A transfer moves, it does not add
Ali and Ben share 48 stickers in the ratio 5 : 3. Then Ali gives Ben some of his stickers, and now the ratio of Ali's stickers to Ben's is 4 : 4.
Both shares changed: Ali has fewer stickers and Ben has more. But no sticker came in from outside and no sticker was thrown away. The stickers only moved from one boy to the other, so the total is still 48. In every transfer problem, the total is the amount that stays the same.
Ali's bar gets shorter and Ben's gets longer by the same amount, so the two bars together are as long as before.
The same parts before and after
Count the parts in each ratio. Before the gift there are 5 + 3 = 8 parts. After the gift there are 4 + 4 = 8 parts. Both ratios cut the same 48 stickers into 8 equal parts, so one part is the same number of stickers in both: 48 ÷ 8 = 6.
Now every share can be found. Before the gift, Ali had 5 × 6 = 30 stickers and Ben had 3 × 6 = 18. Check: 30 + 18 = 48.
48 ÷ 8 = 6 in every part, so before the gift Ali's 5 parts are 30 stickers and Ben's 3 parts are 18.
Count the parts that moved
Ali went from 5 parts to 4 parts, so 1 part moved across to Ben, and Ben went from 3 parts to 4. One part is 6 stickers, so Ali gave Ben 6 stickers.
After the gift, each boy has 4 × 6 = 24 stickers. Check it against the story: Ali had 30 and gave away 6, and 30 − 6 = 24. Ben had 18 and received 6, and 18 + 6 = 24.
After the gift, both boys hold 4 parts of 6 stickers, which is 24 each. There are still 8 parts and 48 stickers.
When the totals have different numbers of parts
Sometimes the two ratios do not have the same number of parts. Suppose two amounts are in the ratio 2 : 5 before a transfer and 4 : 5 after it. That is 7 parts before and 9 parts after, so a part cannot be the same size in both, and parts cannot be counted across from one ratio to the other.
The total still did not change, and that decides the size of a part. In the drawing below, the top bar is 2 : 5 with 9 in every part: 18 and 45, which is 63 altogether. You choose the size of a part in the lower bar, which is 4 : 5. Only one size keeps its total at 63.
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 until its total is 63, the same as the top bar. At 7, the shares are 28 and 35: the second share gave 45 − 35 = 10 to the first, which rose from 18 to 28.
The usual mistakes
Giving the giver's amount instead of the gift. When 48 is shared 5 : 3 and becomes 4 : 4, Ali had 30 and keeps 24. The gift is the difference, 30 − 24 = 6.
Treating the ratios as amounts. The ratio 5 : 3 does not mean Ali has 5 stickers. The numbers in a ratio count parts, and a part here is 6 stickers.
Worked example: A Transfer That Makes Two Amounts Equal
Question The ratio of the number of cards Devi has to the number of cards Farah has is 7 : 3. After Devi gives Farah 12 cards, the two girls have the same number of cards. (a) How many cards did Devi have at first? (b) How many cards does each girl have in the end?
1.Draw Devi's cards as 7 units and Farah's cards as 3 units. The cards only move between the two girls, so the total stays at 7 + 3 = 10 units.
Devi has 7 units and Farah has 3. The total of 10 units does not change. 2.In the end the two girls have the same number of cards, so each has 10 ÷ 2 = 5 units.
When they are equal, each girl has 10 ÷ 2 = 5 units. 3.Devi goes from 7 units to 5 units, so she gives away 2 units. 2 units = 12 cards, and 1 unit = 12 ÷ 2 = 6 cards.
Devi gives away 7 − 5 = 2 units, which are the 12 cards. 1 unit = 6 cards. 4.(a) Devi had 7 × 6 = 42 cards at first.
(a) Devi had 7 × 6 = 42 cards at first. 5.(b) In the end each girl has 5 × 6 = 30 cards. Check: 42 − 12 = 30 and 18 + 12 = 30.
(b) Each girl ends with 5 × 6 = 30 cards.
Answer: (a) 42 cards; (b) 30 cards
Common mistakes
- Writing 7 − 3 = 4 units = 12 cards. Devi gives away only half of the difference. If she gave all 4 units, Farah would then have more cards than Devi.
- Giving 42 as the answer to part (b). 42 is what Devi had at first. In the end she has 12 fewer cards.
Three people and two transfers
The total stays the same however many people pass things between them. In the next problem, the ratio before the transfers has 12 parts and the ratio after has 6 parts. Multiplying both numbers of a ratio by the same number gives an equivalent ratio, so doubling every number in the after ratio makes it 12 parts too. Then follow the person who only gave and never received, because one change in parts is easier to read than two.
Worked example: Three-Party Closed System Internal Transfer
Question Initially, the ratio of Alan's tokens to Bryan's tokens to Colin's tokens was 3 : 4 : 5. Alan gave 20 tokens to Bryan, and Colin gave 30 tokens to Alan. In the end, the ratio of Alan's tokens to Bryan's tokens to Colin's tokens became 2 : 3 : 1. How many tokens did the three of them have altogether?
1.Before model: Alan (3u), Bryan (4u), Colin (5u) → Total = 12u.
Before: 3:4:5, a total of 12 units. 2.After model: Alan (2p), Bryan (3p), Colin (p) → Total = 6p.
After: 2:3:1, a total of 6 parts. Tokens only changed hands, so both bars are the same length. 3.Standardize After bar into 12 units by multiplying each part by 2:
Make the totals agree: double every part, 6p = 12u. 4.After: Alan (4u), Bryan (6u), Colin (2u).
After, in units: Alan 4u, Bryan 6u, Colin 2u. 5.Inspect Colin's bar: dropped from 5u to 2u (a loss of 3u).
Colin is the clean one: he only gave, never received. From 5u to 2u is a loss of 3u. 6.Since Colin only gave away 30 tokens: 3u = 30 ⟹ u = 10.
He gave exactly 30, so 3u = 30 and u = 10. 7.Total tokens = 12u = 12 × 10 = 120.
Altogether: 12u = 120 tokens.
Answer: 120 tokens
Common mistakes
- Attempting to track Alan first, which involves two separate transactions (+30 and -20), increasing the likelihood of arithmetic errors.
- Failing to standardize the total units before comparing individual changes.