You generally find multiplication more effortless as compared to division, right? So to mentally solve division, simply convert the division into multiplication. Yes, you heard me right. Hold on to see how this makes things relatively easier for you.
In this article, I will share with you the trick to divide any number by 5, 25 and 125 by converting these into single digit multiplications. You can refer to an earlier article where certain Vedic Mathematics trick are used for division of more difficult numbers.
Believe me, you will be amazed at the simplicity of this method. So stay tuned and read ahead.
To divide any number by 5, you don’t actually need to perform any division. You will get your answer in 2 simple steps.
First step, you simply need to multiply the given number by 2.
Let us take some examples to understand this further.
2323/5 = 2323×2 = 4646
Second step, put place a decimal just before one digit from the right. So in this case the decimal will come before 6. Hence the answer would be 464.6
Thus, 2323/5 = 464.6
Isn’t that simple? You just need to know multiplication or table of 2.
Another example,
Step 1: 23487/5 = 23487×2 = 46974
Step 2: placing the decimal before one digit from right: 4697.4
Did you ever think it could be this easy?
To divide any number by 25, you simply need to multiply the given number by 4 and then place a decimal before 2 digits from the right.
Let’s try with an example, 5830/25
Step 1: multiply the given number by 4
5830×4 = 23320
Step 2: placing a decimal point before 2 digits from the right
Another example, 719835/25
Step 1: Multiplying by 4:
719835*4 = 2879340
Adjusting the decimal by 2 places
28793.40, that’s the answer.
This is fun.
To divide any number by 125, multiply the number by 8 and then put a decimal point before 3 digits from the right. Let’s take an example, 2650/125
Step 1: Multiplying by 8:
2650*8 = 21200
Adjusting the decimal by 3 places
21.200 = 21.2, that’s the answer.
Another example, 348847/125
Step 1: Multiplying by 8:
348847*8 = 2790776
Adjusting the decimal by 3 places
2790.776, that’s the answer.
Don’t you love this?
For those who’re interested in understanding what’s going on here, you can read further.
Let us first understand for 5.
x/5 = x*2/10 = (x/10) * 2 = 0.x * 2
thus, to divide any number by 5, we can multiply the same number by 2 and adjust the decimal point by one place.
Similarly, we can understand shortcut we adopted above for multiplication by 25.
x/25 = x * 4/100 = (x/100) * 4 = 0.0x * 4
Thus, to divide any number by 25, we can multiply the same number by 4 and adjust the decimal point by 2 places.
Stretching the method further, to divide anything by 125 we multiply the number by 8 because –
x/125 = x*8/1000 = (x/1000) * 8 = 0.00x * 8
Once you understand the logic explained above, you don’t have to memorize the steps in these tricks. In fact, by using the same logic you can do quick division by numbers like 50, 250, 500 and so on. So please don’t memorize, please understand and then apply.
If you’ve slightest doubt, ask your questions in the comments below.
]]>A set of 30 questions in each subject, Mathematics, Chemistry and Physics in a time period of 180 minutes (3 hours) is not a task that can be counted easy. One has to be fast and accurate while solving questions, since any mistake can lead to negative marking.
The marking scheme is quite simple for the exam, as every correct answer awards 4 marks and every incorrect deducts 1 from his aggregate total. So it is clear that one is being evaluated in a maximum score of 360.
JEE Main is one of the exam that is based on basic concepts and understanding of different principles/theorems that are being taught in 11^{th} and 12^{th} standard. The syllabus of the exam includes all the topics that are included in NCERT books.
The exam has all types of questions, some of them are based on principles (like Archimedes, Bernoulli, etc.), theorems (Pythagoras, Theorems for triangles, etc.) and applications of theory (Projectile, Equations of motion, etc.). It has both theoretical and numerical type of problems, with 4 multiple choices. Among the 4 choices, only one choice is most appropriate or correct.
One can follow multiple books to practice and learn the concepts that are required to crack the exam. The most recommended books for understanding of basic concepts for JEE Main is NCERT books (for all the three subjects). It is counted as a baseline material for JEE Main and learning all the principles, formulas and theorems. It is one of the best book for theoretical questions as some of the problems in JEE Main are manipulated from NCERT textbooks.
Once an aspirant has completed the theoretical study from NCERT books, he/she can refer below-mentioned books for practice and improving on their concepts.
