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: \[ r_{{eff}} = 1.1038 - 1 = 0.1038 \quad {or} \quad 10.38\%. \]
In number theory, it is often important to find factors of an integer \( N \). The number \( N \) has two trivial factors, namely 1 and \( N \). Any other factor, if it exists, is called a non-trivial factor of \( N \). Naresh has plotted a graph of some constraints (linear inequations) with points \( A(0, 50) \), \( B(20, 40) \), \( C(50, 100) \), \( D(0, 200) \), and \( E(100, 0) \). This graph is constructed using three non-trivial constraints and two trivial constraints. One of the non-trivial constraints is \( x + 2y \geq 100 \).
Based on the above information, answer the following questions:
On her birthday, Prema decides to donate some money to children of an orphanage home.
If there are 8 children less, everyone gets ₹ 10 more. However, if there are 16 children more, everyone gets ₹ 10 less. Let the number of children in the orphanage home be \( x \) and the amount to be donated to each child be \( y \).
Based on the above information, answer the following questions:
Let \( X \) denote the number of hours a Class 12 student studies during a randomly selected school day. The probability that \( X \) can take the values \( x_i \), for an unknown constant \( k \):
\[ P(X = x_i) = \begin{cases} 0.1, & {if } x_i = 0, \\ kx_i, & {if } x_i = 1 { or } 2, \\ k(5 - x_i), & {if } x_i = 3 { or } 4. \end{cases} \]A recent accounting graduate opened a new business and installed a computer system that costs ₹ 45,200. The computer system will be depreciated linearly over 3 years and will have a scrap value of ₹ 0.
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]