The document provides frequently asked questions about aptitude tests, including data sufficiency and reasoning questions. It discusses topics like geometry, ratios, percentages, time/speed/distance problems, and series questions. The questions are meant to help students prepare for competitive exams.
The document provides information about the Management Aptitude Test (MAT) Afterschool Centre for Social Entrepreneurship and its PGPSE (Post Graduate Programme in Social Entrepreneurship) program. The 3-year integrated PGPSE program can be done along with civil service exams and provides an option to do a part-time job while studying. The 18-month PGPSE is available in both regular and distance learning modes and focuses on developing social entrepreneurs. Workshops on social entrepreneurship are also conducted across India.
(8) Lesson 5.7 - Distance on the Coordinate Planewzuri
This document provides examples for applying algebraic concepts to geometry problems. It demonstrates using the Pythagorean theorem and distance formula to calculate distances between points on a coordinate plane. Sample problems are worked through step-by-step and answers are provided. Common Core standards addressed include applying the Pythagorean theorem to find distances between points and using square root and cube root symbols to represent solutions to equations.
This document provides examples for applying the Pythagorean theorem and algebraic concepts to solve geometric problems involving right triangles. It includes:
1. Five multi-step examples that write and solve equations using the Pythagorean theorem (a^2 + b^2 = c^2) to find missing side lengths of right triangles in real-world contexts.
2. Information on the Common Core State Standards for applying the Pythagorean theorem to determine unknown side lengths in two and three dimensional figures.
3. A question about how using the Pythagorean theorem in this lesson connects to finding the distance between two points on a coordinate plane in the next lesson.
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
The document contains a collection of quantitative aptitude questions often asked in competitive exams. It provides the questions, explains the solutions, and includes links to additional practice resources on quantitative aptitude and entrepreneurship.
This is a collection of fifty questions from important topics in Aptitude where students should pay more attention and practice. Questions taken from various net sources. Some of the answers were edited. This presentation could be run only in office 2010 or latest.
The document provides frequently asked questions related to aptitude tests. It contains 15 questions covering topics like percentages, ratios, time/work/speed problems involving trains, and geometry questions related to circles, spheres, and tethered animals grazing fields. The questions are meant to help students prepare for competitive exams.
Here is a possible table with a constant rate of change:
Hours Worked | Money Earned
1 | $10
2 | $20
3 | $30
4 | $40
The constant rate of change is $10 per hour. The money earned increases by $10 for every 1 hour worked.
The document discusses an afterschool management aptitude test and social entrepreneurship program. It provides contact information for Dr. T.K. Jain of the Centre for Social Entrepreneurship in Bikaner and mentions that the program is open and free for all. It also includes sample practice questions and solutions for the test.
The document provides information about the Management Aptitude Test (MAT) Afterschool Centre for Social Entrepreneurship and its PGPSE (Post Graduate Programme in Social Entrepreneurship) program. The 3-year integrated PGPSE program can be done along with civil service exams and provides an option to do a part-time job while studying. The 18-month PGPSE is available in both regular and distance learning modes and focuses on developing social entrepreneurs. Workshops on social entrepreneurship are also conducted across India.
(8) Lesson 5.7 - Distance on the Coordinate Planewzuri
This document provides examples for applying algebraic concepts to geometry problems. It demonstrates using the Pythagorean theorem and distance formula to calculate distances between points on a coordinate plane. Sample problems are worked through step-by-step and answers are provided. Common Core standards addressed include applying the Pythagorean theorem to find distances between points and using square root and cube root symbols to represent solutions to equations.
This document provides examples for applying the Pythagorean theorem and algebraic concepts to solve geometric problems involving right triangles. It includes:
1. Five multi-step examples that write and solve equations using the Pythagorean theorem (a^2 + b^2 = c^2) to find missing side lengths of right triangles in real-world contexts.
2. Information on the Common Core State Standards for applying the Pythagorean theorem to determine unknown side lengths in two and three dimensional figures.
3. A question about how using the Pythagorean theorem in this lesson connects to finding the distance between two points on a coordinate plane in the next lesson.
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
The document contains a collection of quantitative aptitude questions often asked in competitive exams. It provides the questions, explains the solutions, and includes links to additional practice resources on quantitative aptitude and entrepreneurship.
This is a collection of fifty questions from important topics in Aptitude where students should pay more attention and practice. Questions taken from various net sources. Some of the answers were edited. This presentation could be run only in office 2010 or latest.
The document provides frequently asked questions related to aptitude tests. It contains 15 questions covering topics like percentages, ratios, time/work/speed problems involving trains, and geometry questions related to circles, spheres, and tethered animals grazing fields. The questions are meant to help students prepare for competitive exams.
Here is a possible table with a constant rate of change:
Hours Worked | Money Earned
1 | $10
2 | $20
3 | $30
4 | $40
The constant rate of change is $10 per hour. The money earned increases by $10 for every 1 hour worked.
The document discusses an afterschool management aptitude test and social entrepreneurship program. It provides contact information for Dr. T.K. Jain of the Centre for Social Entrepreneurship in Bikaner and mentions that the program is open and free for all. It also includes sample practice questions and solutions for the test.
1. The document provides 20 multiple choice questions from a past IB ACIO exam paper from 2015.
2. It tests knowledge on topics like spices, human anatomy, world geography, science, history and more.
3. For each question there are 4 answer options and explanations are provided for some questions.
This module introduces ratio, proportion, and the Basic Proportionality Theorem. Students will learn about ratios, proportions, and how to use the fundamental law of proportions to solve problems involving triangles. The module is designed to teach students to apply the definition of proportion of segments to find unknown lengths and illustrate and verify the Basic Proportionality Theorem and its Converse. Examples are provided to demonstrate how to express ratios in simplest form, find missing values in proportions, determine if ratios form proportions, and solve problems involving angles and segments in triangles using ratios and proportions.
This module covers similarity and the Pythagorean theorem as they relate to right triangles. It discusses how the altitude to the hypotenuse of a right triangle divides it into two smaller right triangles that are similar to each other and the original triangle. It also explains how the altitude is the geometric mean of the hypotenuse segments. Special right triangles like 45-45-90 and 30-60-90 triangles are examined, relating side lengths through their properties. The Pythagorean theorem is derived and used to solve for missing sides of right triangles. Students work through examples and multi-step problems applying these concepts.
This document provides a mark scheme for the January 2015 International GCSE Physics exam. It begins by outlining the qualifications awarded by Pearson, the exam board, and their commitment to education. The document then provides general guidance for examiners on how to apply the mark scheme consistently. It emphasizes rewarding what students can show they know and avoiding penalizing for omissions. Several sample exam questions and answers are then provided, with notes explaining the marks awarded for each part.
This document provides information about Module 17 on similar triangles. The key points covered are:
1. The module discusses the definition of similar triangles, similarity theorems, and how to determine if two triangles are similar or find missing lengths using properties of similar triangles.
2. Students are expected to learn how to apply the definition of similar triangles, verify the AAA, SAS, and SSS similarity theorems, and use proportionality theorems to calculate lengths of line segments.
