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The equal-parts method: 25 illustrated problems from primary to lower secondary school

Solve word problems using equal parts and bar models: clear explanations, 25 progressive exercises, diagrams and worked solutions revealed with a click.

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The equal-parts method: 25 illustrated problems from primary to lower secondary school

30 min

A problem becomes easier when you can see it. The equal-parts method, also called the bar-model method, represents quantities as groups of equal rectangles. First find the value of one rectangle, then rebuild the required quantities: often no equations or trial and error are needed.

Illustration for the Arithmetic section
Section: Arithmetic

Source document: The equal-parts method, from primary to lower secondary school (PDF in Italian). This course preserves the explanation and all 12 problems from the supplied document, adds 13 new problems and develops every solution. Problems are reordered by difficulty; fractional ages and amounts that cannot be paid exactly in cents receive an explicit explanation.

1. The idea: draw a part, not every object

Marco and Luca have 40 cards altogether. Marco has three times as many as Luca. How many does each have? If Luca occupies 1 part, Marco occupies 3 parts equal to Luca’s part. This does not mean Luca has 1 card and Marco 3: we do not yet know the value of a part.

The total number of parts is 1 + 3 = 4. One part therefore contains 40 : 4 = 10 cards. Luca has 1 × 10 = 10 cards and Marco 3 × 10 = 30. Check: 10 + 30 = 40 and 30 = 3 × 10.

Luca has one part of 10 cards; Marco has three parts of 10, giving 40 altogether.
One part for Luca and three for Marco: four equal parts altogether.

The key question is always: which given quantity corresponds to how many parts? Divide only after answering this.

2. The three basic cases

You know the sum: add the parts

Two brothers have €56 altogether. The older has three times as much money as the younger. Their parts are 3 and 1: 56 : (3 + 1) = 14 euros per part. The older has 3 × 14 = 42 euros and the younger 14. The total, not the difference, represents four parts.

Three parts of 14 euros for the older brother and one for the younger: 42 and 14 euros.
Known sum: divide by 3 + 1.

You know the difference: subtract the parts

One segment is five times as long as another. Their difference is 32 cm. The parts are 5 and 1; the extra parts are 5 − 1 = 4. Thus 32 : (5 − 1) = 8 cm per part. The segments measure 5 × 8 = 40 cm and 8 cm. Check: 40 − 8 = 32.

The long segment has five 8-centimetre parts and the short one a single part: difference 32 centimetres.
The actual difference corresponds only to the four extra parts.

You know one quantity: use its own parts

In a class, the number of girls is 4/3 of the number of boys, and there are 12 boys. Girls occupy 4 parts and boys 3. One part is 12 : 3 = 4 pupils; there are 4 × 4 = 16 girls. Do not divide by 7: 12 is not the total class size.

Girls: four parts of 4 pupils. Boys: three parts of 4 pupils.
The known 12 pupils correspond to the boys’ three parts.

3. At secondary school: a fraction is a ratio of parts

Writing AC = (7/5) CB does not mean AC measures 7 cm and CB 5 cm. It means AC contains 7 parts and CB 5 parts of the same size. If AC exceeds CB by 10.8 cm, those two extra parts are exactly 10.8 cm.

(7 − 5) parts = 10.8 cm; therefore 1 part = 10.8 : 2 = 5.4 cm. Rebuild the lengths: AC = 7 × 5.4 = 37.8 cm, CB = 5 × 5.4 = 27 cm. Check: 37.8 − 27 = 10.8 and 37.8 : 27 = 7 : 5.

AC contains seven parts of 5.4 centimetres and CB five: 37.8 and 27 centimetres.
The ratio 7 : 5 gives the numbers of parts, not lengths in centimetres.

For quantities A and B in the ratio a : b, this table gives the value of one part:

Known quantityValue of one part
Sum SS : (a + b)
Difference D, with a > bD : (a − b)
First quantity AA : a
Second quantity BB : b

Once you know one part, multiply by a and b. With three quantities, for example 1, 2 and 6 parts, count 1 + 2 + 6 parts in the sum. First make sure all parts really represent the same amount.

