Ratio, proportion and rates of change

Explanation

• A conversion is multiplication or division by a fixed factor, and the direction is checked by size: $8000$ cm$^3$ is a small volume, so it must become $0.008$ m$^3$, not $8\,000\,000$. Lengths use $1$ m $=100$ cm and $1$ km $=1000$ m; areas square the factor and volumes cube it, so $1$ m$^3=100^3=1\,000\,000$ cm$^3$ and $1$ litre $=1000$ cm$^3$.
• Time converts through $60$: $1$ hour $=3600$ seconds. A compound unit converts one part at a time: $180$ km/h becomes m/s by multiplying by $1000$ and dividing by $3600$, giving $50$; each conversion is a method mark.
• Before substituting into a formula, put every quantity into the unit the answer needs, converting millimetres and centimetres to metres for a density in kg/m$^3$.
• In algebraic contexts the same factor multiplies the whole expression.
• Write the conversion line down; the mark is for seeing it.

Explanation

• A scale written $1:n$ compares a diagram length with the corresponding real length in the same units. Multiply a diagram length by $n$ to obtain the real length, or divide the real length by $n$ to obtain the diagram length.
• Convert units only after establishing which direction the scale factor acts.
• For areas, square the linear scale factor; for volumes, cube it.
• On a map, measure between the stated points accurately and use any given scale bar or numerical scale.
• Examiners award method for applying the scale in the correct direction and then converting units correctly.

Explanation

• To express quantity $A$ as a fraction of quantity $B$, form $\dfrac{A}{B}$ and simplify.
• The order of the wording fixes the order of the fraction: the quantity before “as a fraction of” is the numerator.
• Quantities must first be written in the same units, so that the units cancel.
• The result may be less than $1$, equal to $1$, or greater than $1$; an improper fraction is valid when the first quantity is larger.
• Examiners expect a simplified exact fraction unless the question asks for another form.

Explanation

• Ratio notation compares quantities in a fixed order, for example $a:b$. Before simplifying, express every quantity in the same units.
• An equivalent ratio is produced by multiplying or dividing every part by the same non-zero value.
• For an integer ratio, divide all parts by their highest common factor; when decimals occur, first scale them to integers if helpful.
• Ratios can have more than two parts, and a shared part can be matched using a common multiple.
• Examiners require the stated order to be preserved and usually expect the simplest integer form.

Explanation

• In a part:part ratio $a:b$ the whole is $a+b$ equal shares, so divide the quantity by $a+b$ to find one share and multiply by each part; dividing by $a$ and by $b$ separately is the error every scheme lists. Check that the parts add back to the total.
• A part:whole statement is read differently: if one part is $\dfrac{3}{8}$ of the whole, the other is $\dfrac{5}{8}$ and the ratio is $3:5$. To express a division as a ratio, write the two amounts in order and simplify by their highest common factor.
• Multi-step problems combine a ratio with fractions or percentages of the same total: remove the fraction and percentage shares first, then split the remainder in the ratio.
• For mixing or concentration, track the actual amount of the ingredient, not an average of percentages.
• Label which part is which in the final answer.

Explanation

• A multiplicative relationship compares quantities by division rather than subtraction. If $A=kB$, then $\dfrac{A}{B}=k$, so $A:B=k:1$ and $A$ is the fraction $k$ of $B$.
• Convert decimal multipliers to fractions when an integer ratio is required; for example, $1.75=\dfrac{7}{4}$ gives $A:B=7:4$.
• For a chain of relationships, substitute one equation into another or match the shared ratio part.
• Check that a multiplier greater than $1$ corresponds to the larger quantity.
• Examiners expect the direction of the comparison to agree with the wording, because reversing it produces the reciprocal.

Explanation

• Proportion means that two corresponding ratios are equal. Keep the quantities in the same order on both sides, then solve by scaling, finding a unit value, or cross-multiplying.
• If $\dfrac{a}{b}=\dfrac{c}{d}$ with non-zero denominators, then $ad=bc$.
• A unitary method is often clearest in context: find the amount for one unit and scale to the required number.
• Decimal answers involving money must be interpreted in pounds and pence.
• Examiners award method for a correct proportional relationship even if a later arithmetic slip occurs, so show the equation or unit value.

