Fractions · applications

Applications: Fractions

21 problems · Grade 5

PSLE

01

Two Children's Savings, Where Equal Fractions of Them Match

heuristicMake the Numerators the Same, Then Compare the Whole Bars

23 of Lucas's savings is equal to 47 of Nathan's savings. Nathan has $48 more than Lucas. (a) How much money does Lucas have? (b) How much money do they have altogether?

Lucas2 of 3Nathan4 of 7
Two shaded parts of three, against four of seven. The shaded amounts are equal, so the parts are not the same size.
Draw Lucas's savings as 3 parts with 2 shaded, and Nathan's as 7 parts with 4 shaded. The shaded amounts are equal, but Lucas's 2 parts match Nathan's 4 parts, so the parts are not the same size yet.
step 1 of 5

The shaded parts of the two bars are equal amounts. Make the numerators the same, so that both shaded parts have the same number of units. Then every unit in both bars is the same size, and the two whole bars can be compared.

All steps
  1. Draw Lucas's savings as 3 parts with 2 shaded, and Nathan's as 7 parts with 4 shaded. The shaded amounts are equal, but Lucas's 2 parts match Nathan's 4 parts, so the parts are not the same size yet.
  2. Make the numerators the same: 23 = 46. Cut each of Lucas's thirds in half. Now 4 of Lucas's 6 units equal 4 of Nathan's 7 units, so all the units are the same size.
  3. Lucas has 6 units and Nathan has 7 units. Nathan has 7 − 6 = 1 unit more, and that is $48. So 1 unit is $48.
  4. (a) Lucas has 6 × $48 = $288.
  5. (b) Together they have 6 + 7 = 13 units, which is 13 × $48 = $624. Check: Nathan has 7 × $48 = $336, and 23 of $288 and 47 of $336 are both $192.

answer(a) $288; (b) $624

techniqueEquivalent Fractions · Same Top Number

examsPSLE

Common pitfalls

  • Making the denominators the same instead of the numerators. The amounts that are equal are the shaded parts, so it is the shaded parts that must have the same number of units: make both numerators 4, and then every unit is the same size.
  • Taking 4 − 2 = 2 units as the $48. The $48 is the difference between the whole savings, and once 23 is written as 46 the whole savings are 6 units and 7 units, so the difference is 7 − 6 = 1 unit.
02

Money Spent on Groceries, Then on Transport from What Was Left

heuristicCut What Is Left into the Pieces the Next Fraction Needs

Mrs. Tan had some money. She spent 14 of it on groceries and 25 of the remaining money on transport. She had $108 left. (a) How much money did Mrs. Tan have at first? (b) How much did she spend on transport?

Moneygroceries3 parts left
Her money as 4 equal parts. Groceries take 1 part, and 3 parts are left.
Draw her money as a bar of 4 equal parts. Groceries take 1 part, so 3 parts are left.
step 1 of 6

Draw her money as one bar cut into quarters. The money left after groceries must then be cut into fifths, so cut every quarter into 5 equal units. Then each fraction is a number of units.

All steps
  1. Draw her money as a bar of 4 equal parts. Groceries take 1 part, so 3 parts are left.
  2. Transport is 25 of the 3 parts left, so they must split into fifths. Cut every part into 5 units: the bar has 4 × 5 = 20 units, and 3 × 5 = 15 units are left after groceries.
  3. Transport takes 25 of 15 units: 15 ÷ 5 = 3 and 2 × 3 = 6 units. So 15 − 6 = 9 units are left.
  4. The 9 units left are $108, so 1 unit is $108 ÷ 9 = $12.
  5. (a) She had 20 × $12 = $240 at first.
  6. (b) She spent 6 × $12 = $72 on transport. Check: groceries cost $240 ÷ 4 = $60, and $240 − $60 − $72 = $108.

answer(a) $240; (b) $72

techniqueA Fraction of an Amount · Fractions of a Changing Whole

examsPSLE · GCSE Higher

Common pitfalls

  • Adding 14 + 25 = 1320 as the fraction she spent. The 25 is a fraction of the money left after groceries, not of all her money, so transport is only 25 of 34, which is 620.
  • Taking $108 as 35 of all her money. The $108 is 35 of the money left after groceries, and that is 35 of 34, which is 920 of all her money.
03

Tarts Sold as a Fraction and a Few More, Morning and Afternoon

heuristicWork Backwards from the End / Add Back the Extra Tarts Before Undoing the Fraction

A baker made some tarts. In the morning he sold 13 of them and 6 more. In the afternoon he sold 12 of the tarts that were left and 4 more. He then had 20 tarts left. (a) How many tarts were left after the morning? (b) How many tarts did he make altogether?

After morninghalf sold420 left
Start from the end. What was left after the morning is 2 halves: one half and 4 more tarts were sold, and 20 were left.
Draw the tarts left after the morning as a bar of 2 equal halves. In the afternoon he sold one half and 4 more tarts, and 20 tarts were left.
step 1 of 6

Start from the end and work backwards. Draw the tarts left after the morning, then all the tarts, and add back each fixed number of tarts before undoing the fraction.

