Ratio and proportion · applications

Applications: Ratio and Proportion

25 problems

O-Level · SAT · GCSE Higher

01

An Art Club of Boys and Girls: The Members, Then the Paintbrushes Given Out

heuristicFind One Unit from the Difference, Then Count the Units

The ratio of the number of boys to the number of girls in an art club was 4 : 7. There were 36 more girls than boys. (a) How many members were there in the art club altogether? (b) Each girl was given 3 paintbrushes and each boy was given 2 paintbrushes. How many paintbrushes were given out altogether?

Boysuuuu4uGirlsuuuuuuu7u
Boys : girls = 4 : 7. The boys are 4 units and the girls are 7 units of the same size.
Draw the boys as 4 equal units and the girls as 7 units of the same size.
step 1 of 5

Draw the boys as 4 units and the girls as 7 units of the same size. The girls' bar is 3 units longer, and those 3 units are the 36 more girls. Find one unit, then count the units.

All steps
  1. Draw the boys as 4 equal units and the girls as 7 units of the same size.
  2. The girls' bar is 7 − 4 = 3 units longer than the boys' bar, and those 3 units are the 36 more girls. So 1 unit is 36 ÷ 3 = 12 children.
  3. (a) There are 4 + 7 = 11 units altogether, so the club has 11 × 12 = 132 members.
  4. The boys are 4 × 12 = 48 children and the girls are 7 × 12 = 84 children.
  5. (b) The boys get 48 × 2 = 96 paintbrushes and the girls get 84 × 3 = 252, so 96 + 252 = 348 paintbrushes were given out. Check: 84 − 48 = 36 more girls than boys.

answer(a) 132 members; (b) 348 paintbrushes

techniquePart-to-Whole Ratios · The Unitary Method

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Dividing the 36 by all 11 units. The 36 is how many more girls there are than boys, which is only the 3 units the girls' bar has beyond the boys' bar, so 1 unit is 36 ÷ 3 = 12.
  • Multiplying the 132 members by 2 + 3 = 5 paintbrushes, or by one of the two numbers. A boy gets 2 paintbrushes and a girl gets 3, so count the boys and the girls on their own: 48 × 2 + 84 × 3 = 348.
02

Stickers Shared by Three Girls, in Two Ratios That Both Include Brenda

heuristicMake Brenda the Same Number of Units in Both Ratios, Then Find One Unit

The ratio of Alicia's stickers to Brenda's stickers was 3 : 5. The ratio of Brenda's stickers to Clara's stickers was 4 : 7. Alicia had 92 fewer stickers than Clara. (a) Find the ratio of Alicia's stickers to Brenda's stickers to Clara's stickers. (b) How many stickers did the three girls have altogether?

Alicia3 unitsBrenda5 units3 : 5 rowBrenda4 units4 : 7 rowClara7 units
Alicia : Brenda = 3 : 5 and Brenda : Clara = 4 : 7. Brenda's bar is the same length in both rows, but it is 5 units in one and 4 units in the other.
Draw Alicia : Brenda = 3 : 5 and Brenda : Clara = 4 : 7. Brenda's bar is the same length in both rows, but it is 5 units in one and 4 units in the other, so the units are different sizes.
step 1 of 5

Brenda is in both ratios, but she is 5 units in one and 4 units in the other. Cut the units smaller until Brenda has the same number in both. Then one unit is the same number of stickers for all three girls.

All steps
  1. Draw Alicia : Brenda = 3 : 5 and Brenda : Clara = 4 : 7. Brenda's bar is the same length in both rows, but it is 5 units in one and 4 units in the other, so the units are different sizes.
  2. The least common multiple of 5 and 4 is 20, so cut Brenda into 20 units in both rows. Multiply the first ratio by 4: 3 : 5 = 12 : 20. Multiply the second ratio by 5: 4 : 7 = 20 : 35.
  3. (a) Now Brenda is 20 units in both rows, so Alicia : Brenda : Clara = 12 : 20 : 35.
  4. Clara has 35 − 12 = 23 units more than Alicia, and that is 92 stickers. So 1 unit is 92 ÷ 23 = 4 stickers.
  5. (b) Altogether there are 12 + 20 + 35 = 67 units, which is 67 × 4 = 268 stickers. Check: Alicia has 12 × 4 = 48 and Clara has 35 × 4 = 140, and 140 − 48 = 92.

answer(a) 12 : 20 : 35; (b) 268 stickers

techniqueCombining Two Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Joining the two ratios as 3 : 5 : 7. Brenda is 5 units in the first ratio and 4 units in the second, so a unit is a different number of stickers in each. Make Brenda the same number of units in both before joining them.
  • Subtracting Alicia's 3 units from Clara's 7 units and calling the 7 − 3 = 4 units the 92 stickers. The 7 and the 3 come from ratios with different units. Once Brenda is matched, Clara has 35 − 12 = 23 units more than Alicia.
03

Marbles Passed from Daryl to Evan, with the Total Unchanged

heuristicMake the Totals Match, Then Follow Daryl's Units

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. (a) How many marbles did Daryl have at first? (b) How many marbles did Evan have after the gift?

BeforeDaryl 7uEvan 5u12uAfterDaryl 1 partEvan 2 parts3 parts
Before, Daryl : Evan = 7 : 5, which is 12 units. After, it is 1 : 2, only 3 parts, but the bar is the same length: no marbles are added or taken away.
Before, Daryl : Evan = 7 : 5, so the total is 7 + 5 = 12 units. After, Daryl : Evan = 1 : 2, which is 1 + 2 = 3 parts.
step 1 of 5

The marbles only move from Daryl to Evan, so the total does not change. Rewrite the ratio after the gift with the same total number of units as before. Then one unit is the same number of marbles before and after.

All steps
  1. Before, Daryl : Evan = 7 : 5, so the total is 7 + 5 = 12 units. After, Daryl : Evan = 1 : 2, which is 1 + 2 = 3 parts.
  2. No marbles are added or taken away, so the total bar is the same length before and after. Multiply the ratio after by 4 so that it also has 12 units: 1 : 2 = 4 : 8.
  3. Daryl goes from 7 units to 4 units, so he gave away 7 − 4 = 3 units. He gave 18 marbles, so 1 unit is 18 ÷ 3 = 6 marbles.
  4. (a) Daryl had 7 × 6 = 42 marbles at first.
  5. (b) After the gift, Evan has 8 units, which is 8 × 6 = 48 marbles. Check: Daryl is left with 42 − 18 = 24 marbles, and 24 : 48 = 1 : 2.

answer(a) 42 marbles; (b) 48 marbles

techniqueRatios with a Constant Total

Common pitfalls

  • Taking 7 − 1 = 6 units as Daryl's 18 marbles. The ratio before has 12 units and the ratio after has only 3 parts, so a part after is not the same size as a unit before. Make both totals 12 units first; then Daryl goes from 7 units to 4 units.
  • Adding the 18 marbles to Daryl, or taking them from Evan. Daryl gave the marbles to Evan, so Daryl ends with 18 fewer and Evan with 18 more: 42 − 18 = 24 and 30 + 18 = 48.
04

Non-Fiction Books Added to a Shelf, with the Fiction Unchanged

heuristicMake the Fiction Units Match, Then Compare the Non-Fiction

The ratio of the number of fiction books to the number of non-fiction books on a shelf was 5 : 3. The librarian then added 54 non-fiction books and no fiction books, and the ratio became 2 : 3. (a) How many fiction books were on the shelf? (b) How many non-fiction books were on the shelf after the 54 were added?