3. Several examples and exercises are provided to help students practice determining if triangles are similar, citing the appropriate similarity theorem, finding missing lengths, and applying properties of similar triangles.
The document is a lesson on adding and subtracting integers:
1. It provides examples of adding and subtracting integers and finding the sums and differences.
2. It explains that to subtract an integer, you add its additive inverse, such as 4 - 9 = 4 + (-9) = -5.
3. It contains practice problems evaluating integer expressions and solving real-world problems involving adding and subtracting integers.
The document discusses various methods to calculate averages and related concepts:
- It defines an average as the sum of a data set divided by the number of data points. Examples are given of calculating averages for test scores and batting averages.
- Shortcut formulas are provided for calculating averages when data points are added or removed. These allow finding missing data like a player's score or person's age.
- Methods are outlined for finding averages of series using patterns in the number of data points.
- Formulas are presented for calculating average speeds for journeys with multiple distances traveled at different rates. Worked examples demonstrate applying the formulas.
This document provides a mathematics module with 31 questions covering topics like arithmetic, geometry, fractions, percentages, and conversions. For each question there are 3 multiple choice answers and an explanation of the correct answer. The questions assess skills in calculating speeds, volumes, ratios, expressions with exponents, area of shapes, conversions between units, and more.
1. The document presents three systems of equations related to word problems involving Justin and Karin's allowance, costs for recreation programs, and ages of Dewan and Adrianne.
2. Graphs are helpful for solving systems of equations because they allow you to visually check solutions obtained through algebraic methods. Graphing the equations shows their point of intersection, which indicates the solution.
3. Learning to solve systems of equations algebraically and using graphs helps answer word problems leading to two linear equations in two variables, as required by the Common Core standards.
1. The document provides examples for writing and solving systems of linear equations by graphing. It explains how to write equations in slope-intercept form, graph the lines on the same coordinate plane, and determine if the system has no solution, one solution, or an infinite number of solutions based on where the lines intersect.
2. Key steps shown include writing equations in y=mx+b form, finding the slope and y-intercept, graphing the lines and analyzing where they intersect to determine the number of solutions.
3. Graphing systems of equations allows you to visually see where lines intersect, satisfying both equations simultaneously and showing the solution to the system.
This document is a lesson on the Pythagorean theorem from a Holt Geometry textbook. It begins with examples of using the Pythagorean theorem to solve for missing sides of right triangles. It then discusses how to identify if a set of three sides forms a Pythagorean triple. The lesson continues with examples of using the converse of the Pythagorean theorem and Pythagorean inequalities to classify triangles as right, acute, or obtuse based on their side lengths. It concludes with a two-part lesson quiz reviewing the material.
Exact Solutions of Convection Diffusion Equation by Modified F-Expansion MethodIJMER
In this paper, the modified F-expansion method is proposed for constructing more than one
exact solutions of nonlinear convection diffusion equation with the aid of symbolic computation
Mathematica. By using this method, some new exact travelling wave solutions of the convection diffusion
equation are successfully obtained. These exact solutions include the soliton-like solutions, trigonometric
function solutions and rational solutions. Also, it is shown that the proposed method is efficient for
solving nonlinear partial differential equations arising in mathematical physics.
The document discusses paragraph proofs and two-column proofs in geometry. A paragraph proof states the given information, what is to be proven, then uses a logical chain of statements and reasons to justify the conclusion. A two-column proof separates the statements from the reasons into two columns, with the given information and conclusion stated. Both proof styles use definitions, properties, and theorems to logically justify each step. The document provides examples of each type of proof to illustrate the process.
The document provides information about an afterschool management aptitude test and social entrepreneurship program. It includes sample questions to test analytical ability covering topics like averages, percentages, profit/loss calculations, and data interpretation. Solutions to the example questions are also provided.
(8) Lesson 2.1 - Solve Equations with Rational Coefficientswzuri
This document provides instruction on solving linear equations with one variable. It begins by defining key terms like multiplicative inverse and coefficient. It then provides examples of solving various equations by multiplying or dividing both sides of the equation by the same term to isolate the variable. These examples include equations with fractions, decimals, and percentages. The document emphasizes using the inverse operation property to solve equations and checking solutions. It aims to help students solve equations that may require expanding and combining like terms.
This document summarizes a module on rational exponents and radicals that was presented at a 2014 mid-year inset for secondary mathematics teachers. The module covered lessons on zero, negative integral and rational exponents, radicals, and solving radical equations. It provided examples of simplifying expressions using laws of exponents and radicals. Recommended teaching strategies included problem-solving activities and a group brainstorming activity to discuss critical content areas and difficulties from teacher and student perspectives.
1. The document provides examples of finding slopes of lines from given points and determining if points form a parallelogram. It also discusses direct variation relationships and using equations, tables, and graphs to represent and compare them.
2. A constant of variation in a direct variation relationship is the constant that relates the ratio of the output quantity to the input quantity. It represents the unit rate or slope in the direct variation equation and graph.
3. Graphs are helpful for comparing different proportional relationships represented in different ways and finding the unit rate between quantities in a relationship from its graphical representation.
- Mauldin Middle School wants to make $4,740 from yearbook sales. Print yearbooks cost $60 each and digital yearbooks cost $15 each.
- This relationship can be represented by the equation 60x + 15y = 4,740.
- Setting y = 0 in the equation yields an x-intercept of 79, meaning 79 print yearbooks would earn the school $4,740.
- Setting x = 0 yields a y-intercept of 316, meaning 316 digital yearbooks would earn the school $4,740.
The document provides a collection of aptitude test questions and their solutions. Some key questions covered include: calculating percentages in mixtures, work problems, interest rate problems, sets and Venn diagrams, time and work problems related to trains, and geometry problems involving circles, spheres, cylinders. The questions are meant to help students prepare for competitive exams.
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
The document contains a collection of quantitative aptitude questions often asked in competitive exams. It provides the questions, explains the solutions, and includes links to additional practice resources on quantitative aptitude and entrepreneurship.
Questions of quantitative aptitude tests for competitive examinations Dr. Trilok Kumar Jain
This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship PGPSE is for those who want to transform the world. It is different from MBA, BBA, CFA, CA,CS,ICWA and other traditional programmes. It is based on self certification and based on self learning and guidance by mentors. It is for those who want to be entrepreneurs and social changers. Let us work together. Our basic idea is that KNOWLEDGE IS FREE & AND SHARE IT WITH THE WORLD
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
This document provides a collection of quantitative aptitude questions often asked in competitive exams, along with explanations and links to additional practice resources. The questions cover a range of topics including number series, time, ratios, mixtures, averages, profit and loss, time and work, and data interpretation. Solutions to sample puzzles are also included.
