You wake up in a room with two floors, each having 3 puzzles. To escape, you must find every answer in the order: Red, Yellow, Green, Cyan, Navy, Pink, and combine them together.

**Part 1:**
The sequence is: 0, 1, 2, 4, 10, 34, Red. This appears to be a pattern sequence.

**Part 2:**
The sequence is: 1, 4, 7, 11, 15, 19, 24, Yellow. This also seems to be a pattern sequence.

**Part 3:**
Given sets are: [8, 2, 0, 8, 11], [12, 3, 3, 81, 21], [17, 4, 20, 1024, 1211], Green. You need to find the next set, add all the numbers, and if there's a fraction, take the denominator.

**Part 4:**
The sequence is: 0, 1, 5, 16, 41, 92, 189, Cyan.

**Part 5:**
The sequence is: 1, 3, 9, 28, 276, 867, Navy.

**Part 6:**
This part consists of math questions:
1. What is the largest possible number that has no "N" in it?
2. If all Roman numerals were arranged alphabetically, what would be the last?
3. What is the highest number that has no repeated letters (English)?
4. What is the highest whole number consisting entirely of alternating consonants and vowels?
5. What is the smallest natural number that uses a, e, I, o, and u in order?
6. Add all 3-lettered numbers. What is the result?
7. What is the smallest natural number whose name is spelled with 15 letters?
8. Add all the numbers that only use "E" as a vowel. What is the result?
9. What’s the first number that uses "A" as a vowel?

After answering these questions, you should get a 5-digit number.

**Question:** What is the number for every puzzle, and what is the code to escape?


To determine the limiting distance for a tree line, enter the diameter at breast height (DBH) and the plot radius factor (PRF) into a calculator.

## Limiting Distance Formula

The limiting distance (LD) is calculated using the equation: LD = DBH * PRF, where:
- LD is the limiting distance in feet (ft)
- DBH is the diameter at breast height of the tree in inches (in)
- PRF is the plot radius factor in feet per inch (ft/in)

## Calculating Limiting Distance

To calculate the limiting distance, follow these steps:
1. Determine the DBH of the tree.
2. Determine the PRF.
3. Use the formula LD = DBH * PRF to calculate the limiting distance.

Example:
- DBH: 12.5 inches
- PRF: 2.5 ft/in (note: the unit for PRF should be ft/in, not inches/ft, to be consistent with the formula)
- LD = 12.5 * 2.5
- LD = 31.25 ft


**Pulse Code Modulation**

Pulse code modulation (PCM) is a method of converting discrete signals into digital form. It is widely used in modern communication systems and is essential for digital communication. In PCM, each sampled value is converted into its binary equivalent and coded as a series of zeros and ones.

**Working Principle**

To convert a discrete signal into digital form, each discrete value must be represented as its binary equivalent. This binary number can be of any number of bits, with a greater number of bits providing better accuracy but also increasing memory and bandwidth usage. In this project, 3-bit binary numbers are used, representing 8 values (0 to 7). To ensure the signal values fall within this range, the signal is normalized by dividing it by its maximum amplitude and then multiplying it by 7.

**Types of Pulses**

There are three types of pulses used in this project: Return to Zero (RZ), Not Returning to Zero (NRZ), and Manchester.

*   **Return to Zero Pulse**: For a digital '1', the pulse is on for T/2 seconds and off for T/2 seconds. For a digital '0', the pulse is off for T seconds.
*   **Not Returning to Zero (NRZ) Pulse**: For a digital '1', the pulse is on for T seconds. For a digital '0', the pulse is off for T seconds.
*   **Manchester Pulse**: For a digital '1', the pulse is on for the first T/2 seconds and off for the next T/2 seconds. For a digital '0', the pulse is off for the first T/2 seconds and on for the next T/2 seconds. This pulse has the advantage of not having a DC component in the modulated signal but requires twice the bandwidth of the NRZ pulse.

**Algorithm and Code Explanation**

The project involves the following steps:

1.  Clear all existing windows and variables.
2.  Take the input signal from the user and store it in the variable 'signal'.
3.  Normalize the signal by dividing it by its maximum amplitude and multiplying it by 7.
4.  Define empty vectors 'binsignal' and 'pulsesignal' to represent the binary value and pulse output of each signal value, respectively.
5.  Convert each signal value to its binary equivalent using the dec2bin function.
6.  Convert the binary string to a matrix of zeros and ones by subtracting the ASCII value of '0' (48) from the binary string.
7.  Concatenate the binary matrix with the existing 'binsignal' and repeat the process until the entire signal is converted into binary values.
8.  Take input from the user to determine which type of pulse to use.
9.  Define the pulses for RZ, NRZ, and Manchester types.
10. Define an empty vector 'pulsesignal'.
11. Read 'binsignal' and concatenate the appropriate pulse with the 'pulsesignal' matrix.
12. Plot the output 'pulsesignal' using the plot() function.

**Output**

The project outputs the modulated signal in the form of a plot, showcasing the pulse code modulation of the input signal using the chosen pulse type.

**Project Source Code**

```matlab
clc; clear all; close all;
signal = input('Enter signal: ');
normsignal = (signal ./ max(signal)) * 7;
binsignal = [];
pulsesignal = [];

for i = 1:length(signal)
    binvalue = dec2bin(signal(i), 3);
    binvalue = binvalue - 48;
    binsignal = [binsignal binvalue];
end

pulse = input('Choose pulse: 1 = Return to Zero, 2 = Not return to Zero, 3 = Manchester');

if pulse == 1
    for i = 1:length(binsignal)
        if binsignal(i) == 1
            pulsesignal = [pulsesignal ones(1, 10) zeros(1, 10)];
        end
        if binsignal(i) == 0
            pulsesignal = [pulsesignal zeros(1, 20)];
        end
    end
end

if pulse == 2
    for i = 1:length(binsignal)
        if binsignal(i) == 1
            pulsesignal = [pulsesignal ones(1, 20)];
        end
        if binsignal(i) == 0
            pulsesignal = [pulsesignal zeros(1, 20)];
        end
    end
end

if pulse == 3
    for i = 1:length(binsignal)
        if binsignal(i) == 1
            pulsesignal = [pulsesignal ones(1, 10) (-1 * ones(1, 10))];
        end
        if binsignal(i) == 0
            pulsesignal = [pulsesignal (-1 * ones(1, 10)) ones(1, 10)];
        end
    end
end

plot(pulsesignal);
```

**Questions Related to This Article**

1.  What is pulse code modulation, and how is it used in modern communication systems?
2.  What are the different types of pulses used in pulse code modulation, and what are their advantages and disadvantages?
3.  How is the signal normalized in the project, and why is it necessary?
4.  What is the purpose of the dec2bin function in the project, and how is it used?
5.  How does the Manchester pulse differ from the Return to Zero and Not Returning to Zero pulses, and what are its advantages?

**Multiple Choice Questions**

1.  What is the primary purpose of pulse code modulation?
    *   A) To convert analog signals to digital signals
    *   B) To convert digital signals to analog signals
    *   C) To modulate digital signals using different pulse types
    *   D) To demodulate digital signals
2.  Which of the following pulses has no DC component in the modulated signal?
    *   A) Return to Zero pulse
    *   B) Not Returning to Zero pulse
    *   C) Manchester pulse
    *   D) None of the above
3.  What is the purpose of normalizing the signal in the project?
    *   A) To increase the signal amplitude
    *   B) To decrease the signal amplitude
    *   C) To ensure the signal values fall within a specific range
    *   D) To remove noise from the signal

**Answers**

1.  C) To modulate digital signals using different pulse types
2.  C) Manchester pulse
3.  C) To ensure the signal values fall within a specific range


We can represent a group of friends by drawing a graph, where each node represents a person, and an edge joins two nodes if those two people are friends.

**Problem 1:**
A graph shows a group of friends. Using the clues below, determine who's who:
- Alan has 3 friends: Barney, Charlie, and Daniel.
- Barney and Ed are both friends with Charlie.
- Ed is Frank's only friend.

**Problem 2:**
A second network of friends is given. Use the clues below to figure out who's who:
- Bella and Ciara are friends.
- Emily and Ciara are not friends.
- Bella is Fiona's only friend.
- Anna has more friends than anyone else.
- Daphne has three friends.
- Gill and Daphne are not friends.
- Emily has two friends.

**Questions to Consider:**
1. Did each problem have a unique solution?
2. Were there any clues you didn't need to use?
3. If you label each node with the number of friends the person has and add together all the numbers, what can you say about the answer? Can you explain why?
4. Can you design a puzzle with five friends, where some people have more than two friends, with a unique solution?

**Solution Approach:**
To solve these puzzles, list the known information and use it to start building the friendship chain. Then, apply the Guess and Check strategy to place the other people, regularly referring back to the known information to ensure consistency.

**Key Observation:**
When labeling each node with the number of friends and adding these numbers together, the sum can be related to the structure of the graph. In the case of a graph with a triangular pattern, the sum of the numbers can be derived from the number of nodes and the pattern's properties. For instance, in a graph with 7 nodes arranged in a triangular pattern, the expected sum would be close to the number of nodes multiplied by the number of connections per node, adjusted for any incomplete patterns.


In engineering, mathematics, and computer science, the integral is a powerful tool. A single-variable integral can yield the area under a curve, and a double-variable integral can provide an efficient way to compute the volume under a surface. Integrals also allow us to determine accumulated changes. For example, if we have a function f(t) and want to evaluate it at some positive t = b, then we can take f(0) and add on the change in f that happens between 0 and b. Mathematically, this can be expressed as:

f(b) = f(0) + ∫_{[0, b]} (df/dx) dx.

The integral used throughout this discussion is the Riemann integral. To construct the Riemann integral, we need to understand the concept of infimum and supremum of a set. The infimum of a set is its greatest lower bound, while the supremum is its least upper bound. Notationally, we write "sup A" to denote the supremum of A and "inf A" to denote the infimum of A.

Consider the set E of all numbers of the form 1/n, where n is an integer greater than or equal to 1. As n becomes larger, 1/n becomes closer to 0, but 0 itself is not in E. Therefore, E has no minimum element, but it has an infimum, namely 0. The supremum of E is 1, which is an element of E.

To construct the Riemann integral, we start with a single-variable function f(x) defined on a closed interval [a, b]. We assume that f(x) is bounded on [a, b] and continuous almost everywhere. We subdivide [a, b] into many pieces by defining a partition P, where P = {x_{0}, x_{1}, x_{2}, …, x_{n}}, with x_{0} = a and x_{n} = b.