Beforefiction 5 parts3 parts5 : 3Afterfiction 2 partsnon-fiction 3 parts2 : 3
Before, fiction : non-fiction = 5 : 3; after, 2 : 3. No fiction book is added, so the fiction bar is the same length in both, cut into 5 parts and into 2.
Before, fiction : non-fiction = 5 : 3. After, fiction : non-fiction = 2 : 3. The fiction books do not change, so the fiction bar is the same length in both.
step 1 of 5

No fiction books are added or taken away, so the fiction bar is the same length before and after. Rewrite both ratios so the fiction has the same number of units. Then one unit is the same number of books in both, and the growth in the non-fiction is the 54 books.

All steps
  1. Before, fiction : non-fiction = 5 : 3. After, fiction : non-fiction = 2 : 3. The fiction books do not change, so the fiction bar is the same length in both.
  2. The least common multiple of 5 and 2 is 10, so make the fiction 10 units in both. Multiply the first ratio by 2: 5 : 3 = 10 : 6. Multiply the second ratio by 5: 2 : 3 = 10 : 15.
  3. The non-fiction grows from 6 units to 15 units, so the 54 books added are 15 − 6 = 9 units. 1 unit is 54 ÷ 9 = 6 books.
  4. (a) The fiction is 10 units, which is 10 × 6 = 60 books.
  5. (b) After the 54 are added, the non-fiction is 15 units, which is 15 × 6 = 90 books. Check: before, there were 6 × 6 = 36 non-fiction books, and 60 : 36 = 5 : 3; after, 60 : 90 = 2 : 3.

answer(a) 60 fiction books; (b) 90 non-fiction books

techniqueRatios with a Constant Part

examsPSLE · SAT

Common pitfalls

  • Thinking the non-fiction did not change because it is 3 in both ratios. The fiction is 5 units before and 2 parts after, so a unit is a different number of books in each ratio. Once the fiction is matched, the non-fiction goes from 6 units to 15 units.
  • Taking the fiction from 5 down to 2 and calling the 3 lost units the 54 books. No fiction books were taken away; the 54 books were added to the non-fiction. Match the fiction first, and then the growth in the non-fiction, 9 units, is the 54 books.
05

Mr. Wong and His Son: Their Ages Five Years Ago and in Ten Years' Time

heuristicThe Age Difference Never Changes, So Give It the Same Number of Units in Both Ratios

Five years ago, the ratio of Mr. Wong's age to his son's age was 7 : 2. In 10 years' time, the ratio of Mr. Wong's age to his son's age will be 2 : 1. (a) How old is Mr. Wong's son now? (b) What is the ratio of Mr. Wong's age to his son's age now, in its simplest form?

5 yrs agoMr. Wong 7uson 2udifference 5uIn 10 yrsMr. Wong 2 partsson 1 partdifference 1 part
Five years ago, Mr. Wong was 7 units and his son 2, a difference of 5 units. In 10 years' time the ratio is 2 : 1, a difference of 1 part.
Five years ago, Mr. Wong was 7 units and his son was 2 units, so the difference was 7 − 2 = 5 units. In 10 years' time, the ratio is 2 : 1, so the difference is 2 − 1 = 1 part.
step 1 of 6

Mr. Wong and his son grow older by the same number of years, so the difference between their ages never changes. Rewrite the second ratio so the difference has the same number of units as in the first. Then one unit is the same number of years in both.

All steps
  1. Five years ago, Mr. Wong was 7 units and his son was 2 units, so the difference was 7 − 2 = 5 units. In 10 years' time, the ratio is 2 : 1, so the difference is 2 − 1 = 1 part.
  2. The difference between their ages never changes, so make it 5 units in both. Multiply the second ratio by 5: 2 : 1 = 10 : 5.
  3. The son grows from 2 units to 5 units, and Mr. Wong grows from 7 units to 10 units. Each of them gains 3 units.
  4. From 5 years ago to 10 years from now is 5 + 10 = 15 years. So 3 units are 15 years, and 1 unit is 15 ÷ 3 = 5 years.
  5. (a) Five years ago the son was 2 × 5 = 10 years old, so now he is 10 + 5 = 15 years old.
  6. (b) Five years ago Mr. Wong was 7 × 5 = 35, so now he is 35 + 5 = 40. The ratio now is 40 : 15, and dividing both by 5 gives 8 : 3. Check: in 10 years' time they will be 50 and 25, and 50 : 25 = 2 : 1.

answer(a) 15 years old; (b) 8 : 3

techniqueRatios with a Constant Difference

Common pitfalls

  • Counting the years between the two ratios as 10 − 5 = 5. One ratio is 5 years before now and the other is 10 years after now, so they are 5 + 10 = 15 years apart.
  • Giving the son's age five years ago, 10, as his age now. The units give the ages five years ago, so add 5 years to find each age now: the son is 15 and Mr. Wong is 40.
06

Laptops, Tablets and Smartphones at a Fair, in Two Brands

heuristicSplit Each Kind's Units by Brand, Then Compare the Two Brands

At an electronics fair, the ratio of the number of laptops to the number of tablets to the number of smartphones was 3 : 4 : 8. One third of the laptops were Brand X and the rest were Brand Y. Half of the tablets were Brand X and the rest were Brand Y. None of the smartphones were Brand X. There were 180 more Brand Y devices than Brand X devices. (a) What fraction of all the devices were Brand X? (b) How many devices were at the fair altogether?

Laptops3 unitsTablets4 unitsSmartphones8 units
The laptops, tablets and smartphones are 3, 4 and 8 units, all the same size.
Draw the laptops as 3 units, the tablets as 4 units and the smartphones as 8 units, all the same size.
step 1 of 6

Draw each kind of device with its units from the ratio 3 : 4 : 8. Each fraction is taken of its own kind, so split each bar into Brand X units and Brand Y units. Then count the units of each brand.

All steps
  1. Draw the laptops as 3 units, the tablets as 4 units and the smartphones as 8 units, all the same size.
  2. 13 of the 3 laptop units is 1 unit, so 1 unit is Brand X and 2 units are Brand Y. Half of the 4 tablet units is 2 units, so 2 are Brand X and 2 are Brand Y. All 8 smartphone units are Brand Y.
  3. Brand X has 1 + 2 = 3 units and Brand Y has 2 + 2 + 8 = 12 units, out of 3 + 4 + 8 = 15 units altogether.
  4. (a) Brand X is 3 of the 15 units, which is 315 = 15 of the devices.
  5. Brand Y has 12 − 3 = 9 units more than Brand X, and that is 180 devices. So 1 unit is 180 ÷ 9 = 20 devices.
  6. (b) There were 15 × 20 = 300 devices at the fair. Check: Brand X has 3 × 20 = 60, Brand Y has 12 × 20 = 240, and 240 − 60 = 180.

answer(a) 15 of the devices; (b) 300 devices

techniqueThree-Part Ratios · Combining Two Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking 13 and 12 of all 15 units. The 13 is of the laptops only and the 12 is of the tablets only, so Brand X is 13 of 3 units and 12 of 4 units, which is 3 units.
  • Leaving the smartphones out of Brand Y. None of the smartphones is Brand X, so all 8 of their units are Brand Y, and Brand Y is 2 + 2 + 8 = 12 units, not 4.
07

Notes of Three Values in a Charity Box

heuristicMake One Set of Notes from the Two Ratios, Then Count the Sets

In a charity box, the ratio of the number of $2 notes to the number of $5 notes was 5 : 3, and the ratio of the number of $5 notes to the number of $10 notes was 2 : 1. The notes in the box were worth $1280 altogether. (a) How many $5 notes were in the box? (b) What was the total value of the $10 notes?