1. The document provides 20 multiple choice questions from a past IB ACIO exam paper from 2015.
2. It tests knowledge on topics like spices, human anatomy, world geography, science, history and more.
3. For each question there are 4 answer options and explanations are provided for some questions.
This module introduces ratio, proportion, and the Basic Proportionality Theorem. Students will learn about ratios, proportions, and how to use the fundamental law of proportions to solve problems involving triangles. The module is designed to teach students to apply the definition of proportion of segments to find unknown lengths and illustrate and verify the Basic Proportionality Theorem and its Converse. Examples are provided to demonstrate how to express ratios in simplest form, find missing values in proportions, determine if ratios form proportions, and solve problems involving angles and segments in triangles using ratios and proportions.
This module covers similarity and the Pythagorean theorem as they relate to right triangles. It discusses how the altitude to the hypotenuse of a right triangle divides it into two smaller right triangles that are similar to each other and the original triangle. It also explains how the altitude is the geometric mean of the hypotenuse segments. Special right triangles like 45-45-90 and 30-60-90 triangles are examined, relating side lengths through their properties. The Pythagorean theorem is derived and used to solve for missing sides of right triangles. Students work through examples and multi-step problems applying these concepts.
This document provides a mark scheme for the January 2015 International GCSE Physics exam. It begins by outlining the qualifications awarded by Pearson, the exam board, and their commitment to education. The document then provides general guidance for examiners on how to apply the mark scheme consistently. It emphasizes rewarding what students can show they know and avoiding penalizing for omissions. Several sample exam questions and answers are then provided, with notes explaining the marks awarded for each part.
This document provides information about Module 17 on similar triangles. The key points covered are:
1. The module discusses the definition of similar triangles, similarity theorems, and how to determine if two triangles are similar or find missing lengths using properties of similar triangles.
2. Students are expected to learn how to apply the definition of similar triangles, verify the AAA, SAS, and SSS similarity theorems, and use proportionality theorems to calculate lengths of line segments.
3. Several examples and exercises are provided to help students practice determining if triangles are similar, citing the appropriate similarity theorem, finding missing lengths, and applying properties of similar triangles.
The document is a lesson on adding and subtracting integers:
1. It provides examples of adding and subtracting integers and finding the sums and differences.
2. It explains that to subtract an integer, you add its additive inverse, such as 4 - 9 = 4 + (-9) = -5.
3. It contains practice problems evaluating integer expressions and solving real-world problems involving adding and subtracting integers.
The document discusses various methods to calculate averages and related concepts:
- It defines an average as the sum of a data set divided by the number of data points. Examples are given of calculating averages for test scores and batting averages.
- Shortcut formulas are provided for calculating averages when data points are added or removed. These allow finding missing data like a player's score or person's age.
- Methods are outlined for finding averages of series using patterns in the number of data points.
- Formulas are presented for calculating average speeds for journeys with multiple distances traveled at different rates. Worked examples demonstrate applying the formulas.
This document provides a mathematics module with 31 questions covering topics like arithmetic, geometry, fractions, percentages, and conversions. For each question there are 3 multiple choice answers and an explanation of the correct answer. The questions assess skills in calculating speeds, volumes, ratios, expressions with exponents, area of shapes, conversions between units, and more.
1. The document presents three systems of equations related to word problems involving Justin and Karin's allowance, costs for recreation programs, and ages of Dewan and Adrianne.
2. Graphs are helpful for solving systems of equations because they allow you to visually check solutions obtained through algebraic methods. Graphing the equations shows their point of intersection, which indicates the solution.
3. Learning to solve systems of equations algebraically and using graphs helps answer word problems leading to two linear equations in two variables, as required by the Common Core standards.
1. The document provides examples for writing and solving systems of linear equations by graphing. It explains how to write equations in slope-intercept form, graph the lines on the same coordinate plane, and determine if the system has no solution, one solution, or an infinite number of solutions based on where the lines intersect.
2. Key steps shown include writing equations in y=mx+b form, finding the slope and y-intercept, graphing the lines and analyzing where they intersect to determine the number of solutions.
3. Graphing systems of equations allows you to visually see where lines intersect, satisfying both equations simultaneously and showing the solution to the system.
This document is a lesson on the Pythagorean theorem from a Holt Geometry textbook. It begins with examples of using the Pythagorean theorem to solve for missing sides of right triangles. It then discusses how to identify if a set of three sides forms a Pythagorean triple. The lesson continues with examples of using the converse of the Pythagorean theorem and Pythagorean inequalities to classify triangles as right, acute, or obtuse based on their side lengths. It concludes with a two-part lesson quiz reviewing the material.
Exact Solutions of Convection Diffusion Equation by Modified F-Expansion MethodIJMER
In this paper, the modified F-expansion method is proposed for constructing more than one
exact solutions of nonlinear convection diffusion equation with the aid of symbolic computation
Mathematica. By using this method, some new exact travelling wave solutions of the convection diffusion
equation are successfully obtained. These exact solutions include the soliton-like solutions, trigonometric
function solutions and rational solutions. Also, it is shown that the proposed method is efficient for
solving nonlinear partial differential equations arising in mathematical physics.
The document discusses paragraph proofs and two-column proofs in geometry. A paragraph proof states the given information, what is to be proven, then uses a logical chain of statements and reasons to justify the conclusion. A two-column proof separates the statements from the reasons into two columns, with the given information and conclusion stated. Both proof styles use definitions, properties, and theorems to logically justify each step. The document provides examples of each type of proof to illustrate the process.
The document provides information about an afterschool management aptitude test and social entrepreneurship program. It includes sample questions to test analytical ability covering topics like averages, percentages, profit/loss calculations, and data interpretation. Solutions to the example questions are also provided.
(8) Lesson 2.1 - Solve Equations with Rational Coefficientswzuri
This document provides instruction on solving linear equations with one variable. It begins by defining key terms like multiplicative inverse and coefficient. It then provides examples of solving various equations by multiplying or dividing both sides of the equation by the same term to isolate the variable. These examples include equations with fractions, decimals, and percentages. The document emphasizes using the inverse operation property to solve equations and checking solutions. It aims to help students solve equations that may require expanding and combining like terms.
This document summarizes a module on rational exponents and radicals that was presented at a 2014 mid-year inset for secondary mathematics teachers. The module covered lessons on zero, negative integral and rational exponents, radicals, and solving radical equations. It provided examples of simplifying expressions using laws of exponents and radicals. Recommended teaching strategies included problem-solving activities and a group brainstorming activity to discuss critical content areas and difficulties from teacher and student perspectives.
1. The document provides examples of finding slopes of lines from given points and determining if points form a parallelogram. It also discusses direct variation relationships and using equations, tables, and graphs to represent and compare them.
2. A constant of variation in a direct variation relationship is the constant that relates the ratio of the output quantity to the input quantity. It represents the unit rate or slope in the direct variation equation and graph.
3. Graphs are helpful for comparing different proportional relationships represented in different ways and finding the unit rate between quantities in a relationship from its graphical representation.
- Mauldin Middle School wants to make $4,740 from yearbook sales. Print yearbooks cost $60 each and digital yearbooks cost $15 each.
- This relationship can be represented by the equation 60x + 15y = 4,740.
- Setting y = 0 in the equation yields an x-intercept of 79, meaning 79 print yearbooks would earn the school $4,740.