4. A five-step procedure

  1. Read: what is given and what must you find? Underline the units.
  2. Draw: translate double, triple or a fraction into equal parts.
  3. Match: is the given number a sum, a difference or one quantity?
  4. Calculate: find one part, then every required quantity.
  5. Check: verify both the numerical condition and the stated ratio. Give an answer with the correct units.

Watch the wording: “three times as much” means multiplication by three, whereas “3 more” means a difference of 3 units. For a rectangle, perimeter = 2 × (length + width): before adding the length and width parts, divide the perimeter by 2. With coins of different denominations, counting coins is not the same as adding euros.

The ratio may not hold at the start. When both people age or receive the same gift, their difference stays unchanged. When objects move from one group to another, their total stays unchanged. First identify what does not change, then apply the method to the requested situation.

25 problems with diagrams and expandable solutions

Read the question, examine the diagram and try it in your notebook. Within a model, rectangles representing the same part have the same value. Colours distinguish quantities, not the value of a part. A question mark does not mean zero. Click Show the solution and completed diagram after trying; click again to close it. These are models, not drawings to measure with a ruler.

Level 1 - Primary: equal parts and simple information

  1. The sweets

    Anna and Pietro have 48 sweets altogether. Anna has twice as many as Pietro. How many does each have?

    Anna occupies two equal parts and Pietro one; together they have 48 sweets.
    Two parts plus one: the total represents three parts.
    Show the solution and completed diagram

    The parts are 2 for Anna and 1 for Pietro: 2 + 1 = 3. One part is 48 : 3 = 16 sweets.

    Anna: 2 × 16 = 32 sweets. Pietro: 1 × 16 = 16. Answer: Anna 32, Pietro 16.

    Check: 32 + 16 = 48 and 32 = 2 × 16.

    One part is 16 sweets: Anna 32, Pietro 16.
    The completed model shows the value of each part and the required quantities.

  2. Bags of biscuits

    Two bags contain 35 biscuits altogether. The large bag contains four times as many as the small bag. How many are in each?

    The large bag occupies four parts, the small one a single part; total 35 biscuits.
    The small bag is the comparison unit.
    Show the solution and completed diagram

    There are 4 + 1 = 5 parts. One part is 35 : 5 = 7 biscuits.

    The large bag contains 4 × 7 = 28; the small one 7. Answer: 28 and 7 biscuits.

    Check: 28 + 7 = 35 and 28 : 7 = 4.

    One part is 7 biscuits: large 28, small 7.
    The completed model shows the value of each part and the required quantities.

  3. Savings

    A boy has €24 more than his sister and three times her savings. How much does each have?

    Three parts for the boy and one for his sister: the extra two parts are 24 euros.
    The given amount is the difference, not the total.
    Show the solution and completed diagram

    There are 3 − 1 = 2 extra parts. One part is therefore 24 : 2 = 12 euros.

    The boy has 3 × 12 = 36 euros and his sister 12. Answer: €36 and €12.

    Check: 36 − 12 = 24 and 36 = 3 × 12.

    One part is 12 euros: boy 36 euros, sister 12 euros.
    The completed model shows the value of each part and the required quantities.

  4. Giulia and Leo’s books

    Giulia has 12 books, three times as many as Leo. How many books does Leo have?

    Giulia’s 12 books represent three parts; Leo has one part.
    The known quantity belongs to Giulia and represents three parts.
    Show the solution and completed diagram

    The 12 books represent Giulia’s 3 parts, not both children’s total. One part is 12 : 3 = 4 books.

    Leo has one part: 4 books. Check: 3 × 4 = 12.

    One part is 4 books: Giulia 12, Leo 4.
    The completed model shows the value of each part and the required quantities.

  5. The trees

    A garden has 72 trees, all pines or olive trees. There are five times as many pines as olive trees. How many of each?

    Five parts of pines and one of olive trees, with 72 trees altogether.
    Six equal parts make up the garden.
    Show the solution and completed diagram

    The total number of parts is 5 + 1 = 6. One part is 72 : 6 = 12 trees.

    Pines: 5 × 12 = 60. Olive trees: 12. Answer: 60 pines and 12 olive trees.

    Check: 60 + 12 = 72 and 60 = 5 × 12.