$2$2$2$2$2$2$2$2$2$210 × $2$5$5$5$5$5$56 × $5$10$10$103 × $105 : 3 = 10 : 6 and 2 : 1 = 6 : 3
The $5 notes are 3 in one ratio and 2 in the other. Make them 6 in both: 5 : 3 = 10 : 6 and 2 : 1 = 6 : 3.
The $5 notes are 3 in the first ratio and 2 in the second. The least common multiple of 3 and 2 is 6, so make them 6 in both: 5 : 3 = 10 : 6 and 2 : 1 = 6 : 3.
step 1 of 6

The $5 notes are in both ratios, so make their number the same in both. That gives one set of notes in the right ratio. Find what one set is worth, and how many sets make $1280.

All steps
  1. The $5 notes are 3 in the first ratio and 2 in the second. The least common multiple of 3 and 2 is 6, so make them 6 in both: 5 : 3 = 10 : 6 and 2 : 1 = 6 : 3.
  2. So the notes can be sorted into equal sets, each holding ten $2 notes, six $5 notes and three $10 notes.
  3. One set is worth 10 × $2 + 6 × $5 + 3 × $10 = $20 + $30 + $30 = $80.
  4. The box holds $1280, so there are 1280 ÷ 80 = 16 sets.
  5. (a) Each set has six $5 notes, so there are 16 × 6 = 96 five-dollar notes.
  6. (b) Each set has $30 in $10 notes, so the $10 notes are worth 16 × $30 = $480. Check: the $2 and $5 notes are worth 16 × $50 = $800, and $800 + $480 = $1280.

answer(a) 96 notes; (b) $480

techniqueThe Unitary Method · Bar Models for Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing $1280 by 10 + 6 + 3 = 19 units. The 19 counts notes, but the $1280 is money, and each kind of note is worth a different amount. Find what one set of notes is worth, $80, and divide by that.
  • Giving 16 as the number of $10 notes, so that they are worth $160. 16 is the number of sets, and each set has three $10 notes, so there are 16 × 3 = 48 of them, worth $480.
08

Zachary and Hannah Spend Different Amounts from Their Savings

heuristicWrite the Amounts After as Parts, Then Make the Parts Match

The ratio of Zachary's savings to Hannah's savings was 4 : 5. Then Zachary spent $45 and Hannah spent $100, and the ratio of Zachary's savings to Hannah's savings became 3 : 2. (a) How much money did Zachary have at first? (b) How much money did Hannah have left?

Zachary4uHannah5uZachary afterppp3pHannah afterpp2p
Before, Zachary has 4 units and Hannah has 5. After spending, Zachary has 3 parts and Hannah has 2. The bars are drawn to scale, and a part is not the same size as a unit.
Before, Zachary has 4 units and Hannah has 5 units. After spending, Zachary has 3 parts and Hannah has 2 parts.
step 1 of 6

Both children spend, and they spend different amounts, so neither the total nor the difference stays the same. Draw the savings before as units and the savings after as parts. Then make the parts match, and compare the units.

All steps
  1. Before, Zachary has 4 units and Hannah has 5 units. After spending, Zachary has 3 parts and Hannah has 2 parts.
  2. Zachary's 4 units take away $45 leave 3 parts. Hannah's 5 units take away $100 leave 2 parts.
  3. Make the parts match. Twice Zachary's money after is 6 parts, which is 8 units take away $90. Three times Hannah's money after is also 6 parts, which is 15 units take away $300.
  4. So 15 units take away $300 is the same as 8 units take away $90. Add $300 to both: 15 units are 8 units and $210. The 7 extra units are $210, so 1 unit is $210 ÷ 7 = $30.
  5. (a) Zachary had 4 × $30 = $120 at first.
  6. (b) Hannah had 5 × $30 = $150 at first, so she has $150 − $100 = $50 left. Check: Zachary has $120 − $45 = $75 left, and 75 : 50 = 3 : 2.

answer(a) $120; (b) $50

techniqueBefore-and-After Ratio Problems · Simultaneous by Elimination

examsSAT · GCSE Higher

Common pitfalls

  • Subtracting the ratio numbers: 4 − 3 = 1 unit for Zachary's $45 and 5 − 2 = 3 units for Hannah's $100. The units before and the parts after are different sizes, so the numbers of the two ratios cannot be subtracted. Write the savings after as 3 parts and 2 parts, and make the parts match.
  • Writing 7 units = $300 + $90 = $390. The 15 units lose $300 and the 8 units lose only $90, yet both leave 6 parts, so the 7 extra units make up the difference: $300 − $90 = $210.
09

Tokens Passed Among Three Boys, with the Total Unchanged

heuristicMake the Totals Match, Then Follow the Boy Who Only Gave

At first, the ratio of Alan's tokens to Bryan's tokens to Colin's tokens was 3 : 4 : 5. Then 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 was 2 : 3 : 1. (a) How many tokens did the three boys have altogether? (b) How many tokens did Alan have in the end?

At firstAlan 3uBryan 4uColin 5u12uEndAlan 2 partsBryan 3 parts1 part6 parts
At first, 3 : 4 : 5 is 12 units. In the end, 2 : 3 : 1 is only 6 parts, but the bar is the same length: the tokens only pass between the boys.
At first, Alan : Bryan : Colin = 3 : 4 : 5, a total of 3 + 4 + 5 = 12 units. In the end, the ratio is 2 : 3 : 1, a total of 2 + 3 + 1 = 6 parts.
step 1 of 5

The tokens only pass between the three boys, so the total never changes. Rewrite the ratio at the end with the same total number of units as at first. Then follow Colin, who only gave tokens, because his change is a single move.

All steps
  1. At first, Alan : Bryan : Colin = 3 : 4 : 5, a total of 3 + 4 + 5 = 12 units. In the end, the ratio is 2 : 3 : 1, a total of 2 + 3 + 1 = 6 parts.
  2. No tokens are added or taken away, so the total stays the same. Double every number in the ratio at the end to make 12 units: 2 : 3 : 1 = 4 : 6 : 2.
  3. Colin only gave tokens and never received any. He goes from 5 units to 2 units, so the 30 tokens he gave are 5 − 2 = 3 units. 1 unit is 30 ÷ 3 = 10 tokens.
  4. (a) Altogether the boys have 12 × 10 = 120 tokens.
  5. (b) In the end, Alan has 4 units, which is 4 × 10 = 40 tokens. Check: Alan started with 3 × 10 = 30, gave 20 and received 30, and 30 − 20 + 30 = 40.

answer(a) 120 tokens; (b) 40 tokens

techniqueRatios After a Transfer · Three-Part Ratios

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Comparing the ratios before the totals match, so that Colin goes from 5 to 1 and the 4 units lost are the 30 tokens. The first ratio has 12 units and the second has only 6 parts, so each part at the end is 2 units. Double the second ratio first: Colin goes from 5 units to 2 units.
  • Following Alan and calling his gain of 4 − 3 = 1 unit the 30 tokens he received. Alan also gave 20 tokens away, so his 1 unit is 30 − 20 = 10 tokens. Colin made only one move, so his change is simpler to read.
10

Milk Cartons Sent Out and Juice Cartons Delivered at a Warehouse

heuristicWrite the Amounts at the End as Parts, Then Make the Parts Match

At a warehouse, the ratio of the number of cartons of milk to the number of cartons of juice was 5 : 2. Then 35 cartons of milk were sent out and 10 cartons of juice were delivered, and the ratio of milk cartons to juice cartons became 3 : 2. (a) How many cartons of milk were there at first? (b) How many cartons of juice were there in the end?