All steps
  1. Draw the tarts left after the morning as a bar of 2 equal halves. In the afternoon he sold one half and 4 more tarts, and 20 tarts were left.
  2. The other half is the 20 tarts left and the 4 extra tarts sold, so one half is 20 + 4 = 24 tarts.
  3. (a) The bar is 2 halves, so 2 × 24 = 48 tarts were left after the morning.
  4. Now draw all the tarts he made as a bar of 3 equal units. In the morning he sold 1 unit and 6 more tarts, so the 48 tarts left are 2 units less 6 tarts.
  5. Add the 6 tarts back: 2 units are 48 + 6 = 54 tarts, so 1 unit is 54 ÷ 2 = 27 tarts.
  6. (b) He made 3 × 27 = 81 tarts. Check: in the morning he sold 27 + 6 = 33, which leaves 81 − 33 = 48, and in the afternoon he sold 24 + 4 = 28, which leaves 48 − 28 = 20.

answer(a) 48 tarts; (b) 81 tarts

techniqueFractions of a Changing Whole · Multiplying and Dividing Undo Each Other

Common pitfalls

  • Taking 20 − 4 = 16 as half of the tarts left after the morning. The 4 tarts were sold, so working backwards they are added back: one half is 20 + 4 = 24.
  • Taking 13 of the 48 tarts as the morning sale. The 13 is a fraction of all the tarts he made, not of the 48 left: the 48 tarts are 23 of them less 6.
04

Children Come into a Library While the Adults Stay

heuristicThe Adults Do Not Change / Make the Adults the Same Number of Units

At a library, 38 of the visitors were children and the rest were adults. Then 45 more children came in, and no adults came or left. After that, 35 of the visitors were children. (a) How many adults were in the library? (b) How many visitors were in the library in the end?

At firstchildren 3adults 53 : 5
At first 38 are children and 58 adults: children : adults = 3 : 5. Gold is the children.
At first 38 of the visitors are children, so the other 58 are adults: children : adults = 3 : 5.
step 1 of 6

Turn each fraction into a ratio of children to adults. The adults do not change, so make the adults the same number of units in both ratios. Then the children's units can be compared.

All steps
  1. At first 38 of the visitors are children, so the other 58 are adults: children : adults = 3 : 5.
  2. In the end 35 of the visitors are children and 25 are adults: children : adults = 3 : 2.
  3. The adults are the same people both times, so make their units the same. The LCM of 5 and 2 is 10: 3 : 5 times 2 is 6 : 10, and 3 : 2 times 5 is 15 : 10.
  4. The children went from 6 units to 15 units, which is 15 − 6 = 9 units more. That is the 45 children who came in, so 1 unit is 45 ÷ 9 = 5.
  5. (a) There were 10 × 5 = 50 adults.
  6. (b) In the end there were 15 + 10 = 25 units, which is 25 × 5 = 125 visitors. Check: at first there were 30 children among 80 visitors, and 3080 = 38; in the end there were 75 among 125, and 75125 = 35.

answer(a) 50 adults; (b) 125 visitors

techniqueRatios with a Constant Part

examsPSLE · SAT

Common pitfalls

  • Comparing the children's 3 units at first with their 3 units in the end, as if nothing had changed. The two 3s belong to different ratios, 3 : 5 and 3 : 2, and their units are not the same size until the adults are made the same number of units.
  • Taking 35 − 38 of the visitors as the 45 children. The total grew by 45 when the children came in, so the two fractions are fractions of different totals; the adults are the part that stays the same.
05

Oil Poured from One Container into the Other

heuristicThe Total Does Not Change / Draw the Total in Twelfths

Container A and Container B hold 720 ml of oil altogether. At first, Container A holds 512 of the oil. Then 80 ml of oil is poured from Container B into Container A. (a) How much oil was in Container B at first? (b) What fraction of the oil is in Container B after the oil is poured?

All the oil720 ml1 unit = 60 ml
All the oil as 12 equal units, because A's fraction is in twelfths: 720 ÷ 12 = 60 ml in each.
Draw all the oil as a bar of 12 equal units. The 12 units are 720 ml, so 1 unit is 720 ÷ 12 = 60 ml.
step 1 of 5

Draw the 720 ml of oil as one bar of 12 equal units, because A's fraction is in twelfths. Pouring moves oil from B to A, but it does not change the length of the bar.

All steps
  1. Draw all the oil as a bar of 12 equal units. The 12 units are 720 ml, so 1 unit is 720 ÷ 12 = 60 ml.
  2. At first A holds 5 units, which is 5 × 60 = 300 ml.
  3. (a) B holds the other 12 − 5 = 7 units, which is 7 × 60 = 420 ml.
  4. Move 80 ml from B's part of the bar to A's. The bar stays the same length, because the oil only moved: the total is still 720 ml, and B now holds 420 − 80 = 340 ml.
  5. (b) B holds 340720 of the oil, and dividing both numbers by 20 gives 1736. Check: A now holds 300 + 80 = 380 ml, and 380 + 340 = 720 ml.

answer(a) 420 ml; (b) 1736 of the oil

techniqueRatios with a Constant Total

Common pitfalls

  • Giving 300 ml as the oil in Container B. 512 of the oil is in Container A; B holds the other 712, which is 420 ml.
  • Writing B's share as 340640, as if the 80 ml had left the containers. The oil was poured from B into A, so the total is still 720 ml, and B holds 340720 of it.
06

Two Children Spend the Same Amount, and Their Fraction Changes

heuristicThe Difference Does Not Change / Compare the Units of the Difference

Ethan had 25 as much pocket money as Fiona. Then each of them spent $18 on stationery. After that, Ethan had 14 as much money as Fiona. (a) How much money did Fiona have at first? (b) How much money did Fiona have left?

Ethan2 unitsFiona5 unitsdifference: 3 units
Ethan : Fiona = 2 : 5, and the difference is 5 − 2 = 3 units.
At first Ethan had 25 as much as Fiona: Ethan : Fiona = 2 : 5, and the difference is 5 − 2 = 3 units.
step 1 of 6

Both of them spend the same amount, so the difference between their money does not change. Write both ratios and compare the units of the difference.