    One part is 12 trees: 60 pines and 12 olive trees.
    The completed model shows the value of each part and the required quantities.

  6. Two ribbons

    One ribbon is four times as long as another. The long ribbon exceeds the short one by 18 cm. How long is each?

    Four parts and one part: the three extra parts measure 18 centimetres.
    The difference corresponds to three parts.
    Show the solution and completed diagram

    The difference in parts is 4 − 1 = 3. One part measures 18 : 3 = 6 cm.

    The long ribbon measures 4 × 6 = 24 cm and the short one 6 cm. Answer: 24 cm and 6 cm.

    Check: 24 − 6 = 18 and 24 = 4 × 6.

    One part measures 6 centimetres: the ribbons are 24 and 6 centimetres long.
    The completed model shows the value of each part and the required quantities.

  7. Three boxes of pencils

    Three boxes contain 60 pencils altogether. The second contains twice as many as the first, and the third three times as many as the first. How many are in each box?

    The boxes occupy one, two and three parts; total 60 pencils.
    Both comparisons refer to the first box.
    Show the solution and completed diagram

    The parts are 1, 2 and 3: 1 + 2 + 3 = 6. One part is 60 : 6 = 10 pencils.

    The boxes contain 10, 20 and 30 pencils. Check: 10 + 20 + 30 = 60; the second has double and the third triple the first.

    One part is 10 pencils: first 10, second 20, third 30.
    The completed model shows the value of each part and the required quantities.

  8. Red marbles

    A bag contains 40 marbles, all red or blue. 3/5 of them are red. How many are there of each colour?

    The whole is divided into five parts: three red and two blue.
    If the whole has five parts, the non-red parts are 5 − 3.
    Show the solution and completed diagram

    The denominator 5 tells us how many parts make up the whole. One part is 40 : 5 = 8 marbles.

    Red: 3 × 8 = 24. Blue: (5 − 3) × 8 = 16. Answer: 24 red and 16 blue.

    Check: 24 + 16 = 40 and 24/40 = 3/5.

    One part is 8 marbles: 24 red and 16 blue.
    The completed model shows the value of each part and the required quantities.

  9. Cards: sum and difference

    Sara and Paolo have 54 cards altogether. Sara has 14 more than Paolo. How many does each have?

    Both have one unknown equal part; Sara also has 14 extra cards.
    The extra rectangle of 14 is a known amount, not another equal part.
    Show the solution and completed diagram

    Set aside Sara’s 14 extra cards. The remaining 54 − 14 = 40 cards are divided into two equal quantities.

    One part is 40 : 2 = 20. Paolo has 20; Sara has 20 + 14 = 34. Answer: Sara 34, Paolo 20.

    Check: 34 + 20 = 54 and 34 − 20 = 14.

    The two equal parts are 20 cards each; Sara also has 14, giving 34 and 20.
    The completed model shows the value of each part and the required quantities.

  10. Three trays

    Distribute 45 small cakes between three trays in the ratio 2 : 3 : 4. How many cakes go on each tray?

    The trays occupy two, three and four parts of the same value.
    The ratio divides the total into nine equal parts.
    Show the solution and completed diagram

    The total number of parts is 2 + 3 + 4 = 9. One part is 45 : 9 = 5 cakes.

    The trays receive 2 × 5 = 10, 3 × 5 = 15 and 4 × 5 = 20 cakes. Answer: 10, 15 and 20.

    Check: 10 + 15 + 20 = 45 and 10 : 15 : 20 = 2 : 3 : 4.

    One part is 5 cakes: the trays contain 10, 15 and 20.
    The completed model shows the value of each part and the required quantities.

Level 2 - Lower secondary: fractions, ratios and geometry

  1. Two segments and their sum

    Two segments have a combined length of 84 cm. The first is 4/3 of the second. Find their lengths.

    The first segment has four parts and the second three; their sum is 84 centimetres.
    4/3 of the second means four parts against three, not four centimetres against three.
    Show the solution and completed diagram

    The parts are 4 and 3. One part measures 84 : (4 + 3) = 12 cm.