We then define m_{j} = inf {f(t): x_{j-1} ≤ t ≤ x_{j}} and M_{j} = sup {f(t): x_{j-1} ≤ t ≤ x_{j}}. These represent the greatest lower bound and least upper bound of the set of all values of f(t) where t ranges between the endpoints of each subinterval defined by our partition P.

Let ∆x_{j} = x_{j} – x_{j-1}, which represents the width of each subinterval. We can approximate the area under the graph of f(x) using infinitely many infinitely thin rectangles. The upper Riemann sum of f with respect to the partition P is U(f, P) = ∑ M_{j} ∆x_{j}, and the lower Riemann sum of f with respect to the partition P is L(f, P) = ∑ m_{j} ∆x_{j}.

U(f, P) overestimates the area under the graph of f(x), while L(f, P) underestimates it. If the infimum of U(f, P) over all possible partitions of [a, b] is equal to the supremum of L(f, P) over all possible partitions of [a, b], then we denote that common limiting value as ∫_{[a, b] }f(x) dx, which is the Riemann integral of f over [a, b].

As we subdivide [a, b] into finer and finer partitions P, U(f, P) gets smaller and smaller, while L(f, P) gets bigger and bigger, both approaching the true value of ∫_{[a, b] }f(x) dx. Therefore, the Riemann integral is defined as the common value that U(f, P) and L(f, P) approach as we take partitions that become infinitely fine.


The half-angle identity calculator is an online tool used to find the value of half-angle using the value of the full angle. Users can choose between different units of angles. Half-angle refers to the half of an angle, represented as Sin (θ/2), Cos (θ/2), and Tan (θ/2).

The formulas used for half-angle in trigonometry are:
- Sin (θ/2) = ± √((1-cosθ)/2)
- Cos (θ/2) = ± √((1+cosθ)/2)
- Tan (θ/2) = ± Sin(θ) / (1 + Cos(θ))

To find half-angle, follow these steps:
1. Substitute the angle in the formula.
2. Calculate the result.

For example, find the half angle of sin 30 degrees:
Sin (θ/2) = ± √((1-cos(30))/2)
= ± √((1-0.866)/2)
= ± √(0.134/2)
= ± √0.067
= ± 0.2588

Note: A double angle calculator can be used to find the double angle of trigonometric identities.


### Problem Definition

A three-dimensional steel truss structure with 260 members and 76 nodes is given. The goal is to perform sizing and shape optimization of this truss. The topology of the truss is fixed, and the structure is designed under three independent load cases:
- Horizontal loads of 5kN in the positive x-direction
- Horizontal loads of 1kN in the positive y-direction
- Vertical loads of 5kN in the negative z-direction

The displacements of all nodes in x, y, and z directions are limited to ±25 mm. The material properties are:
- Density: 7.85 ton/m^3
- Elastic modulus: 200 GPa
- Yield stress: 248.2 MPa

The structure is designed according to AISC-LRFD 1994 regulations, considering the limit states of tensile yielding and compressive buckling.

### Design Variables and Constraints

The problem has 260 sizing variables (cross-sections of the truss members) and 10 shape variables (z-coordinates of the bottom nodes). The total number of design variables is 270. Each pair of nodes with the same x-coordinate will have the same height. The sizing variables can take integer values from 1 to 37, representing a section ID from a database of 37 pipe sections. The shape variables can take integer values from -25000 to 3500 mm.

### Optimization Objective and Constraints

The objective is to minimize the weight of the structure while satisfying the given constraints. The maximum number of objective function evaluations allowed is 200,000 times. Any optimization technique can be used, but commercial optimization software is not allowed. The optimization process must start from a random point in the design space.

### Structural Analysis and Design Function

A MATLAB function, ISCSO_2019, is provided for structural analysis and design. The function takes three input variables:
- Sections: a row vector (1×260) containing the sizing variables
- Coordinates: a row vector (1×10) containing the shape variables
- Flag: either 0 or 1, indicating whether graphical output is requested

The function returns three output variables:
- Weight: the net weight of the proposed structure
- Const_Vio_Stress: the relative amount of constraint violation based on stress constraints
- Const_Vio_Disp: the relative amount of constraint violation based on displacement constraints

### Submission Requirements

The final optimum design with all 270 design variables, a brief description of the optimization approach, and the code must be submitted in a single PDF file. The problem must be solved in the given units, and the submission must be made before the deadline.


To solve the system of equations, we first need to clarify the equations. The original equations are:
y = (3/5)x - 2
y = (x/10) + 1

These can be rewritten using the slope-intercept form (y = mx + b), where the slope of the first equation is 3/5 and the slope of the second equation is 1/10.

To avoid confusion, it's helpful to use grouping symbols or the three-line method:
y = 3x - 2 --- 5
y = x + 1 -- 10

To solve the system, we can set the two equations equal to each other:
(3/5)x - 2 = (x/10) + 1

Multiplying through by 10 to eliminate the fractions:
6x - 20 = x + 10

Subtracting x from both sides:
5x - 20 = 10

Adding 20 to both sides:
5x = 30

Dividing both sides by 5:
x = 6

Now, we can substitute x back into one of the original equations to find y:
y = (3/5)(6) - 2
y = 18/5 - 10/5
y = 8/5
y = 1.6

Therefore, the solution to the system of equations is x = 6 and y = 1.6 or 8/5.


To find the area of an irregular pentagon, several approaches can be taken depending on the information available.

1. **Given Side Lengths and Angles**: If the lengths of all sides and the angles between them are known, the area can be calculated using the formula for the area of a polygon: \(A = \frac{1}{2} \times \text{apothem} \times \text{perimeter}\), or by dividing the polygon into simpler shapes like triangles and rectangles.

2. **Given Parallel Sides**: If two sides are parallel, the pentagon can be divided into a rectangle and a triangle. The area of the rectangle is found by multiplying its length by its width, and the area of the triangle is found using the formula \(\frac{1}{2} \times \text{base} \times \text{height}\).

3. **Given Certain Angles and Side Lengths**: Knowing certain angles (like right angles) and side lengths can help in determining the area by breaking down the pentagon into simpler geometric shapes whose area formulas are known.

4. **No Known Angles or Limited Information**: Without specific angles or with limited side length information, determining the exact area may require additional geometric principles or assumptions about the shape.

**Specific Examples**:

- **Dale's Problem**: An irregular pentagon with sides of lengths 67ft, 39ft, 18ft, 35ft, and 49ft, where the 39ft and 18ft sides are parallel, and angles approximately 80 degrees and 100 degrees at the ends of the 67ft side. The area can be approached by dividing the pentagon into simpler shapes.

- **Joe's Barn Roof**: A potential irregular pentagon with a 12' base and other sides of 4' each. Feasibility and area calculation depend on the specific geometry and angles involved.

- **Amber's Hexagon**: Given vertices of a hexagon, the area can be found by dividing the hexagon into triangles and summing their areas using the formula \(A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\) for each triangle.

- **Richie's Fish Tank**: An irregular pentagon with sides 24", 24", 8", 8", and 32", and three right angles. The area can be calculated by breaking the shape into simpler figures like rectangles and triangles.

- **Steven's and Grace's Problems**: For irregular pentagons with known side lengths and some angles, the area can be calculated using trigonometry and the formula for the area of a polygon, or by dividing the pentagon into simpler shapes.

- **Kelly's and Jason's Problems**: Without specific angles or with incomplete side length information, additional geometric or trigonometric principles must be applied to find the area or missing side lengths.

**Calculating Area**:

- **With Known Angles and Sides**: Use \(A = \frac{1}{2}ab\sin(C)\) for each pair of sides and angle between them, summing these areas for the total.

- **Without Known Angles**: Consider the polygon's properties, such as dividing it into known shapes or using the Shoelace formula for the area of a simple polygon whose vertices are given.

Each problem's unique characteristics dictate the approach to finding the area of an irregular pentagon.