- Setting x = 0 yields a y-intercept of 316, meaning 316 digital yearbooks would earn the school $4,740.
The document provides a collection of aptitude test questions and their solutions. Some key questions covered include: calculating percentages in mixtures, work problems, interest rate problems, sets and Venn diagrams, time and work problems related to trains, and geometry problems involving circles, spheres, cylinders. The questions are meant to help students prepare for competitive exams.
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
The document contains a collection of quantitative aptitude questions often asked in competitive exams. It provides the questions, explains the solutions, and includes links to additional practice resources on quantitative aptitude and entrepreneurship.
Questions of quantitative aptitude tests for competitive examinations Dr. Trilok Kumar Jain
This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship PGPSE is for those who want to transform the world. It is different from MBA, BBA, CFA, CA,CS,ICWA and other traditional programmes. It is based on self certification and based on self learning and guidance by mentors. It is for those who want to be entrepreneurs and social changers. Let us work together. Our basic idea is that KNOWLEDGE IS FREE & AND SHARE IT WITH THE WORLD
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
This document provides a collection of quantitative aptitude questions often asked in competitive exams, along with explanations and links to additional practice resources. The questions cover a range of topics including number series, time, ratios, mixtures, averages, profit and loss, time and work, and data interpretation. Solutions to sample puzzles are also included.
Questions Of Quantitative Aptitude Tests For Competitive ExaminationsDr. Trilok Kumar Jain
This document provides a collection of quantitative aptitude questions often asked in competitive exams, along with explanations and links to additional practice resources. The questions cover a range of topics including number series, time, ratios, mixtures, averages, profit and loss, time and work, and data interpretation. Solutions to sample puzzles are also included.
Actuarial Science (ACET) Mock Test Paper II By Sourav Sir's ClassesSOURAV DAS
Actuarial Science (ACET) Mock Test Paper II With Solution By Sourav Sir's Classes, Kolkata, New Delhi.
Contact Us For Any Query About Mock Tests & Solutions.
Call : 9836793076
Accounting And Bookkeeping For Business And Management 13 OctoberDr. Trilok Kumar Jain
The document advertises the PGPSE program run by AFTERSCHO☺OL, which is a free online program focused on social entrepreneurship. It is described as the world's most comprehensive program in this field. The summary provides details on program components, branches, workshops conducted, placement support, and ability to pursue other courses concurrently.
Actuarial Science (ACET) Mock Test Paper III By Sourav Sir's ClassesSOURAV DAS
Actuarial Science (ACET) Preparation
Actuarial Science (ACET) Mock Test Paper III With Solution By Sourav Sir's Classes, Kolkata, New Delhi.
Contact Us For Any Query About Mock Tests & Solutions.
Call : 9836793076
The document contains information about SAT exams, scoring, and target scores for students at ATHS. It provides 10 tips for scoring higher on the SAT and states that the minimum number of correct answers needed in each section to reach the target score is 10 questions. It also includes a sample SAT worksheet with 20 multiple choice questions testing various math and reasoning skills.
This document provides practice questions and tips in business mathematics. It contains multiple choice questions related to topics like ratios, percentages, profit and loss, time and work, averages, simple and compound interest, discounts, and permutations and combinations. The questions are intended to help students prepare for competitive exams in subjects like commerce and management.
This document provides practice questions and tips in business mathematics. It contains multiple choice questions related to topics like ratios, percentages, profit and loss, time and work, averages, simple and compound interest, discounts, and permutations and combinations. The questions are intended to help students prepare for competitive exams in subjects like commerce and management.
1. The document contains 10 math problems with solutions. The problems cover topics like arithmetic progressions, rates of change, probability, and geometry.
2. One problem involves finding the value of n given that the sum of even numbers between 1 and n is a specific value. The solution uses the formula for the sum of an arithmetic progression.
3. Another problem asks what fraction of a solution must be replaced if the original solution was 40% and replaced with 25% solution to get a final concentration of 35%. The solution sets up an equation to solve for the fraction replaced.
The document contains solutions to 18 math and probability problems. Some key details:
- Problem 1 involves finding an odd number n such that the sum of even numbers between 1 and n equals 79*80.
- Problem 2 calculates the price at which a bushel of corn costs the same as a peck of wheat, given changing prices.
- Problem 3 determines the minimum number of people needed to have over a 50% chance that one was born in a leap year.
The document provides information about a management aptitude test and social entrepreneurship program. It discusses developing change makers and offers a free, comprehensive program in social and spiritual entrepreneurship open to all. It then provides examples of math and reasoning questions along with solutions.
This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship PGPSE is for those who want to transform the world. It is different from MBA, BBA, CFA, CA,CS,ICWA and other traditional programmes. It is based on self certification and based on self learning and guidance by mentors. It is for those who want to be entrepreneurs and social changers. Let us work together. Our basic idea is that KNOWLEDGE IS FREE & AND SHARE IT WITH THE WORLD
The document provides 14 questions related to mensuration, area, volume, and other concepts tested in aptitude tests. It includes questions on finding the area of different shapes like circles, rectangles, triangles, as well as word problems involving distances, speeds, and rates.
The document contains 44 math word problems. It provides the problems, solutions, and answers. The problems cover a range of math topics including ratios, percentages, averages, algebra, geometry, and more. For each problem, the full question is stated, the steps to solve the problem are shown, and the final numerical answer is provided.
Aptitude Questions With Answers For Time Distance ProblemsRichard Hogue
The document provides examples and solutions for various time and distance aptitude questions. It includes 10 sample questions covering topics like speed, ratio of speeds, time calculations for journeys, and averages. The questions are multi-step word problems requiring setting up equations to arrive at the solution.
This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship
Examination reforms are essential to transform the education system according to the document. The current examination system focuses only on rote memorization but needs to evaluate creativity and problem-solving. The document outlines steps to reform examinations including setting goals based on program and course objectives, evaluating whether objectives are achieved through direct and indirect methods, using continuous evaluations, and adopting open book exams and multiple evaluation methods.
leewayhertz.com-AI in predictive maintenance Use cases technologies benefits ...alexjohnson7307
Predictive maintenance is a proactive approach that anticipates equipment failures before they happen. At the forefront of this innovative strategy is Artificial Intelligence (AI), which brings unprecedented precision and efficiency. AI in predictive maintenance is transforming industries by reducing downtime, minimizing costs, and enhancing productivity.
Nunit vs XUnit vs MSTest Differences Between These Unit Testing Frameworks.pdfflufftailshop
When it comes to unit testing in the .NET ecosystem, developers have a wide range of options available. Among the most popular choices are NUnit, XUnit, and MSTest. These unit testing frameworks provide essential tools and features to help ensure the quality and reliability of code. However, understanding the differences between these frameworks is crucial for selecting the most suitable one for your projects.
TrustArc Webinar - 2024 Global Privacy SurveyTrustArc
How does your privacy program stack up against your peers? What challenges are privacy teams tackling and prioritizing in 2024?
