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GCSE Maths · EdexcelChapter 15 · Ratio and proportion
GCSE Maths · Ratio and Proportion · Higher · Lesson 15.9

GCSE Maths: Multi-stage ratio problems with algebra (Higher)

Part of the GCSE Maths ratio and proportion chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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15.9

Multi-stage ratio problems with algebra

Edexcel 1MA1: R5, R6, R7, N11
Grade 6-9

In harder problems the numbers change: items are added or removed, money is given away, or people get older. The original numbers are in one ratio and the new numbers are in another. We cannot just divide the total, because we do not know it. Instead we use algebra.

The method. (1) If the original ratio is 5 : 7, write the original amounts as 5x and 7x, where x is the value of one part. This keeps the original ratio exactly right for every value of x. (2) Write expressions for the amounts after the change: for example 6 more of the first gives 5x + 6, and 7x stays the same. (3) Use the new ratio, say 6 : 7, to form an equation: 5x + 67x = 67. (4) Cross-multiply: 7(5x + 6) = 6 × 7x, expand and solve for x. (5) Put x back to find the amounts the question asks for, and check that the new amounts really are in the new ratio.

Transfers. If money is passed from one person to another, subtract from one and add to the other: 5x − 4 and 3x + 4. The total does not change. Ages. In 6 years’ time everyone is 6 years older, so add 6 to every age. Expressions in a ratio. If x : y = 3 : 5, write x = 3k and y = 5k. Then any expression in x and y becomes an expression in k, and k cancels from the ratio. Equations from ratios. A ratio with letters in it, such as (3x + 1) : (2x − 1) = 5 : 3, is solved by cross-multiplying.

Boys: 5 partsGirls: 7 partsratio 5 : 7
Boys : girls = 5 : 7. Write the amounts as 5x and 7x, where x is the value of one part.

Key ideas

  • Original ratio p : q means the amounts are px and qx.
  • Write expressions for the new amounts, then form an equation from the new ratio by cross-multiplying.
  • Transfers: subtract from one amount, add to the other. The total stays the same.
  • Ages: add the same number of years to every age.
  • If x : y = p : q, write x = pk and y = qk. Check the answer by testing the new ratio.
Worked example 15.9.1
The ratio of boys to girls in a club is 5 : 7. Six more boys join the club and the ratio of boys to girls becomes 6 : 7. How many boys and girls were in the club before the new boys joined?

Key words. An equation says two things are equal. We form an equation from the information in the question, then solve it to find x. Cross-multiplying means multiplying the top of each fraction by the bottom of the other.

Method. Write the amounts as multiples of one letter (the value of one part). Use the change to write new expressions, then form and solve an equation from the new ratio.

  1. Use a letter for the value of one part. The ratio 5 : 7 means the boys is 5 equal parts and the girls is 7 equal parts. Let one part be x.
    Boys = 5x
    Girls = 7x
  2. Write expressions for the amounts after the change. Six boys join, so add 6 to the number of boys. The number of girls does not change.
    Boys = 5x + 6
    Girls = 7x
  3. Write the ratio as a fraction (or use cross-multiplying). Equal ratios give equal fractions: the first quantity divided by the second equals r over s. Then we can multiply both sides to remove the fractions.
    5x + 67x = 67
  4. Cross-multiply. Multiply the top of each fraction by the bottom of the other. (This is the same as multiplying both sides by 7 and by 7x.) Write each product in brackets.
    7 × (5x + 6) = 6 × 7x
  5. Expand the brackets. Multiply every term inside each bracket by the number outside.
    35x + 42 = 42x
  6. Collect the x terms on one side and the numbers on the other. Do the same to both sides so that the equation stays balanced.
    35x + 42 = 42x
    Subtract 35x from both sides: 42 = 7x
  7. Divide both sides by 7. 7x means 7 multiplied by x, so the inverse is dividing by 7.
    x = 42 ÷ 7 = 6
  8. Find the amounts the question asks for. Put the value of one part back into the expressions.
    Original boys: 5 × 6 = 30
    Original girls: 7 × 6 = 42

Check. After the change: boys = 36 and girls = 42. 36 : 42 simplifies (÷ 6) to 6 : 7, which is the ratio 6 : 7 we were told. ✓

Watch out. The ratio is only true for the amounts after the change. Before the change the amounts are in the original ratio. Do not mix the two.