All steps
  1. At first Ethan had 25 as much as Fiona: Ethan : Fiona = 2 : 5, and the difference is 5 − 2 = 3 units.
  2. Afterward Ethan had 14 as much as Fiona: Ethan : Fiona = 1 : 4, and the difference is 4 − 1 = 3 units.
  3. They spent the same amount, so the difference in dollars did not change. It is 3 units both times, so a unit before is the same size as a unit after.
  4. Ethan went from 2 units to 1 unit, so he spent 1 unit. He spent $18, so 1 unit is $18.
  5. (a) Fiona had 5 × $18 = $90 at first.
  6. (b) Fiona had 4 × $18 = $72 left. Check: Ethan had 2 × $18 = $36 and was left with $36 − $18 = $18, which is 14 of $72.

answer(a) $90; (b) $72

techniqueRatios with a Constant Difference

Common pitfalls

  • Taking 25 as Ethan's share of all the money, so that Ethan : Fiona = 2 : 3. 'Ethan had 25 as much as Fiona' compares Ethan with Fiona, so Ethan is 2 units when Fiona is 5 units.
  • Comparing Ethan's 2 units with his 1 unit without checking the differences first. The units are the same size only because both ratios give a difference of 3 units; when the differences are not the same, scale the ratios until they are.
07

Laptops Sold at Three Stores, Each Compared with Store B

heuristicDraw the Store Both Fractions Are Of / Eight Units Hold Quarters and Eighths

Store A sold 14 fewer laptops than Store B. Store C sold 38 more laptops than Store B. Store C sold 180 more laptops than Store A. (a) How many laptops did Store B sell? (b) How many laptops did the three stores sell altogether?

Store B8 units
Both fractions are fractions of Store B, so Store B is drawn first: 8 units, which quarters and eighths both divide.
Draw Store B's sales as a bar of 8 equal units, because 8 is a multiple of both denominators, 4 and 8.
step 1 of 6

Both fractions compare a store with Store B, so draw Store B first. Quarters and eighths of it are both whole units when Store B is 8 units.

All steps
  1. Draw Store B's sales as a bar of 8 equal units, because 8 is a multiple of both denominators, 4 and 8.
  2. Store A sold 14 fewer than Store B. 14 of 8 units is 2 units, so Store A sold 8 − 2 = 6 units.
  3. Store C sold 38 more than Store B. 38 of 8 units is 3 units, so Store C sold 8 + 3 = 11 units.
  4. Store C's bar is 11 − 6 = 5 units longer than Store A's, and that is 180 laptops, so 1 unit is 180 ÷ 5 = 36 laptops.
  5. (a) Store B sold 8 × 36 = 288 laptops.
  6. (b) The three stores sold 6 + 8 + 11 = 25 units, which is 25 × 36 = 900 laptops. Check: Store A sold 6 × 36 = 216 and Store C sold 11 × 36 = 396, and 396 − 216 = 180.

answer(a) 288 laptops; (b) 900 laptops

techniqueA Fraction of an Amount · Ratio, Fraction, Percent

examsPSLE · GCSE Higher

Common pitfalls

  • Drawing Store A as 14 of Store B, which is 2 units. '14 fewer than Store B' means Store B's amount less a quarter of it: 1 − 14 = 34 of Store B, which is 6 units.
  • Taking 38 of Store A's sales as Store C's extra. Both fractions are fractions of Store B, so Store C's extra is 38 of Store B's 8 units, which is 3 units.
08

Red and Blue Pens, Both Sold in Different Numbers

heuristicUnits Before and Parts After / Make the Parts the Same in Both Equations

In a shop, the number of red pens was 34 of the number of blue pens. Then 30 red pens and 10 blue pens were sold. After that, the number of red pens was 23 of the number of blue pens. (a) How many blue pens were there at first? (b) How many red pens were left?

Red3uBlue4u
At first red is 3 units and blue is 4 units.
At first red is 34 of blue: red is 3 units and blue is 4 units. Write u for 1 unit.
step 1 of 6

Both colours were sold, by different amounts, so nothing stays the same. Use units for the pens at first and parts for the pens afterward, write one equation for each colour, and make the parts match.

All steps
  1. At first red is 34 of blue: red is 3 units and blue is 4 units. Write u for 1 unit.
  2. Afterward red is 23 of blue: red is 2 parts and blue is 3 parts. Different numbers of red and blue pens were sold, so a part is not the same size as a unit. Write p for 1 part.
  3. Red lost 30 pens and blue lost 10 pens, so 3u − 30 = 2p and 4u − 10 = 3p.
  4. Make the parts the same: multiply the red equation by 3 and the blue equation by 2. Then 9u − 90 = 6p and 8u − 20 = 6p, so 9u − 90 = 8u − 20.
  5. (a) Take 8u from both sides and add 90: u = 70. There were 4 × 70 = 280 blue pens at first.
  6. (b) There were 3 × 70 = 210 red pens at first, so 210 − 30 = 180 were left. Check: 280 − 10 = 270 blue pens were left, and 23 of 270 is 180.

answer(a) 280 blue pens; (b) 180 red pens

techniqueBefore-and-After Ratio Problems · Simultaneous by Elimination

examsSAT · GCSE Higher

Common pitfalls

  • Keeping the difference between the blue and red pens at 1 unit, as if it did not change. 30 red pens and only 10 blue pens were sold, so the difference changed, and nothing in the question stays the same.
  • Writing 3u − 30 = 2u, as if a part were a unit. The parts come from a different ratio after a different sale, so they need their own letter until the two equations show how u and p are related.
09

A Piggy Bank of 50-Cent and 20-Cent Coins

heuristicGroup the Coins in Fives / Divide the Total by the Value of One Group

In a piggy bank, 35 of the coins are 50-cent coins and the rest are 20-cent coins. The coins are worth $57 altogether. (a) How many coins are in the piggy bank? (b) How much are the 20-cent coins worth altogether?

50¢50¢50¢20¢20¢1 group = 5 coins
35 of the coins are 50-cent coins: 3 of every 5. One group is 3 fifty-cent coins and 2 twenty-cent coins.
35 of the coins are 50-cent coins, so 3 of every 5 coins are 50-cent coins and the other 2 are 20-cent coins. Take these 5 coins as one group.
step 1 of 5

3 of every 5 coins are 50-cent coins. Put the coins into groups of 5, each with 3 fifty-cent coins and 2 twenty-cent coins, and find how many groups make $57.