Milk5uJuice2uMilk in endppp3pJuice in endpp2p
At first, the milk is 5 units and the juice is 2 units. In the end, the milk is 3 parts and the juice is 2 parts. The bars are drawn to scale.
At first, the milk is 5 units and the juice is 2 units. In the end, the milk is 3 parts and the juice is 2 parts.
step 1 of 6

The milk goes down and the juice goes up, so neither the total nor the difference stays the same. Draw the cartons at first as units and the cartons in the end as parts. Then make the parts match, and compare the units.

All steps
  1. At first, the milk is 5 units and the juice is 2 units. In the end, the milk is 3 parts and the juice is 2 parts.
  2. The milk's 5 units take away 35 cartons leave 3 parts. The juice's 2 units and 10 more cartons make 2 parts.
  3. Make the parts match. Twice the milk in the end is 6 parts, which is 10 units take away 70 cartons. Three times the juice in the end is also 6 parts, which is 6 units and 30 cartons.
  4. So 10 units take away 70 is the same as 6 units and 30. Add 70 to both: 10 units are 6 units and 100 cartons. The 4 extra units are 100 cartons, so 1 unit is 100 ÷ 4 = 25 cartons.
  5. (a) There were 5 × 25 = 125 cartons of milk at first.
  6. (b) The juice was 2 × 25 = 50 cartons, and 10 more were delivered, so there are 50 + 10 = 60 cartons of juice in the end. Check: the milk left is 125 − 35 = 90 cartons, and 90 : 60 = 3 : 2.

answer(a) 125 cartons of milk; (b) 60 cartons of juice

techniqueBefore-and-After Ratio Problems

examsSAT · GCSE Higher

Common pitfalls

  • Taking 10 cartons away from the juice as well, and writing 2 units take away 10. The 10 cartons of juice were delivered, so the juice goes up to 2 units and 10; only the milk goes down.
  • Subtracting the ratio numbers: the milk goes from 5 to 3, so 2 units are the 35 cartons. The total and the difference both change, so the units at first and the parts in the end are different sizes. Write the cartons in the end as 3 parts and 2 parts, and make the parts match.
11

Lucas and Marcus Each Spend a Different Fraction of Their Money

heuristicCut Every Unit into 5, So Each Fraction Is a Whole Number of Units

Lucas and Marcus had pocket money in the ratio 7 : 4. Lucas spent 35 of his money, and Marcus spent 12 of his money. Then Lucas had $48 more left than Marcus. (a) How much money did Lucas have at first? (b) What was the ratio of the money the two boys spent altogether to the money they had left altogether?

Lucas35 units7 units × 5Marcus20 units4 units × 5
Lucas : Marcus = 7 : 4. Lucas spends fifths, so cut every unit into 5: Lucas has 35 small units and Marcus has 20.
Lucas has 7 units and Marcus has 4 units. Lucas spends fifths of his money, so cut every unit into 5 smaller units: Lucas has 7 × 5 = 35 small units and Marcus has 4 × 5 = 20.
step 1 of 5

Lucas spends fifths of his money, so cut every unit into 5 smaller units. Cut Marcus's units the same way, so that all the small units are the same size. Then find what each boy keeps, and compare.

All steps
  1. Lucas has 7 units and Marcus has 4 units. Lucas spends fifths of his money, so cut every unit into 5 smaller units: Lucas has 7 × 5 = 35 small units and Marcus has 4 × 5 = 20.
  2. Lucas spends 35 of 35 units, which is 21 units, and keeps 35 − 21 = 14 units. Marcus spends 12 of 20 units, which is 10 units, and keeps the other 10 units.
  3. Lucas has 14 − 10 = 4 units more left than Marcus, and that is $48. So 1 unit is $48 ÷ 4 = $12.
  4. (a) Lucas had 35 × $12 = $420 at first.
  5. (b) Together the boys spent 21 + 10 = 31 units and kept 14 + 10 = 24 units, so the ratio of the money spent to the money left is 31 : 24. Check: Lucas keeps 14 × $12 = $168 and Marcus keeps 10 × $12 = $120, and $168 − $120 = $48.

answer(a) $420; (b) 31 : 24

techniqueRatio, Fraction, Percent · Combining Two Ratios

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Subtracting the fractions, 35 − 12 = 110, and calling that the $48. The 35 is of Lucas's money and the 12 is of Marcus's money, which are different amounts, so the fractions cannot be subtracted. Find what each boy keeps in units first.
  • Comparing what the boys spent, 21 − 10 = 11 units, with the $48. The $48 is how much more Lucas has left, so compare the units left: 14 − 10 = 4 units.
12

Water Poured from Container X into Container Y, Then Some Poured Back

heuristicChoose Units That Both Pours Divide, Then Follow the Water

Container X and Container Y held water in the ratio 5 : 3. First, 14 of the water in Container X was poured into Container Y. Then 15 of the water now in Container Y was poured back into Container X. In the end, Container X had 144 ml more water than Container Y. (a) How much water was in Container X at first? (b) How much water was in Container Y in the end?

At firstX 100uY 60u160u
Water only moves between X and Y, so the total never changes. Multiply 5 : 3 by 20: X is 100 units and Y is 60, so both pours give whole units.
X : Y = 5 : 3. A quarter of X is poured first, and then a fifth of Y, so multiply the ratio by 4 × 5 = 20 to make both pours whole numbers of units: X has 100 units and Y has 60 units, 160 units in all.
step 1 of 6

Water only moves between the two containers, so the total never changes. Choose units small enough that a quarter of X, and then a fifth of the new amount in Y, are whole numbers of units. Then follow the water through both pours.

All steps
  1. X : Y = 5 : 3. A quarter of X is poured first, and then a fifth of Y, so multiply the ratio by 4 × 5 = 20 to make both pours whole numbers of units: X has 100 units and Y has 60 units, 160 units in all.
  2. First pour: 14 of 100 units is 25 units. X has 100 − 25 = 75 units and Y has 60 + 25 = 85 units.
  3. Second pour: 15 of 85 units is 17 units. X has 75 + 17 = 92 units and Y has 85 − 17 = 68 units, still 160 units in all.
  4. In the end, X has 92 − 68 = 24 units more than Y, and that is 144 ml. So 1 unit is 144 ÷ 24 = 6 ml.
  5. (a) X had 100 × 6 = 600 ml at first.
  6. (b) In the end, Y has 68 × 6 = 408 ml. Check: X has 92 × 6 = 552 ml in the end, and 552 − 408 = 144.

answer(a) 600 ml; (b) 408 ml

techniqueMultiplying and Dividing Undo Each Other · Ratios After a Transfer

Common pitfalls

  • Taking 15 of the water Y held at first, 60 units, which is 12 units. The second pour is 15 of the water in Y after the first pour, 85 units, so it is 17 units.
  • Taking the 144 ml as the difference between the two pours, 25 − 17 = 8 units. The 144 ml compares the two containers in the end, so it is 92 − 68 = 24 units.
13

Three Gears in a Row: Their Teeth and Their Turns

heuristicEvery Gear Passes the Same Number of Teeth, So Turns Times Teeth Stays the Same

Gear A, Gear B and Gear C are in a row. The teeth of Gear A fit into the teeth of Gear B, and the teeth of Gear B fit into the teeth of Gear C. The ratio of the number of teeth on Gear A to the number on Gear B is 2 : 3, and the ratio of the number of teeth on Gear B to the number on Gear C is 4 : 5. (a) Find the ratio of the number of teeth on Gear A to the number of teeth on Gear C. (b) Gear A makes 90 complete turns. How many complete turns does Gear C make?