There is one well known book by a professor in IIT Kanpur, naming “Concepts of physics by H.C Verma”. The book comes in 2 volumes(Concepts of Physics Vol.1 and Vol.2) and are the most recommended books for JEE aspirants. This book includes short answer type questions to sharpen one’s basic understanding. Also this book includes, single choice and multiple choice questions that can be very helpful to build up your concepts and apply whatever you have learnt in your theory. At the end of every topic, you can find exercise problems, which can be really fruitful to score good marks in JEE Main.
That single book can surely make you crack JEE Main with a very good score along with NCERT Physics book. If you feel you still have time, you can check DC Pandey (Arihant Publications) and Resnic Halliday.
Concentrate on NCERT for theoretical questions and learning of concepts including chemical reactions and formulas. NCERT chemistry can be referred for solving theoretical problems, but it doesn’t have good number of problems. For practice of problems and extending your knowledge towards different topics, you can refer these books-
For Organic chemistry-
For Inorganic chemistry–
For Physical chemistry-
Practice as many problems (numerical as well as theoretical) from the mentioned books when you are through with NCERT books and you surely can score really well in Chemistry.
Mathematics is considered as one of the subjects that play a vital role in cracking IIT JEE Main and scoring a good rank. For learning concepts and theorems, one can refer Mathematics NCERT book, but it doesn’t have very good level problems for practice.
For practicing Mathematical problems and to get idea of implementation of theorems/principles, you can refer these books-
Once you are through with the above mentioned books, you must practice mock test papers and solve previous year exam papers. Some of the mock test papers are released by JEE organising committee (CBSE earlier, NTA from 2019 onwards) and some are available from different publications. Some of the publications are-
Mock tests can help you to get an idea of where you stand and a rough estimate of how much you can score. The last two to three weeks before the exam must be dedicated to solving previous year’s problems. You should take last 15 year papers of IIT JEE Main and solve them. That will get you ready for your IIT JEE Main exam and you will be surely score really good.
Along with theory, numerical questions will help you boost your marks and get you a very good rank. As it is said for competitive exams-
“More you practice, closer you move towards scoring good marks”.
Don’t forget to download JEE Main preparation app on Google Play Store. It provides free notes, topic-wise study material, quizzes, mock test, question papers, and all the updates related JEE Mains/Advanced exam.
]]>I decided to write this post to ease this struggle to the best of my capacity.
For ease of understanding I’ve made categories based on the type of decisions we need to make.
All the examples given below are from the items I picked up from the grocery section of the supermarket.
1^{st} comparison
To compare these options, first thing you do is round off the numbers.
After that, you need to compare the price by making the quantity same. Ideal way is to scale up the lower quantity.
In this case if you multiply smaller quantity, i.e. 24 by 3 you get 72.
Hence 24×3 mangoes in 1^{st} case will be Rs.700×3 = Rs.2100
Where in 2^{nd} case, 72 mangoes are available for Rs.1900
So you know, if you can consume 72 mangoes, it’s available at a bargain. However, we need to consider the perishable nature of mangoes and the quantity we can consume in a reasonable time till which mangoes can remain fresh.
2^{nd} Comparison
Again, 1^{st} thing is to round off the prices.
Next, easier thing to do in this case will be to scale it up and see.
If we multiply the smaller quantity by 2, we get
85×2 = 170 gm for Rs.150 x 2 = Rs.300
Comparing once again,
In 1^{st} case, we get 170 gm for Rs.300
In 2^{nd} case, we get 200 gm of Rs.300
Thus clearly 2^{nd} case is more beneficial.
3^{rd} comparison
We need to scale up the smaller quantity to match the larger quantity. Then the price comparison will be apple to apple. That too of the same size.
In the above case you saw that just simple multiplication was good enough to do that. However, here you need to do some quick thinking.
You remember HCF taught in school? Don’t panic. No worries even if you don’t.
HCF or highest common factor means finding a number which divides both the numbers.
Let me explain this for the above example. If we have smaller packets of 500 gm each, for buying 1.5 kg you need 3 packets and for buying 4 kg you need 8 packets.
So now, if 1.5kg is at Rs.153, 1 packet of 0.5kg is for Rs.153/3 = Rs.51
Therefore, 8 packets (i.e. 4kgs) will be for Rs.51*8 = Rs.408.
So, if you ignore the free bucket available you’re better off buying the smaller packet i.e. of 1.5 kg compared to 4 kg pack.