All steps
  1. 35 of the coins are 50-cent coins, so 3 of every 5 coins are 50-cent coins and the other 2 are 20-cent coins. Take these 5 coins as one group.
  2. One group is worth 3 × 50 + 2 × 20 = 150 + 40 = 190 cents, which is $1.90.
  3. The coins are worth $57, which is 5700 cents, so there are 5700 ÷ 190 = 30 groups.
  4. (a) Each group has 5 coins, so there are 30 × 5 = 150 coins.
  5. (b) There are 30 × 2 = 60 twenty-cent coins, worth 60 × 20 = 1200 cents, which is $12. Check: the 30 × 3 = 90 fifty-cent coins are worth $45, and $45 + $12 = $57.

answer(a) 150 coins; (b) $12

techniqueA Fraction of a Group · The Unitary Method

examsPSLE · GCSE Higher

Common pitfalls

  • Dividing $57 by 5, as if every coin were worth the same. The coins in a group have different values, so divide by the value of one whole group, $1.90, to count the groups.
  • Mixing dollars and cents, as in 190k = 57. Write every value in the same unit: 5700 cents and 190 cents a group, or $57 and $1.90 a group.
10

A Rectangle and a Triangle That Overlap

heuristicCall the Overlap 1 Unit / Count the Overlap Once in the Whole Figure

A rectangle and a triangle overlap, as shown, and the overlap is shaded. The unshaded part of the rectangle is 45 of the rectangle's area, and the unshaded part of the triangle is 67 of the triangle's area. The whole figure has an area of 176 cm2. (a) What is the area of the shaded overlap? (b) What is the area of the triangle?

rectangletriangleRectangle14 unshaded5 units
The unshaded part of the rectangle is 45, so the shaded overlap is 15: call it 1 unit, and the rectangle is 5 units.
The unshaded part of the rectangle is 45 of it, so the shaded overlap is the other 15. Call the overlap 1 unit: the rectangle is 5 units, and 4 of them are unshaded.
step 1 of 5

The overlap is part of both shapes. Call it 1 unit, and write each shape as a number of these units.

All steps
  1. The unshaded part of the rectangle is 45 of it, so the shaded overlap is the other 15. Call the overlap 1 unit: the rectangle is 5 units, and 4 of them are unshaded.
  2. The unshaded part of the triangle is 67 of it, so the overlap is 17 of the triangle. The triangle is 7 units, and 6 of them are unshaded.
  3. The whole figure is the unshaded part of the rectangle, the overlap and the unshaded part of the triangle, with the overlap counted once: 4 + 1 + 6 = 11 units.
  4. (a) The 11 units are 176 cm2, so the overlap, 1 unit, is 176 ÷ 11 = 16 cm2.
  5. (b) The triangle is 7 units: 7 × 16 = 112 cm2. Check: the rectangle is 5 × 16 = 80 cm2, and 80 + 112 − 16 = 176 cm2.

answer(a) 16 cm2; (b) 112 cm2

techniqueA Fraction of an Amount · Area of a Composite Figure

examsPSLE · GCSE Higher

Common pitfalls

  • Taking the whole figure as 5 + 7 = 12 units. The overlap belongs to both shapes, so adding them counts it twice; the whole figure is 12 − 1 = 11 units.
  • Taking the overlap as 45 of the rectangle. The 45 is the unshaded part; the shaded overlap is what is left, 1 − 45 = 15 of the rectangle.
11

Two Children Spend in Two Stages, and Only the Totals Are Known

heuristicFind Each Child's Fraction Spent / Compare the Bar Spent with the Bar Left

Chloe and Daniel each had some money. Chloe spent 13 of her money on clothes and 14 of the remaining money on makeup. Daniel spent 15 of his money on games and 12 of the remaining money on books. Together they spent $300, and together they had $260 left. (a) How much money did Chloe have at first? (b) How much money did Daniel have at first?

Chloeunit spentunit left2 units
Chloe spent 13 and then 14 of the 23 left: 13 + 16 = 12. Her money is 2 units, 1 spent and 1 left.
Chloe: the clothes take 13 of her money, and the makeup takes 14 of the 23 left, which is 16. She spent 13 + 16 = 12, so draw her money as 2 units: 1 unit spent and 1 unit left.
step 1 of 6

Work out what fraction of their own money each child spent. They started with different amounts, so Chloe's bar is cut into units and Daniel's into parts of a different size. Then compare the bar of money spent with the bar of money left.

All steps
  1. Chloe: the clothes take 13 of her money, and the makeup takes 14 of the 23 left, which is 16. She spent 13 + 16 = 12, so draw her money as 2 units: 1 unit spent and 1 unit left.
  2. Daniel: the games take 15 of his money, and the books take 12 of the 45 left, which is 25. He spent 35, so draw his money as 5 parts: 3 parts spent and 2 parts left.
  3. Put what they spent in one bar, 1 unit and 3 parts, which is $300. Put what they kept in another bar, 1 unit and 2 parts, which is $260.
  4. The bar spent is the bar kept and 1 more part, so 1 part is $300 − $260 = $40.
  5. (a) The bar kept is 1 unit and 2 parts, $260, so 1 unit is $260 − 2 × $40 = $180. Chloe had 2 units: 2 × $180 = $360.
  6. (b) Daniel had 5 parts: 5 × $40 = $200. Check: Chloe spent $120 + $60 = $180 and Daniel spent $40 + $80 = $120, which is $300; they kept $180 + $80 = $260.

answer(a) $360; (b) $200

techniqueFractions of a Changing Whole · Simultaneous by Elimination

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Drawing Chloe's and Daniel's bars with units of the same size. They started with different amounts, so a unit of Chloe's money and a part of Daniel's are different sizes and need different letters, u and p.
  • Taking the makeup as 14 of all Chloe's money, so that she spent 13 + 14 = 712. The makeup is 14 of the money left after the clothes, 14 of 23, which is 16.
12

Game Cards Given Away as a Fraction and a Few More

heuristicThe Total Does Not Change / Share the Same Units Before and After

Afiq had 35 as many game cards as Bala. Bala gave 14 of his cards and 14 more cards to Afiq. Then Afiq had 3 times as many cards as Bala. (a) How many cards did Afiq have at first? (b) How many cards did Bala have left?