A8 teethB12 teethC15 teeth
Gear B is 3 in one ratio and 4 in the other. Make it 12 in both: A : B : C = 8 : 12 : 15.
Gear B is 3 in the first ratio and 4 in the second. The least common multiple of 3 and 4 is 12, so make B 12 in both: 2 : 3 = 8 : 12 and 4 : 5 = 12 : 15.
step 1 of 5

Where two gears meet, their teeth pass the meeting point one for one, so every gear in the row passes the same number of teeth. Write the teeth as units, count the units of teeth that Gear A passes, and share them among Gear C's units.

All steps
  1. Gear B is 3 in the first ratio and 4 in the second. The least common multiple of 3 and 4 is 12, so make B 12 in both: 2 : 3 = 8 : 12 and 4 : 5 = 12 : 15.
  2. (a) So A : B : C = 8 : 12 : 15, and the ratio of the teeth on A to the teeth on C is 8 : 15.
  3. Where A meets B, one tooth of A passes for each tooth of B, and where B meets C, one tooth of B passes for each tooth of C. So A, B and C all pass the same number of teeth.
  4. Gear A has 8 units of teeth. In 90 turns it passes 90 × 8 = 720 units of teeth, so Gear C passes 720 units of teeth too.
  5. (b) Gear C has 15 units of teeth, so it makes 720 ÷ 15 = 48 complete turns. Check: 48 × 15 = 720, and 90 × 8 = 720.

answer(a) 8 : 15; (b) 48 complete turns

techniqueInverse Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Thinking the gear with more teeth turns more times, and working out 90 × 158 = 16834 turns. Gear C has more teeth, so more teeth must pass before it finishes one turn, and it turns fewer times: 90 × 8 ÷ 15 = 48.
  • Joining the ratios as 2 : 3 : 5, so that A : C = 2 : 5. Gear B is 3 in the first ratio and 4 in the second, so first make B the same in both: A : B : C = 8 : 12 : 15.
14

Highlighters, Pens and Markers Sold at Three Prices

heuristicMake One Set from the Two Ratios, Then Compare the Money in a Set

A bookstore sold highlighters, pens and markers. The ratio of the number of highlighters sold to the number of pens sold was 3 : 4, and the ratio of the number of pens sold to the number of markers sold was 2 : 3. A highlighter cost $1.50, a pen cost $2.00 and a marker cost $3.50. The bookstore took in $65 more from selling markers than from selling pens. (a) How many pens were sold? (b) How much money did the bookstore take in from all three items altogether?

1.501.501.503 highlighters2.002.002.002.004 pens3.503.503.503.503.503.506 markers
The pens are 4 in one ratio and 2 in the other. Make them 4 in both: one set is 3 highlighters, 4 pens and 6 markers.
The pens are 4 in the first ratio and 2 in the second, so make them 4 in both: 2 : 3 = 4 : 6. The items were sold in sets of 3 highlighters, 4 pens and 6 markers.
step 1 of 6

The pens are in both ratios, so make their number the same in both. That gives one set of items sold in the right ratio. Find how much more the markers bring in than the pens in one set, and how many sets make the $65.

All steps
  1. The pens are 4 in the first ratio and 2 in the second, so make them 4 in both: 2 : 3 = 4 : 6. The items were sold in sets of 3 highlighters, 4 pens and 6 markers.
  2. In one set, the pens bring in 4 × $2.00 = $8.00 and the markers bring in 6 × $3.50 = $21.00. The markers bring in $21.00 − $8.00 = $13.00 more than the pens.
  3. The markers brought in $65 more altogether, so the bookstore sold 65 ÷ 13 = 5 sets.
  4. (a) Each set has 4 pens, so 5 × 4 = 20 pens were sold.
  5. One set brings in 3 × $1.50 + $8.00 + $21.00 = $4.50 + $8.00 + $21.00 = $33.50.
  6. (b) Five sets bring in 5 × $33.50 = $167.50. Check: the 30 markers bring in $105 and the 20 pens $40, and $105 − $40 = $65.

answer(a) 20 pens; (b) $167.50

techniqueThe Unitary Method · Bar Models for Ratio

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the markers' 6 − 4 = 2 extra units as the $65. The $65 is money, not a number of items, and a marker costs more than a pen. Compare the money from the markers and from the pens in one set: $21.00 − $8.00 = $13.00.
  • Leaving out the highlighters in part (b), so that a set brings in only $8.00 + $21.00 = $29.00. The question asks for the money from all three items, so a set brings in $4.50 + $8.00 + $21.00 = $33.50.
15

Adults and Children at a Funfair, with Boys and Girls Among the Children

heuristicCut the Units into 4 for the Boys and Girls, Then Use Their Ratio at the End

At a funfair, the ratio of the number of adults to the number of children was 3 : 5. Among the children, the ratio of the number of boys to the number of girls was 1 : 3. Later, 33 adults and 7 boys came in, and 6 girls left. Then the ratio of adults to children was 1 : 1, and the ratio of boys to girls was 1 : 2. (a) How many children were at the funfair at first? (b) How many adults were at the funfair in the end?

Adults12uChildren20u
The children split 1 : 3, which is 4 parts, so cut every unit into 4: adults : children = 3 : 5 = 12 : 20.
The children split 1 : 3, which is 4 parts, so cut every unit into 4. Adults : children = 3 : 5 = 12 : 20, so the adults are 12 units and the children are 20 units.
step 1 of 6

Draw the adults and the children, then split the children's bar into boys and girls. Cut the units small enough that the boys and the girls are whole numbers of units. The boys and the girls at the end give one unit.

All steps
  1. The children split 1 : 3, which is 4 parts, so cut every unit into 4. Adults : children = 3 : 5 = 12 : 20, so the adults are 12 units and the children are 20 units.
  2. The boys are 14 of the 20 units, which is 5 units, and the girls are the other 15 units.
  3. Then 7 boys come in and 6 girls leave, so the boys are 5 units and 7, and the girls are 15 units take away 6. There are now twice as many girls as boys, so 15 units take away 6 is the same as 10 units and 14.
  4. Add 6 to both: 15 units are 10 units and 20. The 5 extra units are 20 children, so 1 unit is 20 ÷ 5 = 4 children.
  5. (a) There were 20 × 4 = 80 children at first.
  6. (b) There were 12 × 4 = 48 adults at first, and 33 more came in, so there are 48 + 33 = 81 adults in the end. Check: in the end there are 27 boys and 54 girls, which is 81 children, the same as the adults.

answer(a) 80 children; (b) 81 adults

techniqueRatios with a Constant Part · Combining Two Ratios

examsPSLE · SAT

Common pitfalls

  • Using the 1 : 3 of boys to girls as if it split everyone at the funfair. It splits only the children, so the boys are 14 of the children's 20 units, which is 5 units.
  • Adding the 6 girls who left, as if they came in, and writing 15 units and 6. The girls who leave make the number of girls smaller, so the girls in the end are 15 units take away 6.
16

A Van's Fuel Log: The Fuel Used for Each Kilometer, and How Far the Tank Lasts

heuristicDivide Each Output by Its Input, Then Use the Constant

On three trips, a delivery van used 4.8 liters of fuel to travel 60 km, 12 liters to travel 150 km and 19.2 liters to travel 240 km. (a) Show that the fuel used is in direct proportion to the distance, and find the constant of proportionality in liters per kilometer. (b) The van starts its next trip with 36 liters of fuel in the tank. The driver refuels when 4 liters are left. How far can the van travel on this trip before she must refuel?