However, getting a bucket by paying additional Rs.10 is not a bad deal. So you should go for 4 kg packet.
In an earlier article I’ve touched upon calculations of gross margin, cost and sale price. That article will also help you in calculations of discounts. Have a look.
I hope I helped you to overcome your dilemma during shopping when you are spoilt for choices and one offer seems to be better than the other.
Did you all like my tips? If yes, write in the comments below. Stay tuned for another set of situations on Discount Shopping.
]]>At times, it might look little scary or daunting to understand the properties of these polygons. If you know polygons in and out, this article may not be for you. It is for learners who would like to understand polygons in a simple common sensical manner.
So let us delve deep into the world of polygons. So, here we go with the most obvious question first –
Polygon is any shape made up of 3 or more connecting lines on a flat sheet surface. Hence, polygons are 2-dimensional closed figures made up of straight lines. Thus the intersecting lines have to terminate at the point of intersection.
If you would like to dive deeper into the formal definition and terminology related to polygons, I recommend you check out Chegg’s page on polygons
The angles formed in the interior or inside of a polygon where two pair of sides intersect are called interior angles.
A polygon in which all sides are equal (equilateral) and all angles are equal (equiangular). Otherwise, it’s an irregular polygon.
I can directly give you the formula to calculate the sum of all interior angles of a polygon. But I want you to dive a little deeper and understand how that formula is arrived.
Our goal for today is to figure out how the interior angles of a polygon change as the number of sides of the figure increases. So let us start with the polygon which has smallest number of sides, i.e. triangles.
To understand interior angles of a polygon, we have to keep triangles (type of polygon) as the starting point. Triangle has 3 interior angles and 3 sides. It’s a known fact that sum of these 3 interior angles of a triangle is always 180°.
Little out of context, but the moment we think of triangles Pythagoras Theorem is generally the first things which comes to our mind. In a previous post, I have tried exploring, was Pythagoras Theorem actually proved by Pythagoras?
3 sided polygon, sum of interior angles = 180°
Moving forward and looking at polygons with higher number of sides –
Quadrilateral is polygon with 4 sides and obviously 4 interior angles.
If we connect the opposite corners of the quadrilateral (i.e. the diagonal) divides the quadrilateral into 2 triangles. If we add the sum of interior angles of theses 2 triangles, we get 180°+180°=360°. Thus the sum of interior angles in a 4 sided polygon is always 360°.
Another way of looking at it: square is a polygon with all sides equal, i.e. it’s a regular polygon. We know in a square each of the 4 angles is equal to 90°. Thus sum of all the interior angles is 90°x4 = 360°
4 sided polygon, sum of interior angles = 360°
In a 5 sided polygon, also called pentagon, you can join opposite 2 vertices from any one vertex and you will get 3 triangles. Again, we know that sum of interior angles of a triangle is 180°.
Thus, sum of interior angles of 3 triangles = sum of all interior angles of a pentagon = 180°x3 = 540°
5 sided polygon, sum of interior angles = 540°
Have you noticed that, as we keep increasing the number of sides in a polygon the interior angles keep increasing.
As we move further, i.e. for each increase in number of sides of a polygon, 180° gets added to the sum of interior angles.
So the general rule or formula is,
Sum of Interior Angles = (n-2) x 180°
In a ‘n’ sided polygon, the sum of interior angles of a polygon = (n-2) x 180°
A regular polygon is equi-angular; thus each interior angle will be equal. In a ‘n’ sided or ‘n’ angled polygon,
Each Interior Angle = Sum of Interior Angles / no. of sides = [(n-2) x 180°]/n
For example, in a dodecagon (12 sided figure shown below),
Sum of interior angles
= (n-2) x 180°
= (12-2) x 180°
= 10 x 180°
= 1800°
And in a regular dodecagon, each interior angle
= 1800°/12
= 150°
An exterior angle in any polygon is an angle formed by one side of the polygon and the extension of an adjacent side of the polygon.
Interior and exterior angles are supplementary angles.
Exterior angle = 180° – Interior angle
Thus, if interior angle is X°, exterior angle = 180° – X°
Another important point to keep in mind is that,
Sum of all exterior angles = 360°
So, in a regular polygon of n sides, each
Exterior angle = 360°/number of sides in a polygon
Example, in a hexagon, each exterior angle will be 360°/6 = 60°
The easiest way to finding the number of sides, is to first find out the exterior angle.