Worked example 15.9.2
Sam and Tia have money in the ratio 5 : 3. Sam gives Tia £4. The ratio of their money is now 13 : 11. How much money do they have altogether?

Key words. An equation says two things are equal. We form an equation from the information in the question, then solve it to find x. Cross-multiplying means multiplying the top of each fraction by the bottom of the other.

Method. Write the amounts as multiples of one letter (the value of one part). Use the change to write new expressions, then form and solve an equation from the new ratio.

  1. Use a letter for the value of one part. The ratio 5 : 3 means the Sam is 5 equal parts and the Tia is 3 equal parts. Let one part be x.
    Sam = 5x
    Tia = 3x
  2. Write expressions for the amounts after the gift. Sam gives away £4, so subtract 4 from Sam's amount. Tia receives £4, so add 4 to Tia's amount.
    Sam = 5x − 4
    Tia = 3x + 4
  3. Write the ratio as a fraction (or use cross-multiplying). Equal ratios give equal fractions: the first quantity divided by the second equals r over s. Then we can multiply both sides to remove the fractions.
    5x − 43x + 4 = 1311
  4. Cross-multiply. Multiply the top of each fraction by the bottom of the other. (This is the same as multiplying both sides by 11 and by (3x + 4).) Write each product in brackets.
    11 × (5x − 4) = 13 × (3x + 4)
  5. Expand the brackets. Multiply every term inside each bracket by the number outside.
    55x − 44 = 39x + 52
  6. Collect the x terms on one side and the numbers on the other. Do the same to both sides so that the equation stays balanced.
    55x − 44 = 39x + 52
    Subtract 39x from both sides: 16x − 44 = 52
    Add 44 to both sides: 16x = 96
  7. Divide both sides by 16. 16x means 16 multiplied by x, so the inverse is dividing by 16.
    x = 96 ÷ 16 = 6
  8. Find the amounts the question asks for. Put the value of one part back into the expressions.
    Original Sam: 5 × 6 = £30
    Original Tia: 3 × 6 = £18
  9. Add to find the total. The question asks for the total.
    £30 + £18 = £48

Check. After the gift: Sam = £26 and Tia = £22. £26 : £22 simplifies (÷ 2) to 13 : 11, which is the ratio 13 : 11 we were told. ✓

Watch out. The ratio is only true for the amounts after the change. Before the change the amounts are in the original ratio. Do not mix the two.

Worked example 15.9.3
Solve (3x + 1) : (2x − 1) = 5 : 3.

Key words. An equation says two things are equal. Cross-multiplying turns a = b over c = d into a × d = b × c.

Method. Write the ratio as a fraction, cross-multiply to remove the fractions, expand and solve the linear equation.

  1. Write the ratio as a fraction (or use cross-multiplying). Equal ratios give equal fractions: the first quantity divided by the second equals r over s. Then we can multiply both sides to remove the fractions.
    3x + 12x − 1 = 53
  2. Cross-multiply. Multiply the top of each fraction by the bottom of the other. (This is the same as multiplying both sides by 3 and by (2x − 1).) Write each product in brackets.
    3 × (3x + 1) = 5 × (2x − 1)
  3. Expand the brackets. Multiply every term inside each bracket by the number outside.
    9x + 3 = 10x − 5
  4. Collect the x terms on one side and the numbers on the other. Do the same to both sides so that the equation stays balanced.
    9x + 3 = 10x − 5
    Subtract 9x from both sides: 3 = x − 5
    Add 5 to both sides: 8 = x
  5. Divide both sides by 1. 1x means 1 multiplied by x, so the inverse is dividing by 1.
    x = 8 ÷ 1 = 8

Check. Substitute x = 8: 3x + 1 = 25 and 2x − 1 = 15. 25 : 15 = 5 : 3 ✓

Watch out. Cross-multiply carefully: top of one fraction times bottom of the other. Put the expressions in brackets before you multiply.