At firstAfiq 12 unitsBala 20 units32 units
Afiq : Bala = 3 : 5, times 4 so that a quarter of Bala's cards is whole: 12 units and 20 units, 32 in all.
At first Afiq : Bala = 3 : 5. Bala gives away 14 of his cards, so his share should be a multiple of 4: multiply by 4 to make Afiq 12 units and Bala 20 units, 32 units in all.
step 1 of 6

Cards only move from Bala to Afiq, so the total does not change. Draw the total as the same number of units before and after, and compare Bala's cards.

All steps
  1. At first Afiq : Bala = 3 : 5. Bala gives away 14 of his cards, so his share should be a multiple of 4: multiply by 4 to make Afiq 12 units and Bala 20 units, 32 units in all.
  2. Afterward Afiq : Bala = 3 : 1, which is 4 parts. The total is still 32 units, so each part is 32 ÷ 4 = 8 units: Afiq has 24 units and Bala has 8 units.
  3. Bala went from 20 units to 8 units, so he gave away 20 − 8 = 12 units.
  4. He gave 14 of his cards, which is 14 of 20 units, 5 units, and 14 more cards. So 12 units are 5 units and 14 cards, and the other 12 − 5 = 7 units are the 14 cards.
  5. (a) 1 unit is 14 ÷ 7 = 2 cards, so Afiq had 12 × 2 = 24 cards at first.
  6. (b) Bala had 8 × 2 = 16 cards left. Check: Bala had 40 cards and gave away 10 + 14 = 24, which leaves 16, and Afiq then had 24 + 24 = 48, which is 3 × 16.

answer(a) 24 cards; (b) 16 cards

techniqueRatios After a Transfer · Ratios with a Constant Total

Common pitfalls

  • Giving Afiq the 14 cards but not taking them from Bala. Every card Afiq gets comes from Bala, so Bala has 5u + 14 fewer cards, not just 5u fewer.
  • Drawing the ratio 3 : 1 afterward as 3 units and 1 unit of the size used before. The total of 32 units does not change, so the 4 parts of 3 : 1 share those 32 units, and each part is 8 units.
13

Two Taps Filling Two Tanks of Different Sizes

heuristicDraw Both Tanks in Blocks of One Size / Find How Long Each Tap Takes for One Block

Tank A and Tank B have capacities in the ratio 3 : 4. Tap X fills 12 of Tank A in 6 minutes. Tap Y fills 23 of Tank A in 10 minutes. Both tanks start empty, and at the same moment Tap X is turned on to fill Tank A and Tap Y is turned on to fill Tank B. (a) How many minutes does Tap X take to fill Tank A? (b) What fraction of Tank B is filled when Tank A is full?

Tank A3 blocksTap XTank B4 blocksTap Y
The capacities are in the ratio 3 : 4: 3 blocks against 4 blocks of the same size.
Draw Tank A as 3 equal blocks and Tank B as 4 blocks of the same size, because their capacities are in the ratio 3 : 4.
step 1 of 5

Draw the tanks with blocks of one size, 3 blocks for Tank A and 4 blocks for Tank B, and find how long each tap takes to fill its tank.

All steps
  1. Draw Tank A as 3 equal blocks and Tank B as 4 blocks of the same size, because their capacities are in the ratio 3 : 4.
  2. (a) Tap X fills half of Tank A in 6 minutes, so it fills the whole of Tank A in 6 × 2 = 12 minutes.
  3. Tap Y fills 23 of Tank A, which is 2 blocks, in 10 minutes, so it fills 1 block every 10 ÷ 2 = 5 minutes. The 4 blocks of Tank B take it 4 × 5 = 20 minutes.
  4. The taps start together, and Tank A is full after 12 minutes. By then Tap Y has run for 12 of the 20 minutes that Tank B needs.
  5. (b) Tank B is 1220 = 35 full. Check: in 12 minutes Tap Y fills 12 ÷ 5 = 225 blocks, and 225 blocks out of 4 is 125 ÷ 4 = 35 of Tank B.

answer(a) 12 minutes; (b) 35 of Tank B

techniqueRates · A Fraction of an Amount

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Using 6 minutes as the time Tap X takes to fill Tank A. In 6 minutes it fills only half of Tank A, so the whole tank takes 6 × 2 = 12 minutes, and at 6 minutes Tank B is only 620 = 310 full.
  • Reading Tap Y's 23 as two thirds of Tank B, which gives 45. Tap Y's rate is measured on Tank A: 23 of Tank A is 2 blocks, and Tank B has 4 blocks, so Tap Y takes 20 minutes, not 15, to fill Tank B.
14

Three Overlapping Shapes, Both Overlaps Given as Fractions of the Middle One

heuristicDraw the Middle Shape in Twelfths / Build the Outer Shapes from the Overlaps

Three shapes overlap as shown: A overlaps B, B overlaps C, and A and C do not overlap. The overlap of A and B is 14 of A and 16 of B. The overlap of B and C is 15 of C and 14 of B. The part of B that overlaps neither A nor C has an area of 70 cm2. (a) What is the area of Shape C? (b) What is the area of the whole figure?

ABCShape B12 units
Both overlaps are fractions of B, and 12 is the LCM of 6 and 4, so B is drawn as 12 units.
Sixths and quarters of B must both be whole units, so draw B as 12 units, since 12 is the LCM of 6 and 4.
step 1 of 6

Both overlaps are given as fractions of B, so draw B first, with a number of units that both sixths and quarters divide. Then find each overlap in units, and build A and C from them.