Tata Consultancy Services Placement Question Papers

1. Online Test:
   - Verbal section: Antonyms and Synonyms
   - Verbal reasoning
   - Quantitative aptitude
   - Critical reasoning
   - Analytical Reasoning
2. Interview:
   - Technical + HR
   - MR (Managerial Round)
3. Written Test:
   - Ray writes a two-digit number. He sees that the number exceeds 4 times the sum of its digits by 3. If the number is increased by 18, the result is the same as the number formed by reversing the digits.
     a) 35
     b) 42
     c) 49
     d) 57
   Solution: Let the two-digit number be xy.
             4(x + y) + 3 = 10x + y
             10x + y + 18 = 10y + x
             Solving the equations, we get x = 3 and y = 5
   - a, b, c are non-negative integers such that 28a + 30b + 31c = 365. a + b + c = ?
     a) Greater than 14
     b) Less than or equal to 11
     c) 13
     d) 12
   Solution: Number of months having 28 days = 1
             Number of months having 30 days = 4
             Number of months having 31 days = 7
             28 * 1 + 30 * 4 + 31 * 7 = 365
             Here, a = 1, b = 4, c = 7.
             a + b + c = 12
   - George can do a piece of work in 8 hours. Paul can do the same work in 10 hours, Hari can do the same work in 12 hours. George, Paul, and Hari start the same work at 9 am, while George stops at 11 am, the remaining two complete the work. What time will the work complete?
     a) 11.30 am
     b) 12 noon
     c) 12.30 pm
     d) 1 pm
   Solution: Let the total work = 120 units.
             As George completes this entire work in 8 hours, his capacity is 15 units/hour
             Similarly, the capacity of Paul is 12 units/hour
             the capacity of Hari is 10 units/hour
             All 3 started at 9 am and worked up to 11 am.
             So total work done up to 11 am = 2 * (15 + 12 + 10) = 74
             Remaining work = 120 - 74 = 46
             Now this work is to be done by Paul and Hari.
             46 / (12 + 10) = 2 hours (approx)
             So work gets completed at 1 pm
   - If x^y denotes x raised to the power y, Find the last two digits of (1141^3843) + (1961^4181)
     a) 02
     b) 82
     c) 42
     d) 22
   Solution: Remember 1 raised to any power will give 1 as the unit digit.
             To find the digit in the 10th place, we have to multiply, 10th digit in the base x unit digit in the power.
             So the Last two digits of the given expression = 21 + 61 = 82
   - J can dig a well in 16 days. P can dig a well in 24 days. J, P, H dig in 8 days. H alone can dig the well in how many days?
     a) 32
     b) 48
     c) 96
     d) 24
   Solution: Assume the total work = 48 units.
             Capacity of J = 48 / 16 = 3 units/day
             Capacity of P = 48 / 24 = 2 units/day
             Capacity of J, P, H = 48 / 8 = 6 units/day
             From the above capacity of H = 6 - 2 - 3 = 1
             So H takes 48 / 1 days = 48 days to dig the well
   - If a lemon and apple together cost Rs.12, tomato and a lemon cost Rs.4 and an apple costs Rs.8 more than a lemon. What is the cost of lemon?
     Solution: Let's denote the cost of lemon as L, apple as A, and tomato as T.
               L + A = 12
               T + L = 4
               L + 8 = A
               Taking 1 and 3, we get A = 10 and L = 2
   - 3 mangoes and 4 apples cost Rs.85. 5 apples and 6 peaches cost 122. 6 mangoes and 2 peaches cost Rs.144. What is the combined price of 1 apple, 1 peach, and 1 mango?
     a) 37
     b) 39
     c) 35
     d) 36
   Solution: 3m + 4a = 85
             5a + 6p = 122
             6m + 2p = 144
             (1) x 2 => 6m + 8a = 170
             4 x (3) => 6m + 2p = 144
             Solving we get 8a - 2p = 56
             (2) => 5a + 6p = 122
             3 x (4) = 24a - 6p = 168
             Solving we get a = 10, p = 12, m = 15
             So a + p + m = 37
   - An organization has 3 committees, only 2 persons are members of all 3 committees but every pair of committees has 3 members in common. what is the least possible number of members on any one committee?
     a) 4
     b) 5
     c) 6
     d) 1
   Solution: Total 4 members minimum required to serve only on one committee.
   - There are 5 sweets - Jammun, kaju, Peda, Ladu, Jilebi which can be consumed in 5 consecutive days. Monday to Friday.
     A person eats one sweet a day, based on the following constraints.
     (i) Ladu not eaten on Monday
     (ii) If Jamun is eaten on Monday, Ladu should be eaten on Friday.
     (iii) Peda is eaten the day following the day of eating Jilebi
     (iv) If Ladu eaten on Tuesday, kaju should be eaten on Monday
     based on above, peda can be eaten on any day except
     a) Tuesday
     b) Monday
     c) Wednesday
     d) Friday
   Solution: From the (iii) clue, peda must be eaten after jilebi. so Peda should not be eaten on Monday.
   - If YWVSQ is 25 - 23 - 21 - 19 - 17, Then MKIGF
     a) 13 - 11 - 8 - 7 - 6
     b) 1 - 2-3-5-7
     c) 9 - 8 - 7 - 6 - 5
     d) 7 - 8 - 5 - 3
   Solution: MKIGF = 13 - 11 - 9 - 7 - 6
   - Addition of 641 + 852 + 973 = 2456 is incorrect. What is the largest digit that can be changed to make the addition correct?
     a) 5
     b) 6
     c) 4
     d) 7
   Solution: 641
             852
             963
             ------
             2466
             Largest among tens place is 7, so 7 should be replaced by 6 to get 2456
   - Value of a scooter depreciates in such a way that its value at the end of each year is 3/4th of its value at the beginning of the same year. If the initial value of the scooter is 40,000, what is the value of the scooter at the end of 3 years.
     a) 23125
     b) 19000
     c) 13435
     d) 16875
   Solution: value of the scooter at the end of the year = 40000*(3/4)^3 = 16875
   - At the end of 1994, R was half as old as his grandmother. The sum of the years in which they were born is 3844. How old R was at the end of 1999
     a) 48
     b) 55
     c) 49
     d) 53
   Solution: In 1994, Assume the ages of GM and R = 2k, k then their birth years are 1994 - 2k, 1994 - k.
             But given that sum of these years is 3844.
             So 1994 - 2k + 1994 - k = 3844
             K = 48
             In 1999, the age of R is 48 + 5 = 53
   - When numbers are written in base b, we have 12 * 25 = 333, the value of b is?
     a) 8
     b) 6
     c) None
     d) 7
   Solution: Let the base = b
             So, (b+2)(2b+5) = (b+2)(2b+5)=3b^2+3b+3
             2b^2+9b+10=3b^2+3b+3
             b^2−6b−7=0
             Solving we get b = 7 or -1
             So b = 7
   - How many polynomials of degree >=1 satisfy f(x^2)=[f(x)]^2=f(f(x))
     a) more than 2
     b) 2
     c) 0
     d) 1
   Solution: Let f(x) = x^2
             f(x^2)=[x^2]^2=x^4
             (f(x))^2=[x^2]^2=x^4
             f(f(x))=f(x^2)=[x^2]^2=x^4
             Only 1
   - Figure shows an equilateral triangle of side of length 5 which is divided into several unit triangles. A valid path is a path from the triangle in the top row to the middle triangle in the bottom row such that the adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example is given below. How many such valid paths are there?
     a) 120
     b) 16
     c) 23
     d) 24
   Solution: Number of valid paths = (n-1) ! = (5-1)! = 24
   - In the question, A^B means, A raised to power B. If x*y^2*z < 0, then which one of the following statements must be true?
     (i) xz < 0
     (ii) z < 0
     (iii) xyz < 0
     a) (i) and (iii)
     b) (iii) only
     c) None
     d) (i) only
   Solution: As y^2 is always positive, x*y^2*z < 0 is possible only when xz < 0. Option d is correct.
   - The marked price of a coat was 40% less than the suggested retail price. Eesha purchased the coat for half the marked price at the fiftieth anniversary sale. What percentage less than the suggested retail price did Eesha pay?
     a) 60
     b) 20
     c) 70
     d) 30
   Solution: Let the retail price is Rs.100. then market price is (100-40) % of 100 = 60.
             Eesha purchased the coat for half of this price. i.e., 30 only.
             which is 70 less than the retail price. So Option C is correct.

Sample Questions:

