GCSE Maths

Numbers
37 Topics | 1 Quiz
Integers
Prime Numbers and Composite Numbers
Negative Numbers
Comparing Numbers Using Symbols
The Square Numbers
The Laws of Indices
Negative Indices
The Cube Numbers
Triangular Numbers
The Fibonacci Sequence
Factors
Multiples
The Highest Common Factor
Fractions
Adding and Subtracting Fractions
Multiplication of Fractions
Division of Fractions
Fractional Indices
Decimal Places
Fractions and Decimals
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Vectors
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Vectors
Probability
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Introduction to probability
Finding the sum of probabilities
Conditional probability
Set Notation
Venn diagrams
The Complement of a Set
Statistics
2 Topics
Population and samples
Representing data
Algebra
16 Topics | 1 Quiz
Algebraic Notation
Algebraic Vocabulary
Collecting Like Terms
Expanding Brackets and Simplifying
Expanding Double Brackets
The Difference of Two Squares
Substituting into Formulae
Subject of Formula
Factorisation
Factorising Quadratic Expressions
More on Factorisation
Equations
Quadratic Equations
Mathematical Formulae
Algebraic Equivalence and Proof
Functions
Algebra
Sequences
2 Topics | 1 Quiz
Generating sequences
Nth Term Sequences
Sequences
Ratio, Proportion and Rates of Change
8 Topics | 1 Quiz
Inverse Proportion
Conversion
Scale Factors
Percentages
Direct Proportion
Rate, Ratio and Proportion Examples
Ratios
Ratios
Ratio, Proportion and Rates of Change
Geometry and Measures
9 Topics
Loci and constructions
Properties of angles
Properties of polygons
Transformations
Enlargement
Circle theorems
Plane figure properties
Plans and elevations
Areas and Volumes in Map Scale Problems
Mensuration and Calculation
10 Topics
Map scales
Bearings
Area
Volume
Circles: arcs and sectors
Congruence and similarity
Pythagoras’ theorem
Sine and cosine rule
Area of a triangle
Standard units of measure
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Rate, Ratio and Proportion Examples

GCSE Maths Ratio, Proportion and Rates of Change Rate, Ratio and Proportion Examples

Example

  • John can finish a large chocolate bar in three hours
  • Michael eats the same chocolate bar in two hours
  • Nathaniel can eat the chocolate bar in 5 hours

Assuming that their rate of eating doesn’t change and that they start eating at the same time, how long does it take all three of them to finish one large chocolate bar?

Observation: If Michael can finish the chocolate bar in two hours, then all three of them should finish the chocolate bar in less than two hours.

We find the least common multiple of 2,3 and 5. This value is 2 \times 3 \times 5 = 30.

So, given 30 hours, John eats 10 chocolate bars. Given 30 hours, Michael eats 15 chocolate bars. Also given 30 hours, Nathaniel eats 6 chocolate bars.

So, given 30 hours, John, Michael and Nathaniel eat a total of 10+15+6 = 31 chocolate bars.

To eat 31 chocolate bars takes 30 hours and to eat 1 goat takes \dfrac{30}{31} = 58 mins, 4 secs, correct to the nearest second.

Example

A sum of money is divided in the ratio 2:3:5. Given that the smallest share is £45. Find the largest share of money.

Let x = sum of money, so we are given:

\dfrac{2}{10}\times x=£45

Solving for x gives x = £45\times\dfrac{10}{2}

= £225.

Largest share = \dfrac{5}{10}\times£225

= £112.50

Example

What is the ratio of £50 to £125?

