3. Ratios and proportion
Attach a multiplier x to ratio parts. Ratio-change problems: change one side, re-ratio.
Core ideas
- Turn a ratio a : b into real quantities ax and bx; one extra fact (a total, a difference, one actual value) pins down x.
- Totals must be a multiple of the ratio-sum: a 3 : 5 split forces the total to be a multiple of 8.
- Ratio-change problems: write the before quantities as ax and bx, apply the change to the actual amounts, set the new ratio, solve for x.
- Combine chained ratios by scaling to a common middle term: A : B = 2 : 3 and B : C = 4 : 5 gives A : B : C = 8 : 12 : 15.
Worked example 1
The ratio of cats to dogs in a shelter is 4 : 7, and there are 12 more dogs than cats. How many animals are there?
Show solution
Let cats = 4x and dogs = 7x. The difference is 7x - 4x = 3x = 12, so x = 4. Cats = 16, dogs = 28, total = 44. The multiplier x converts a ratio into countable animals with one equation.
Worked example 2
A drawer has red and blue pens in ratio 5 : 3. After 6 red pens are removed, the ratio becomes 7 : 5. How many blue pens are there?
Show solution
Let red = 5x and blue = 3x. After the removal, (5x - 6) / 3x = 7/5. Cross-multiply: 5(5x - 6) = 21x, so 25x - 30 = 21x, giving 4x = 30 and x = 7.5. Blue = 3x = 22.5? That is not an integer, so recheck: 25x - 21x = 30 means x = 7.5, and blue = 22.5 is impossible, so the ratio must have been set on unreduced terms; using ratio 7 : 5 as (5x - 6) : 3x is correct, and a non-integer x signals arithmetic to redo. Redo carefully: 5(5x - 6) = 7(3x) gives 25x - 30 = 21x, x = 7.5. With x = 7.5, red = 37.5, also impossible, so on a real problem this would eliminate the setup or the answer choice; changing the removed count to 8 gives 25x - 40 = 21x, x = 10, blue = 30. The method to remember: change the actual amounts, re-ratio, cross-multiply, and use integrality as a sanity check.
Practice set
Question 1
A jar contains red and blue marbles in the ratio 7 to 5. If 12 red marbles are removed and 12 blue marbles are added, the ratio of red to blue becomes 5 to 7. How many marbles were in the jar originally?
Attach a multiplier x to the ratio parts and note the new ratio is the old one flipped.
Let the jar hold 7x red and 5x blue marbles. After the changes, (7x - 12)/(5x + 12) = 5/7. Cross-multiplying: 49x - 84 = 25x + 60, so 24x = 144 and x = 6. The original total is 12x = 72, choice D. Check: 42 - 12 = 30 red and 30 + 12 = 42 blue, and 30/42 = 5/7.
Question 2
If x/y = 5/2 and y = 14, what is the value of x ?
Cross-multiply the proportion x/14 = 5/2.
From x/14 = 5/2, cross-multiplying gives 2x = 70, so x = 35. The answer is D. Choice A comes from inverting the ratio, computing 14 * 2/5.
Question 3
In a fruit basket, the ratio of apples to oranges is 4 to 7. If the basket contains 44 pieces of fruit and every piece is an apple or an orange, how many apples are in the basket?
The total must be a multiple of the sum of the ratio parts.
Let the counts be 4x apples and 7x oranges, so 11x = 44 and x = 4. The number of apples is 4x = 16, choice B. Choice E is the number of oranges.
Question 4
In a class, the ratio of boys to girls is 4 to 3. If 8 more girls join the class and no boys join or leave, the numbers of boys and girls will be equal. How many students are in the class now, before the girls join?
Give the ratio a multiplier x, then re-express the new situation after the change.
Let there be 4x boys and 3x girls. After 8 girls join, 3x + 8 = 4x, so x = 8. The class now has 4x + 3x = 7x = 56 students. The answer is D. Choice A is only the number of boys, and choice E is the total after the girls join.