1. If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?
   Solution: log 0.317=0.3332 and log 0.318=0.3364, then log 0.319=log0.318+(log0.318-log0.317) = 0.3396
2. A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
   Solution: x= 2 kg Packs
             y= 1 kg packs
             x + y = 150 .......... Eqn 1
             2x + y = 264 .......... Eqn 2
             Solve the Simultaneous equation; x = 114 so, y = 36
             ANS : Number of 2 kg Packs = 114.
3. My flight takes off at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
   a) 6:00 am
   b) 6:40am
   c) 7:40
   d) 7:00
   e) 8:00
   Solution: The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
             When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
             Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM
4. A plane moves from 9°N40°E to 9°N40°W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?
   Solution: Since it is moving from east to west longitude we need to add both i.e.,40+40=80
             multiply the ans by 4 =>80*4=320min
             convert this min to hours i.e., 5hrs 33min
             It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
             So the ans is 10am+2hr 30 min =>ans is 12:30 it will reach
5. The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent?
   Solution: see it doesn't matter that what the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....
6. A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes.
   Solution: (n/1000)*(n/10)*10+(n/100)
7. A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week
   Solution: Go to river catch fish
             4*9=36
             7*1=7
             2*5=10
             36+7+10=53...
             take what is given
             53 good days means --- 9 fishes so 53/9=4(remainder=17)
             if u assume 5 then there is no chance for bad days.
             fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10)
             if u assume 2 then there is no chance for bad days.
             bad days means -------5 fishes so remaining 10---10/5=2days.
             Ans: 4 good, 1 fair, 2 bad.
             ==== total 7 days.
             x+y+z=7--------- eq1
             9*x+7*y+5*z=53 -------eq2
             multiply eq 1 by 9, 9*x+9*y+9*z=35 -------------eq3
             from eq2 and eq3
             2*y+4*z=10-----eq4
             since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3
             then z = 2 or 1
             from eq 4
             for first y=1,z=2 then from eq1 x= 4 so 9*4+1*7+2*5=53.... satisfied
             now for second y=3 z=1 then from eq1 x=3 so 9*3+3*7+1*5=53 ......satisfied
             so finally there are two solution of this question (x,y,z)=(4,1,2) and (3,3,1)...
8. Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then the number of fishes caught by X?
   Solution: Let no. of fish x catches=p
             no. caught by y =r
             r=5p.
             r+p=42
             then p=7,r=35
9. Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
   Solution: suppose total income is 100
             so amount x is getting is 80
             y is 70
             z =60
             total=210
             but total money is 300
             300-210=90
             so they are getting 90 rs less
             90 is 30% of 300
             so they r getting 30% discount
10. The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
    Solution: incomes:3:4
              expenditures:4:5
              3x-4y=1/4(3x)
              12x-16y=3x
              9x=16y
              y=9x/16
              (3x-4(9x/16))/((4x-5(9x/16)))
              ans:12/19
11. If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?
    Solution: g(2)=g(1)-g(0)=1+1=2
              g(3)=1
              g(4)=-1
              g(5)=-2
              g(6)=-1
12. If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages?
    Solution: A can copy 50 pages in 10 hrs.
             A can copy 5 pages in 1hr.(50/10)
             now A & B can copy 70 pages in 10hrs.
             thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x--> no. of pages B can copy in 10 hrs.]
             so, B can copy 9 pages in 1hr.
             therefore, to copy 26 pages B will need almost 3hrs.
             since in 3hrs B can copy 27 pages.
13. A, B and C are 8 bit no's. They are as follows:
    A -> 1 1 0 0 0 1 0 1
    B -> 0 0 1 1 0 0 1 1
    C -> 0 0 1 1 1 0 1 0
    ( - =minus, u=union)
    Find ((A - C) u B) =?
    Solution: To find A-C, We will find 2's compliment of C and them add it with A, That will give us (A-C)
              2's compliment of C=1's compliment of C+1 =11000101+1=11000110
              A-C=11000101+11000110 =10001001
              Now (A-C) U B is .OR. logic operation on (A-C) and B
              10001001 .OR . 00110011
              The answer is = 10111011, Whose decimal equivalent is 187.
14. One circular array is given(means memory allocation takes place in circular fashion) dimension(9X7) and starting add. is 3000, What is the address of (2,3)?
    Solution: it's a 9x7 int array so it requires a 126 bytes for storing.b'ze integer value need 2 bytes of memory allocation.
             and starting add is 3000 so starting add of 2x3 will be 3012.
15. In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).
    Solution: initial x (1,1) = 3000
             u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elements
             u need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 -1) * 4 = 212 -----( -1 is to deduct the 1 element )
             3000 + 212 = 3212
16. Which of the following is a power of 3
    a) 2345
    b) 9875
    c) 6504
    d) 9833
17. The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ?
    Solution: M=sqrt(100N)
             N is increased by 1% therefore new value of N=N + (N/100) =101N/100
             M=sqrt(100 * (101N/100) )
             Hence, we get M=sqrt(101 * N)
18. 1)SCOOTER --------- AUTOMOBILE--- A. PART OF
    2.OXYGEN----------- WATER ------- B. A Type of
    3.SHOP STAFF------- FITTERS------ C. NOT A TYPE OF
    4. BUG -------------REPTILE------ D. A SUPERSET OF
    1)B 2)A 3)D 4)C
19. A bus started from the bus stand at 8.00 am and after 30 min staying at the destination, it returned back to the bus stand. the destination is 27 miles from the bus stand. the speed of the bus 50 percent fast speed. at what time it returns to the bus stand
    Solution: a bus cover 27 miles with 18 mph in =27/18= 1 hour 30 min.
             and it wait at stand =30 min.
             after this speed of return increase by 50% so 50%of 18 mph=9mph
             Total speed of returning=18+9=27
             Then in return it takes 27/27=1 hour
             then total time in journey=1+1:30+00:30 =3 hour
             so it will come at 8+3 hour=11 a.m.
             So Ans==11 a.m
20. In two-dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of the first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?
    Solution: Here, the address of the first element x[1][1] is 1258 and also 2 byte of memory is given.
             now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338
             so the answer is 1338.
21. The temperature at Mumbai is given by the function: -t2/6+4t+12 where t is the elapsed time since midnight. What is the percentage rise (or fall) in temperature between 5.00PM and 8.00PM?
22. Low temperature at the night in a city is 1/3 more than 1/2 high as higher temperature in a day. Sum of the low temperature and highest temp. is 100 degrees. Then what is the low temp?
    Solution: Let highest temp be x so low temp=1/3 of x of 1/2 of x plus x/2
             i.e. x/6+x/2
             total temp=x+x/6+x/2=100
             therefore, x=60
             Lowest temp is 40
23. In Madras, temperature at noon varies according to -t^2/2 + 8t + 3, where t is elapsed time. Find how much temperature more or less in 4pm to 9pm.
    Solution: In equation first put t=9, we will get 34.5...........................(1)
             now put t=4, we will get 27..............................(2)
             so ans=34.5-27 =7.5
24. A person had to multiply two numbers. Instead of multiplying by 35, he multiplied by 53 and the product went up by 540. What was the raised product?
    a) 780
    b) 1040
    c) 1590
    d) 1720
    Solution: x*53-x*35=540=> x=30
             therefore, 53*30=1590
             Ans
25. How many positive integer solutions does the equation 2x+3y = 100 have?
    a) 50
    b) 33
    c) 16
    d) 35
    Solution: There is a simple way to answer this kind of Q's
             given 2x+3y=100, take l.c.m of 'x' coeff and 'y' coeff
             i.e. l.c.m of 2,3 ==6
             then divide 100 with 6 , which turns out 16
             hence answer is 16
             short cut formula--- constant / (l.cm of x coeff and y coeff)
26. The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs.70 per head when there are 25 boarders and Rs.60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
    a) 65
    b) 55
    c) 50
    d) 45
    Solution: Let a = fixed cost and k = variable cost and n = number of boarders
             total cost when 25 boarders c = 25*70 = 1750
             i.e. 1750 = a + 25k
             total cost when 50 boarders c = 50*60 = 3000
             i.e. 3000 = a + 50k
             solving above 2 eqns, 3000-1750 = 25k
             i.e. 1250 = 25k
             i.e. k = 50
             therefore, substituting this value of k in either of above 2 eqns we get a = 500
             (a = 3000-50*50 = 500 or a = 1750 - 25*50 = 500)
             so total cost when 100 boarders = c = a + 100k = 500 + 100*50 = 5500
             so cost per head = 5500/100 = 55
27. Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
    a) 37.5%
    b) 62.5%
    c) 50%
    d) None of these
    Solution: Let, 5 pens + 7 pencils + 4 erasers = x rupees
             so 10 pens + 14 pencils + 8 erasers = 2*x rupees
             also mentioned, 6 pens + 14 pencils + 8 erasers = 1.5*x rupees
             so (10-6) = 4 pens = (2-1.5)x rupees
             so 4 pens = 0.5x rupees
             => 8 pens = x rupees
             so 5 pens = 5x/8 rupees = 5/8 of total (note x rupees is total amt paid by Amal)
             i.e 5/8 = 500/8% = 62.5%
             is the answer
28. I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend lost Rs.4 more than me in the second race. What is the amount lost by my friend in the second race?
    Solution: x + x+6 = rs 68
             2x + 6 = 68
             2x = 68-6
             2x = 62
             x=31
             x is the amt lost in I race
             x+ 6 = 31+6=37 is lost in second race
             then my friend lost 37 + 4 = 41 Rs
29. Ten boxes are there. Each ball weighs 100 gms. One ball is weighing 90 gms.
    i) If there are 3 balls (n=3) in each box, how many times will it take to find 90 gms ball?
    ii) Same question with n=10
    iii) Same question with n=9
    Solution: to me the chances are
             when n=3 (i) nC1= 3C1 =3 for 10 boxes .. 10*3=30
             (ii) 10C1=10 for 10 boxes ....10*10=100
             (iii)9C1=9 for 10 boxes .....10*9=90
30. (1-1/6) (1-1/7).... (1- (1/ (n+4))) (1-(1/ (n+5))) = ?
    Solution: leaving the first numerator and last denominator, all the numerator and denominator will cancel out one another.
             Ans. 5/(n+5)
31. A face of the clock is divided into three parts. First part hours total is equal to the sum of the second and third part. What is the total of hours in the bigger part?
    Solution: the clock normally has 12 hr
             three parts x,y,z
             x+y+z=12
             x=y+z
             2x=12
             x=6
             so the largest part is 6 hrs
32. With 4/5 full tank vehicle travels 12 miles, with 1/3 full tank how much distance travels
    Solution: 4/5 full tank= 12 mile
             1 full tank= 12/(4/5)
             1/3 full tank= 12/(4/5)*(1/3)= 5 miles
33. wind blows 160 miles in 330min.for 80 miles how much time required
    Solution: 160 miles= 330 min
             1 mile = 330/160
             80 miles=(330*80)/160=165 min.
34. A person was fined for exceeding the speed limit by 10mph.another person was also fined for exceeding the same speed limit by twice the same if the second person was traveling at a speed of 35 mph. find the speed limit
    Solution: (x+10)=(x+35)/2
             solving the eqn we get x=15
35. A sales person multiplied a number and get the answer is 3 instead of that number divided by 3. what is the answer he actually has to get.
    Solution: Assume 1
             1* 3 = 3
             1*1/3=1/3
             so he has to got 1/3
             this is the exact answer
36. A person who decided to go weekend trip should not exceed 8 hours driving in a day
             average speed of forward journey is 40 mph
             due to traffic in Sundays the return journey average speed is 30 mph.
             How far he can select a picnic spot.
37. Low temperature at the night in a city is 1/3 more than 1/2 hinge as higher temperature in a day. Sum of the low temp and high temp is 100 c. then what is the low temp.
    Solution: let x be the highest temp.
             then, x+x/2+x/6=100.
             therefore, x=60
             which is the highest temp and 100-x=40
             which is the lowest temp.
38. car is filled with four and half gallons of oil for full round trip. Fuel is taken 1/4 gallons more in going than coming. What is the fuel consumed in coming up.
    Solution: let feul consumed in coming up is x.
             thus equation is: x+1.25x=4.5
             ans:2gallons
39. A work is done by the people in 24 min. One of them can do this work alone in 40 min. How much time required to do the same work for the second person
    Solution: Two people work together in 24 mins.
             So, their one day work is (1/A)+(1/B)=(1/24)
             One man can complete the work in 40mins
             one man's one day work (1/B)= (1/40)
             Now, (1/A)=(1/24)-(1/40)
             (1/A)=(1/60)
             So, A can complete the work in 60 mins.
40. In a company 30% are supervisors and 40% employees are male if 60% of supervisors are male. What is the probability?
             That a randomly chosen employee is a male or female?
    Solution: 40% employees are male
             if 60% of supervisors are male so for 100% is 26.4%
             so the probability is 0.264
41. In 80 coins one coin is counterfeit what is minimum number of weighing to find out counterfeit coin
    Solution: the minimum number of weighings needed is just 5.as shown below
             (1) 80->30-30
             (2) 15-15
             (3) 7-7
             (4) 3-3
             (5) 1-1
42. 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas, and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples?
    Solution: 2x+3y+4z=15
             3x+2y+z=10
             adding
             5x+5y+5z=25
             x+y+z=5
             that is for 1 orange, 1 banana and 1 apple requires 5Rs.
             so for 3 orange, 3 banana and 3 apple requires 15Rs.
             i.e. 3x+3y+3z=15
43. In 8*8 chess board what is the total number of squares refers
    Solution: odele discovered that there are 204 squares on the board
             We found that you would add the different squares - 1 + 4 + 9 + 16+ 25 + 36 + 49 + 64.
             Also in 3*3 tic tac toe board what is the total no of squares
             Ans 14
             ie 9+4(bigger ones)+1 (biggest one)
             If you get 100*100 board just use the formula
             the formula for the sum of the first n perfect squares is n x (n + 1) x (2n + 1)
             ______________________ 6
             if in this formula if you put n=8 you get your answer 204
44. One fast typist type some matter in 2hr and another slow typist type the same matter in 3hr. If both do combinely in how much time they will finish.
    Solution: Faster one can do 1/2 of work in one hours
             lower one can do 1/3 of work in one hour
             both they do (1/2+1/3=5/6) th work in one hour.
             so work will b finished in 6/5=1.2 hour
             i e 1 hour 12 min.
45. If Rs20/- is available to pay for typing a research report & typist A produces 42 pages and typist B produces 28 pages. How much should typist A receive?
    Solution: Find of 42 % of 20 rs with respect to 70 (i.e 28 + 42)
             ==> (42 * 20 )/70
             ==> 12 Rs
46. An officer kept files on his table at various times in the order 1,2,3,4,5,6. Typist can take file from top whenever she has time and type it.What order she can't type.?
47. In some game 139 members have participated every time one fellow will get bye what is the number of matches to choose the champion to be held?
    Solution: since one player gets a bye in each round,he will reach the finals of the tournament without playing a match.
             therefore 137 matches should be played to determine the second finalist from the remaining 138 players(excluding the 1st player)
             therefore to determine the winner 138 matches shd be played.
48. One rectangular plate with length 8inches, breadth 11 inches and 2 inches thickness is there. What is the length of the circular rod with diameter 8 inches and equal to volume of rectangular plate?
    Solution: Vol. of rect. plate= 8*11*2=176
             area of rod=(22/7)*(8/2)*(8/2)=(352/7)
             vol. of rod=area*length=vol. of plate
             so length of rod= vol of plate/area=176/(352/7)=3.5
49. One tank will fill in 6 minutes at the rate of 3cu ft /min, length of tank is 4 ft and the width is 1/2 of length, what is the depth of the tank?
    Solution: 3 ft 7.5 inches
50. A man has to get air-mail. He starts to go to the airport on his motorbike. Plane comes early and the mail is sent by a horse-cart. The man meets the cart in the middle after half an hour. He takes the mail and returns back, by doing so, he saves twenty minutes. How early did the plane arrive?
    Solution: assume he started at 1:00,so at 1:30 he met cart.
             He returned home at 2:00.so it took him 1 hour for the total journey.
             by doing this he saved 20 min.so the actual time if the plane is not late is 1 hour and 20 min.
             so the actual time of plane is at 1:40.
             The cart traveled a time of 10 min before it met him.so the plane is 10 min early.


