^2} - {1 \over {{P^2}}}.

In following question, two equations are given, you have to solve them and choose the correct option: x2– 3x – 4 = 0, y2– 4 = 0 In following question, two equations are given, you have to solve them and chose the correct option: x2+ 11x + 30 = 0, y2+ 7y + 12 = 0 In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘l’ and ‘m’ and mark correct answer. I. (125)1/3l – (1331)1/3= 4 × (729)1/6+ (1024)1/10II. (m)2– (15625)1/3= 0 Direction: In the following question two equations numbered I and II are given. You have to solve both the equations and give answers:I: x - √121 = 0 II: y2- 121 = 0 The number of solutions of x2+ 4|x| + 5 = 0 is

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