In the fifth annual Global Privacy Benchmarks Survey, we asked over 1,800 global privacy professionals and business executives to share their perspectives on the current state of privacy inside and outside of their organizations. This year’s report focused on emerging areas of importance for privacy and compliance professionals, including considerations and implications of Artificial Intelligence (AI) technologies, building brand trust, and different approaches for achieving higher privacy competence scores.
See how organizational priorities and strategic approaches to data security and privacy are evolving around the globe.
This webinar will review:
- The top 10 privacy insights from the fifth annual Global Privacy Benchmarks Survey
- The top challenges for privacy leaders, practitioners, and organizations in 2024
- Key themes to consider in developing and maintaining your privacy program
HCL Notes and Domino License Cost Reduction in the World of DLAUpanagenda
Webinar Recording: https://www.panagenda.com/webinars/hcl-notes-and-domino-license-cost-reduction-in-the-world-of-dlau/
The introduction of DLAU and the CCB & CCX licensing model caused quite a stir in the HCL community. As a Notes and Domino customer, you may have faced challenges with unexpected user counts and license costs. You probably have questions on how this new licensing approach works and how to benefit from it. Most importantly, you likely have budget constraints and want to save money where possible. Don’t worry, we can help with all of this!
We’ll show you how to fix common misconfigurations that cause higher-than-expected user counts, and how to identify accounts which you can deactivate to save money. There are also frequent patterns that can cause unnecessary cost, like using a person document instead of a mail-in for shared mailboxes. We’ll provide examples and solutions for those as well. And naturally we’ll explain the new licensing model.
Join HCL Ambassador Marc Thomas in this webinar with a special guest appearance from Franz Walder. It will give you the tools and know-how to stay on top of what is going on with Domino licensing. You will be able lower your cost through an optimized configuration and keep it low going forward.
These topics will be covered
- Reducing license cost by finding and fixing misconfigurations and superfluous accounts
- How do CCB and CCX licenses really work?
- Understanding the DLAU tool and how to best utilize it
- Tips for common problem areas, like team mailboxes, functional/test users, etc
- Practical examples and best practices to implement right away
How to Interpret Trends in the Kalyan Rajdhani Mix Chart.pdfChart Kalyan
A Mix Chart displays historical data of numbers in a graphical or tabular form. The Kalyan Rajdhani Mix Chart specifically shows the results of a sequence of numbers over different periods.
Dive into the realm of operating systems (OS) with Pravash Chandra Das, a seasoned Digital Forensic Analyst, as your guide. 🚀 This comprehensive presentation illuminates the core concepts, types, and evolution of OS, essential for understanding modern computing landscapes.
Beginning with the foundational definition, Das clarifies the pivotal role of OS as system software orchestrating hardware resources, software applications, and user interactions. Through succinct descriptions, he delineates the diverse types of OS, from single-user, single-task environments like early MS-DOS iterations, to multi-user, multi-tasking systems exemplified by modern Linux distributions.
Crucial components like the kernel and shell are dissected, highlighting their indispensable functions in resource management and user interface interaction. Das elucidates how the kernel acts as the central nervous system, orchestrating process scheduling, memory allocation, and device management. Meanwhile, the shell serves as the gateway for user commands, bridging the gap between human input and machine execution. 💻
The narrative then shifts to a captivating exploration of prominent desktop OSs, Windows, macOS, and Linux. Windows, with its globally ubiquitous presence and user-friendly interface, emerges as a cornerstone in personal computing history. macOS, lauded for its sleek design and seamless integration with Apple's ecosystem, stands as a beacon of stability and creativity. Linux, an open-source marvel, offers unparalleled flexibility and security, revolutionizing the computing landscape. 🖥️
Moving to the realm of mobile devices, Das unravels the dominance of Android and iOS. Android's open-source ethos fosters a vibrant ecosystem of customization and innovation, while iOS boasts a seamless user experience and robust security infrastructure. Meanwhile, discontinued platforms like Symbian and Palm OS evoke nostalgia for their pioneering roles in the smartphone revolution.
The journey concludes with a reflection on the ever-evolving landscape of OS, underscored by the emergence of real-time operating systems (RTOS) and the persistent quest for innovation and efficiency. As technology continues to shape our world, understanding the foundations and evolution of operating systems remains paramount. Join Pravash Chandra Das on this illuminating journey through the heart of computing. 🌟
GraphRAG for Life Science to increase LLM accuracyTomaz Bratanic
GraphRAG for life science domain, where you retriever information from biomedical knowledge graphs using LLMs to increase the accuracy and performance of generated answers
Ocean lotus Threat actors project by John Sitima 2024 (1).pptxSitimaJohn
Ocean Lotus cyber threat actors represent a sophisticated, persistent, and politically motivated group that poses a significant risk to organizations and individuals in the Southeast Asian region. Their continuous evolution and adaptability underscore the need for robust cybersecurity measures and international cooperation to identify and mitigate the threats posed by such advanced persistent threat groups.
Generating privacy-protected synthetic data using Secludy and MilvusZilliz
During this demo, the founders of Secludy will demonstrate how their system utilizes Milvus to store and manipulate embeddings for generating privacy-protected synthetic data. Their approach not only maintains the confidentiality of the original data but also enhances the utility and scalability of LLMs under privacy constraints. Attendees, including machine learning engineers, data scientists, and data managers, will witness first-hand how Secludy's integration with Milvus empowers organizations to harness the power of LLMs securely and efficiently.
Programming Foundation Models with DSPy - Meetup SlidesZilliz
Prompting language models is hard, while programming language models is easy. In this talk, I will discuss the state-of-the-art framework DSPy for programming foundation models with its powerful optimizers and runtime constraint system.
HCL Notes und Domino Lizenzkostenreduzierung in der Welt von DLAUpanagenda
Webinar Recording: https://www.panagenda.com/webinars/hcl-notes-und-domino-lizenzkostenreduzierung-in-der-welt-von-dlau/
DLAU und die Lizenzen nach dem CCB- und CCX-Modell sind für viele in der HCL-Community seit letztem Jahr ein heißes Thema. Als Notes- oder Domino-Kunde haben Sie vielleicht mit unerwartet hohen Benutzerzahlen und Lizenzgebühren zu kämpfen. Sie fragen sich vielleicht, wie diese neue Art der Lizenzierung funktioniert und welchen Nutzen sie Ihnen bringt. Vor allem wollen Sie sicherlich Ihr Budget einhalten und Kosten sparen, wo immer möglich. Das verstehen wir und wir möchten Ihnen dabei helfen!
Wir erklären Ihnen, wie Sie häufige Konfigurationsprobleme lösen können, die dazu führen können, dass mehr Benutzer gezählt werden als nötig, und wie Sie überflüssige oder ungenutzte Konten identifizieren und entfernen können, um Geld zu sparen. Es gibt auch einige Ansätze, die zu unnötigen Ausgaben führen können, z. B. wenn ein Personendokument anstelle eines Mail-Ins für geteilte Mailboxen verwendet wird. Wir zeigen Ihnen solche Fälle und deren Lösungen. Und natürlich erklären wir Ihnen das neue Lizenzmodell.
