1. Direct and inverse proportion problems
Grade 5
Direct and inverse proportion problems
Solve problems involving direct and inverse proportion.
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2. Lesson Plan Lesson Overview Progression of Learning
Objective(s)
Solve problems involving direct and inverse proportion
Grade 6
Prior Knowledge
Solving equations
Rearranging equations
Substitution
Duration
60 minutes
Resources
Print slides: 4, 6, 8, 11
Equipment
Calculator
Progression of Learning
What are the students learning?
How are the students learning? (Activities & Differentiation)
How to write a direct proportion equation and use this
Give students slide 4 printed. Tell them that you are going to model each step. The instructions are given in the left hand column. Read question in 2nd column and reveal each step slowly with students copying down the answer. Once complete move onto 2nd example in the 3rd column. Show students how the equation for step 1 is slightly different – relate back to the text in red (x can change depending on the information given) 15
How to write a inverse proportion equation and use this
Give students slide 6 printed. Repeat the exact same steps as above. Tell the students that the steps are identical however the equation changes as inverse. 10
Applying method of direct and inverse proportion to contextualised problems
Give students slide 8 printed. Again repeat that the steps are the same. The key to success is to ”find” the numbers from the information given in the question and to get them into the correct places. Allow students to work on this problem independently before teacher input. Then reveal the answers using slide 9 and 10. The slides have been animated and coloured to show how to interpret the context of the question and to extract the numbers from the question.
Solve problems involving direct and inverse proportion in OCR exam questions (from specimen papers)
Give students slide 11. This includes an questions from specimen papers. Students to complete and for answers to be discussed using the mark schemes. 20
Next Steps
Assessment
PLC/Reformed Specification/Target 6/Ratio, Proportion and Rates of change/Direct & Inverse Proportion

3. Key Vocabulary
Inverse Direct Proportional Equation Constant

4. Direct Proportion Student Sheet 1 y is directly proportional to x.
y = 20 when x = 4.
By writing an equation of proportionality, work out the value of y when x = 7
y is directly proportional to square of x.
y = 150 when x = 5.
By writing an equation of proportionality, work out the value of x when y = 54
Write y = kx
x can change depending on the information given
Substitute y = and x = into the equation
Solve to find value of k
Re write original equation with value for k substituted
Substitute value given in question into equation to find other value
Student Sheet 1

5. Direct Proportion y = kx y = kx2 150 = k × 52 20 = k × 4 150 = k × 25
y is directly proportional to x.
y = 20 when x = 4.
By writing an equation of proportionality, work out the value of y when x = 7
y is directly proportional to square of x.
y = 150 when x = 5.
By writing an equation of proportionality, work out the value of x when y = 54
Write y = kx
x can change depending on the information given
Substitute y = and x = into the equation
Solve to find value of k
Re write original equation with value for k substituted
Substitute value given in question into equation to find other value
y = kx
y = kx2
150 = k × 52
20 = k × 4
150 = k × 25
5 = k
6 = k
y = 5x
y = 6x2
y = 5 × 7 = 35
54 = 6x2
9 = x2
y = 35
54 = x2 6
3 = x

6. Inverse Proportion Student Sheet 2 y is inversely proportional to x.
y = 6 when x = 2.
By writing an equation of proportionality, work out the value of x when y = 0.4
y is inversely proportional to cube of x.
y = 0.5 when x = 4.
By writing an equation of proportionality, work out the value of y when x = 2
Write y = k x
x can change depending on the information given
Substitute y = and x = into the equation
Solve to find value of k
Re write original equation with value for k substituted
Substitute value given in question into equation to find other value
Student Sheet 2

7. Inverse Proportion y = k x y = k x3 6 = k 2 0.5 = k 43 12 = k 32 = k
y is inversely proportional to x.
y = 6 when x = 2.
By writing an equation of proportionality, work out the value of x when y = 0.4
y is inversely proportional to cube of x.
y = 0.5 when x = 4.
By writing an equation of proportionality, work out the value of y when x = 2
Write y = k x
x can change depending on the information given
Substitute y = and x = into the equation
Solve to find value of k
Re write original equation with value for k substituted
Substitute value given in question into equation to find other value
y = k x
y = k x3
6 = k 2
0.5 = k 43
12 = k
32 = k
y = 12 x
y = 32 x3
0.4 = 12 x
0.4x = 12
y = 32 23
y = 32 8
x = 12
0.4
y = 4
x = 30

8. Problem solving and reasoning
The attractive force between 2 objects is inversely proportional to the square of their distance apart.
The force is 0.96 Newtons when the distance is 10 miles.
What will the force be when the distance is 12 miles?
The distance it is possible to see on a clear day, d miles, varies directly with the square root of the height h metres somebody is stood above sea level.
Standing on a pier 9 metres above sea level it is possible to see 21 miles.
Standing on top of a cliff it is possible to see the coast 36 miles away. How high are the cliffs to the nearest metre?
Student Sheet 3

9. Problem solving and reasoning
The attractive force between 2 objects is inversely proportional to the square of their distance apart. The force is 0.96 Newtons when the distance is 10 miles.
What will the force be when the distance is 12 miles?
Force ∝ 1/Distance2
F = k/D2
Substitute F = 0.96 and D = 10 into F = k/D2
0.96 = k/102
0.96 × 100 = k
96 = k
F = 96/D2
Substitute D = 12 into the equation above
F = 96/122
F = 96/144
F = 2/3

10. Problem solving and reasoning
The distance it is possible to see on a clear day, d miles, varies directly with the square root of the height h metres somebody is stood above sea level.
Standing on a pier 9 metres above sea level it is possible to see 21 miles.
Standing on top of a cliff it is possible to see the coast 36 miles away. How high are the cliffs to the nearest metre?
Distance ∝ √Height
d = k √h
Substitute d = 21 and h = 9
21 = k √9
21 = 3k
7 = k
d = 7 √h
Substitute d = 36
36 = 7 √h
5.143 = √h
26.45 = h
h = 26 metres to the nearest metre

11. Exam Question – Specimen Papers
Student Sheet 4

12. Exam Question – Specimen Papers