So let’s say the interior angle given is 144°, how to find the number of sides in the polygon?
First step is to calculate the exterior angle, which in this case = 180° – 144° = 36°
Now since it’s known that sum of all exterior angles is 360°, thus the formula is
Number of sides in a polygon = 360°/exterior angle
In above case, number of sides = 360°/36° = 10 sides. Thus it’s a decagon.
For your further reading, you can check this previous post where I’ve explained the ratio of area and volume derived from ratio of sides.
Objective of QuickerMaths is to make mathematics fun, quick and simple. I would love to hear from you in comments below.
]]>First Egg Riddle
If one and a half hens lay one and a half eggs in one and a half days, how many eggs does one hen lay in one day?
Leave your answers below.
Second Egg Riddle
2 fathers and 2 sons sat on the table to eat eggs for breakfast. They ate exactly three eggs, each person had an egg. Now you need to explain how that’s possible?
Third Egg Riddle
Let say you started selling a basket of eggs.
First customer buys one-half of your eggs plus one-half of an egg. Second customer buys one-half of your eggs plus one-half of an egg.
Third customer buys one-half of your eggs plus one half an egg.
At this point you have sold all of your eggs, and you never broke an egg. How many eggs did you start with?
]]>Get all information for GMAT in French here
Correct answers will result in points being awarded and difficulty being raised for the next question on-screen, while wrong answers will result in similar difficulty level and no points awarded for that question. So, those planning to appear for the exam must have a look below to get an idea of the best ways to master the section.
The test evaluates four key skills of a student, i.e. analytical writing, quantitative analysis, verbal skills, and reading skills. The language of the exam is English, with emphasis on grammar, algebra, geometry, and arithmetic. The exam also assesses analytical writing and problem-solving abilities of the students. GMAC believes that data sufficiency, logic, and critical reasoning skills are extremely vital to businesses in the real world.
GMAT 2018 Paper Pattern
To prepare for any exam, you must first be familiar with the structure of that exam. There are four sections in a GMAT exam. These are analytical writing assessment, integrated reasoning section, quantitative section, and verbal section. The exam is 3 hours and 7 minutes long, with time divided unequally amongst all the three sections.
GMAT Test Section | Questions | Question Types | Timing |
Analytical Writing Assessment | 1 Topic | Analysis of Argument | 30 Minutes |
Integrated Reasoning | 12 Questions | Multi-Source Reasoning Graphics Interpretation Two-Part Analysis Table Analysis |
30 Minutes |
Quantitative | 31 Questions | Data Sufficiency Problem Solving |
62 Minutes |
Verbal | 36 Questions | Reading Comprehension Critical Reasoning Sentence Correction |
65 Minutes |
As you can see in the table above, there are two types of questions in the Quantitative section:
The Problem Solving section has the same multiple choice format that is popular for every standardized test in today day and age. There are five options, out of which only one can be the correct answer.
The other format, Data Sufficiency, is unique to the GMAT exam, with unique rules too, which require different strategies as to all other tests. This section, thus, requires the most amount of practice and work.
7 Best Tips for GMAT Quantitative Ability preparation
If you are preparing for the GMAT exam, you will need all the help you can get to clear the exam. Provided below are a few tips you can follow in order to better prepare for the Quantitative section of the exam.
Strengthen your Basics
The mathematical concepts tested in GMAT are extremely simple, consisting of basic arithmetic, algebra, and geometry. The only problem is that students tend to forget the basics as time moves on. Your GMAT preparation should first and foremost cover the basics, and only after completing those should you think about going ahead with further preparations. The best way to remember formulae is to create flashcards and stick them around in your room. That way, every time you walk by a formula, your eyes will tend to hover over the flashcard.
Practise tests and mock tests
It should be obvious that practise will make you better eventually. The more you practise, the easier you will find the test to be. Thankfully, practise tests don’t have to be very expensive. Several online resources provide free practice content to use.
After every iteration of a practice or a mock test, you would also need to analyse your performance. Review the results and note the questions that you have answered incorrectly. Improve upon these particular areas identify your area of weakness. Majority of the GMAT exam questions revolve around students’ familiarity with different types of questions and avoiding common mistakes. With ample practice, you will be able to realise which questions are trick questions, thus also saving you a lot of time. One practice every week should be a comfortable place to start.
