MYP 3 · Maths

REAL NUMBERS AND RATIO

Operations with fractions

QUESTION 1 [7 marks] — Criterion B Medium
Investigate the pattern formed when adding two consecutive unit fractions, $\frac{1}{n} + \frac{1}{n+1}$.
a. Calculate $\frac{1}{2}+\frac{1}{3}$, $\frac{1}{3}+\frac{1}{4}$, and $\frac{1}{4}+\frac{1}{5}$, leaving each answer as a single fraction (not simplified further unless needed).
[3]
Show Solution
$\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$. $\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$. $\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$.
b. Look at the numerators (5, 7, 9) and denominators (6, 12, 20) of your three answers. Describe the pattern in each.
[2]
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The numerators increase by 2 each time (5, 7, 9 — an arithmetic sequence). The denominators are each the product of the two consecutive numbers used ($2\times3=6$, $3\times4=12$, $4\times5=20$).
c. Using your pattern, predict $\frac{1}{6}+\frac{1}{7}$ without doing the full calculation, then check your prediction by calculating it directly.
[2]
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Prediction: numerator continues the pattern to 13, denominator $=6\times7=42$, so $\frac{13}{42}$. Checking directly: $\frac{1}{6}+\frac{1}{7} = \frac{7}{42}+\frac{6}{42} = \frac{13}{42}$ — matches the prediction.
QUESTION 2 [6 marks] — Criterion C Medium
A recipe requires $2\frac{1}{4}$ cups of flour. A baker is making $1\frac{1}{3}$ batches of the recipe.
a. Write a clear, step-by-step solution showing how much flour is needed in total, converting all mixed numbers to improper fractions first and showing every step.
[4]
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Convert to improper fractions: $2\frac{1}{4}=\frac{9}{4}$, $1\frac{1}{3}=\frac{4}{3}$. Multiply: $\frac{9}{4}\times\frac{4}{3}=\frac{36}{12}=3$. So exactly 3 cups of flour are needed.
b. Explain why converting to improper fractions is a more reliable method than trying to multiply mixed numbers directly.
[2]
Show Solution
Multiplying mixed numbers directly (e.g. multiplying whole number parts and fraction parts separately) does not give the correct product, since a mixed number represents an addition, not a simple combination — converting to a single improper fraction avoids this common error.
QUESTION 3 [7 marks] — Criterion D Medium
A water tank starts full. Over one day, $\frac{2}{5}$ of the water is used for irrigation in the morning, then $\frac{1}{3}$ of the remaining water is used for cleaning in the afternoon.
a. Find what fraction of the original full tank remains at the end of the day.
[3]
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After morning: remaining $=1-\frac{2}{5}=\frac{3}{5}$. Afternoon usage $=\frac{1}{3}\times\frac{3}{5}=\frac{1}{5}$. Remaining at end of day $=\frac{3}{5}-\frac{1}{5}=\frac{2}{5}$.
b. If the tank holds 4500 litres when full, how many litres remain at the end of the day?
[2]
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$$\frac{2}{5} \times 4500 = 1800 \text{ litres}$$
c. The tank needs at least 1500 litres remaining to supply the household overnight. Based on your answer, is there enough water? Explain.
[2]
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Yes — 1800 litres remain, which is more than the required 1500 litres, so there is enough water for the household overnight.