    First segment: 4 × 12 = 48 cm. Second: 3 × 12 = 36 cm. Answer: 48 cm and 36 cm.

    Check: 48 + 36 = 84 and 48/36 = 4/3.

    One part is 12 centimetres: first 48 centimetres, second 36.
    The completed model shows the value of each part and the required quantities.

  2. Two numbers and their difference

    Two numbers differ by 45. The larger is 8/5 of the smaller. Find the numbers.

    Eight parts and five parts: the three-part difference is 45.
    The given number represents only the extra parts, not all thirteen parts.
    Show the solution and completed diagram

    The difference represents 8 − 5 = 3 parts. One part is 45 : 3 = 15.

    Larger: 8 × 15 = 120. Smaller: 5 × 15 = 75. Answer: 120 and 75.

    Check: 120 − 75 = 45 and 120/75 = 8/5.

    One part is 15: the numbers are 120 and 75.
    The completed model shows the value of each part and the required quantities.

  3. The class

    In a class, the number of girls is 5/4 of the number of boys. There are 27 pupils altogether. How many boys and girls are there?

    Five parts of girls and four of boys make 27 pupils.
    Nine parts represent the whole class.
    Show the solution and completed diagram

    One part is 27 : (5 + 4) = 3 pupils. Girls: 5 × 3 = 15; boys: 4 × 3 = 12.

    Answer: 12 boys and 15 girls. Check: 12 + 15 = 27 and 15/12 = 5/4.

    Each part is 3 pupils: 15 girls and 12 boys.
    The completed model shows the value of each part and the required quantities.

  4. One segment is known

    Segment A is 5/3 of segment B. B measures 18 cm. How long is A?

    A has five parts; B’s 18 centimetres represent three parts.
    Divide 18 by the number of B’s parts.
    Show the solution and completed diagram

    B’s 18 cm represent 3 parts: 18 : 3 = 6 cm per part.

    A measures 5 × 6 = 30 cm. Answer: 30 cm. Check: 30/18 = 5/3.

    One part is 6 centimetres: A 30 centimetres and B 18.
    The completed model shows the value of each part and the required quantities.

  5. The rectangle with perimeter 84

    A rectangle has perimeter 84 cm. Its length is 4/3 of its width. Find length, width and area.

    Length four parts and width three; the perimeter counts both twice.
    To work with length + width, use half the perimeter.
    Show the solution and completed diagram

    First halve the perimeter: 84 : 2 = 42 cm is length plus width. There are 4 + 3 = 7 parts.

    One part is 42 : 7 = 6 cm. Length: 4 × 6 = 24 cm; width: 3 × 6 = 18 cm.

    Area: 24 × 18 = 432 cm². Answer: length 24 cm, width 18 cm, area 432 cm². Check: 2 × (24 + 18) = 84.

    One part measures 6 centimetres: length 24, width 18; a rectangle in ratio four to three.
    The completed model shows the value of each part and the required quantities.

  6. The plot of land

    The length of a rectangular plot exceeds its width by 18 m. Its length is 8/5 of its width. Find perimeter and area.

    Length eight parts, width five; three parts correspond to 18 metres.
    Here the known quantity is the difference of the dimensions, not the perimeter.
    Show the solution and completed diagram

    The extra parts are 8 − 5 = 3. One part is 18 : 3 = 6 m. Length: 8 × 6 = 48 m; width: 5 × 6 = 30 m.

    Perimeter: 2 × (48 + 30) = 156 m. Area: 48 × 30 = 1440 m². Answer: perimeter 156 m, area 1440 m².

    Check: 48 − 30 = 18 and 48/30 = 8/5.

    One part is 6 metres: length 48 and width 30; rectangle in ratio eight to five.
    The completed model shows the value of each part and the required quantities.

  7. A classroom percentage

    In a class of 30 pupils, 20% travel to school by scooter. The others do not. How many pupils are in each group?

    Twenty per cent is one fifth: one part uses scooters and four do not.
    A percentage can be rewritten as a fraction of the whole.
    Show the solution and completed diagram

    Convert the percentage to a fraction: 20% = 20/100 = 1/5. The whole has 5 parts; one part is 30 : 5 = 6 pupils.