Nehmen Sie an diesem Webinar teil, bei dem HCL-Ambassador Marc Thomas und Gastredner Franz Walder Ihnen diese neue Welt näherbringen. Es vermittelt Ihnen die Tools und das Know-how, um den Überblick zu bewahren. Sie werden in der Lage sein, Ihre Kosten durch eine optimierte Domino-Konfiguration zu reduzieren und auch in Zukunft gering zu halten.
Diese Themen werden behandelt
- Reduzierung der Lizenzkosten durch Auffinden und Beheben von Fehlkonfigurationen und überflüssigen Konten
- Wie funktionieren CCB- und CCX-Lizenzen wirklich?
- Verstehen des DLAU-Tools und wie man es am besten nutzt
- Tipps für häufige Problembereiche, wie z. B. Team-Postfächer, Funktions-/Testbenutzer usw.
- Praxisbeispiele und Best Practices zum sofortigen Umsetzen
Trusted Execution Environment for Decentralized Process MiningLucaBarbaro3
Presentation of the paper "Trusted Execution Environment for Decentralized Process Mining" given during the CAiSE 2024 Conference in Cyprus on June 7, 2024.
Skybuffer SAM4U tool for SAP license adoptionTatiana Kojar
Manage and optimize your license adoption and consumption with SAM4U, an SAP free customer software asset management tool.
SAM4U, an SAP complimentary software asset management tool for customers, delivers a detailed and well-structured overview of license inventory and usage with a user-friendly interface. We offer a hosted, cost-effective, and performance-optimized SAM4U setup in the Skybuffer Cloud environment. You retain ownership of the system and data, while we manage the ABAP 7.58 infrastructure, ensuring fixed Total Cost of Ownership (TCO) and exceptional services through the SAP Fiori interface.
This presentation provides valuable insights into effective cost-saving techniques on AWS. Learn how to optimize your AWS resources by rightsizing, increasing elasticity, picking the right storage class, and choosing the best pricing model. Additionally, discover essential governance mechanisms to ensure continuous cost efficiency. Whether you are new to AWS or an experienced user, this presentation provides clear and practical tips to help you reduce your cloud costs and get the most out of your budget.
Building Production Ready Search Pipelines with Spark and MilvusZilliz
Spark is the widely used ETL tool for processing, indexing and ingesting data to serving stack for search. Milvus is the production-ready open-source vector database. In this talk we will show how to use Spark to process unstructured data to extract vector representations, and push the vectors to Milvus vector database for search serving.
Building Production Ready Search Pipelines with Spark and Milvus
Aptitude test papers
1. APTITUDE TEST PAPERS - FREQUENTLY ASKED QUESTIONS by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – CSE & PGPSE STUDENTS (CSE & PGPSE are free online programmes open for all, free for all) mobile : 91+9414430763
2. My words..... My purpose here is to give a few questions, which are often asked in aptitude tests and competitive examinations. Please prepare well for your examinations. Please pass this presentation to all those who might need it. Let us spread knowledge as widely as possible. I welcome your suggestions. I also request you to help me in spreading social entrepreneurship across the globe – for which I need support of you people – not of any VIP. With your help, I can spread the ideas – for which we stand....
3. Data sufficiency question ... Are two triangles congruent? a. Both triangles are right angled. b. Both are of same perimeter. Answer : this information is indequate and we cant get the answer on the basis of such limited information
4. Data sufficiency question ? In what time the cylinder completes one rotation? a. Radius of cylinder is given. b. Speed (angular) is given. Answer : BOTH THE STATEMENTS ARE REQUIRED (we get anwer using both)
5. Data sufficiency question ... PQR is an isosceles triangle. What's the length of PQ? a. The length of QR is 7. b. The angle of PQR is 90 degree. ANSWER : BOTH THE STATEMENTS ARE REQUIRED
6. What's the ratio of length of the line segment PQ to the length of arc PQ, when PQ forms the arc of a circle? a. Area of circle is168. b. The segment PQ forms the diameter of circle. Answer : both the statements are required
7. In a 1 km race,A beats B by 28 m or 7sec.Find A's time over the course? Speed of B = 28/7 = 4 m per second total time by B = 1000/4 = 250 seconds total time by A = 250-7 = 243 seconds answer
8. A runs 1 3/4 times as fast as B.If A gives B a start of 84 m,how far must be winning post be so that A and B might reach it at the same time ? Options : 220, 196, 190, 144 solve from options : start with 196 196 – 84 = 112 196/112 = 1.75 this is given in question : 1 3/4 ratio 1 is equal to 1.75 answer
9. A can run 1km in 3 min ,10sec and B can cover same distance in 3 min 20 sec. By what distance A beat B? 1 km = 1000 meters distance covered by B when A finishes : 3.16 / 3.33 * 1000 = 948.9 or 949 meters thus A defeats B by 51 meters. Answer
10. In a 100m race,A runs at 8km per hour.If A gives b a start of 4 m and still beats him by 15 sec,what is the speed of B? Speed of A = 8 * 5/18 = 2.22 meters per second time taken by A = 100/2.22 =45.05 time taken by B = 45.05 +15 = 60.05 seconds distance covered by B = 100-4 = 96 meters speed of B = 96/60.05 =1.6 meters per second speed of B in km / h = 1.6 * 18/5 = 5.76 km per hour answer
11. A and B take part in 100m race .A runs at 5km per hour. A gives B a start of 8 m and still beats him by 8 sec. What is the speed of B? Speed of A = 5 * 5/18 = 1.39 time taken by A = 100/1.39 =71.94 seconds time taken by B 71.94 + 8 = 79.94 sec. Distance covered by B = 100 -8 = 92 meters speed of B -92/79.94 =1.15 speed of B 1.15 * 18/5 = 4.14 km per hour. Answer
12. If length and breadth of a rectangle are increased by 20% each, what is the change in area ? Let us assume L = 20 and B = 10 area = 200 increase : L =20+4 = 24 and B = 12 new area = 288 increase in area is 44% shortcut = X +X +(X*X%) =20+20+4 = 44 %
13. If length and breadth of a rectangle are increased by 30% each, what is the change in area ? Let us assume L = 20 and B = 10 area = 200 increase : L =20+6 = 26 and B = 13 new area = 338 increase in area is 69% shortcut = X +X +(X*X%) =30+30+9 = 69 %
14. If out of length and breadth of a rectangle length is increased by 30% and the other is decreased by 30%, what is the change in area ? Let us assume L = 20 and B = 10 area = 200 increase : L =20+6 = 26 and B = 7 new area = 182 decrease in area is 9% shortcut = decrease of (X*X%) =30*30% = 9% decrease note : it doesnt matter whether length is increased or width is increased, the answer is same in such questions.