Pay special attention to Data Sufficiency questions
The Quantitative section is the most difficult section of GMAT primarily due to the Data Sufficiency portion. They would require you to think a little differently, but the more you practice them, the easier they become. There are several key points to remember when working on Data Sufficiency questions. Read the provided statements individually, and very carefully. Only after carefully evaluating the statements, make your answer choice.
Data sufficiency requires only sufficiency, not the actual answer, which means that if a problem states if the value of a variable can be determined, you only have to see whether it can be or cannot be determined, without actually solving for the value. You are just trying to find out if there’s enough information to answer the question, but you don’t actually have to find the answer.
Memorize the Five Answer Choices
There are always the same five answer choices for every Data Sufficiency Question. These answer choices are:
If you were to memorise these statements, you could save precious few seconds for every problem you attempt. You would only need to read the statement provided alongside the question and place judgement based on them, by clicking on the answer choices.
Be careful with Graphs, Charts, and Tables
A lot of questions in GMAT quantitative section will require you to read and interpret information provided on charts, graphs, and tables. It is extremely important that you read the axes, the key, units of measurement, etc. correctly so that you don’t misinterpret the data.
Use the rough paper in exam
Even if you feel like the GMAT quantitative section is too easy for you, it would only benefit you to use a paper for calculations as much as possible. Writing down your calculations will help you notice any mistake you might have made before you press the answer and move on to the next question. Writing also forces you to make sure you’re thinking through progress in steps instead of leaps, which can help reduce mistakes further. Remember, use of a calculator is forbidden in the GMAT exam.
Read the Questions carefully
Last but not the least, this is the most crucial yet the most ignored piece of advice that one can offer to any student aspiring to crack any national level exam. It is simply because students tend to make more mistakes when they are fatigued.
One of the most common mistakes on the GMAT exam is to misinterpret or read the question incorrectly. The GMAT exam purposefully throws in questions with difficult language, or questions that can mistakenly be read differently.
Instead of asking “Which of the following may be false?” GMAT will present the question as “Which of the following may not be true?” which might be misunderstood as “Which of the following may be true?” While this may sound a bit far-fetched right now, the atmosphere of pressure inside the hall and the fatigue during the exam itself could easily lead to such mistakes.
Thus, make sure that you read every question carefully so you can save yourself from these easily avoidable mistakes.
]]>
It is an institute which is of national importance that has been established and recognized under the act of Indian parliament in the year 1959. It has been established in the year 1931 which is a public university and is considered one of the oldest and prestigious universities which focus on the study of stastics and mathematics.
It is one of the leading training and research institute in the fields of computer science, quantitative aptitude, statics, and related sciences and so on. The headquarters of the university is located in West Bengal, Kolkata.
Courses offered
It is a deemed to be university in Mulana in the state of Haryana. It was established in the year 1993 in the name of Maharishi Markandeshwar and was founded by Tarsem Garg. The institute is the leading symbol of education in terms of technical, medical and other professional streams. It has been accredited by the NAAC team with an A grade.
The university is committed to excel in research and innovation and the skill development of the students. It has developed an industry oriented education system which helps the students in order to make the students leaders in the professional world.
Courses offered
It is an institute of central excellence which has been formed for the teaching and research in the mathematical sciences which is established in the year 1989 with the motto of bridging the gap between the teaching and the research in the fields of mathematics and various other allied subjects.
It has been recognized by the government of India under the section 3 of the UGC act, 1956. The objective of the institute is to provide the excellent quality education to the students in the best possible way.
Courses offered
It is one of the autonomous public institutes of higher education situated in India. It is governed by the Institute of technology Act, 1961 which has been declared as the institution of national importance in India. There are various branches of the institute which has been established in various other cities of India to impart excellent education to the students in the fields of technology, science and mathematics. Every year, a large number of students gear up to take admission in this university.
Courses offered
It is a private university situated in Jalandhar, Punjab. It was established in the year 2005 by the lovely international trust under the lovely professional university act, 2005 which started its operation in the year 2006. It has the largest single campus in terms of the private university in India. The campus spreads over 600 acres of land on the outskirts of Jalandhar and accommodates more than 24,000 students.
It has been recognized by the UGC, NCTE and AIU. It was also ranked 18th as India’s best college in the year 2017. It offers various courses in Mathematics which are spread over 3 years across six semesters. In order to get admissions in such courses, the candidates should have scored 60% and above in the secondary examination or any other degree equivalent to it. The candidates also need to clear the LPUNEST Exam 2018-2019 in order to take admission in the examination.