All steps
  1. Sixths and quarters of B must both be whole units, so draw B as 12 units, since 12 is the LCM of 6 and 4.
  2. The overlap of A and B is 16 of 12 units, which is 2 units. The overlap of B and C is 14 of 12 units, which is 3 units.
  3. The part of B that overlaps neither shape is 12 − 2 − 3 = 7 units, and that is 70 cm2, so 1 unit is 70 ÷ 7 = 10 cm2.
  4. (a) The overlap of B and C, 3 units, is 15 of C, so C is 5 × 3 = 15 units, which is 15 × 10 = 150 cm2.
  5. The overlap of A and B, 2 units, is 14 of A, so A is 4 × 2 = 8 units.
  6. (b) Adding the three shapes counts each overlap twice, so the whole figure is 8 + 12 + 15 − 2 − 3 = 30 units, which is 30 × 10 = 300 cm2. Check: A, B and C are 80, 120 and 150 cm2, and 80 + 120 + 150 − 20 − 30 = 300.

answer(a) 150 cm2; (b) 300 cm2

techniqueEquivalent Fractions · A Common Denominator

examsPSLE

Common pitfalls

  • Taking the part of B that overlaps nothing as 1 − 16 = 56 of B. B overlaps C as well as A, so both overlaps come off: 1 − 212 − 312 = 712 of B.
  • Adding the three shapes, 80 + 120 + 150 = 350 cm2, for the whole figure. Each overlap lies inside two shapes, so it has been counted twice, and the overlaps of 20 and 30 cm2 are each taken away once.
15

Egg Tarts and Cream Puffs, Each Sold in Its Own Fraction

heuristicSplit the Whole First / Take Each Fraction of Its Own Kind of Pastry

A baker made 240 pastries. 38 of them were egg tarts and the rest were cream puffs. He sold 56 of the egg tarts at $2 each and 45 of the cream puffs at $3 each. (a) How many pastries did he sell? (b) How much money did he collect?

240 pastries30303030303030301 unit = 30
The 240 pastries as 8 equal units: 240 ÷ 8 = 30 in each.
Draw the 240 pastries as a bar of 8 equal units. One unit is 240 ÷ 8 = 30 pastries.
step 1 of 6

Cut the 240 pastries into eighths to find the egg tarts and the cream puffs. Then take each fraction sold of its own kind of pastry, not of all 240.

All steps
  1. Draw the 240 pastries as a bar of 8 equal units. One unit is 240 ÷ 8 = 30 pastries.
  2. The egg tarts are 3 units, 3 × 30 = 90, and the cream puffs are the other 5 units, 5 × 30 = 150.
  3. He sold 56 of the 90 egg tarts: 90 ÷ 6 = 15, and 5 × 15 = 75 egg tarts.
  4. He sold 45 of the 150 cream puffs: 150 ÷ 5 = 30, and 4 × 30 = 120 cream puffs.
  5. (a) He sold 75 + 120 = 195 pastries.
  6. (b) He collected 75 × $2 + 120 × $3 = $150 + $360 = $510. Check: he did not sell 90 − 75 = 15 egg tarts and 150 − 120 = 30 cream puffs, and 195 + 15 + 30 = 240.

answer(a) 195 pastries; (b) $510

techniqueA Fraction of an Amount · The Unitary Method

examsPSLE · GCSE Higher

Common pitfalls

  • Taking 56 of all 240 pastries as the egg tarts sold. The 56 is a fraction of the 90 egg tarts only, so he sold 56 × 90 = 75 of them.
  • Pricing the pastries left instead of the pastries sold, 15 × $2 + 30 × $3. Only the pastries sold bring in money: 75 egg tarts and 120 cream puffs.
16

Slices of Cake Eaten at a Party, as Whole Cakes and What They Cost

heuristicGroup the Slices into Wholes / Improper Fraction to Mixed Number

At a class party, every cake is cut into 8 equal slices. The children eat 43 slices, which is 438 cakes. (a) How many cakes do the children eat? Give the answer as a mixed number. (b) Each cake cost $16. What was the cost of the cake that the children ate?

Cakes 1-3cake 1cake 2cake 3Cakes 4-6cake 4cake 5cake 6
Six cakes of 8 equal slices each. One whole cake is 8 slices, 88.
Draw each cake as a bar cut into 8 equal slices. One whole cake is 8 slices, which is 88.
step 1 of 6

Draw the cakes as bars of 8 slices each and fill them 8 slices at a time. The full bars give the whole number and the slices in the last bar give the fraction. Then find the cost of the whole cakes and of the extra slices separately.

All steps
  1. Draw each cake as a bar cut into 8 equal slices. One whole cake is 8 slices, which is 88.
  2. Fill the bars 8 slices at a time. Five bars take 5 × 8 = 40 slices, and the other 43 − 40 = 3 slices go into a sixth bar.
  3. (a) Five bars are full and the sixth bar has 3 of its 8 slices filled, so the children eat 538 cakes.
  4. The 5 whole cakes cost 5 × $16 = $80.
  5. One slice is 18 of a cake, so it costs $16 ÷ 8 = $2, and the 3 slices cost 3 × $2 = $6.
  6. (b) The cake the children ate cost $80 + $6 = $86. Check: 43 slices at $2 each cost 43 × $2 = $86.

answer(a) 538 cakes; (b) $86

techniqueImproper Fractions · Mixed Numbers and Improper Fractions · A Fraction of an Amount

examsPSLE

Common pitfalls

  • Writing 438 as 438 or as 535. The whole number is how many times 8 goes into 43, which is 5 times, and the remainder is still counted in eighths: 3 slices out of 8 is 38.
  • Answering (b) with 5 × $16 = $80, the cost of the whole cakes only. The children also ate 38 of a sixth cake, and that part costs 38 of $16, which is $6 more.
17

A Ribbon Cut into Quarter-Metre Pieces, and the Length Not Used

heuristicCount Unit Pieces / Mixed Number to Improper Fraction and Back

Mrs. Diaz has a ribbon 534 m long. She cuts all of it into pieces that are each 14 m long. (a) How many pieces does she cut? (b) She uses 10 of the pieces to tie bows on gift boxes. What is the total length of the pieces that she does not use? Give the answer as a mixed number.