Explanation

• For parts in ratio $a:b$, the whole has $a+b$ equal parts, so their fractions of the whole are $\dfrac{a}{a+b}$ and $\dfrac{b}{a+b}$. A constant ratio between variables can also define a linear function.
• If $y:x=m:1$, then $\dfrac{y}{x}=m$ and $y=mx$.
• Its graph is a straight line through the origin, with gradient $m$.
• A line with a non-zero intercept does not represent a constant ratio.
• Examiners may ask you to move between ratio, fraction, equation and graph, so always state the link explicitly.

Explanation

• A percentage is a number of parts per hundred, so $p\%=\dfrac{p}{100}$ and the multiplier for a change of $p\%$ is $1\pm\dfrac{p}{100}$. Percentage change is always measured against the original value: $\text{percentage change}=\dfrac{\text{change}}{\text{original}}\times100$, so a rise from $245$ to $883$ is $\dfrac{638}{245}\times100\approx260\%$, above $100\%$ because the value more than doubled.
• Percentage profit uses the cost price as the denominator. For an original value problem, the given amount is the multiplier times the original, so divide by the multiplier: a price of $\pounds46.75$ after a $15\%$ reduction was $46.75\div0.85$.
• Applying the percentage to the reduced price is the standard error and scores nothing.
• Simple interest is the same amount each year, calculated from the original principal only.
• Keep unrounded values until the comparison or final answer.

Explanation

• Write the relationship as an equation with a constant before doing anything numerical: $y=kx$, $y=\dfrac{k}{x}$, or at Higher tier $y=kx^2$, $y=k\sqrt{x}$, $y=\dfrac{k}{x^2}$. Substitute the given pair to find $k$, write the formula out, then substitute the new value.
• For inverse proportion the product $xy$ is constant, so the same problem can be solved by $xy=k$; for direct proportion the ratio $\dfrac{y}{x}$ is constant and doubling $x$ doubles $y$, while for $y=kx^3$ doubling $x$ multiplies $y$ by $8$, not $6$.
• The graph of direct proportion is a straight line through the origin; inverse proportion is a decreasing curve approaching both axes.
• In a practical problem such as seed for a lawn, scale by the ratio of areas and compare with what is available, and finish with a supported yes or no.
• Writing $k$ on the wrong side, $y=kx^2$ for $y=\dfrac{k}{x^2}$, loses every mark.

Explanation

• Write the formula before substituting: $\text{speed}=\dfrac{\text{distance}}{\text{time}}$, $\text{density}=\dfrac{\text{mass}}{\text{volume}}$, $\text{pressure}=\dfrac{\text{force}}{\text{area}}$. Rates of pay and unit prices are a total divided by the hours or the number of items.
• Make the units match the unit you must answer in before you calculate: $180$ km per hour becomes metres per second by multiplying by $1000$ and dividing by $3600$. Rearrange when the unknown is not the subject; mass is density multiplied by volume, and time is distance divided by speed.
• Multi-step problems chain these: mass and density give a volume, subtracting a known volume gives the volume of the remaining material, and dividing its mass by that volume gives its density.
• The unit in the question tells you which quantity is divided by which.
• State the unit with the value; some schemes award it a separate mark.

Explanation

• Similar shapes have equal corresponding angles and proportional corresponding lengths. If the linear scale factor from one shape to another is $k$, the length ratio is $k$, the area ratio is $k^2$, and the volume ratio is $k^3$.
• Match corresponding measurements and keep the comparison order consistent.
• To recover a length factor from an area ratio take a square root; from a volume ratio take a cube root.
• Trigonometric ratios remain constant in similar right-angled triangles.
• Examiners expect you to identify whether the measurements are lengths, areas or volumes before applying the scale factor.

Explanation

• If $X$ is inversely proportional to $Y$, then $X\propto\dfrac{1}{Y}$ and $X=\dfrac{k}{Y}$ for a constant $k$; equivalently, $XY=k$. Foundation questions can require interpreting equations that describe direct and inverse proportion.
• Higher tier: construct equations involving a power, such as $y=kx^2$ or $y=\dfrac{k}{x^n}$.
• Substitute a known pair to find $k$, write the complete equation, then use it to find an unknown.
• Check any stated restrictions, such as a positive length, before choosing a root.
• Examiners usually award separate method marks for the correct proportional form and for finding the constant.