Ratio Problems – Pearson Edexcel International GCSE (9–1) Mathematics A Higher Tier

These ratio problems are based on questions from the Revise Pearson Edexcel International GCSE (9–1) Mathematics A – Higher Tier Revision Guide. The solutions below use an algebraic ratio method in which each part of the ratio is represented as a multiple of x.

Question 1 – Sharing Money in a Ratio

Andre, Becky and Makito share money in the ratio 3 : 6 : 7. Andre and Becky receive £207 altogether. Work out how much Makito receives.

Working

A : B : M = 3x : 6x : 7x

3x + 6x = 207
9x = 207
x = 207/9

7 × (207/9) = 161

Makito receives £161.
Answer: £161

Question 2(a) – Ages in a Ratio

Amir and Petra's ages are in the ratio 3 : 7. Amir is 9 years old. Work out Petra's age.

Working

A : P = 3x : 7x

3x = 9
x = 3

P = 7 × 3 = 21

Petra is 21 years old.
Answer: 21 years old

Question 2(b) – Finding an Age from the Total

Karl and Yasmina's ages are in the ratio 5 : 2. The sum of their ages is 84. Work out Yasmina's age.

Working

K : Y = 5x : 2x

5x + 2x = 84
7x = 84
x = 84/7 = 12

Y = 2 × 12 = 24

Yasmina is 24 years old.
Answer: 24 years old

Algebraic Ratio Method

A ratio such as 3 : 7 can be represented algebraically as 3x : 7x. The variable x represents the value of one ratio part.

Once x has been found from the information in the question, each quantity can be calculated by multiplying x by its corresponding ratio number.

Source: Questions adapted from Harry Smith, Revise Pearson Edexcel International GCSE (9–1) Mathematics A – Higher Tier Revision Guide, Pearson.

Solutions and workings: Tiago Hands.

Popular posts from this blog

A Geometric Way to Visualise sin(x + y) and cos(x + y)

Image
The angle–addition identities for sine and cosine often appear as algebraic formulas, but they can also be understood by combining two right triangles in a simple geometric construction. The calculations for the side lengths follow directly from the definitions of sine and cosine. sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y − sin x sin y Start with a right triangle of angle y and hypotenuse 1. From basic trigonometry, its horizontal and vertical sides are: cos y and sin y. Next, attach a second right triangle with angle x . Its hypotenuse is the side of length cos y from the first triangle, so its adjacent and opposite sides become: adjacent = cos x · cos y opposite = sin x · cos y Likewise, if the first triangle's vertical side sin y is used as a hypotenuse in a similar way, it contributes: adjacent = cos x · sin y opposite = sin x · sin y When the horizontal components are combined, they give the expression for cos(x + y...
Read more

The Method of Differences — A Clean Proof of the Sum of Cubes

The Method of Differences — A Clean Proof of the Sum of Cubes The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r) , the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum . 1) Telescoping Sums Assume the general term u r can be written as: u r = f(r+1) − f(r). Then the finite sum from r = 1 to r = n becomes: Σ u r = Σ ( f(r+1) − f(r) ). To see what happens, write out a few terms: u₁ = f(2) − f(1) u₂ = f(3) − f(2) u₃ = f(4) − f(3) ⋮ uₙ = f(n+1) − f(n) When these are added, everything cancels except the first and last pieces: Σ u r = f(n+1) − f(1). This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states. 2) A Classic Application — The Sum of Cubes We will use this technique to prove the well-known ...
Read more

The Shortest Distance Between Two Skew Lines in ℝ³

The Shortest Distance Between Two Skew Lines in ℝ³ This post derives, from first principles, a vector formula for the shortest distance between two skew lines in ℝ³. The argument uses only the definitions and basic properties of the dot product and cross product; no higher results are assumed. 1. Vector Equations of the Lines Let the two lines be given in vector form by r = a + λb r = c + μd where: a , c are position vectors of fixed points on each line, b , d are non-zero direction vectors, λ, μ ∈ ℝ are parameters. The lines are skew if they are neither parallel nor intersecting. Our goal is to find a closed-form expression for the minimum distance between them. 2. Direction of the Common Perpendicular The segment that realises the shortest distance lies along a line perpendicular to both direction vectors b and d . A vector perpendicular to both is given by their cross product: b × d. Assuming b and d are not parallel, b × d ≠ 0. A unit ...
Read more