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Class 10th- Maths Periodic Test Marathon 🔥 | Part 2 | Shobhit Nirwan

Shobhit Nirwan・2 minutes read

The text discusses solving linear equations in two variables using graphical and algebraic methods, emphasizing the importance of understanding and applying these techniques to solve mathematical problems. It also covers scenarios involving boat speeds in still and flowing water, as well as methods for solving quadratic equations, such as splitting the middle term and the discriminant method.

Insights

Linear equations in two variables can be solved using graphical, substitution, and elimination methods, with each technique offering a distinct approach to finding solutions.
Understanding and applying algebraic methods like substitution and elimination are crucial for solving mathematical problems effectively, emphasizing the importance of mastering these techniques.
Quadratic equations can be solved using methods like splitting the middle term and the discriminant method, each providing a structured approach to finding roots and solutions.
Real and distinct roots in quadratic equations are determined by the discriminant, showcasing how mathematical formulas and methods can help identify the nature of solutions in algebraic problems.

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Recent questions

How are linear equations solved?
By graphical and algebraic methods.
What is the importance of understanding linear equations?
Essential for solving mathematical problems.
What are the methods to solve quadratic equations?
Splitting the middle term and discriminant method.
How can boat speeds in still and flowing water be calculated?
By considering upstream and downstream scenarios.
Why is perseverance important in education?
To overcome challenges and succeed.

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Summary

00:00

"Mathematical Methods for Solving Linear Equations"

The text begins with greetings and introductions among individuals.
It mentions starting a class and continuing a marathon that was previously interrupted.
The focus shifts to studying mathematics, specifically part two of a marathon.
The text delves into solving linear equations in two variables.
It discusses methods for solving linear equations, including graphical and algebraic methods.
The algebraic method is further broken down into substitution and elimination techniques.
Substitution involves replacing the value of one variable in another equation.
Elimination requires making the coefficients of one variable equal in both equations.
The text emphasizes the importance of understanding and applying these methods in solving mathematical problems.
It concludes with a reference to the challenges faced in education and the need for perseverance and dedication.

14:11

Solving Equations: Graphical, Substitution, Elimination

The equation given is 2x + 3y = 4
To eliminate 3y, add both equations resulting in -3y + 3y = 0
The value of x is found to be 13/5
To find y, substitute x = 13/5 into one of the equations
The value of y is calculated to be -2/5
Graphing the equations shows the intersection point at x = 13/5, y = -2/5
Three methods to solve equations are mentioned: graphical, substitution, and elimination
Graphical method is rarely asked unless specified in the question
Substitution involves replacing variables in one equation with values from another
Elimination method focuses on eliminating a variable to solve the equations

27:51

"Number Puzzle: Solving Equations for Fractions"

The difference between the numbers is 27, 63, and 17, with 7, 2, 6, and 3 being single digits.
The sum of the digits of a two-digit number is 8, with one digit being x and the other y.
Interchanging the digits of a number results in a number that exceeds the original by 18.
The difference between the original number and the reversed number is 18, leading to x - y = 2.
Solving the equations x - y = 2 results in x = 5 and y = 7.
The number formed by x = 5 and y = 7 is 57.
A fraction becomes 1/3 when the numerator is x and the denominator is y.
The fraction becomes 1/2 when 3x - 2 = y.
Cross-multiplying the fraction equation simplifies to 3x - 6 = y.
The solution to the fraction equation is x = 5 and y = 7, resulting in the fraction being 5/7.

43:00

"Equations, Fractions, and Hostel Charges Solved"

The first equation is 3x - y = 6, solved by taking LCM and cross multiplying.
The second equation is 2x - y = -1, derived from the previous equation.
The values of x and y are found to be 7 and 1 respectively.
A fraction is determined to be 9/1, representing the fraction.
Adding two to both the numerator and denominator changes the fraction to 5/6.
The first equation is 3x - y = 6, while the second equation is 2x - y = -1.
Solving both equations involves cross multiplying to get the answer of 7.
The fixed charge for a hostel entry is x, with a daily charge of y.
Swati's equation is x + 20y = 3000, while Mansi's equation is x + 25y = 3500.
Solving both equations using substitution method yields x = 1000 and y = 100, indicating the fixed and daily charges.