15. The ratio of the flow of water in pipes varies inversely as the square of the radius of the pipe, what is the ratio of the rates of flow in two pipes of diameter 2 &4 ? (1/2)^2 and (1/4)^2 ¼ : 1/16 or 4:1 or take radius : 1^2 : (1/2)^2 =1:1/4 or 4:1 answer
16. 4 friends A,B,C,D are studing together in class 12. A and B are good in Hindi but poor in English. A and C are good in Sanskrit, but poor in geography. D and B are good in Maths as well as in Sanskrit. Who is not good in maths but good in Hindi?: A g In H & S p in e, g B G in H & M & S p in e C g in S p in g D g in M,S ANSWER = A
17. A man buys 12 lts of liquid which contain 20% of the liquid and the rest is water. He then mixes it with 10lts of another mixture with 30% of liquid. What is the % of water in the new mixture? Total quantity of mixture = 12+10 =22 total water : 9.6+7 = 16.6 =16.6/22 =75% answer
18. SERIES : b3c, c4e,e6h,h9l,?? L13q answer in the first case, one is added to b, in 2 nd case we add 2 to e , in 3 rd we add 3 to e ... so we will add 5 to l numbers also increase : 3 to 4 (addition of 1), 4 to 6 (addition of 2), 6 to 9 (addition of 3) and 9 to 13 (addition of 4) answer
19. In a class of 50 students 23 speak English 15 speak Hindi and 18 speak Punjabi 3 speak only english and hindi, 6 speak english and punjabi 9 can speak only english, how many speak all the 3 languages? 23 - ( 9+6+3) = 5 answer
20. In which of the system, decimal number 194 is equal to 1234? Options : 3,5,8,16 try with options : try for 5 1*(5^3)+(2*5^2)+(3*5^1)+(4*5^0) =125+50+15+4 =194 so answer is 5 answer
21. If TAFJHH is coded as RBEKGI then RBDJK can be coded as --------- ANS: PCCKJ
22. Find the result of the following expression if, M denotes modulus operation, R denotes round-off, T denotes truncation: M(373,5)+R(3.4)+T(7.7)+R(5.8) 373 / 5 = remainder is 3 3.4 rounded off to 3 7.7 to 7 5.8 to 6 total : = 19 ANS:19
23. A certain number of men can finish a piece of work in 10 days. If however there were 10 men less it will take 10days more for the work to be finished. How many men were there originally? Number of men =x total work =10X 10X = (X-10)(20) 10X=20X-200 X=20 answer
24. In simple interest what sum amount of Rs. 1120/- in 4 years and Rs.1200/- in 5 years? Interest = 1200 – 1120 = 80 the total interest for 5 years = 400 principal = 1200-400 = 800 answer
25. There are total 15 people. 7 speaks french and 8 speaks spanish. 3 do not speak any language. Which part of total people speaks both languages. 7+8 = 15 BUT IT SHOULD BE 12 (because 3 dont speak any language). So there are 3 persons who speak both the languages. Answer
26. A jogger wants to save 1/4 th of his jogging time. He should increase his speed by how much Let us assume that the jogger takes 4 ours for 12 KM. Now he will save 1/4 th so he will take 3 hours for 12 KM. His earlier speed was 12/4 = 3, but his new speed is 12/3 = 4. he increased his speed from 3 to 4 or 33% increase in his speed. Answer (you may assume other figures also but the answer will be the same).
27. A is an integer. Dividing 89 & 125 by A gives remainders 4 & 6 respectively. Find a ? Find HCF of (89-4) and (125-6) HCF of 85 and 119 is 17 answer = 17 answer
28. In a office work is distributed between p persons. If 1/8 members are absent then work increased for each person by what %. Suppose there are 560 units of work among 8 persons. Now there are only 7 persons so each person is doing 80 units of work (instead of 70 units of work). The work increased by : 10 on 70 10/70*100 = 14.29% answer
29. 120, 315, 300, 345, ? We can write these numbers as multiples of 15 : 8, 21,20,23 the next digit should be :
30.
31. RULE : a^0 = 1 IF ZERO IS PUT AS POWER OF ANY NUMBER, IT WILL BECOME 1. THUS IF YOU PUT POWER OF ZERO ON 100000, IT WILL BE EQUAL TO 1.
32. RULE : (a/b)^n = a^n / b^n IF YOU TAKE UP POWER OVER A FRACTION, YOU CAN PUT IT SEPARATELY ALSO. EXAMPLE : (2/3) ^3 = 8/27 ANSWER
33. (27)^2/3 =? 1. FIRST WE PUT 27 AS 3^3 BECAUSE WE KNOW 3*3*3 = 27 = (3^3)2/3 WE CAN CANCEL POWR OF 3 3^2 = 9. ANSWER
34. (1024)^-4/5 =? 1. FIRST PUT 1024 AS 2^X, WE KNOW THAT 2*2*2*2*2..... (TILL 10) = 1024 SO WE CAN WRITE : 1024 = 2^10 = 4^5 (4^5)-4/5 (NOW CANCEL 5) = (4)-4= 1/(4)4 (ANY POWER IN NEGATIVE WILL PUT THE NUMERATOR INTO DENOMINATOR) = 1/256.
35. If 2^(x-1)+ 2^(x+1) = 256 then find the value of x .(X-1) * (X+1) = (X^2 – 1) So we can write it as : =2^(x^2-1) = 256 now factorise 256 = 2*2*2*2*2*2*2*2=2^8 2^(x^2 -1) = 2^8 or X^2 = 9 X = 3 answer
36. Simplify : 125^(2/3) * 25^(1/2)*5*5^(1/2) 125^(2/3) = (5^3)^(2/3) 3 gets cancelled so : 5^2 = 25 =5*5 we can write it as : 5*5*5*5*5^(1/2) =5^4 * 5^ (½) (add the two powers, when there is multiplication of their base) =5^(9/2) answer
37. What is the compounded ratio of : 4:9, duplicate ratio of 3:4, triplicate ratio of 2:3 and 9:7? Duplicate ratio of 3:4 =9:16 (square of 3 and 4 respectively) triplicate ratio of 2:3 = 8:27 (take cube of 2 & 3 respectively). Now let us compound (multiply all the X and Y individuallY ) them : X =4*9*8*9 = 2592 Y = 9*16*27*7=27216 their ratio is 1: 10.5 or 2:21 answer
38. A train 140 m long running at 72 kmph.In how much time will it pass a platform 260m long. Total distance to cover = 140+260 = 400 time = distance / speed speed in meters per second = 72 *5/18 = 20 meters per second time = 400/20 = 20 seconds answer
39. A man is standing on a railway bridge which is 180 m.He finds that a train crosses the bridge in 20 seconds but himself in 8 sec. Find the length of the train and its sppeed Let us assume the length of train = X the speed of train = x meters/8 180 meters are covered by train in (20-8) seconds 180/12 = 15 meters per second. Thus the length of train is : 15*8 = 120 meters. Answer
40. A train 150m long is running with a speed of 54 Km per hour. In what time will it pass a man who is running at a speed of 9 km ph in the same direction in which the train is going Speed of train = 54 * 5/18 = 15 meters per second speed of mann = 9 * 5/18 = 2.5 meters per second distance to cover = 150 time = 150/(15-2.5) =12 seconds answer
41. A train 220m long is running with a speed of 59 k m ph ..In what time will it pass a man who is running at 4 kmph in the direction opposite to that in which train is going. Distance to cover = 220 meters speed = 59+4 = 63 km per hour. In meters per second = 63*5/18 = 17.5 meters per second. Time required : 220/17.5 =12.57 seconds. Answer
42. Two trains 137m and 163m in length are running towards each other on parallel lines,one at the rate of 42kmph & another at 48 mph.In wht time will they be clear of each other from the moment they meet. Distance to cover 137+163 = 300 meters speed = 42+48 = 90 km per hour speed in meters per second = 90 * 5/18 = 25 meters per second time required = 300/25 = 12 seconds answer
43. A train running at 54 kmph takes 20 sec to pass a platform. Next it takes 12 sec to pass a man walking at 6kmph in the same direction in which the train is going.Find length of the train and length of platform Solution : Train v/s man speed = 54 -6 = 48 km per hour speed in m/s =48 * 5/18 = 13.33 m / s length of train = 12*13.33 = 159.6 meters speed for platform =54*5/18 = 15 m / s length of platform+ train = 20*15 = 300 length of platform = 300 – 159 = 140 meters approx.