Choosing mathematics has its own advantages such as it requires more of application part and zero mugging. Secondly not many UPSC aspirants take up maths as the optional subjects, so there is less competition. Thirdly, the study material is easily available. And lastly, this subject has a limited syllabus unlike other subjects in which current affairs are also involved. Here are few easy tips to get high marks in the exam.
During the exam, if you are unable to solve the question in two attempts, leave it. You will not be able to solve that. Focus on other questions. If a problem is taking too much time, then hold on, you may be following a wrong approach to solve it.
Practice some good tips and tricks to solve the problems. The more problems you solve, you efficiency will automatically be increased. Try to adopt Vedic Maths techniques. This will help you to solve even the biggest of equations within fractions of seconds.
A few tips and tricks using Vedic Maths :
Step 1:
Let us consider multiplication of three digit numbers 208 x 206.
Step 2:
Now, deduct the last digit from the respective numerals.
208 – 8 = 200
206 – 6 = 200
Step 3:
Pick any one number and add it with the unit digit of another number.
208 + 6 = 214
Step 4:
Now, multiply the result obtained in step 1 and step 2.
214 * 200 = 42800
Step 5:
Then, multiply the unit digits of the given numbers.
8 * 6 = 48
Step 6:
Add the values obtained in step 4 and step 5.
42800 + 48 = 42848
208 x 206 = 42848
Maths is one such subject where a single silly mistake can deviate you from the actual answer. Be careful of the sign the number is carrying. Practice, practice and practice till you become perfect.
Make a chart (or two) of all the formulas and paste them in your study room. Don’t be hesitant in looking through the formula in case you forget. After some practice, the formulas will automatically be memorized by you.
Rather than just reading any solution, better take a pen and paper and try solving it. There is a huge difference between solving in mind and solving on paper.
It can be illustrated through an example :
Some new topics have been introduced such as Mechanics and Fluid dynamics. Try to understand the concepts in the beginning of your preparation period. After reading all the solved examples, then start off with previous year question papers. This will give you an idea what and how much to study for the prelims.
If you are writing short answers then give one or two lines for introduction and then move straight to the point. In case of long answers, give introduction and explain wherever necessary and make a perfect conclusion at the end.
Joining test series is an effective way to improve your time management and self evaluation skills. If not daily, then give a mock test at least twice a week for a sound preparation.
You cannot solve any mathematics problem if you lack focus or concentration. Keep all the distractions away from you and be cool & relaxed. This way you will not only be able to understand the problem but will also hit the right answer too.
Mathematics is the first love of science. If studied and applied properly, mathematics will fetch you good marks and ultimately a good UPSC score.
Hiring a tutor or enrolling a child in a tutorial program, in general, is a popular option among parents. It is something that many parents decide to do to help their kids excel at or meet the demands of a difficult subject (or more than one). It is one of the best ways parents can show their love and support for the children who are having a hard time keeping up with their schoolwork.
As a parent, your role in your child’s academic progress won’t stop or be diminished if you decide to enroll your child in an after-school math tutorial program. Your youngster will experience more benefits from a tutorial program if you have a good working relationship with his tutor.
To make sure you and your child’s tutor are on the same page regarding his improvements and continuous progress in math, follow these helpful tips:
Communication is one of the vital pillars of a tutoring process. From the start, ask questions if you have any, share your feedback and raise any concerns you might have about the program. Make sure the tutor has all the necessary information regarding the key needs and preferences of your child as well.
Before scheduling a talk with the tutor or learning center, come up with a list of questions regarding their procedures and policies. Get details about their assessment process and their teaching methodologies as well.
Before the formal tutorial program starts, create a list of goals that you want your child to achieve. Discuss these goals with the tutor. Ensure you highlight the particular concepts or areas your kid needs help with.
When coming up with your list of goals, talk to your child. Also, keep in mind that a tutor isn’t just a person you hire to help your child do his math assignments or review for an upcoming test. If your child needs help with this subject, work with the tutor to ensure that the whole problem will be tackled and not just one area will be solved or managed.
One goal that you should never forget to list is to see a change in your child’s attitude towards math. Aside from getting higher grades, your tutor should also play an important role in changing your youngster’s negative perceptions about math, and in the process inspire the student to actively participate in studying it.
For your child to benefit more from the tutorial program, inform his teacher about this after-school activity. It is important that the teacher and tutor work together so that they are on the same page in terms of the lessons and goals.