    There are 6 scooter users; the others number 4 × 6 = 24. Answer: 6 and 24 pupils.

    Check, converting the fraction to a percentage: (6/30) × 100% = 20%; also 6 + 24 = 30.

    One part is 6 pupils: 6 use scooters and 24 do not.
    The completed model shows the value of each part and the required quantities.

  8. Pages still to read

    Elena has read 3/7 of a book. She has 48 pages left. How many pages does the book have, and how many has she read?

    Three parts have been read; the remaining four parts represent 48 pages.
    The given 48 refers to the unread fraction, not the whole book.
    Show the solution and completed diagram

    The unread parts are 7 − 3 = 4. One part is 48 : 4 = 12 pages.

    The book has 7 × 12 = 84 pages; Elena has read 3 × 12 = 36. Answer: 84 pages altogether, 36 already read.

    Check: 84 − 36 = 48 and 36/84 = 3/7.

    One part is 12 pages: 36 read and 48 unread, total 84.
    The completed model shows the value of each part and the required quantities.

  9. Three ribbons, two ratios

    Three ribbons A, B and C have lengths in these ratios: A : B = 2 : 3 and B : C = 2 : 5. Their combined length is 100 cm. Find each length.

    Two separate comparisons: A two parts and B three; B two parts and C five. The parts in the two comparisons are not yet the same size.
    B has different part counts in the comparisons: first make the ratios compatible.
    Show the solution and completed diagram

    Do not add 2 + 3 + 5: B has different numbers of parts in the two ratios. Give B 6 parts: A : B = 4 : 6 and B : C = 6 : 15.

    The common ratio is A : B : C = 4 : 6 : 15. Total parts: 4 + 6 + 15 = 25; one part is 100 : 25 = 4 cm.

    Answer: A = 16 cm, B = 24 cm, C = 60 cm. Check: 16 + 24 + 60 = 100, 16/24 = 2/3 and 24/60 = 2/5.

    Common ratio A:B:C is 4:6:15; one part is 4 centimetres, giving 16, 24 and 60 centimetres.
    The completed model shows the value of each part and the required quantities.

  10. The PDF’s rectangle: exact results

    A rectangle has perimeter 90 cm. Its length is 7/4 of its width. Find length, width and area, retaining exact fractions.

    Length seven parts and width four; together they equal the half-perimeter of 45 centimetres.
    A part may have a fractional size: the method remains the same.
    Show the solution and completed diagram

    Half the perimeter is 90 : 2 = 45 cm. There are 7 + 4 = 11 parts: one part measures 45/11 cm.

    Length: 7 × (45/11) = 315/11 cm, approximately 28.64 cm. Width: 4 × (45/11) = 180/11 cm, approximately 16.36 cm.

    Exact area: (315/11) × (180/11) = 56700/121 cm², approximately 468.60 cm². Do not multiply already-rounded dimensions if you want to retain the exact answer.

    Check: 2 × (315/11 + 180/11) = 90 and (315/11) : (180/11) = 7/4.

    One part is 45/11 centimetres; length 315/11 and width 180/11 centimetres.
    The completed model shows the value of each part and the required quantities.

Level 3 - One step further: values, ages and transformations

  1. Three brothers and a theoretical division

    Three brothers share €240. The second receives twice the first’s share, and the third three times the second’s. How much does each receive?

    The shares are one, two and six parts: the third is triple the second, not the first.
    The third share occupies 3 × 2 = 6 parts.
    Show the solution and completed diagram

    The first has 1 part, the second 2, the third 3 × 2 = 6. Total parts: 1 + 2 + 6 = 9.

    One part is 240 : 9 = 80/3 euros. The exact shares are 80/3 euros, 160/3 euros and 160 euros. Check: 80/3 + 160/3 + 160 = 240; the second is double the first and the third triple the second.

    A note about money: 80/3 euros and 160/3 euros have recurring decimals. They are exact mathematical shares but cannot be paid exactly in cents; rounding cannot preserve all ratios. For a practical division without rounding, a version with €270 gives a €30 part and shares of €30, €60 and €180.

    One part is 80/3 euros: shares 80/3, 160/3 and 160 euros.
    The completed model shows the value of each part and the required quantities.