60 km6 blocks150 km15 blocks240 km24 blocks
Each distance is a whole number of 10 km blocks: 60 km is 6 blocks, 150 km is 15 and 240 km is 24.
Every distance is a whole number of 10 km blocks: 60 km is 6 blocks, 150 km is 15 blocks and 240 km is 24 blocks.
step 1 of 5

Cut every distance into blocks of 10 km. If each block uses the same amount of fuel on every trip, the fuel is in direct proportion to the distance, and that amount gives the constant.

All steps
  1. Every distance is a whole number of 10 km blocks: 60 km is 6 blocks, 150 km is 15 blocks and 240 km is 24 blocks.
  2. Share each trip's fuel over its blocks: 4.8 ÷ 6 = 0.8, 12 ÷ 15 = 0.8 and 19.2 ÷ 24 = 0.8 liters. Every 10 km block uses the same 0.8 liters, so the fuel is in direct proportion to the distance.
  3. (a) One kilometer uses 0.8 ÷ 10 = 0.08 liters. The constant of proportionality is 0.08 liters per kilometer.
  4. (b) The driver keeps 4 liters in the tank, which leaves 36 − 4 = 32 liters to use.
  5. 32 liters is 32 ÷ 0.8 = 40 blocks of 10 km, so the van can travel 40 × 10 = 400 km. Check: 400 × 0.08 = 32 liters.

answer(a) Fuel divided by distance is 0.08 for every trip, so the fuel is in direct proportion to the distance and the constant is 0.08 liters per km; (b) 400 km

techniqueThe Constant of Proportionality · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing for one trip only. Any single pair of numbers gives a quotient; the fuel is in direct proportion to the distance only because all three trips give the same one, so every trip must be checked.
  • Dividing the whole 36 liters by 0.08 and answering 450 km. The 4 liters the driver keeps in the tank are not used on the trip, so only 32 liters are shared out.
17

Apple Pickers in an Orchard: A Bigger Team, Then a Bigger Order

heuristicFind What One Picker Fills in One Hour, Then Scale

In an orchard, 6 pickers fill 90 crates of apples in 3 hours. Every picker works at the same steady rate. (a) How many crates do 9 pickers fill in 3 hours? (b) The orchard has an order for 400 crates, which must be filled in 5 hours. How many pickers are needed?

6 pickers, 3 hours: 90 crates
One box for each picker in each hour: 6 × 3 = 18 boxes, holding 90 crates between them.
Draw one box for each picker in each hour: 6 pickers for 3 hours make 6 × 3 = 18 boxes, and together they hold 90 crates.
step 1 of 5

Draw one box for each picker in each hour. The boxes all hold the same number of crates, so find what one box holds and count boxes.

All steps
  1. Draw one box for each picker in each hour: 6 pickers for 3 hours make 6 × 3 = 18 boxes, and together they hold 90 crates.
  2. One box, one picker for one hour, holds 90 ÷ 18 = 5 crates.
  3. (a) 9 pickers for 3 hours make 9 × 3 = 27 boxes, so they fill 27 × 5 = 135 crates.
  4. (b) 400 crates need 400 ÷ 5 = 80 boxes.
  5. The 80 boxes are spread over 5 hours, so each hour needs 80 ÷ 5 = 16 boxes, one for each picker: 16 pickers are needed. Check: 16 × 5 × 5 = 400 crates.

answer(a) 135 crates; (b) 16 pickers

techniqueProportion by Unit Value · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Scaling (a) by the time as well as by the team. The time is still 3 hours; only the number of pickers changes, from 6 to 9, so the crates change in the same way: 90 × 96 = 135.
  • Treating a longer time as needing more pickers. Each picker fills more crates in 5 hours than in 3, so a longer time needs fewer pickers; dividing the 80 picker-hours by the 5 hours is what gives the team.
18

Paint Mixed Too Dark: White Added Until the Shade Is Lighter

heuristicRewrite Both Ratios So the Unchanged Share Matches

A painter mixes 14 liters of paint from blue paint and white paint in the ratio 3 : 4. The shade is too dark, so she adds white paint only, until the ratio of blue paint to white paint is 2 : 5. (a) How many liters of white paint does she add? (b) White paint is sold only in 2-liter cans, at $9.50 a can. White paint is sold only in whole cans. How much does she pay for the cans of white paint she needs?

Beforeblue 3 partswhite 4 parts14 LAfterblue 2 partswhite 5 parts
Before, blue : white = 3 : 4; after, 2 : 5. No blue is added, so the blue bar is the same length in both.
No blue paint is added, so the blue share is the same before and after. Before, blue : white = 3 : 4; after, blue : white = 2 : 5.
step 1 of 5

The blue share is the same paint before and after, so rewrite both ratios with the same number of units for blue. Then one unit means the same amount of paint in both, and the white units can be compared.

All steps
  1. No blue paint is added, so the blue share is the same before and after. Before, blue : white = 3 : 4; after, blue : white = 2 : 5.
  2. Rewrite both ratios so the blue share matches. The least common multiple of 3 and 2 is 6: 3 : 4 = 6 : 8 and 2 : 5 = 6 : 15.
  3. Before, the 14 liters are 6 + 8 = 14 units, so one unit is 14 ÷ 14 = 1 liter.
  4. (a) The white goes from 8 units to 15 units, so she adds 15 − 8 = 7 units, which is 7 liters.
  5. (b) 3 cans hold only 6 liters, so she buys 4 cans and pays 4 × $9.50 = $38.00. Check: blue 6 liters, white 15 liters, and 6 : 15 = 2 : 5.

answer(a) 7 liters; (b) $38.00

techniqueRewriting a Ratio to Match a Share · Ratios with a Constant Part

examsPSLE · O-Level

Common pitfalls

  • Adding 5 − 4 = 1 unit of white, 2 liters. The blue share is 3 units before and 2 units after, so a unit is a different amount of paint in the two ratios; they must be rewritten so the blue shares match before the white shares are compared.
  • Paying for 3.5 cans, or for 3 cans. Paint is sold only in whole cans, and 3 cans hold 6 liters, 1 liter short of the 7 liters she needs.
19

A Hiking Trail on Two Maps with Different Scales

heuristicMap to Ground by One Scale, Then Ground to Map by the Other

A hiking map is drawn to the scale 1 : 25000. On it, the trail from a parking lot to a waterfall measures 18.4 cm along its bends. (a) How long is the trail on the ground, in kilometers? (b) The same trail is drawn on a regional map on which 1 cm stands for 2 km. How long is the trail on the regional map, in centimeters?

Hiking map18.4 cm1 cm is 250 m
On the hiking map, 1 cm stands for 25000 cm, which is 250 m on the ground.
On the hiking map, 1 cm stands for 25000 cm on the ground, which is 250 m.
step 1 of 5

Find what one centimeter on the hiking map stands for on the ground, count the centimeters of trail, then share the ground distance into the regional map's centimeters.