Worked example 15.9.4
x : y = 3 : 5. Work out the ratio (2x + y) : (x + 3y) in its simplest form.

Key words. A ratio compares the sizes of two or more quantities, written with a colon, for example 2 : 3 (read “2 to 3”). The numbers in a ratio are called parts. The simplest form of a ratio has no whole number (other than 1) that divides into every part.

Method. Write x and y as multiples of one letter k, substitute, then cancel k.

  1. Write x and y using one letter. The ratio 3 : 5 means x is 3 parts and y is 5 parts. If one part is k, then x = 3k and y = 5k.
    x = 3k
    y = 5k
  2. Substitute into the first expression. Replace x and y by their values in k.
    2x + y = 2(3k) + (1)(5k) = 6k + (5)k = 11k
  3. Substitute into the second expression. Do the same for the second expression.
    x + 3y = 1(3k) + (3)(5k) = 3k + (15)k = 18k
  4. Write the ratio and cancel k. Both terms contain k, so k divides out. The ratio of the numbers is what remains.
    11k : 18k = 11 : 18

Check. Choose numbers in the ratio, e.g. x = 9, y = 15: 2x + y = 33 and x + 3y = 54; the ratio 33 : 54 simplifies to 11 : 18. ✓

Watch out. Cancel k only at the end, when it is a factor of both parts.

Worked example 15.9.5
The ages of Ali and Bea are in the ratio 3 : 5. In 6 years’ time the ratio of their ages will be 5 : 7. Work out their ages now.

Key words. An equation says two things are equal. We form an equation from the information in the question, then solve it to find x. Cross-multiplying means multiplying the top of each fraction by the bottom of the other.

Method. Write the amounts as multiples of one letter (the value of one part). Use the change to write new expressions, then form and solve an equation from the new ratio.

  1. Use a letter for the value of one part. The ratio 3 : 5 means the Ali is 3 equal parts and the Bea is 5 equal parts. Let one part be x.
    Ali = 3x
    Bea = 5x
  2. Write expressions for the amounts in 6 years. In 6 years each person is 6 years older, so add 6 to both ages.
    Ali = 3x + 6
    Bea = 5x + 6
  3. Write the ratio as a fraction (or use cross-multiplying). Equal ratios give equal fractions: the first quantity divided by the second equals r over s. Then we can multiply both sides to remove the fractions.
    3x + 65x + 6 = 57
  4. Cross-multiply. Multiply the top of each fraction by the bottom of the other. (This is the same as multiplying both sides by 7 and by (5x + 6).) Write each product in brackets.
    7 × (3x + 6) = 5 × (5x + 6)
  5. Expand the brackets. Multiply every term inside each bracket by the number outside.
    21x + 42 = 25x + 30
  6. Collect the x terms on one side and the numbers on the other. Do the same to both sides so that the equation stays balanced.
    21x + 42 = 25x + 30
    Subtract 21x from both sides: 42 = 4x + 30
    Subtract 30 from both sides: 12 = 4x
  7. Divide both sides by 4. 4x means 4 multiplied by x, so the inverse is dividing by 4.
    x = 12 ÷ 4 = 3
  8. Find the amounts the question asks for. Put the value of one part back into the expressions.
    Original Ali: 3 × 3 = 9 years
    Original Bea: 5 × 3 = 15 years

Check. In 6 years: Ali = 15 years and Bea = 21 years. 15 years : 21 years simplifies (÷ 3) to 5 : 7, which is the ratio 5 : 7 we were told. ✓

Watch out. The ratio is only true for the amounts after the change. Before the change the amounts are in the original ratio. Do not mix the two.