01:01:44

Father's age, children's ages, boat speeds explained.

Children consider x and y as the first step.
The age of the father is twice the sum of the ages of the children.
After 20 years, the father's age will be equal to the sum of the ages of his two children.
The age of the father after 20 years will be 40 years.
The age of the children after 5 years will be x + 5 and y + 5.
The father's present age is 45 years.
Upstream and downstream questions involve calculating boat speeds in still and flowing water.
In upstream scenarios, the boat moves against the stream's flow.
The net speed of the boat in upstream conditions is the difference between its speed and the stream's speed.
These types of questions are typically solved using quadratic equations.

01:19:53

Boat Speeds in Stream Scenarios

The text discusses a scenario involving a stream and a boat, with mentions of speeds and distances.
The stream is described as running at a certain speed, while the boat is moving in relation to the stream.
Various speeds are mentioned, including the boat's speed and the stream's speed.
Calculations are made regarding the net speed of the boat in different scenarios.
The text delves into the concept of up stream and down stream movements of the boat.
A specific example is provided where a man rows a boat down stream and up stream, with distances and times mentioned.
The speed of rowing in still water is a key focus, with equations and calculations involved.
The text transitions to discussing quadratic equations and methods to solve them.
Two methods are highlighted: splitting the middle term and the discriminant method.
The process of splitting the middle term is explained, emphasizing the need to find two numbers that add up and multiply to specific values.

01:48:14

Solving Quadratic Equations with Splitting Method

To solve the equation, add four to four to get eight and then four more for a total of 16.
The process involves determining the value of q, which is 4x multiplied by 4x.
Choose whether to use a minus sign by placing B minus before -4x or a plus sign by placing it before 4x.
The goal is to achieve a result of -8 by combining terms, ensuring that -4 -4 equals -8.
Split the equation into two numbers that, when multiplied, equal 16, such as -4 and -4.
The equation simplifies to x - 4, with x - 4 = 0, leading to x = 4 as the answer.
Utilize the Splitting the Middle Term Method to solve the Quadratic Equation for x in the equation 100x s - 20x + 1 = 0.
Identify two numbers that, when added, equal 10 and when multiplied, equal 100, such as 10 and 10.
The equation simplifies to 10x - 1, with x = 1 or x = 10 as the solutions.
Apply the same method to solve the Quadratic Equation √3 x 10x x 7 √3 x 0, resulting in x = -√3 or x = -7/√3 as the answers.

02:02:13

Solving Equations with Roots: Step-by-Step Guide

The question involves solving a complex mathematical equation with roots and variables.
The method to solve such questions with roots is explained step by step.
The process involves isolating the root part of the equation and squaring both sides.
By applying the formula for a - b, the equation is simplified.
The final quartic equation is derived after simplification.
The discriminant method is introduced as an alternative to solving equations with roots.
The discriminant method helps determine the nature of roots in a quadratic equation.
Real and distinct roots are identified based on the sign of the discriminant.
The formula for calculating the roots of a quadratic equation is detailed.
An example equation is provided to demonstrate the application of the quadratic formula in finding roots.

02:20:16

Solving Quadratic Equations: Process and Formula

Real and equal solutions occur when the discriminant (d) equals zero, which is when b^2 - 4ac = 0.
The formula for X is c - a + 4 in a.
The value of S in C is A - B.
The process involves applying mathematics to solve the equation.
The formula for solving the equation involves specific steps like multiplying terms and rearranging them.
The equation is manipulated to simplify it and reach the solution.
The formula for the equation is a + b + c = aS + a + b + cS + 2ab + 2bc + 2ca.
The process involves identifying common terms and rearranging them to simplify the equation.
The equation is further simplified to reach the final solution.
The final solution involves solving a quadratic equation to determine the value of x.

02:34:58

Determining Flight Duration with Speed Calculations

If R is reduced by 200 km, the speed will be affected.
To reach x km on time, consider the speed and distance.
In case of deteriorating weather, return within 200 km.
The time taken to travel x km from R is crucial.
A delay of 30 minutes affects the original flight duration.
The original duration of the flight needs to be determined.
Calculations involving speed, distance, and time are essential for finding the original flight duration.

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