Let us suppose the number be x. The phrase "The sum of \(8\frac{3}{4}\)" means add the number x to \(8\frac{3}{4}\). The following algebraic expression will be formed: \(x + 8\frac{3}{4}\). This expression represents adding x to \(8\frac{3}{4}\), resulting in the algebraic expression \(x + 8\frac{3}{4}\). The algebraic expression is \(x + 8\frac{3}{4}\).


# Grade 4 Math

## Course Overview
The Acellus Grade 4 Math course covers whole numbers, decimals, and fractions through composition, decomposition, and conversion. It also explores addition, subtraction, data representation, money, 2D shapes, and measurements. The course is taught by Acellus Instructor Mark Rogers and was developed by the International Academy of Science.

## Scope and Sequence
The course consists of 15 units:

1. **Place Value of Whole Numbers**: Students examine digits, values, and periods of whole numbers, compose and decompose numbers, and explore rounding.
2. **Place Value of Decimals**: Learners study digits, values, and periods of decimal numbers, compose and decompose decimals, and work with number lines.
3. **Decimals and Fraction Comparison**: Students learn to convert decimals to fractions, model decimal and fraction mixed numbers, and compare numbers.
4. **Addition and Subtraction**: This unit covers standard algorithms for addition and subtraction, regrouping, and decimal addition and subtraction.
5. **Products**: Learners multiply by tens and hundreds, multiply two-, three-, and four-digit numbers, and explore two-digit by two-digit multiplication.
6. **Division**: Students represent quotients with arrays and area models, learn divisibility rules, and study standard algorithms for division.
7. **Unknown Quantities**: Learners explore variables, represent sums and differences with equations, and discover multi-step operations.
8. **Sequences and Tables**: Students examine numeric sequences, explore addition, multiplication, subtraction, and division sequences, and extend sequences to solve for the nth term.
9. **Comparing Fractions**: Learners compose and decompose fractions, explore equivalent fractions, and compare fractions with different numerators and denominators.
10. **Operations on Fractions**: Students discover benchmarks of fractions, add and subtract fractions, and multiply fractions by whole numbers.
11. **Visual Representations of Data**: Learners explore frequency tables, dot plots, and stem-and-leaf plots, and solve two-step problems.
12. **Constructing and Measuring Angles**: Students discover points, lines, line segments, rays, and angles, and learn to measure angles with a protractor.
13. **2D Shapes**: Learners examine lines of symmetry, classify triangles, and find perimeters of rectangles and squares.
14. **Measurement**: Students explore relative sizes of length, mass, and volume, convert units, and study measurement applications.
15. **Money and Time**: The final unit covers converting hours to minutes, exploring elapsed time, and studying money, including change, sales, expenses, and profit.

The course concludes with a Final Review and Exam.


# The Basics of Annuity Calculators

The true value of an annuity is determined by factors such as the amount invested, type of annuity (guaranteed or insured), and interest rates over the contract length. When shopping for annuities, it's essential to compare them side-by-side, considering the payment structure, terms, and other details.

The Present Value of an Annuity is the current value of a stream of future annuity payments at a given interest rate or discount rate. A higher discount rate generally results in a lower annuity value. For example, a regular monthly payment of a dollar amount for 30 years will yield a total value of $1,000 at the end of the period, or $120 per month if paid monthly.

Annuity calculations involve comparing the value of different annuities based on their payment structure, terms, and other details. When making a lump-sum payment, it may be more affordable to buy an aligned annuity in exchange for a higher present value. The Annual Percentage Rate (APR) and Fixed Rate Annuity Payment Formula are crucial in determining the annuity's value. Parties can agree on a structured payment structure, and the APRs and rates can be set accordingly.

The Present Value of the Annuity Due is the amount of money a person is expected to receive upon death. For a fixed annuity due, this amount is determined after all payments have been received, while for an indexed annuity due, it's calculated using premiums paid and life expectancy. This information is typically found in Life Insurance Company periodic reports.

The actual amount received from an annuity depends on the plan's terms and type of annuity invested in. While future payments are uncertain, individuals have some control over them. For variable rate annuities, the present value formula can be used to determine a locked-in rate. Key factors influencing annuity value include:

* Amount invested
* Type of annuity (guaranteed or insured)
* Interest rates over the contract length
* Payment structure
* Terms of the agreement
* Annual Percentage Rate (APR)
* Life expectancy (for indexed annuity due)

Understanding these factors and using annuity calculators can help individuals make informed decisions when purchasing an annuity.


**Kinematics in Two Dimensions**

This chapter covers the kinematics of a particle in two dimensions, including displacement, velocity, and acceleration. The four kinematic equations are still applicable, but in vector form. The superposition principle allows us to treat vector components independently.

**Key Concepts**

* Displacement: the straight line between the final and initial position of a particle, represented by the vector difference between the final and initial position.
* Average Velocity: displacement divided by time interval.
* Instantaneous Velocity: the limit of average velocity as time approaches zero, equal to the derivative of the position vector with respect to time.
* Average Acceleration: change in velocity divided by time interval.
* Instantaneous Acceleration: the limiting value of the ratio of change in velocity to time as time approaches zero, equal to the derivative of the velocity vector with respect to time.

**Two-Dimensional Motion with Constant Acceleration**

The equations of motion are still valid, but in vector form. The superposition principle allows us to break down vector equations into their x and y components and calculate them independently.

**Projectile Motion**

Projectile motion assumes free fall acceleration (g) is constant and air resistance is negligible. The path of a projectile is a parabola. The velocity vector changes in magnitude and direction, with acceleration in the y-direction being g and acceleration in the x-direction being 0.

**Quick Quizzes**

1. You throw two rocks straight up in the air. One reaches a height of 10m, the other a height of 20m. Which one will be in the air longer?
A. 10m
B. 20m
C. The same time
D. Need more information
E. Depends on weight

Answer: C. The same time

2. Is it possible that a particle with constant speed experiences an acceleration?
A. Yes
B. No

Answer: A. Yes

3. A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship gets hit first?
A. A
B. B
C. Both hit at the same time
D. Need more information

Answer: C. Both hit at the same time

4. You shoot a bullet with 600 m/s straight at a target that is 600 m away. By how much does the bullet ‘drop’ during its flight?
A. 0 m
B. 0.49 m
C. 0.98 m
D. 4.9 m
E. 9.8 m

Answer: B. 0.49 m

**Examples**

1. A melon truck brakes right before a ravine and loses a few melons. The melons skit over the edge with an initial velocity of vx = 10.0 m/s.
* Determine the x and y coordinates of the particle at any time t and the position vector r at any time t.
* Determine the x and y components of the velocity at any time and the total velocity at any time.
* Calculate the impact angle, the velocity, and the speed of the melons as they hit the bottom of the ravine.

2. A rescue plane drops a package to stranded explorers. The plane is traveling horizontally at 40.0 m/s and is 100 m above the ground.
* Where does the package strike the ground relative to the point at which it was released?
* What are the horizontal and vertical components of the velocity of the package just before it hits the ground? What is the speed of the package as it hits the ground?
* Where is the plane when the package hits the ground? (Assume that the plane does not change its speed or course.)

**Reading Assignment**

Chapter 4.1, 4.2, 4.3
4.4 and 4.5 will be combined with chapter 6, skip 4.6

**Homework**

Due Thursday, Sept. 20: QQ3, AE1, AE4, 3, 5, 7, 9, 16, 21, 62

Remember: Homework 3.1 is due today (Sept. 13) and Homework 3.2 is due Tuesday (Sept. 18)


The flexible budget variance for admin expenses is a crucial calculation in budgeting and controlling costs for any business. It is the difference between the actual expenses and the flexible budgeted expenses. To calculate the variance, subtract the flexible budget expenses from the actual expenses, using the formula: (Actual Admin Expenses – Flexible Budget Admin Expenses). A positive result indicates that actual expenses were lower than the budgeted amount, while a negative result indicates that actual expenses were higher.

Calculating the flexible budget variance provides valuable insights into how well a company is managing its resources and controlling costs. It helps identify areas where expenses are over budget and require attention from management. By monitoring these expenses, businesses can make informed decisions and take corrective actions to bring expenses in line with their budget goals.

Flexible budget variance is a measure of the difference between actual costs and flexible budgeted costs. This variance helps companies analyze and understand the impact of changing sales levels and costs on the overall budget. A flexible budget is designed to adjust to changes in activity levels, using a formula to calculate costs that should have been incurred on a given level of activity. Fixed costs remain the same, while variable costs change with sales level.

To calculate the flexible budget variance, compare the actual costs with the flexible budgeted costs for the same level of activity. The variance is then divided into price variance and quantity variance. Price variance arises due to changes in unit cost, while quantity variance is due to changes in the level of activity.

Given the actual admin expenses, budgeted expenses for the level of activity achieved, and standard expenses for the same level of activity, the flexible budget variance for admin expenses can be calculated. For example, if the actual expenses are $45,000 and the flexible budget amount is $40,000, the flexible budget variance is $5,000. A positive variance indicates that actual expenses were lower than budgeted or standard expenses, while a negative variance implies that more expenses were incurred than originally planned.

In conclusion, calculating flexible budget variance for admin expenses requires comparing actual, budgeted, and standard expenses. The variance can be broken down into quantity and price variances to understand the impact of changes in activity levels and unit cost.

The flexible budget is the budget that adjusts according to the actual level of activity. For instance, if the actual level of activity is 4,000 units, the flexible budget amount for admin expenses is $40,000. If the actual admin expenses incurred are $45,000, the flexible budget variance is $5,000, indicating that actual admin expenses exceeded the budgeted amount.