Watch out

The new ratio only applies to the amounts after the change. Do not write the new ratio as 5x : 7x with the new numbers; use the new expressions such as (5x + 6) and 7x. And always check that the answer fits the new ratio.

Exercise 15I  Multi-stage ratio problems

Non-calculator
1
PSSolve each problem by forming an equation.
(a)The ratio of red counters to blue counters is 3 : 4. Six more red counters are added and the ratio becomes 3 : 2. Find the original numbers of red and blue counters.
(b)The ratio of boys to girls is 5 : 7. Six more boys join and the ratio becomes 6 : 7. How many boys were there at first?
(c)The ratio of cats to dogs in a rescue centre is 7 : 4. Nine cats are rehomed and the ratio becomes 11 : 8. Find the original numbers of cats and dogs.
(d)Two numbers A and B are in the ratio 2 : 5. When 10 is added to each number the ratio becomes 1 : 2. Find A and B.
2
PSMoney and marbles are passed between two people. Find the amount asked for.
(a)Ali and Ben have marbles in the ratio 7 : 3. Ali gives 8 marbles to Ben, and now they have the ratio 1 : 1. How many marbles did Ali have at first?
(b)Ana and Bo have money in the ratio 4 : 5. Ana gives £3 to Bo and the ratio becomes 7 : 11. How much money do they have altogether?
3
PSA shop has pens and pencils in the ratio 4 : 7. It sells 12 pencils and the ratio of pens to pencils is now 4 : 3. How many pencils were there at the start?
4
FSolve each equation.
(a)(x + 4) : (x + 10) = 2 : 3
(b)(3x + 1) : (2x − 1) = 5 : 3
(c)(4x − 2) : (x + 7) = 2 : 1
(d)(2x + 5) : (5x − 3) = 13 : 17
(e)(3x − 1) : (x + 9) = 1 : 1
5
Fx : y is given. Work out the second ratio in its simplest form.
(a)x : y = 3 : 5. Find (2x + y) : (x + 3y)
(b)x : y = 4 : 3. Find (x + y) : (x − y)
(c)x : y = 2 : 5. Find (3x + 2y) : (x + y)
(d)x : y = 2 : 3. Find (5x − y) : (x + 2y)
6
Fa : b = 2 : 3 and b : c = 4 : 7. Work out the ratio a : c in its simplest form.
7
PSThe ages of Ali and Bea are in the ratio 3 : 5. In 6 years’ time the ratio of their ages will be 5 : 7. Work out their ages now.
8
PSA rectangle has sides in the ratio 3 : 5. The shorter side is increased by 4 cm and the longer side is decreased by 2 cm. The new shape is a square. Find the lengths of the sides of the original rectangle.
9
PSContainer A holds 30 litres of squash and water in the ratio 1 : 4. Container B holds 20 litres of squash and water in the ratio 2 : 3. They are poured into one tank. Write the ratio squash : water in the tank in its simplest form.
10
Ra : b = 3 : 4. Show that (a + b) : (b − a) = 7 : 1.
11
RShow that adding 5 to both parts of the ratio 2 : 3 can never give the ratio 2 : 3 again.
12
PSMarbles in a bag are red and blue in the ratio 3 : 5. Eight red marbles are removed and twelve blue marbles are added. The ratio of red to blue is now 1 : 3. How many marbles were in the bag at the start?
13
PSIn a club, the ratio of girls to boys is 3 : 2. Seven girls leave and then the number of girls equals the number of boys. How many girls and boys were in the club at first?
14
RTwo numbers A and B are in the ratio 3 : 4. When 12 is added to each number, the ratio becomes 5 : 6. Find A and B. Then show that the new ratio is closer to 1 : 1 than the old ratio.

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