The Lumenary · Grades 6–8

Ratios, rates and proportional reasoning

Name: ______________________

Date: ______________________

You will learn to

Solve ratio and rate problems with tables, double number lines and unit rates
Recognise proportional relationships $y = kx$ in tables, graphs and situations, and find the constant $k$
Prove that the graph of a proportional relationship is a straight line through the origin, and that not every straight line is proportional
Compute percentages and percentage changes, and prove that a 10% rise followed by a 10% fall does not return to the start

1.
Packets of tea in a shop in Kampala: a $250$ g packet costs $5000$ shillings and a $400$ g packet costs $7600$ shillings. Which is the better buy?
- A.The 400 g packet
- B.The 250 g packet
- C.They are equally good value
2.
A recipe for $6$ people uses $4$ cups of millet flour. How many cups are needed for $15$ people?
3.
Mr Short is $4$ buttons tall and $6$ paper clips tall. Mr Tall is $10$ buttons tall. How many paper clips tall is Mr Tall?
4.
Which of these relationships are proportional? (Choose all that apply.)
- A.The cost of pens at 25 rupees each, against the number of pens
- B.A taxi fare of 200 plus 150 per km, against the distance
- C.The perimeter of a square, against its side length
- D.The area of a square, against its side length
- E.CFA francs against euros at the fixed rate
5.
In a proportional relationship, $y = 7$ when $x = 2$ , and $y = 17.5$ when $x = 5$ . What is $y$ when $x = 12$ ?
6.
A bicycle costs $200$ dollars. The price rises by $10\%$ , and the next month the new price falls by $10\%$ . What is the final price, in dollars?
7.
In a sale everything is $20\%$ off. A jacket now costs $240$ reais. What was its price before the sale?
8.
The population of a town rises by $25\%$ . By what percentage must it then fall to return to its original size?
9.
Which Lean definition says correctly that $f$ is proportional (there is one constant $k$ with $f(x) = kx$ for every $x$ )?
- A.∃ k : Rat, ∀ x, f x = k * x
- B.∀ x, ∃ k : Rat, f x = k * x
- C.∃ k b : Rat, ∀ x, f x = k * x + b
- D.∀ x, f x = x
10.
Write constant_of_proportionality(table). The table is a list of pairs (x, y). Return the constant $k$ (as a Fraction) if $y = kx$ for every row, and None otherwise. Rows with x == 0 must have y == 0, and there must be at least one row with x != 0. Use Fraction so that decimals such as \"0.5\" are exact.
Computational — Practice

Answer key — Ratios, rates and proportional reasoning

1.
Unit prices: $5000 \div 250 = 20$ shillings per gram and $7600 \div 400 = 19$ shillings per gram. The $400$ g packet is the better buy.
2.
The unit rate is $\tfrac46 = \tfrac23$ cup per person, so $15$ people need $15 \times \tfrac23 = 10$ cups. (Or: $15$ people is $2.5$ times $6$ people, and $2.5 \times 4 = 10$ .)
3.
$4$ buttons match $6$ clips, so $1$ button matches $1.5$ clips and $10$ buttons match $15$ clips. The ratio $4 : 6$ is equivalent to $10 : 15$ (check: $4 \cdot 15 = 60 = 6 \cdot 10$ ).
4.
Proportional: pens ( $y = 25x$ ), perimeter ( $P = 4s$ ) and currency ( $y = 655.957x$ ). The taxi fare $y = 150x + 200$ is linear but not proportional, and the area $A = s^2$ is not even linear.
5.
$k = \tfrac{7}{2} = \tfrac{17.5}{5} = 3.5$ , so $y = 3.5x$ and at $x = 12$ , $y = 42$ .
6.
$200 \times 1.1 = 220$ , then $220 \times 0.9 = 198$ . Overall the price is multiplied by $1.1 \times 0.9 = 0.99$ : it ends $1\%$ lower.
7.
$0.8P = 240$ , so $P = 240 \div 0.8 = 300$ reais. Check: $20\%$ of $300$ is $60$ , and $300 - 60 = 240$ .
8.
We need $1.25 \times f = 1$ , so $f = 0.8$ : a fall of $20\%$ . In general a rise of $p\%$ is undone by a fall of $\tfrac{100p}{100 + p}\%$ ; for $p = 25$ that is $\tfrac{2500}{125} = 20$ .
9.
∃ k : Rat, ∀ x, f x = k * x: one constant for all $x$ . Swapping the quantifiers lets $k$ depend on $x$ , which says almost nothing; adding $b$ describes all straight lines, not just proportional ones.

10.

from fractions import Fraction

def constant_of_proportionality(table):
    pairs = [(Fraction(x), Fraction(y)) for x, y in table]
    ratios = {y / x for x, y in pairs if x != 0}
    if len(ratios) != 1:
        return None
    if any(y != 0 for x, y in pairs if x == 0):
        return None
    return ratios.pop()

This is Theorem 2 as a program: $y$ is proportional to $x$ exactly when $y = 0$ at $x = 0$ and $\tfrac yx$ is constant. Of course a table only samples the relationship — it cannot tell what happens between its rows.