The following table summarizes the information:

| Category | Amount |
| --- | --- |
| Flexible Budget for Admin Expenses (4,000 units) | $40,000 |
| Actual Admin Expenses | $45,000 |
| Flexible Budget Variance (FBV) | -$5,000 (Unfavorable/Adverse) |

By calculating the FBV, companies can understand how well they are controlling expenses and identify areas that need improvement or adjustments in the future.

The interpretation of the results is crucial in making informed decisions. For example, if the flexible budget variance for admin expenses is $15,000, it indicates that actual expenses exceeded the original budget. However, when considering the actual volume produced, the flexible budget variance can provide a more accurate picture of the company's performance.

To calculate the flexible budget variance, use the formula: Flexible Budget Variance = Actual Expense – Flexible Budget. The flexible budget is calculated as: Flexible Budget = Original Budget * (Actual Volume/Static Budget). For instance, if the original budget is $50,000 and the actual volume is 20,000 units, the flexible budget is $40,000. The flexible budget variance can then be calculated as $55,000 – $40,000 = $15,000.

In conclusion, the flexible budget variance for admin expenses provides useful insights into a company's financial performance. This information can help management make better decisions about budgeting and resource allocation. However, it is crucial to use this information with caution and consider other factors that could impact the organization's success.

What is the flexible budget variance for admin expenses?
A) Actual Admin Expenses + Flexible Budget Admin Expenses
B) Actual Admin Expenses – Flexible Budget Admin Expenses
C) Flexible Budget Admin Expenses – Actual Admin Expenses
D) Actual Admin Expenses * Flexible Budget Admin Expenses

Answer: B) Actual Admin Expenses – Flexible Budget Admin Expenses

What does a positive flexible budget variance indicate?
A) Actual expenses were higher than the budgeted amount
B) Actual expenses were lower than the budgeted amount
C) Actual expenses were equal to the budgeted amount
D) Actual expenses were not related to the budgeted amount

Answer: B) Actual expenses were lower than the budgeted amount

What is the purpose of calculating the flexible budget variance?
A) To identify areas where expenses are under budget
B) To identify areas where expenses are over budget
C) To understand how well a company is managing its resources and controlling costs
D) To determine the original budget

Answer: C) To understand how well a company is managing its resources and controlling costs


The formula for continuous interest can be derived by taking the limit as the number of compounding periods approaches infinity. The effective annual rate for an interest rate, r, compounded n periods per year for t years, is given by the equation. To show that the effective annual rate under continuously compounded interest is e^r - 1, we need to derive the equation that equates the limit of the EAR as the compounding periods approach infinity to the formula for the continuously compounded rate.

The derivation starts with the definition of the effective annual rate. Suppose there are n compounding periods per year and r is the interest rate. The effective annual rate can be calculated using the formula. To derive the formula for continuous interest, we take the limit as n approaches infinity.

One way to derive the formula is to use a substitution to avoid using L'Hopital's rule or the definition of a difference quotient. The derivation involves taking the natural logarithm of both sides of the equation, which allows us to use the property of logarithms that the log of a product is the sum of the logs.

Using this property, the right-hand side of the equation can be split into a sum, and the first term no longer depends on the limit. The resulting equation involves logs on both sides and can be simplified using the fact that e and the natural log are inverses.

Euler's number, e, is a fundamental constant in mathematics, approximately equal to 2.71828. It can be defined as the positive real number such that the integral from 1 to e of the function 1/t is equal to 1. Alternatively, e can be defined as the number such that the area under the curve of 1/t from 1 to e is equal to 1.

The natural logarithm, ln(x), is the inverse function of e^x. It is defined as the logarithm of x to the base e. The natural logarithm has several important properties, including the fact that ln(1) = 0 and ln(e) = 1. The graph of the natural logarithm function shows that it crosses the x-axis at x = 1 and increases slowly as x increases.

To derive the formula for continuous interest, we can use the following steps:

1. Start with the definition of the effective annual rate:
2. Take the limit as n approaches infinity:
3. Use the property of logarithms to simplify the equation:
4. Simplify the resulting equation using the fact that e and the natural log are inverses.

The final result is the formula for continuous interest: e^r - 1. This formula gives the effective annual rate for an interest rate r compounded continuously.


## NCERT Solutions For Class 9 Science Chapter 8 Motion

### Exercise Questions

**Q. 1. An athlete completes one round of circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?**

Given: Time taken = 2 minutes 20 s = 140 s, Radius = 100 m.
In 40 s, the athlete completes one round. So, in 140 s, the athlete will complete 140 ÷ 40 = 3.5 rounds.
At the end of his motion, the athlete will be in the diametrically opposite position.
Displacement = Diameter = 200 m

**Q. 2. Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging: (a) from A to B and (b) from A to C?**

(a) Given: Distance covered = 300 m, Displacement = 300 m, Time taken = 150 seconds.
(b) Given: Distance covered = 300 + 100 = 400 m, Displacement = 300 - 100 = 200 m, Time taken = 2.5 + 1 = 3.5 mins = 3.5 x 60 = 210 s

**Q. 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed of Abdul’s trip?**

Let one side distance = x km.

**Q. 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time?**

Given: u = 0 m/s, a = 3ms–2, t = 8s

**Q. 5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops after 5 seconds. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for two cars. Which of the two cars travelled farther after the brakes were applied?**

In the following graph, AB and CD are the time graphs for the two cars whose initial speeds are 52 km/h (14.4 m/s) and 3 km/h (0.83 m/s), respectively.
Thus, the first car travels farther than the second car after they applied the brakes.

**Q. 6. Following figure shows the distance–time graph of three objects A, B and C. Study the graph and answer the following questions:**
**(a) Which of the three is travelling the fastest?**
**(b) Are all three ever at the same point on the road?**
**(c) How far has C travelled when B passes A?**
**(d) How far has B travelled by the time it passes C?**

According to the graph:
(a) B is travelling fastest as he is taking less time to cover more distance.
(b) All three are never at the same point on the road.
(c) Approximately 6 km [as 8 – 2 = 6]
(d) Approximately 7 km [as 7 – 0 = 7]

**Q. 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s–2, with what velocity will it strike the ground? After what time will it strike the ground?**

Given: u = 0 m/s, s = 20 m, a = 10 ms–2

**Q. 8. The speed–time graph for a car is shown in Figure:**
**(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.**
**(b) Which part of the graph represents uniform motion of the car?**

(a) Distance covered = Area under speed = Time graph
Shaded area representing the distance travelled is as follows:
(b) After 6 seconds the car moves in uniform motion (at a speed of 6 m/s).

**Q. 9. State which of the following situations are possible and give an example of each of the following:**
**(a) an object with a constant acceleration but with zero velocity.**
**(b) an object moving in a certain direction with an acceleration in the perpendicular direction.**

(a) Yes, a body can have acceleration even when its velocity is zero. When a body is thrown up, the highest point of its velocity is zero but it has acceleration which is equal to acceleration due to gravity.
(b) Yes, an object moving horizontally is acted upon by acceleration due to gravity that acts vertically downwards.

**Q. 10. An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.**

Given: r = 42,250 km = 4,22,50,000 m, T = 24 hours = 24 x 60 x 60 seconds
Using speed, v = 2πr ÷ T
v = (2 x 3.14 x 4,22,50,000) ÷ (24 × 60 × 60)
= 3070.9 ms–1
= 3.07 km/s

### Intext Questions

**Page 100**

**Q. 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.**

Yes, zero displacement is possible if an object has moved through a distance.
Suppose a body is moving in a circular path and starts moving from point A and it returns back at the same point A after completing one revolution, then the distance will be equal to its circumference while displacement will be zero.

**Q.2. A farmer moves along the boundary of a square field of side 10 m in 40 seconds. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?**

Given: Side of the square field = 10 m, Therefore, perimeter = 10 m x 4 = 40 m
Farmer moves along the boundary in 40 seconds
Time = 2 minutes 20 seconds = 2 x 60 seconds + 20 seconds = 140 seconds
Since, in 40 seconds farmer moves 40 m, Therefore, in 1 second distance covered by farmer = 40 ÷ 40 = 1 m.
in 140 seconds distance covered by farmer = 1 x 140 m = 140 m
Now, number of rotations to cover 140 along the boundary = Total distance / Perimeter = 140 m ÷ 40 m = 3.5 round
Thus after 3.5 round farmer will at point C (diagonally opposite to his initial position) of the field.
Thus, after 2 minutes 20 seconds, the displacement of farmer will be equal to 10√2 m north–east from the initial position.

**Q. 3. Which of the following is true for displacement?**
**(a) It cannot be zero.**
**(b) Its magnitude is greater than the distance travelled by an object.**

(a) Not true, because if the initial and final positions of the object is the same displacement can become zero.
(b) Not true, because, displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object, but it can be smaller or equal to the distance travelled by an object.

**Page 102**

**Q. 4. Distinguish between speed and velocity.**

Differences between speed and velocity are:

**Q. 5. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?**

As we know, The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity in a straight–line motion.

**Q. 6. What does the odometer of an automobile measure?**

Odometer is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

**Q. 7. What does the path of an object look like when it is in uniform motion?**

Uniform motion is a motion in which an object covers equal distances in equal intervals of time and moves along a straight line.
Thus, in the case of uniform motion, the path is a straight line in uniform motion.

**Q. 8. During an experiment, a signal from a spaceship reached the ground station in 5 minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 x 10^8 m s–1.**

Given: Speed = 3 x 10^8 m s–1, Time = 5 minutes = 5 x 60 seconds = 300 seconds
Using, Distance = Speed x Time
Distance = 3 x 10^8 x 300 m
= 900 × 10^8 m
= 9.0 × 10^10 m

**Page 103**

**Q.9. When will you say a body is in: (i) Uniform acceleration? (ii) Non–uniform acceleration?**

(a) A body is said to be in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitudes in equal intervals of time.
(b) A body is said to be in non–uniform acceleration when its velocity changes by unequal magnitudes in equal intervals of time.

**Q. 10. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.**

Given: u = 80 km/h = [80 x 1000] / 3600 ms–1 = 200/9 m s–1, v = 60 km/h = [60 × 1000] / 3600 ms–1 = 150 / 9 m s–1, t = 5s
Therefore, acceleration is -1.1 ms–2

**Q.11 A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.**

Given: Initial velocity, u = 0 m/s, Final velocity, v = 40 km/h = [40 x 1000] / 3600 ms–1, Time (t) = 10 minute = 60 x 10 = 600 s

**Page 107**

**Q. 12. What is the nature of the distance–time graphs for uniform and non–uniform motion of an object?**

(i) The distance–time graph for an object in uniform motion is straight line, inclined with the time axis.
(ii) The distance–time graph for an object in non uniform motion is a curve.

