TMUA: Ratio, proportion and percentages
Ratio, proportion and percentages.
- Section 1 Part 2 · M1-M7 GCSE-level knowledge (Part 2) · spec M3
- Papers 1 and 2
- 7 practice questions
- No calculator
What the specification covers
- Ratio problems, direct and inverse proportion
- Percentage change, reverse percentages, growth and decay
Key ideas
- A percentage increase of \(p\%\) multiplies by \(1+\tfrac p{100}\); a decrease by \(1-\tfrac p{100}\). Successive changes multiply.
- Reverse percentages: divide by the multiplier.
- \(y\) directly proportional to \(x^n\): \(y=kx^n\). Inversely: \(y=\tfrac k{x^n}\).
Common mistakes
- A 25% rise then a 25% fall is not back to the start (\(1.25\times0.75=0.9375\)).
- Ratios \(a:b\) split a total into \(\tfrac a{a+b}\) and \(\tfrac b{a+b}\).
Exam tip
Work with multipliers, never with the percentages themselves.
Worked example
Worked example
A price is increased by 25% and the new price is then decreased by \(p\)%. The final price is 10% below the original price. What is \(p\)?
- \(28\)
- \(35\)
- \(30\)
- \(15\)
- \(25\)
Answer: A: \(28\)
\(1.25\left(1-\tfrac{p}{100}\right)=0.9\), so \(1-\tfrac p{100}=0.72\) and \(p=28\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
A price is increased by 25% and the new price is then decreased by \(p\)%. The final price is 20% above the original price. What is \(p\)?
- \(4\)
- \(9\)
- \(45\)
- \(5\)
- \(96\)
Show the answer and solution
Answer: A: \(4\)
\(\frac{5}{4}\left(1-\tfrac{p}{100}\right)=\frac{6}{5}\), so \(1-\tfrac{p}{100}=\frac{24}{25}\) and \(p=4\).
Question 2
A price is increased by 80% and the new price is then decreased by \(p\)%. The final price is 8% above the original price. What is \(p\)?
- \(40\)
- \(72\)
- \(60\)
- \(88\)
- \(45\)
Show the answer and solution
Answer: A: \(40\)
\(\frac{9}{5}\left(1-\tfrac{p}{100}\right)=\frac{27}{25}\), so \(1-\tfrac{p}{100}=\frac{3}{5}\) and \(p=40\).
Question 3
A price is increased by 50% and the new price is then decreased by \(p\)%. The final price is 4% below the original price. What is \(p\)?
- \(36\)
- \(41\)
- \(46\)
- \(64\)
- \(54\)
Show the answer and solution
Answer: A: \(36\)
\(\frac{3}{2}\left(1-\tfrac{p}{100}\right)=\frac{24}{25}\), so \(1-\tfrac{p}{100}=\frac{16}{25}\) and \(p=36\).
Question 4
A price is increased by 50% and the new price is then decreased by \(p\)%. The final price is 20% above the original price. What is \(p\)?
- \(80\)
- \(25\)
- \(20\)
- \(30\)
- \(70\)
Show the answer and solution
Answer: C: \(20\)
\(\frac{3}{2}\left(1-\tfrac{p}{100}\right)=\frac{6}{5}\), so \(1-\tfrac{p}{100}=\frac{4}{5}\) and \(p=20\).
Question 5
A price is increased by 20% and the new price is then decreased by \(p\)%. The final price is 4% below the original price. What is \(p\)?
- \(25\)
- \(20\)
- \(80\)
- \(16\)
- \(24\)
Show the answer and solution
Answer: B: \(20\)
\(\frac{6}{5}\left(1-\tfrac{p}{100}\right)=\frac{24}{25}\), so \(1-\tfrac{p}{100}=\frac{4}{5}\) and \(p=20\).
Question 6
A price is increased by 25% and the new price is then decreased by \(p\)%. The final price is 5% above the original price. What is \(p\)?
- \(20\)
- \(16\)
- \(30\)
- \(21\)
- \(84\)
Show the answer and solution
Answer: B: \(16\)
\(\frac{5}{4}\left(1-\tfrac{p}{100}\right)=\frac{21}{20}\), so \(1-\tfrac{p}{100}=\frac{21}{25}\) and \(p=16\).
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