44. A man sitting in a train which is travelling at 50mph observes that a goods train travelling in opposite direction takes 9 sec to pass him .If the goods train is 150m long find its speed Solution : - Distance travelled = 150 speed =150/9 = 16.66 meters per second or 16.66 * 18/5 = 60 km per hour approx. Thus the speed of goods train is 60-50 = 10 km per hour. Answer
45. Two trains are moving in the same direction at 65kmph and 47kmph. The faster train crosses a man in slower train in18sec.the length of the faster train is Solution : = When the trains are going in same direction, we take difference of their speed. 65-47 =18 km per hour or 5 meters per second distance travelled =(time 18 seconds * speed 5 meters per second) = 18 * 5 = 90 meters. Thus the length of faster train is 90 meters. Answer
46. A train overtakes two persons who are walking in the same direction in which the train is going at the rate of 2kmph and 4kmph and passes them completely in 9 sec and 10 sec respectively. The length of train is Solution : - let us assume the speed of train to be X. (X-2) * 9/3600 = (X-4) *10/3600 9X – 18 = 10X – 40 X=22 km per hour. thus distance travelled = (22-4) * 5/18 = 5 m/s time=10 seconds so length of train = 5*10 = 50 meters. Answer
47. Two stations A & B are 110 km apart on a straight line. One train starts from A at 7am and travels towards B at 20kmph. Another train starts from B at 8am an travels toward A at a speed of 25kmph.At what time will they meet From 7 am to 8 am only A is travelling. It would travel 20 km. Now 90 km is to be covered. 90 / (20+25) =2 hours so at 10 am they will meet.
48. A train travelling at 48kmph completely crosses another train having half its length an travelling inopposite direction at 42kmph in12 sec.It also passes a railway platform in 45sec.the length of platform is Distance by 2 trains = (48+42) = 90 or 25 m/s 25 * 12=300 meters. So the length of train is 200 meters. Platform : 48 * 5/18 =13.33 m/ s 45*13.33 = 600 so length of platform = 600-200=400 meters. Answer
49. Find the time taken by a train 180m long,running at 72kmph in crossing an electric pole Speed of train = 72 * 5/18 = 20 m / s time required = 180/20 = 6 seconds. Answer
50. Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. Solution: Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 352/7 radius = 8 Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 528/7 radius = 12 thus width of ring = 12-8 = 4 answer
51. Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is Solution : area of one circle = (22/7) * 7 * 7 =154 square of on one circle = 14*14 = 196 difference of area : 42 one side 42/4 =10.5 we have 4 cirlces, each has 10.5 cm of space enclosed, so total space enclosed is 42 sq. cm. Answer
52. A semicircular shaped window has diameter of 63cm. Its perimeter equals Circumference of circle = diameter * 22/7 = 63 * 22/7 = 198 it is semicircle so divide by 2 = 99 add diameter also – to denote one side : 99+63 = 162 cm answer
53. The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Total area = 5.5 * 3.75 =20.63 multiply it by 800 =Rs.16500
54. The no of revolutions a wheel of diameter 49 cm makes in traveling a distance of 176m is Solution : circumference = 22/7 * 49 = 154 17600 / 154 = 114.29 thus the wheel will make 115 revolutions. Answer
55. .A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field? Solution Area 22/7 * 14 * 14 = 616 time required = 616/100 = 6.16 so cow will take little over 6 days to completely graze the whole field. Answer
56. A man runs round a circular field of radius 49m at the speed of 120 m/hr. What is the time taken by the man to take twenty rounds of the field? Solution : circumference = 2* 22/7 * 49 =308 total distance to be travelled = 308 * 20 = 6160 time required = 6160/120 =51.3 hours.
57. .The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec. What is the speed of motorcycle in km/hr? Speed covered in 1 seconds : 4 * 22/7 *70 =880 cm or 8.8 meters. Speed in km per hour : 8.8 * 18/5 = 31.68 km per hour answer
58. A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be Solution : its circumference is : 2 * 22/7 *56 = 352 when you make a circle out of it, one side will be : 352/4 = 88 thus its area will be : 88*88 =7744 answer
59. The area of the largest triangle that can be inscribed in a semicircle of radius 2 is? Area of triangle = Formula = 1/2 * base * height = ½ * (2+2) *2 =4 answer
60. A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed? The total boundary = 2(90+50)= 280 280/5 = 56 so we will need 56 poles
61. The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter? Total fencing = 5300/26.5 = 200 2(L+B) = 200 2(B+20+B) = 200 2B+20=100 or B=40 and L=60 so length is 60 meters answer
62. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required? Length = X bredth = 20 20X =680, or X = 34 boundary =2(34+20)=108 but we will not cover one side =108-20=88 feet. Answer
63. A rectangular paper when folded into two congruent parts had a perimeter of 34cm for each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper? When we fold from mid of length = L+2B=34 When we fold from mid of width = 2L+B=38 add them = 3L+3B=72 L+B=24, where L =14, B=10 so area of paper =14*10 =140 answer
64. A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is? Diagonal = 52*15/60=13 ½ of perimeter=68*15/60=17 X+Y=17 -- 1 st X^2+Y^2 =169 -- 2 nd square of 1 st equation:X^2+Y^2+2XY=289 2XY=120 or XY =60 thus area of field is 60 sq.meter.
65. The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m Boundary = 10.08/.20 = 50.4 one side is 12.6 area : 158.76 area without pavement : (12.6-6)^2= 43.56 pavement = 115.2 cost = 115.2*.5 = Rs. 57.6 answer
66. Aman walked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges? Let us assume that the side of square is 1. had he walked along edges, he would have travelled 2. he walked on diagonal so he walked sqrt(2) = 1.41 thus he has saved (2-1.41) =.59 or we can say that he has saved 29.5% answer
67. .A man walking at the speed of 4 km p.h. crosses a square field diagonally in 30 minutes. The area of the field is The diagonal is 2 km or 2000 meters side of square is : 2000/sqrt(2) =1414 area of field =2000000 sq. m. Or 200 hectares
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