Direct and Inverse ProportionA selfmarking exercise in three levels on solving direct and inverse variation problems. 
This is level 1; Direct proportion. You can earn a trophy if you get at least 9 correct.
InstructionsTry your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help. When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file. 




More Activities: 

Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician? Comment recorded on the 3 October 'Starter of the Day' page by S Mirza, Park High School, Colne: "Very good starters, help pupils settle very well in maths classroom." Comment recorded on the 23 September 'Starter of the Day' page by Judy, Chatsmore CHS: "This triangle starter is excellent. I have used it with all of my ks3 and ks4 classes and they are all totally focused when counting the triangles." 
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AnswersThere are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer. A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves. Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members. If you would like to enjoy adfree access to the thousands of Transum resources, receive our monthly newsletter, unlock the printable worksheets and see our Maths Lesson Finishers then sign up for a subscription now: Subscribe 

Go MathsLearning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school. Maths MapAre you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.  
Teachers  
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows: 
Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes. 
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Level 1  Direct proportion
Level 2  Inverse proportion
Level 3  Mixed nonlinear questions
Unitary Method  Test your understanding of the Unitary Method for solving real life proportion problems with this online, selfmarking quiz.
Exam Style Questions  A collection of problems in the style of GCSE or IB/Alevel exam paper questions (worked solutions are available for Transum subscribers).
More on this topic including lesson Starters, visual aids, investigations and selfmarking exercises.
This video is from Mannel's Maths Music.
If \(a\) varies directly with \(b\) and \(a=24\) when \(b=8\) find \(a\) when \(b=9\)
$$a \propto b$$ $$a = kb$$Where \(k\) is some constant. If \(a=24\) when \(b=8\) then
$$24 = 8k$$ $$k = 3$$so the equation is
$$a = 3b$$If \(b = 9\) then
$$a = 3 \times 9 = 27$$If \(a\) is inversely proportional to \(b\) and \(a=4\) when \(b=6\) find \(a\) when \(b=8\)
$$a \propto \frac{1}{b}$$ $$a = \frac{k}{b}$$Where \(k\) is some constant. If \(a=4\) when \(b=6\) then
$$4 = \frac{k}{6}$$ $$k = 24$$so the equation is
$$a = \frac{24}{b}$$If \(b = 8\) then
$$a = 24 \div 8 = 3$$If \(a\) is directly proportional to the square of \(b\) and \(a=24\) when \(b=2\) find \(a\) when \(b=3\)
$$a \propto b^2$$ $$a = kb^2$$Where \(k\) is some constant. If \(a=24\) when \(b=2\) then
$$24 = 2^2 \times k$$ $$k = 6$$so the equation is
$$a = 6b^2$$If \(b = 3\) then
$$a = 6 \times 3^2 = 54$$Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can doubleclick the 'Check' button to make it float at the bottom of your screen.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
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Claire Longmoor, Twitter
Monday, September 2, 2019