In the following question, two equations numbered I and II are given. You have to solve both equations and give the answer. I. (x - 13)^2 = 0 II. y^2 = 169

x > y
x ≥ y
x < y
x ≤ y

Correct answer: x ≥ y

Solution

From (x - 13)^2 = 0, we get x = 13. From y^2 = 169, y can be 13 or -13. In both cases, x is greater than or equal to y, so the correct relation is x ≥ y.

Related IBPS PO Quantitative Aptitude questions

Directions (141-143): Solve the given equations and answer the question below. I. $Ax^2 + 5x + 6 = 0$ II. $By^2 - 13y + 20 = 0$ III. $u^2 + 7u + 12 = 0$ IV. $v^2 - 9v + 20 = 0$ Note: (i) The sum of the coefficients of $x^2$, $x$, $y^2$, and $y$ is -5. (ii) Roots of all the equations are non-imaginary. (iii) Both A and B are positive integers. Find the difference between the larger root of equation IV and the smaller root of equation II.
Q82. Solve both equations and form a new equation in variable 'z' (reduce to the lowest possible factor) using the roots of equations 1 and 2 as per the instructions given below. 1. $2 - \frac{1}{x} + \frac{9}{x^2} = 0$ 2. $(y - 2)^2 = 2\frac{1}{4}$ What will be the new equation if the roots are the highest root of equation 1 and the lowest root of equation 2?
Directions (66–68): Read the following quadratic equations carefully and answer the questions given below. (i) $x\times x - 3x - \sqrt{4x^2} = -6$ (ii) $y^2 - \sqrt{81y^2} = -4\times 5$ (iii) $\dfrac{z^2}{\sqrt{625z^6}} + (4\times 7) = 39z$ (iv) $p^2 - (3\times 5)p = 7\times (-8)$ Q66. Find the L.C.M. of the larger roots of $x$, $y$, $z$, and $p$. A) 840 B) 650 C) 780 D) 980 E) 1010
P and Q are the correct terms of series A and series B respectively. If $R$ is equal to the square of the larger root of the equation $x^2 - 15x = -9^2 + 5^2$, then which of the following statement(s) is/are correct? (i) $Q + R = P$ (ii) $\frac{P}{3} + 15 = R - Q$ (iii) $\sqrt{R} + P = 7Q - 5$ A) Only (i) & (ii) B) Only (ii) & (iii) C) Only (i) D) Only (ii) E) Only (iii)
Directions (106-108): Solve the quadratic and answer the following questions. A: $(x-2)^2 = (-3x)^2 + 2^2 + 25x - P$ B: $(10y^2 - 3y + \tfrac{2}{3}) \times 3 + 10y = 0$ One root of equation A is 5. Q106. $\frac{7P}{21} \times 0.2P - 89$ is equal to?
In each question, two equations numbered (I) and (II) are given. You have to solve both equations and mark the appropriate answer accordingly. (I) x^2 + 5x + 6 = 0 (II) y^2 - 5y + 6 = 0

⚔️ Practice IBPS PO Quantitative Aptitude free + battle 1v1 →