**Q. 13. What can you say about the motion of an object whose distance–time graph is a straight line parallel to the time axis?**

When the distance–time graph is a straight line parallel to time axis, it means that the object is not changing its position.
So, the object is stationary or at rest.

**Q. 14. What can you say about the motion of an object if its speed–time graph is a straight line parallel to the time axis?**

When the graph of a speed–time graph is a straight line parallel to the time axis, it means that the speed of the object is not changing with time.
So, the object is performing uniform motion.

**Q. 15. What is the quantity which is measured by the area occupied below the velocity–time graph?**

The quantity of displacement is measured by the area occupied below the velocity–time graph.

**Page 109**

**Q. 16. A bus starting from rest moves with a uniform acceleration of 0.1 m s–2 for 2 minutes.**
**Find:**
**(a) the speed acquired.**
**(b) the distance travelled.**

Given: Initial velocity (u) = 0 m/s, Acceleration (a) = 0.1 m s–2, Time (t) = 2 minutes = 2 x 60 seconds = 120 seconds
(a) The speed acquired: v = u + at → v = 0 + 0.1 x 120 m/s → v = 12 m/s
Thus, the bus will acquire a speed of 12 m/s after 2 minutes with the given acceleration.
(b) The distance travelled: s = ut + 1/2 at^2 → s = 0 + 1/2 x 0.1 x (120)^2 → s = 720 m
Thus, bus will travel a distance of 720 m in the given time of 2 minutes.

**Q. 17. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s². Find how far the train will go before it is brought to rest.**

Given: Initial velocity, u = 90 km/h = [90 x 1000] / 3600 ms–1 = 25 ms–1, Final velocity, v = 0 m/s, Acceleration, a = -0.5 ms–2
Using equation of motion: v² = u² + 2as → 0 = (25)² + 2 x (-0.5) x s → s = 625 m
Therefore, train will go 625 m before it is brought to rest.

**Q. 18. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?**

Given: Initial velocity, u = 0 m/s, Acceleration (a) = 2 cm/s² = 0.02 m/s², Time (t) = 3 seconds
We know that, v = u + at → v = 0 + 0.02 x 3 m/s → v = 0.06 m/s
Therefore, the final velocity of trolley will be 0.06 m/s after start.

**Q. 19. A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?**

Given: Acceleration, a = 4 m/s², Initial velocity, u = 0 m/s, Time, t = 10 s
Thus, racing car will cover a distance of 200 m after start in 10 seconds with given acceleration.

**Q. 20. A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?**

Given: Initial velocity (u) = 5 m/s, Final velocity (v) = 0 m/s, Acceleration (a) = -10 m/s²
As we know, v² = u² + 2as → 0 = (5)² + 2 x (-10) x s → s = 1.25 m
Now, we know that, v = u + at → 0 = 5 + (-10) x t → t = 5/10 second → t = ½ second = 0.5 second
Thus, stone will attain a height of 1.25 m and time taken to attain the height is 0.5 second.


# Primality Testing: Miller-Rabin

## Motivation
The existence of Carmichael numbers makes it challenging to create a primality testing algorithm based solely on Fermat tests, prompting the need for alternative methods like the Miller-Rabin algorithm.

## The Miller-Rabin Algorithm
The Miller-Rabin algorithm is a non-deterministic test for compositeness. If the algorithm fails on a number $n$, it indicates that $n$ is a probable prime. The probability of failure is $4^{-k}$, where $k$ is the number of rounds of Miller-Rabin performed. This probability approaches zero rapidly as $k$ increases.

Carmichael numbers that are strongly pseudoprime to multiple bases do exist but are extremely rare.

## Theoretical Foundation
**Theorem 16.1**: Let $p$ be an odd prime, $r = p - 1$, and $r = 2^k \cdot q$, where $k$ and $q$ are integers with $q$ being odd. For any integer $a$ not divisible by $p$, one of the following conditions must be true:
- $a^q \equiv_p 1$
- One of $a^q, a^{2q}, a^{4q}, \dots, a^{2^{k-1}q}$ is congruent to $-1$ modulo $p$.

Given $a^{p-1} \equiv 1 \pmod{p}$ by Fermat's Little Theorem, it follows that $a^{2^{k-1}q} \equiv \pm 1 \pmod{p}$. If $a^{2^lq} \equiv_p 1$ for all $l < k$, then $a^q \equiv_p 1$. Otherwise, there exists an $l$ such that $a^{2^lq} \equiv_p -1$, satisfying the second condition.

## Miller-Rabin Test
To test if an odd composite integer $n$ is prime:
1. Find $k$ and $q$ (with $q$ odd) such that $n - 1 = 2^k \cdot q$.
2. Choose a random integer $a$.
3. Compute $a^q \pmod{n}$. If $a^q \equiv_n \pm 1$, the test fails, indicating $n$ is a probable prime to base $a$.
4. Otherwise, iteratively compute $a^{2^jq} \pmod{n}$ for $j = 1, 2, \dots, k-1$.
   - If $a^{2^jq} \equiv_n 1$ for any $j$, $a$ is a witness to the compositeness of $n$, and the test stops.
   - If $a^{2^jq} \equiv_n -1$ for any $j$, $n$ is a probable prime to base $a$, and the test stops.
5. If the test has not terminated after $k-1$ iterations, $n$ is composite.

The Miller-Rabin test's efficiency and accuracy make it a valuable tool in number theory for determining the primality of large numbers.


### Part 1

**15. Measuring Sound Loudness**

The loudness of sound is measured in units called decibels (dB). The decibel rating, d, of a sound is found using the formula d = 10 log (I / I_{0}), where I is the intensity of the sound and I_{0} is the intensity of the threshold sound. The threshold sound audible to the average person is 1.0 X 10^{-12} W/m^{2}.

Sound level classifications are as follows:
* Moderate: 45-69 dB
* Loud: 70-89 dB
* Very Loud: 90-109 dB
* Deafening: > 110 dB

To classify a sound with intensity 6.3 X 10^{-3} W/m^{2}, use the formula d = 10 log (I / I_{0}). Plug in the values I_{0} = 1.0 X 10^{-12} and I = 6.3 X 10^{-3} to get d = 10 log ( (6.3 X 10^{-3}) / (1.0 X 10^{-12}) ) = 97.99341..., which is between 90 and 109, or "Very Loud".

**16. Ant Farm Population**

Pedro starts with 100 ants and his farm is growing exponentially at a rate of 15% per month. The equation for Pedro's farm is y = 100(1.15)^{x}. Bobby starts with 350 ants and his farm is steadily decreasing by 5 ants per month. The equation for Bobby's farm is y = 350 - 5x.

To find out after how many months they will both have approximately the same number of ants, graph the equations on a calculator and find the intersection point. Using the CALC function, the intersection point is at a fraction past 8 months, where both will have a little less than 310 ants. The correct answer is 8 months.


Math puzzles, such as Tachiawase, are excellent activities for learning and enjoying mathematics. Tachiawase is a popular Japanese puzzle that involves dissecting a geometrical figure into several parts and recombining them to form another geometrical figure. A notable example of Tachiawase, credited to Hikodate Nakane (1743), is a puzzle that requires making two cuts in two different-sized squares to recombine them into a larger square.

To solve this puzzle, one can start by considering the Pythagorean Theorem, as the goal is to form a larger square from two smaller ones. However, determining where to make the cuts can be challenging. A helpful approach is to simplify the problem by using two squares of equal size, which reduces the difficulty significantly. This simplified version can provide insight into where to make the cuts for the side of the square to be formed.

A GeoGebra mathlet, a dynamic math applet, can be created based on the solution to the puzzle. This mathlet can be used to teach the Pythagorean theorem, and a suggested sequence of questions to ask students includes:

- What are the areas of each square in the figure, and how can they be found in at least two ways?
- How are the areas of the squares related, and can this relationship be explored by dragging a point to test conjectures for any size of squares?
- Are there other ways to prove the relationships between the three squares without using side measurements?
- If the smaller squares have sides of length p and q, how can the area of the largest square be expressed in terms of the areas of the smaller squares?
- How can the length of the sides of the largest square be expressed in terms of the sides of the smaller squares?

After exploring these questions, students can be given the original puzzle to solve. Once they have the correct pieces, they can be asked to move the pieces using transformations in the least possible moves, which can be achieved in three moves using rotation.

The original puzzle, attributed to Nakane, involves dividing the squares into parts and making two cuts to create three pieces that can be recombined into a larger square. It is essential to carefully formulate the problem and provide accurate attribution, as this dissection is often attributed to the 17th-century Dutch mathematician Frans van Schooten. Additional references and dissections can be found in G. N. Frederickson's book "Dissections: Plane & Fancy" (Cambridge University Press, 1997).


# Relative Humidity Calculator

The relative humidity calculator is a tool used to calculate the relative humidity, dew point, and air temperature. It can be used to:

* Calculate the relative humidity given the air temperature and dew point
* Calculate the dew point given the air temperature and relative humidity
* Predict the air temperature corresponding to a specific relative humidity and dew point

The calculated relative humidity can be used in the wet-bulb temperature calculator and heat index calculator to determine the influence of relative humidity on the perceived temperature.

## What is Relative Humidity?

Relative humidity measures the amount of water vapor present in the air compared to the total amount of water vapor the air can hold at a given temperature. It is defined as the ratio of water vapor present in the air to the total amount of water vapor the air can hold at a given temperature, ranging from 0% (dry air) to 100% (saturated air).

Mathematically, relative humidity (RH) is expressed as the ratio of vapor pressure (Pv) to the water saturation pressure (Psat@T) at a given temperature (T): RH = Pv / Psat@T.

## How to Find Relative Humidity?

The relative humidity calculator uses the following equation: RH = 100 * (exp((17.625 * Dp) / (243.04 + Dp)) / exp((17.625 * T) / (243.04 + T))), where RH is the relative humidity, Dp is the dew point temperature in °C, and T is the air temperature in °C.

## How to Calculate Wet-Bulb Temperature?

The wet-bulb temperature (Tw) is the lowest temperature a portion of air can reach by evaporative cooling. It can be calculated using the following equation: Tw = T * arctan[0.151977(RH% + 8.313659)^0.5] + arctan(T + RH%) - arctan(RH% - 1.676331) + 0.00391838(RH%)^1.5 * arctan(0.023101 * RH%) - 4.686035, where RH% is the relative humidity expressed as a percentage.