Ribbon1 m1 m1 m1 m1 m3/4 m5 and 3/4 m
The ribbon: 5 whole metres and 34 m. Every metre is marked into quarters, the length of one piece.
Draw the ribbon as 5 whole metres and 34 of a metre. Each piece is 14 m long, so mark every metre into 4 quarters.
step 1 of 6

Draw the ribbon as 5 whole metres and 34 of a metre, and mark every quarter of a metre. Counting the quarters turns the mixed number into an improper fraction; putting the unused quarters back into whole metres turns it into a mixed number again.

All steps
  1. Draw the ribbon as 5 whole metres and 34 of a metre. Each piece is 14 m long, so mark every metre into 4 quarters.
  2. Each whole metre gives 4 pieces: 5 × 4 = 20 pieces. The 34 m gives 3 more pieces.
  3. (a) She cuts 20 + 3 = 23 pieces. This shows that 534 = 234.
  4. She uses 10 pieces, so 23 − 10 = 13 pieces are not used. Together they measure 134 m.
  5. Put the quarters back into whole metres, 4 quarters to a metre: 13 = 3 × 4 + 1, so 13 quarters make 3 whole metres and 1 quarter of a metre.
  6. (b) The unused pieces measure 314 m. Check: the 10 pieces used measure 104 = 224 m, and 534 − 224 = 314.

answer(a) 23 pieces; (b) 314 m

techniqueMixed Numbers and Improper Fractions · Improper Fractions

examsPSLE · GCSE Higher

Common pitfalls

  • Writing 534 as 84 or 154 by adding or multiplying the wrong numbers. Each of the 5 whole metres is 4 quarters, so the whole metres give 5 × 4 = 20 quarters, and the 3 quarters are added to that to make 234.
  • Giving (b) as 13 or as 134. The 13 is a number of pieces, not a length, and the question asks for the length as a mixed number: 134 m is 314 m.
18

Pizzas Shared Equally at Two Picnic Tables, Compared

heuristicA Division Written as a Fraction / Common Denominator to Compare

At a school picnic, all the pizzas are the same size. At the first table, 3 pizzas are shared equally among 4 children. At the second table, 2 pizzas are shared equally among 3 children. (a) What fraction of a pizza does each child at the first table get? (b) Does each child at the first table or each child at the second table get more pizza, and how much more?

Table 14 children
Table 1: three pizzas, each cut into 4 equal parts, one part for each of the 4 children.
First table: cut each of the 3 pizzas into 4 equal parts, one part for each of the 4 children. Each part is 14 of a pizza.
step 1 of 5

Cut every pizza into as many equal parts as there are children, and give each child one part of every pizza. Then draw the two shares on bars of the same length and cut both into twelfths to compare them.

All steps
  1. First table: cut each of the 3 pizzas into 4 equal parts, one part for each of the 4 children. Each part is 14 of a pizza.
  2. (a) Each child takes one part from every pizza, so each child gets 3 quarters: 3 ÷ 4 = 34 of a pizza.
  3. Second table: cut each of the 2 pizzas into 3 equal parts. Each child takes one part from each pizza, so each child gets 2 thirds: 2 ÷ 3 = 23 of a pizza.
  4. Draw the two shares on bars one pizza long and cut both bars into twelfths: 34 = 912 and 23 = 812.
  5. (b) The first share is one twelfth longer, so each child at the first table gets 912 − 812 = 112 of a pizza more. Check: 4 × 34 = 3 pizzas and 3 × 23 = 2 pizzas.

answer(a) 34 of a pizza; (b) each child at the first table, by 112 of a pizza

techniqueFractions as Division · A Common Denominator

examsPSLE

Common pitfalls

  • Writing the share at the first table as 43, children over pizzas. The amount being shared, 3 pizzas, is divided by the number of children, 4, so it goes on top: 3 ÷ 4 = 34. With fewer pizzas than children, each share must be less than one pizza.
  • Deciding that each child at the second table gets more because there are fewer children there. There are also fewer pizzas; only the two shares, written over the same denominator, show which is larger.
19

Smartphones Collected at Three Stations, Two Linked by Equal Fractions

heuristicMake the Numerators Equal / Then Take Three Quarters of One Bar

At an electronics recycling drive, 38 of the smartphones collected by Station A was equal to 512 of the smartphones collected by Station B. Station C collected 34 as many smartphones as Station B. Station A collected 156 more smartphones than Station C. (a) Write 38 and 512 as equivalent fractions with the same numerator. (b) Find the ratio of the numbers of smartphones collected by Station A, Station B and Station C, in its simplest form. (c) How many smartphones did the three stations collect altogether?

Station A1/81/81/81/81/81/81/81/83/8Station B1/121/121/121/121/121/121/121/121/121/121/121/125/12
38 of A and 512 of B are the same number of smartphones. The numerators 3 and 5 have the common multiple 15.
38 of A and 512 of B are the same number of smartphones. To make the shaded parts match unit for unit, make the numerators equal. The common multiple of 3 and 5 is 15.
step 1 of 6

Equal numerators give equal-sized units. With 15 units of A equal to 15 units of B, the bars for A and B can be drawn with units of the same size, and Station C is three quarters of the bar for B.