  2. Coins: count and value

    A money box contains €1 and €2 coins. There are three times as many €1 coins as €2 coins, and their total value is €70. How many of each coin are there?

    The number of one-euro coins has three parts; the number of two-euro coins one part.
    Here parts represent coin counts; their monetary values need weighting.
    Show the solution and completed diagram

    Draw 3 parts of €1 coins and 1 part of €2 coins. A group of three €1 coins and one €2 coin is worth 3 × 1 + 1 × 2 = 5 euros.

    The number of groups is 70 : 5 = 14. There are 3 × 14 = 42 €1 coins and 1 × 14 = 14 €2 coins. Answer: 42 and 14 coins.

    Check: 42 × 1 + 14 × 2 = 70 euros and 42 = 3 × 14. Dividing 70 by 4 would confuse euros with numbers of coins.

    One part contains 14 coins: 42 one-euro coins and 14 two-euro coins.
    The completed model shows the value of each part and the required quantities.

  3. Equal money gifts

    Two friends initially have €30 and €50. Both receive the same amount of money. Afterwards, the second has 3/2 of the first’s money. How much did each receive?

    After the gift the first has two parts and the second three; their difference stays 20 euros.
    The model represents the situation after the gift. The initial difference does not change.
    Show the solution and completed diagram

    Adding the same amount to both preserves the difference: 50 − 30 = 20 euros. In the final ratio the parts are 2 and 3; the difference of 3 − 2 = 1 part is therefore €20.

    After the gift the first has 2 × 20 = 40 euros and the second 3 × 20 = 60. The gift is 40 − 30 = 10 euros, or 60 − 50 = 10 euros.

    Answer: €10 each. Check: 60/40 = 3/2 and both increases are equal.

    One part is 20 euros: after the gift the friends have 40 and 60 euros.
    The completed model shows the value of each part and the required quantities.

  4. Ages: first find what stays unchanged

    A father is 44 years old and his son is 11. In how many years will the father be three times as old as the son?

    At the required time the father has three parts and the son one. Their age difference remains 33 years.
    The parts represent future ages, not the starting ages.
    Show the solution and completed diagram

    The age difference does not change: 44 − 11 = 33 years. At the required time the father occupies 3 parts and the son 1; the two extra parts represent 33 years.

    One part is 33 : (3 − 1) = 16.5 years. The son will be 16.5 and the father 3 × 16.5 = 49.5.

    The waiting time is 16.5 − 11 = 5.5 years. Answer: in 5 years and 6 months. Also 49.5 − 44 = 5.5. Not 5 years: 49 would not be triple 16. We use exact ages in the model, not just whole numbers of birthdays.

    One part is 16.5 years: at the required time father 49.5, son 16.5.
    The completed model shows the value of each part and the required quantities.

  5. A transfer between groups

    Group A has 48 cards and group B has 24. How many cards must A give B so that the groups have equal quantities?

    Before the transfer A has 48 cards, two parts of 24, and B has 24.
    Start with the known quantities. Cards change groups but do not disappear.
    Show the solution and completed diagram

    The total remains 48 + 24 = 72 cards. To be equal, each group must have 72 : 2 = 36 cards.

    A must give away 48 − 36 = 12 cards. B receives 36 − 24 = 12. Answer: A must give B 12 cards.

    Check: 48 − 12 = 24 + 12 = 36. Do not transfer the whole difference of 24: each transferred card reduces the difference by 2, because A loses one and B gains one.

    After the transfer the total of 72 cards is split into two parts of 36; A gives B 12.
    The completed model shows the value of each part and the required quantities.

6. What we have learnt

The equal-parts method is not a shortcut to avoid equations: it helps us understand relationships between quantities. At primary school we start with rectangles and objects; at secondary school we connect the same diagrams to fractions, ratios and proportions. Later, equations give a name to the unknown part.

For the opening example, equal parts give 40 : (1 + 3) = 10; a proportion can be written Luca : 40 = 1 : 4; an equation is x + 3x = 40. The language changes, not the reasoning. In transformation problems first look for what remains unchanged: a difference, a total or the value of a group.