All steps
  1. On the hiking map, 1 cm stands for 25000 cm on the ground, which is 250 m.
  2. The trail is 18.4 cm on the map, so on the ground it is 18.4 × 250 = 4600 m.
  3. (a) 4600 m = 4.6 km, so the trail is 4.6 km long.
  4. (b) On the regional map, every 2 km of ground is 1 cm. The trail's 4.6 km is 4.6 ÷ 2 = 2.3 lots of 2 km.
  5. So the trail is 2.3 cm long on the regional map. Check: 2.3 × 2 = 4.6 km, the same trail.

answer(a) 4.6 km; (b) 2.3 cm

techniqueMap Scales

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Writing 18.4 × 25000 = 460000 and calling it kilometers or meters. A scale written as a ratio has no unit, so the answer is in the map's unit, centimeters, and 100000 cm make one kilometer.
  • Scaling the 18.4 cm up for the regional map. Each centimeter of the regional map stands for more ground than a centimeter of the hiking map, so the same trail must be shorter there; going through the ground distance, 4.6 km, keeps the direction right.
20

An L-Shaped Living Room on a Floor Plan, and the Boxes of Flooring It Needs

heuristicScale Each Length, So the Area Scales by the Square

A floor plan of an apartment is drawn to the scale 1 : 50 on paper ruled in 1 cm squares. The L-shaped living room covers 108 of the squares. (a) What is the floor area of the real living room, in square meters? (b) Flooring is sold in boxes that each cover 2.2 m2, at $64 a box. How much does it cost to buy enough boxes to cover the living room floor? Ignore any waste from cutting.

1 square: 50 cm by 50 cm of floorwhich is 0.5 m by 0.5 m
One 1 cm square of the plan stands for 50 cm by 50 cm of floor: 0.5 m by 0.5 m.
One 1 cm square on the plan stands for a square of floor 50 cm by 50 cm, which is 0.5 m by 0.5 m.
step 1 of 5

Work out what one square of the plan stands for on the floor, as a length each way and then as an area, and count the squares.

All steps
  1. One 1 cm square on the plan stands for a square of floor 50 cm by 50 cm, which is 0.5 m by 0.5 m.
  2. That piece of floor has an area of 0.5 × 0.5 = 0.25 m2. Both of its sides were scaled by 50, so its area was scaled by 50 × 50 = 2500.
  3. (a) The room covers 108 squares, so its area is 108 × 0.25 = 27 m2.
  4. (b) 12 boxes cover 12 × 2.2 = 26.4 m2, which is less than 27 m2, so 13 boxes are needed.
  5. 13 boxes cost 13 × $64 = $832. Check: 13 × 2.2 = 28.6 m2, enough for the 27 m2 room.

answer(a) 27 m2; (b) $832

techniqueScaling Area · Map Scales

examsO-Level · SAT

Common pitfalls

  • Multiplying the plan area by 50 only: 108 × 50 = 5400 cm2, about half a square meter, far too small for a living room. The scale multiplies each length by 50, and an area is two lengths multiplied, so it is multiplied by 50 × 50 = 2500.
  • Rounding 12.27… boxes down to 12. Twelve boxes cover 26.4 m2 and leave 0.6 m2 of floor bare, so a thirteenth box must be bought.
21

A Disinfectant Strengthened: Pine Oil and Ethanol Added, the Water Left Alone

heuristicCombine the Two Ratios Through the Water, Then Hold the Water Fixed

A laboratory makes a disinfectant by mixing pine oil, distilled water and ethanol. At first, the ratio of pine oil to water is 3 : 8, and the ratio of water to ethanol is 4 : 5. (a) Find the ratio of pine oil to water to ethanol at first, in its simplest form. (b) A chemist then adds 180 ml of pine oil and 420 ml of ethanol, but no water. The ratio of pine oil to water becomes 5 : 8. (i) How much water is in the mixture? (ii) What is the ratio of pine oil to ethanol in the end, in its simplest form?

Pine oil3 unitsWater8 unitsEthanol10 units3 : 8 and 4 : 5 = 8 : 10, so 3 : 8 : 10
(a) The water is 8 in one ratio and 4 in the other: 4 : 5 = 8 : 10, so 3 : 8 : 10.
(a) The water is 8 in the first ratio and 4 in the second, so double the second ratio: 4 : 5 = 8 : 10. Pine oil : water : ethanol = 3 : 8 : 10.
step 1 of 5

No water is added, so the water bar is the same length before and after. The water is 8 units in the combined ratio and 8 parts in the new ratio, so a part is a unit, and the pine oil bar shows how many units the 180 ml make.

All steps
  1. (a) The water is 8 in the first ratio and 4 in the second, so double the second ratio: 4 : 5 = 8 : 10. Pine oil : water : ethanol = 3 : 8 : 10.
  2. Before, the pine oil is 3 units and the water is 8 units. After, pine oil : water = 5 : 8, and the water is still the same 8 units, so the pine oil is now 5 units.
  3. The pine oil grows from 3 units to 5 units, so 2 units = 180 ml and 1 unit = 180 ÷ 2 = 90 ml.
  4. (b)(i) The water is 8 × 90 = 720 ml.
  5. (ii) The pine oil ends at 5 × 90 = 450 ml and the ethanol at 10 × 90 + 420 = 1320 ml. The highest common factor of 450 and 1320 is 30, so 450 : 1320 = 15 : 44. Check: 450 : 720 = 5 : 8.

answer(a) 3 : 8 : 10; (b)(i) 720 ml; (ii) 15 : 44

techniqueRatios with a Constant Part · Combining Two Ratios

examsPSLE · SAT

Common pitfalls

  • Joining the ratios as 3 : 8 : 5. The water is 8 units in one ratio and 4 units in the other, so the second ratio must be doubled to 8 : 10 before the three liquids share one set of units.
  • Answering (ii) with 5 : 10 = 1 : 2, as though the ethanol were still 10 units. The 420 ml of ethanol is added as well, and it is not a whole number of 90 ml units, so find both amounts in ml: 450 : 1320 = 15 : 44.
22

A Pancake Recipe's Milk Table: The Constant, Then the Missing Entries

heuristicDivide Each Amount of Milk by Its Pancakes, Then Use the Constant Both Ways

Ana's pancake recipe uses milk in direct proportion to the number of pancakes. Her table shows 200 ml of milk for 8 pancakes, 300 ml for 12 pancakes and 500 ml for 20 pancakes. (a) Find the constant of proportionality, in milliliters of milk for each pancake. (b) Two entries are missing from her table: the milk for 30 pancakes, and the number of pancakes that 1 liter of milk makes. Find both.

8 pancakes200 ml12 pancakes300 ml20 pancakes500 ml
Share each amount of milk among its pancakes: 200 ÷ 8, 300 ÷ 12 and 500 ÷ 20 all give 25.
Share each amount of milk among its pancakes: 200 ÷ 8 = 25, 300 ÷ 12 = 25 and 500 ÷ 20 = 25.
step 1 of 5

Find the milk for one pancake from each entry in the table. If it is the same every time, that amount is the constant of proportionality, and each missing entry follows from it.