All steps
  1. 38 of A and 512 of B are the same number of smartphones. To make the shaded parts match unit for unit, make the numerators equal. The common multiple of 3 and 5 is 15.
  2. (a) 38 = 1540 and 512 = 1536. So 15 units of A equal 15 units of B, the units are the same size, and A is 40 units while B is 36 units.
  3. Station C collected 34 as many as B. Cut the 36 units of B into 4 equal groups of 9 units. C is 3 of the groups: 27 units.
  4. (b) The ratio of A to B to C is 40 : 36 : 27. It is in its simplest form, because no number other than 1 divides 40, 36 and 27.
  5. A has 40 − 27 = 13 units more than C, and that is 156 smartphones, so 1 unit is 156 ÷ 13 = 12 smartphones.
  6. (c) The three stations collected 40 + 36 + 27 = 103 units, which is 103 × 12 = 1236 smartphones. Check: A collected 480 and B 432; 38 × 480 = 180 and 512 × 432 = 180, and C collected 324, which is 480 − 156.

answer(a) 38 = 1540 and 512 = 1536; (b) 40 : 36 : 27; (c) 1236 smartphones

techniqueA Fraction of an Amount · Equivalent Fractions · Same Top Number

examsPSLE · GCSE Higher

Common pitfalls

  • Making the denominators equal, 38 = 924 and 512 = 1024, and taking A : B as 9 : 10. The sentence says that two parts of equal size are the same number, so it is the numerators that are made equal: 15 of A's 40 units match 15 of B's 36 units.
  • Taking the ratio A : B straight from the denominators as 8 : 12. The denominators give the sizes of A and B only when the numerators are the same, and 3 and 5 are not, so the fractions are first written as 1540 and 1536.
20

Laps of a Quarter-Kilometre Track, Counted as a Mixed Number and Back

heuristicFour Quarters Make One Whole / Group the Laps in Fours, or Turn Each Whole into Four Laps

The track at a school is 14 km long all the way around. (a) Leo runs 11 laps of the track. How far does he run? Give the answer as a mixed number. (b) Ana runs 314 km on the same track. How many laps does she run, and how many more laps is that than Leo?

Leo11/4 km11 laps are 11 quarters of a km: 11/4 km
Each lap is 14 km, so Leo's 11 laps are 11 quarters of a kilometre: 114 km.
Each lap is 14 km, so Leo's 11 laps are 11 quarters of a kilometre: 114 km.
step 1 of 5

Draw each lap as one piece of a bar. 4 laps make 1 km, so group the laps in fours to find the whole kilometres. To go back, turn each whole kilometre into 4 laps.

All steps
  1. Each lap is 14 km, so Leo's 11 laps are 11 quarters of a kilometre: 114 km.
  2. 4 laps make 1 km. Group Leo's laps in fours: 2 groups use 2 × 4 = 8 laps, and 11 − 8 = 3 laps are left.
  3. (a) Leo runs 2 whole kilometres and 3 quarters of a kilometre: 114 = 234 km.
  4. Now go the other way for Ana. Each whole kilometre is 4 laps, so 3 km is 3 × 4 = 12 laps, and the 14 km is 1 lap more: 12 + 1 = 13 laps.
  5. (b) Ana runs 13 laps, which is 13 − 11 = 2 laps more than Leo. Check: 314 − 234 = 134 − 114 = 24 km, which is 2 laps.

answer(a) 234 km; (b) 13 laps, which is 2 more than Leo

techniqueMixed Numbers · Mixed Numbers and Improper Fractions

examsPSLE

Common pitfalls

  • Writing 114 km as 334 km. Three whole kilometres would take 3 × 4 = 12 laps, and Leo runs only 11, so he runs 2 whole kilometres and 3 laps more.
  • Counting 314 km as 3 + 1 = 4 laps. Each whole kilometre is 4 laps, so the 3 whole kilometres alone are 12 laps, and the quarter adds 1 more.
21

Blocks of Clay Shared Equally, Less Than One Block Each and Then More

heuristicA Division Written as a Fraction / Give Each Child One Piece of Every Block

Mr. Tan has 3 identical blocks of clay for a group of 4 children, and he shares them out equally. (a) What fraction of a block does each child get? (b) The next group also has 4 children, and there are 9 blocks of clay to share out equally. How much clay does each child get? Give the answer as a mixed number.

3 blocks1234123412344 childreneach block is cut into 4 equal pieces, one for each child
Fewer blocks than children, so every block is cut into 4 equal pieces, one for each child.
There are fewer blocks than children, so no child can have a whole block. Cut each of the 3 blocks into 4 equal pieces, one for each child. Each piece is 14 of a block.
step 1 of 5

Cut every block that has to be shared into as many equal pieces as there are children, and give each child one piece from each block.

All steps
  1. There are fewer blocks than children, so no child can have a whole block. Cut each of the 3 blocks into 4 equal pieces, one for each child. Each piece is 14 of a block.
  2. Each child takes one piece from each of the 3 blocks, which is 3 quarters of a block.
  3. (a) Each child gets 3 ÷ 4 = 34 of a block. Check: 4 children with 3 quarters each have 12 quarters, and 12 quarters fill 3 blocks.
  4. Next group: each child can first have 2 whole blocks, which uses 4 × 2 = 8 blocks. The 9 − 8 = 1 block that is left is cut into 4 equal pieces, one for each child.
  5. (b) Each child gets 2 whole blocks and 14 of a block: 9 ÷ 4 = 214 blocks.

answer(a) 34 of a block; (b) 214 blocks

techniqueFractions as Division · Mixed Numbers

examsPSLE

Common pitfalls

  • Writing each share in (a) as 43, children over blocks. The amount being shared, 3 blocks, goes on top: 3 ÷ 4 = 34. With fewer blocks than children, each share must be less than one block, and 43 is more than one.
  • Writing 9 ÷ 4 = 2 remainder 1 as 2.1 blocks. The block that is left over is shared among the 4 children too, so each child gets 14 of it, not a tenth: the share is 214 blocks.