If possible, schedule parent-teacher conferences and tutor meetings to coincide with the time your child gets his progress report. It is essential that the tutor communicates with your child’s teacher to get feedback on his performance in the classroom. They don’t even have to meet personally; they can correspond by email or private messages. Just make sure your kid’s tutor and teacher coordinate regularly.
Lastly, aside from finding a good tutor and bringing your child to the tutorial center during his sessions, you also need to take a proactive role in the whole process. At the very least, at the end of each session, ask the tutor what topics they covered today and what difficulties your child encountered. Ask the tutor how you can help reinforce what your child learned from the session.
Although this may mean relearning continuity and different ability and other mathematical concepts so that you can give some exercises for your child to work on at home, your efforts will be rewarded in the end.
Your role in helping your child overcome his fear or dislike of math and getting higher grades doesn’t stop at finding a good tutorial center and bringing him here every session. For your child to get more from each tutorial lesson, you must be totally involved in the process and have a good working relationship with the tutor.
AUTHOR BIO
Maloy Burman is the Chief Executive Officer and Managing Director of Premier Genie FZ LLC. He is responsible for driving Premier Genie into a leadership position in STEM (Science, Technology, Engineering and Mathematics) Education space in Asia, Middle East and Africa and building a solid brand value. Premier Genie is currently running 5 centers in Dubai and 5 centers in India with a goal to multiply that over the next 5 years.
Scenario I: When the average age increases then,
Age of the new comer = age of person who left + (no. of persons in the group * increase in average age)
Scenario II: When the average decreases then,
Age of the new comer = age of the person who left – (no. of persons in the group * decrease in average age)
Let us try to understand these with a simple example –
Question: The average age of 45 persons is decreased by 1/9 year when one of them whose age is 60 years is replaced by a new comer. What is the age of the new comer.
Solution:
Age of the new comer = age of the person who left – (no. of persons in the group * decrease in average age)
Age of the new comer = 60 – (45 * 1/9) = 55 years
Isn’t that easy to calculate using the formula I just gave you. However, to intuitively answer such questions you need to keep following things in mind –
Firstly, the average age is reducing, which means an older man is replaced by relatively younger person. Hence the answer has to be lesser than 60. Hence we subtract from 60 and not add to it.
Secondly, average reduction of 1/9 kg for a group of 45 people means, in total there is a wait reduction of 1/9 per person x 45 persons = 5 kg for the entire group. This is result of the replacement of an old person by a relatively younger person in the group. That’s the reason we’re subtracting 5 from 60, to get the age of the new comer.
Scenario I: When the average age increases then,
Age of the incoming person = previous average age of the group + no. of persons including the person who joined * increase in the average age value
Scenario II: When the average age decreases then,
Age of the incoming person = previous average age of the group – no. of present persons including the person who joined * decrease in the average age value
I will explain this further with an example,
Question: The average age of 20 teachers is 45 years which is decreased by 6/7 years when a student joins the group. Then what is the age of the student?
Solution:
Age of the outgoing person = previous average age of the group – new no. of persons including the person who joined * decrease in the average age value
Therefore, age of the student = 45 – 21 * 6/7 = 27 years
Here also you need to observe that since the average age is going down when the new person is joining, the age of new comer will be lesser than the average, hence subtraction.
Moreover, the decrease is equal to 6/7 per person for each person in the new group. That is 21 times 6/7 = 18 kgs, which needs to be subtracted from the overall average to get the age of the new comer.
Scenario I: When the average age increases then,
Age of the outgoing person = previous average age of the group – no. of persons excluding the person who left * increase in the average age value
Scenario II: When the average age decreases then,
Age of the outgoing person = previous average age of the group + no. of present persons excluding the person who left * decrease in the average age value
Hope we’ve covered all scenarios and with the help of the above logical formulas we will be able to do our calculations.
Question 1: In a boat there are 8 men whose average weight is increased by 1 kg when a man of 60 kg is replaced by a new man. What is the weight of the new comer?
Question 2: In a class there are 30 boys whose average weight is decreased by 200 grams when one boy whose weight was 25 kgs leaves the class and a new comer is admitted. Find the weight of the new comer?
Question 3: A cricketer has a certain average for 9 innings. In the 10^{th} innings, he scores 100 runs, thereby increasing his average by 8 runs. His new average is?
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