All steps
  1. Share each amount of milk among its pancakes: 200 ÷ 8 = 25, 300 ÷ 12 = 25 and 500 ÷ 20 = 25.
  2. (a) Every entry gives 25 ml for one pancake. The constant of proportionality is 25 ml of milk for each pancake.
  3. Thirty pancakes need 30 × 25 = 750 ml of milk.
  4. 1 liter is 1000 ml. Each pancake uses 25 ml, so 1000 ml makes 1000 ÷ 25 = 40 pancakes.
  5. (b) The missing entries are 750 ml for 30 pancakes, and 40 pancakes for 1 liter. Check: 750 ÷ 30 = 25 and 1000 ÷ 40 = 25, the same constant.

answer(a) 25 ml of milk for each pancake; (b) 750 ml for 30 pancakes, and 1 liter makes 40 pancakes

techniqueThe Constant of Proportionality · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the pancakes by the milk: 8 ÷ 200 = 0.04. That is the number of pancakes for each milliliter of milk, not the milk for each pancake. To find milliliters for each pancake, divide the milk by the pancakes.
  • Adding instead of multiplying: 30 pancakes are 10 more than 20, so adding 10 ml to the 500 ml gives 510 ml. Each pancake needs 25 ml, so 10 more pancakes need 10 × 25 = 250 ml more: 500 + 250 = 750 ml.
23

A School Printer: The Pages for More Minutes, Then the Minutes for a Bigger Job

heuristicFind the Pages Printed in One Minute, Then Scale

The school printer prints 45 pages in 3 minutes, always at the same rate. (a) How many pages does it print in 7 minutes? (b) A teacher needs 30 copies of a booklet that has 8 pages. How many minutes does the printer take to print them?

3 minutes15151545 pages1 minute1515 pages
Share the 45 pages among the 3 minutes: 15 pages in one minute.
The 45 pages are shared equally among the 3 minutes: 45 ÷ 3 = 15 pages in one minute.
step 1 of 5

Seven minutes is not a whole number of lots of 3 minutes, so go through one minute. Find the pages printed in one minute, then multiply up, or count the minutes that hold the pages needed.

All steps
  1. The 45 pages are shared equally among the 3 minutes: 45 ÷ 3 = 15 pages in one minute.
  2. (a) In 7 minutes the printer prints 7 × 15 = 105 pages. Check: 3 × 15 = 45 pages in 3 minutes.
  3. The booklets have 30 × 8 = 240 pages in all.
  4. Each minute prints 15 pages, so 240 pages take 240 ÷ 15 = 16 minutes.
  5. (b) The printer takes 16 minutes. Check: 16 × 15 = 240 pages.

answer(a) 105 pages; (b) 16 minutes

techniqueProportion by Unit Value · Direct Proportion

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding the extra minutes to the pages: 7 minutes is 4 minutes more than 3, so 45 + 4 = 49 pages. Each extra minute prints 15 pages, not 1, so the 4 extra minutes print 4 × 15 = 60 more pages: 45 + 60 = 105.
  • Dividing the wrong way round: 3 ÷ 45 is the part of a minute that one page takes, not the pages printed in one minute. The pages in one minute are 45 ÷ 3 = 15.
24

A Fish Tank Where Goldfish Are Sold and the Koi Stay

heuristicRewrite Both Ratios So the Unchanged Koi Share Matches

A pet shop has a tank of goldfish and koi. At first, the ratio of goldfish to koi is 5 : 3. The shop sells some of the goldfish and none of the koi, and then the ratio of goldfish to koi is 3 : 4. (a) What fraction of the goldfish were sold? (b) There are 24 koi in the tank. How many goldfish were sold?

Beforegoldfish: 5 partskoi: 3 partsAftergoldfish: 3 partskoi: 4 parts
No koi are sold, so the koi bar is the same length before and after: 3 parts before, 4 parts after.
No koi are sold, so the koi share is the same before and after. Before, goldfish : koi = 5 : 3; after, goldfish : koi = 3 : 4.
step 1 of 5

The koi share does not change, so rewrite both ratios with the same number of units for the koi. Then one unit is the same number of fish before and after, and the goldfish units can be compared.

All steps
  1. No koi are sold, so the koi share is the same before and after. Before, goldfish : koi = 5 : 3; after, goldfish : koi = 3 : 4.
  2. Rewrite both ratios so that the koi shares match. The least common multiple of 3 and 4 is 12: 5 : 3 = 20 : 12 and 3 : 4 = 9 : 12.
  3. The goldfish go from 20 units to 9 units, so 20 − 9 = 11 units were sold. (a) 11 of the 20 units were sold, which is 1120 of the goldfish.
  4. The koi are 12 units, and there are 24 koi, so 1 unit is 24 ÷ 12 = 2 fish.
  5. (b) 11 units were sold: 11 × 2 = 22 goldfish. Check: before, there were 40 goldfish and 40 : 24 = 5 : 3; after, there are 18 and 18 : 24 = 3 : 4.

answer(a) 1120 of the goldfish; (b) 22 goldfish

techniqueRewriting a Ratio to Match a Share · Ratios with a Constant Part

examsPSLE · O-Level

Common pitfalls

  • Taking 5 − 3 = 2 units as the goldfish sold. The koi are 3 units before and 4 units after, so a unit is a different number of fish in the two ratios. They must be rewritten so that the koi shares match before the goldfish shares are compared.
  • Giving 1112 as the fraction of the goldfish sold. The 12 units are the koi. The goldfish at first were 20 units, so the fraction sold is 1120.
25

A Class Flag Copied with Every Length Three Times as Long, and the Paint It Needs

heuristicMultiply Each Length by 3, So the Area Is Multiplied by 3 × 3

A class paints a small flag 30 cm long and 20 cm wide. For the school hall, it makes a big flag of the same shape, with every length 3 times as long. (a) What is the area of each flag? How many times the area of the small flag is the area of the big flag? (b) The small flag used 40 ml of paint. How much paint does the big flag need, painted in the same way?

30 cm20 cmsmall: 3 × 2 = 6 squares, 600 sq cm
The small flag is 3 squares of 10 cm long and 2 squares wide: 6 squares, 600 cm2.
The small flag is 3 squares long and 2 squares wide: 3 × 2 = 6 squares. Its area is 30 × 20 = 600 cm2.
step 1 of 5

Cut both flags into squares 10 cm by 10 cm. The big flag has every length 3 times as long, so it has 3 times as many squares in each row, and 3 times as many rows.

All steps
  1. The small flag is 3 squares long and 2 squares wide: 3 × 2 = 6 squares. Its area is 30 × 20 = 600 cm2.
  2. The big flag is 30 × 3 = 90 cm long and 20 × 3 = 60 cm wide, so it is 9 squares long and 6 squares wide: 9 × 6 = 54 squares. Its area is 90 × 60 = 5400 cm2.
  3. (a) 54 ÷ 6 = 9, so the big flag's area is 9 times the small flag's. The length was multiplied by 3 and the width by 3, and 3 × 3 = 9.
  4. The paint covers the area, so the big flag needs 9 times the paint of the small flag.
  5. (b) The big flag needs 9 × 40 = 360 ml of paint. Check: 40 ml covered 600 cm2, which is 600 ÷ 40 = 15 cm2 for each milliliter, and 5400 ÷ 15 = 360 ml.

answer(a) 600 cm2 and 5400 cm2, so the big flag's area is 9 times the small flag's; (b) 360 ml

techniqueScaling Area

examsO-Level · SAT

Common pitfalls

  • Multiplying the paint by 3 only: 3 × 40 = 120 ml. The big flag is 3 times as long and also 3 times as wide, so its area, and the paint it needs, is 3 × 3 = 9 times as much.
  • Adding the two factors: 3 + 3 = 6 times the area. The length factor multiplies the width factor, so the area is multiplied by 3 × 3 = 9, not by 6.