Find the effective rate which is equivalent to a normal rate of 10% p.a. compounded:
Given:
\[ (1.05)^2 = 1.1025, \quad (1.025)^4 = 1.1038. \]Step 1: For compounding semi-annually, the effective annual rate \( r_{{eff}} \) is calculated using the formula: \[ r_{{eff}} = \left(1 + \frac{r}{n}\right)^n - 1, \] where: - \( r \) is the nominal annual interest rate, - \( n \) is the number of compounding periods per year.
Step 2: Substitute the given values \( r = 0.10 \) (10%) and \( n = 2 \): \[ r_{{eff}} = \left(1 + \frac{0.10}{2}\right)^2 - 1 = (1.05)^2 - 1. \]
Step 3: Compute the value: \[ r_{{eff}} = 1.1025 - 1 = 0.1025 \quad {or} \quad 10.25\%. \]
Step 1: The effective annual rate \( r_{{eff}} \) for a given nominal rate \( r \) compounded \( n \) times per year is given by: \[ r_{{eff}} = \left(1 + \frac{r}{n}\right)^n - 1. \] where: - \( r = 0.10 \) (10% nominal rate), - \( n \) is the number of compounding periods per year.
Step 2: First, compute for semi-annual compounding (\( n = 2 \)): \[ r_{{eff}} = \left(1 + \frac{0.10}{2}\right)^2 - 1 = (1.05)^2 - 1. \]
Step 3: Compute the value: \[ r_{{eff}} = 1.1025 - 1 = 0.1025 \quad {or} \quad 10.25\%. \]
Step 4: Now, compute for quarterly compounding (\( n = 4 \)): \[ r_{{eff}} = \left(1 + \frac{0.10}{4}\right)^4 - 1 = (1.025)^4 - 1. \]
Step 5: Compute the value: \[ r_{{eff}} = 1.1038 - 1 = 0.1038 \quad {or} \quad 10.38\%. \]
Show that \( R \) is an equivalence relation. Also, write the equivalence class \([2]\).
List-I | List-II |
(A) Absolute maximum value | (I) 3 |
(B) Absolute minimum value | (II) 0 |
(C) Point of maxima | (III) -5 |
(D) Point of minima | (IV) 4 |
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
In a Linear Programming Problem (LPP), the objective function $Z = 2x + 5y$ is to be maximized under the following constraints:
\[ x + y \leq 4, \quad 3x + 3y \geq 18, \quad x, y \geq 0. \] Study the graph and select the correct option.