When things only move between two groups
Sometimes nothing is added and nothing is taken away: some of one share simply moves to the other. One friend gives marbles to another, or oil is poured from one container into another. Both shares change, but the total stays the same. The total is the anchor.
Suppose two amounts are in the ratio 3 : 5, which is 8 parts in total. After some of the second amount moves to the first, the ratio is 5 : 3. That is still 8 parts in total, so the two ratios already use parts of the same size: the first share gained 5 − 3 = 2 parts, and the second share lost the same 2 parts.
Before and after, the two bars together have the same length. Nothing entered and nothing left, so the total is still 8 parts.
Make the totals match first
When the two ratios have totals with different numbers of parts, rewrite them so the totals match. If the ratio was 5 : 7 before (12 parts) and 3 : 1 after (4 parts), multiply both numbers in the second ratio by 3: 3 : 1 = 9 : 3, which is also 12 parts. Now one part is the same amount in both ratios. The first share went from 5 parts to 9 parts, a gain of 4 parts, and the second share lost the same 4 parts. The real number in the question then tells you what those 4 parts are worth.
In the drawing below, the top bar is the ratio 2 : 5, which is 18 and 45, or 63 in total. The lower bar is the ratio 4 : 5, and you choose the size of its parts. Only one size keeps the total at 63, and that is the only case where the change could have been a move from one share to the other.
nothing is held here — 2 : 5 → 4 : 5 alone does not fix any amount
Make the total the quantity that did not move.
Drag the size of a part in the lower bar. At 9 the second share stays at 45 while the total grows; at 7 the total stays at 63 while 10 moves from the second share to the first.
Worked example: Constant Total (Internal Transfer)
Question Daryl and Evan had some marbles in the ratio 7 : 5. After Daryl gave 18 marbles to Evan, the ratio of Daryl's marbles to Evan's marbles became 1 : 2. How many marbles did Daryl have at first?
1.Draw Total bar of 12 units: Daryl gets 7 units, Evan gets 5 units.
Daryl : Evan = 7:5, so the total is 12 units. 2.Draw identical Total bar divided into 3 equal blocks: Daryl gets 1 block, Evan gets 2 blocks.
After: 1:2 of the same total. Marbles moved, none left, so the bar keeps its length: 3 blocks. 3.Convert 1 block into units: 12 ÷ 3 = 4 units.
A block is 12 ÷ 3 = 4 units. 4.Daryl's change: from 7 units down to 4 units = 3 units.
Daryl went from 7 units to 4: the 3 units he gave away. 5.3 units = 18 ⟹ 1 unit = 6.
3u = 18, so u = 6. 6.Daryl at first = 7 × 6 = 42 marbles.
Daryl at first: 7u = 42 marbles.
Answer: 42 marbles
Common mistakes
- Comparing the 'Before' units (7) and 'After' units (1) directly (7 − 1 = 6 units) without normalizing total units.
- Adding 18 to Daryl instead of subtracting it from Daryl.
Worked example: Constant Total (Internal Transfer)
Question Container A and Container B held a total of 720 ml of oil. At first, Container A held 512 of the oil. After 80 ml of oil was poured from Container B into Container A, what fraction of the total oil was in Container B?
1.Draw a total bar of 12 units representing 720 ml.
Draw the total as 12 units, because 512 names twelfths. 2.Compute value of u: 12u = 720 ⟹ u = 60 ml.
12u = 720 ml, so u = 60 ml. 3.Before: Container A = 5u = 300 ml, Container B = 7u = 420 ml.
Before: A holds 5u = 300 ml and B holds 7u = 420 ml. 4.Since 80 ml moves from B to A, shift an 80 ml block visually from bar B to bar A.
Pour 80 ml from B into A. The bar does not change length: the total is still 720 ml. 5.New volume of B = 420 − 80 = 340 ml.
B now holds 420 − 80 = 340 ml. 6.New fraction in B = 340720, divide top and bottom by 20 to simplify: 1736.
B holds 340720 = 1736 of the oil.
Answer: 1736 of the oil
Common mistakes
- Assuming the total volume changes because an exchange occurred.
- Adding 80 ml to A while forgetting to subtract 80 ml from B.