The relative humidity equation and its applications are essential in understanding the relationship between air temperature, dew point, and relative humidity.


# One Hot Encoding

**One hot encoding** is a method of encoding categorical variables as binary vectors for machine learning algorithms. These algorithms work with numbers (0 and 1), not categorical values. One hot encoding translates categorical values into numbers that can be interpreted by algorithms.

## Encoding Categorical Values

The process involves:
- Assigning a unique sequential integer value to each possible categorical value.
- Representing each value as a binary vector with a position for each integer value.
- Creating a vector with a value of 1 in the position for its corresponding integer value, and 0 elsewhere.
- Replacing categorical values with the corresponding vector.

Encoded values may be represented as vector data types or expanded as additional columns in the data.

## Example

Original data:

| Category | Value |
| --- | --- |
| A | 24 |
| A | 90 |
| A | 75 |
| A | 36 |
| B | 45 |
| B | 28 |
| B | 62 |
| C | 65 |
| C | 97 |

Assigning integer values to categories:

| Category | Integer Value |
| --- | --- |
| A | 0 |
| B | 1 |
| C | 2 |

Assigning vectors to values:

| Category | Integer Value | Vector |
| --- | --- | --- |
| A | 0 | [1,0,0] |
| B | 1 | [0,1,0] |
| C | 2 | [0,0,1] |

Applying vector values as columns to the original data:

| Category | Value | A | B | C |
| --- | --- | --- | --- | --- |
| A | 24 | 1 | 0 | 0 |
| A | 90 | 1 | 0 | 0 |
| A | 75 | 1 | 0 | 0 |
| A | 36 | 1 | 0 | 0 |
| B | 45 | 0 | 1 | 0 |
| B | 28 | 0 | 1 | 0 |
| B | 62 | 0 | 1 | 0 |
| C | 65 | 0 | 0 | 1 |
| C | 97 | 0 | 0 | 1 |


# The Distance and Power of Aaron Rodgers' Pass

The Green Bay Packers and Detroit Lions played their second game of the 2015 season on December 3rd, which ended with a highly unlikely Hail Mary catch by Richard Rodgers to score the winning touchdown. The league posted a video clip asking "How far exactly did Rodgers' Hail Mary travel?" with an official marking of 61 yards, measured from the line of scrimmage to the front of the end zone line. However, Ben Rohrbach believed the total travel distance of the ball was close to 100 yards, including vertical and horizontal elements.

When a ball is thrown, its vertical profile forms an upside-down parabola due to gravity's constant downward acceleration. The total distance along this segment is known as the arc length, which can be calculated by integrating the changing height across the horizontal distance.

To calculate the actual arc length of Aaron Rodgers' December 3rd pass, we can use the Pythagorean Theorem with length and width coordinate estimates from the back end zone and visual top of the field. The throw starts an estimated 39 yards from the back end zone and ends 103 yards from the same end zone, with a horizontal displacement of 64.195 yards. The vertical displacement can be estimated by calculating the hang time, squaring it, and multiplying by 1.34058, resulting in a vertical displacement of 42.041 yards.

Using these estimates, the following results were calculated:

- Horizontal Displacement: 64.195 yards
- Vertical Displacement: 42.041 yards
- Total Travel Distance: 110.716 yards
- Velocity of Pass: 12.706 yards per second
- Energy Imparted to Football: 21.405 foot-pounds
- Power Output of Aaron Rodgers during pass: 0.216 horsepower

This analysis vindicates Rohrbach's estimate and calculates Aaron Rodgers' throwing technique to be as strong as just over a fifth of a horse. The calculations were performed using the following Excel formulas:

- A6 = SQRT(C2^2 + C4^2)
- A11 = SQRT((A6^2/4) + (4*B9^2)) + (A6^2/(8*B9))*ASINH(4*B9/A6)
- A13 = (3*B11)^2*(0.94799/(2*32.174049))
- B9 = 1.34058*A9^2
- B11 = SQRT((A6/A9)^2 + 10.72467*(A9/2))
- B13 = (A13/B6)/550
- C2 = ABS(B2-A2)
- C4 = ABS(B4-A4)

Note: Air friction is neglected in this discussion. The method used to calculate the arc length can also be applied to other projectiles, such as golf drives, log throws, and rock catapultations.


## Difference between Symmetric Key Encryption and Asymmetric Key Encryption

The two basic techniques for encrypting information are symmetric encryption (secret key encryption) and asymmetric encryption (public key encryption).

### Symmetric Encryption

Symmetric encryption uses a secret key for both encryption and decryption. This key can be a number, word, or string of random letters, and is applied to the message text to change its content. Both the sender and recipient must know the secret key to encrypt and decrypt messages. Examples of symmetric key algorithms include DES, AES, and Blowfish.

### Asymmetric Encryption

Asymmetric encryption uses a key pair, consisting of a public key and a private key. The public key is freely available, while the private key is kept secret. Messages encrypted with the public key can only be decrypted with the matching private key, and vice versa. This method eliminates the need to securely exchange secret keys, but is slower and requires more processing power. Examples of asymmetric key algorithms include RSA, Elliptic Curve Cryptography (ECC), El Gamal, and Diffie-Hellman.

## Multiple Choice Question

Which of the following statements correctly describes the difference between symmetric key encryption and asymmetric key encryption?

- In symmetric key encryption, the same key is used for encryption and decryption, whereas asymmetric key encryption uses a private key for encryption and decryption.
- In symmetric key encryption, the public key is used for encryption and the symmetric key for decryption, whereas in asymmetric key encryption, the public key is used for encryption and the private key is used for decryption.
- In symmetric key encryption, the same key is used for encryption and decryption, whereas in asymmetric key encryption, the public key is used for encryption and the private key is used for decryption.
- Both use a private key for encryption, and the decryption process can be done using a public key.

Correct answer: In symmetric key encryption, the same key is used for encryption and decryption, whereas in asymmetric key encryption, the public key is used for encryption and the private key is used for decryption.


In the embedded world, fixed-point arithmetic is often necessary for cost and performance reasons. However, multiplying by a non-integer value can be challenging. For example, multiplying by 1.2764 can be done using the naive approach:

uint16_t scale(uint8_t x) { uint16_t y; y = (x * 12764) / 10000; return y; }

However, this will fail due to numeric overflow in the expression (x * 12764). To avoid this, expensive casts are necessary:

uint16_t scale(uint8_t x) { uint16_t y; y = ((uint32_t)x * 12764) / 10000; return y; }

A better approach is to use a fraction (a/b) that closely approximates 1.2764, with a numerator that doesn't exceed 255, allowing for 16-bit arithmetic. Continued fractions can be used to find such approximations. Using a calculator, the following convergents are obtained:

1: 1/1 = 1
3: 4/3 = 1.3333333333333333
1: 5/4 = 1.25
1: 9/7 = 1.2857142857142858
1: 14/11 = 1.2727272727272727
1: 23/18 = 1.2777777777777777
1: 37/29 = 1.2758620689655173
1: 60/47 = 1.2765957446808511
1: 97/76 = 1.2763157894736843
1: 157/123 = 1.2764227642276422
2: 411/322 = 1.2763975155279503
1: 1801/1411 = 1.2763997165131113
1: 3191/2500 = 1.2764

The fraction 157/123 is chosen for its high accuracy and numerator less than 255. The refined code becomes:

uint16_t scale(uint8_t x) { uint16_t y; y = ((uint16_t)x * 157) / 123; return y; }

This approach results in an error of less than 0.002% and significantly improved calculation speed, avoiding the need for 32-bit arithmetic. On an ATmega88 processor, the continued fraction approach takes 53,300 cycles, compared to 148,677 cycles for the naive approach. Mathematicians confirm that this technique generates optimal fractions.


To calculate the thermal efficiency of a power plant, we use the equation E = work / Qh, where E is the efficiency, work is the useful energy produced, and Qh is the heat that flows into the power plant. Since we're given units of power instead of work, we can modify the equation to use power, which is work divided by time (delta T). The equation becomes power / Qh, where power is 250 megawatts (MW) and Qh needs to be determined.

We're told that the power plant expels 550 MW out to the surrounding environment, which represents Qc (the cold reservoir). To find Qh, we use the equation W = Qh - Qc, where W is the work produced. Since we're working with units of power, we can express this as 250 MW = Qh - 550 MW. Rearranging the equation gives us Qh = W + Qc = 250 MW + 550 MW = 800 MW.

Using the energy flow diagram, we can see that Qh is 800 MW, Qc is 550 MW, and the work produced (W) is 250 MW. Plugging these values into the efficiency equation gives us E = 250 MW / 800 MW = 0.313, or 31.3% efficiency. This is the final answer.

The key concepts here are the equations for efficiency and energy flow, as well as the ability to work with units of power instead of work. The important numbers are 250 MW (power produced), 550 MW (power expelled), and 800 MW (heat that flows into the power plant), which result in an efficiency of 31.3%.


MPMP: TakeAway Triangles

The challenge proposed in Matt Parker's Maths Puzzles (MPMP) is to find three starting numbers for a takeaway triangle so that eventually each set of the three new numbers generated always adds to 14. The goal is to find the starting numbers that meet this condition.

To solve this problem, a function `takeaway()` is defined to perform "one round" of takeaway on a set of 3 numbers. This function uses the first digit as the left corner, the right digit as the right corner, and the middle digit as the centered corner of the triangles. The operation can be formalized as:

\begin{align} \operatorname{takeaway}: \mathbb{N}^3 &\to \mathbb{N}^3 \\ \{a,b,c\} &\mapsto \{ab, ac, bc\} \\ \end{align}

This function is then used to iteratively perform the takeaways for a set number of steps (50 in this case) while starting off with a random set of integers between 1 and 100. The result is a list of numbers at each step, which can be used to check if the sum of the numbers is 14.

To increase the chances of finding a solution, 100 successive runs are generated, each with random starting numbers. The condition for a "winning" run is defined as:

If the last couple steps (i.e. starting at step 45) are all equal to 14, then it's a winner.

The runs are then checked for this condition, and the runs that meet this condition are filtered and extracted. The starting numbers for these winning runs are:

* 52 + 17 + 94 = 163
* 3 + 66 + 80 = 149
* 46 + 53 + 74 = 173

These starting numbers do not sum to 14, making them valid solutions. A plot of all the runs by their number's sums at each step, highlighting the 3 winners, is also generated.

The approach used to solve this problem is a "computer fast brain slow" approach, where a large number of simulations are run to find a solution. This approach can be further explored by using larger starting numbers and more steps.