A) 3:7

B) 5:6

C) \dfrac{2}{5}

D) 2:5

The ratio of £50 to £125 is:

\dfrac{50}{125}=\dfrac{2}{5} (fraction)

= 2:5 (In ratio form)

The correct answer is D = 2:5

Example

Given that y varies directly as \left( x^{2}+1\right) and that when x=2, y=125. Find the value of y when x is 3.

y\propto x^{2}+1

y=k\left( x^{2}+1\right), where k is a constant

Substituting x=2, y=125 gives:

125=k\left( 2^{2}+1\right)

125=k(5)

k=\dfrac{125}{5}

k=25

Hence, the connecting equation between x and y is:

y=25\left( x^{2}+1\right)

Now when x=3, y=25\left( 3^{2}+1\right)

y=25(10)

=250

Example

  • Anne and Betty can complete a job in four hours
  • Anne and Clara can complete the job in six hours
  • Betty and Clara can complete the job in nine hours

How long does it take if Anne, Betty and Clara work together? Assuming they work at the same rate.

A) 9 hours, 30 minutes

B) 6 hours, 20 minutes

C) 6 hours, 33 minutes

D) 3 hours, 47 minutes

Anne and Betty can complete the job in four hours. The percentage (%) of the job Anne does is 4 hours + the percentage of the job Betty does in 4 hours, which equals 100% or 1.

4(% job Anne does in 1 hour) + 4(% job Betty does in 1 hour) = 1

4(A+B)=1 (i)

Likewise, Anne and Clara can complete the job in six hours. So, the percentage Anne does in 6 hours + the percentage Clara does in 6 hours = 100% or 1.

6(A+C)=1 (ii)

For Betty and Clara, we have 9(B+C)=1 (iiii)

So, we need to find k(A+B+C)=1, for some value k.

The lowest common multiple (LCM) of 4, 6 and 9 is 36.

So, multiply (i), (ii) and (iii) BY 9, 6 and 4, to obtain:

36(A+B)=9

36(A+C)=6

36(B+C)=4

Adding the last three equations, we have:

36(A+B)+36(A+C)+36(B+C)=9+6+4

36(2A+2B+2C)=19

72(A+B+C)=19

72(A+B+C)=1

The time taken for Anne, Betty and Clara working together:

=\dfrac{72}{19} hours

= 3 hours, 47 minutes

So, the answer is D = 3 hours, 47 minutes.

Example

If a:b = 2:5 and b:c=14:15, then a:c equals…

A) 4:5

B) 28:75

C) 75:28

D) 14:75

Rewriting these ratios as fractions, we get:

\dfrac{a}{b}=\dfrac{2}{5}, \dfrac{14}{15}

Now \dfrac{a}{c}=\dfrac{a}{b} \times \dfrac{b}{c}=\dfrac{2}{5} \times \dfrac{14}{15}=\dfrac{28}{75}

As a ratio, this is 28:75, so the answer is B.

Example

Suppose that there are 3 bells:

  • Bell A
  • Bell B
  • Bell C

Bell A rings every three minutes, bell B rings every five minutes and bell C rings every six minutes. They last rang together at 9:00 AM. When will they ring together again?

You can draw a table showing the successive ringing times for each bell, which you can see below.

Bell A Bell B Bell C
9:00 9:00 9:00
9:03 9:05 9:06
9:06 9:10 9:12
9:09 9:15 9:18
9:12 9:20 9:24
9:15 9:25 9:30
9:18 9:30 9:36
9:21 9:35  
9:24    
9:27    
9:30    
9:33

Then we look for the time that is common in all three columns. The least number we can see in all three columns is 9:30 AM. So, the three bells will ring together at 9:30AM.

However, we usually solve this type of problem using least common multiple (LCM).

We find the LCM of 3, 5 and 6 minutes.

LCM = 2 \times 3 \times 5

= 30 minutes

So, the three bells will ring together 30 minutes after 9:00 AM.

9:00 AM + 30 minutes = 9:30 AM

Example

If five people can dig a ditch in 16 days, how long will it take eight people to dig the ditch?

If five people take 16 days, one person will take 5 \times 16 days, which is 80 days.

So, eight people will take \dfrac{80}{8}=10 days to dig the ditch

Example

A book has 540 pages and is 2.7 cm thick, ignoring the covers. How many pages are there in a book that is 3.7 cm thick?

2.7 cm → 540 pages

1 cm → \dfrac{540}{2.7} pages

3.2 cm →\dfrac{540}{2.7} \times 3.2 = 640 pages.

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