eduKatePunggol · Name the quantities, match the parts
Keep the quantity beside the number.
A ratio question can go wrong before the first calculation. You may divide accurately but use the number of parts for the wrong quantity. Or you may find one group correctly and stop when the question asks for the total.
This Mathematics companion gives you 12 original questions on reading a ratio, finding the value of one equal part and checking what the answer counts. Keep a first attempt, discuss the explanations, try fresh work, and return on another day. The later set includes a question about two records that cannot both be correct.
Starting point: you have been taught introductory ratios, can multiply and divide whole numbers, and can compare quantities using “more than” and “altogether”. These tasks revisit upper-primary foundations and can also support an early-secondary learner. All scenarios and family examples are fictional. Use paper and your usual learning support.
Before opening the worked answers
Write the date and task code. Read each question twice: once for the situation and once for the quantities. Put the names beside the ratio numbers. Underline what is known and circle what you have to find.
Show what the known amount represents in equal parts before dividing. Then give the answer with its label or unit. A short line such as “three parts represent 18 red labels” exposes the reasoning that a bare calculation leaves hidden.
Each numbered question describes a separate collection or situation. Do not carry an amount from one question into another. Keep the original attempt, including an unfinished model. Leave the answer panel closed until you have tried the set.
If reading a phrase is the difficulty, ask for it to be explained and record that help. If multiplication or division is still insecure, work on a smaller example with a teacher. The aim is to understand the relationship well enough to choose a calculation yourself.
RA1: The same ratio, different information
For each question, explain which number of equal parts matches the known amount. The boxes below contain only the two colours named.
- RA1a. A box contains blue and red labels in the ratio blue : red = 2 : 3. There are 30 labels altogether. How many labels of each colour are there?
- RA1b. In another box, blue : red = 2 : 3. There are 18 red labels. How many blue labels are there?
- RA1c. In a third box, blue : red = 2 : 3. There are 10 more red labels than blue labels. How many labels of each colour are there?
- RA1d. You are told only that blue : red = 2 : 3. Must there be exactly five labels altogether? Give two possible collections to explain your answer.
Open RA1 worked answers after your attempt
RA1a: 12 blue labels and 18 red labels
The total represents 2 + 3 = 5 equal parts. One part is 30 ÷ 5 = 6 labels. Blue labels occupy two parts, so there are 2 × 6 = 12 blue labels. Red labels occupy three parts, so there are 3 × 6 = 18 red labels.
Check both conditions: 12 + 18 = 30, and 12 : 18 simplifies to 2 : 3. Finding numbers that add to 30 is only one part of the check; the colour relationship must also fit.
RA1b: 12 blue labels
The 18 red labels represent three parts. One part is 18 ÷ 3 = 6 labels. Two parts are 2 × 6 = 12 blue labels. Dividing 18 by five would treat the red labels as the whole collection, which is not the information given.
The answer happens to match the blue amount in RA1a. The starting information is different, so the explanation still matters. Check that 12 blue and 18 red give the required ratio.
RA1c: 20 blue labels and 30 red labels
The extra red labels represent the difference of 3 − 2 = 1 part. That one part is worth 10 labels. Blue labels total 2 × 10 = 20; red labels total 3 × 10 = 30.
Check the difference: 30 − 20 = 10 labels. Check the ratio: 20 : 30 simplifies to 2 : 3. The 10 in the question is neither the blue amount nor the total; it describes the gap between the two amounts.
RA1d: five parts can represent different totals
There need not be exactly five labels. Two blue and three red give a total of five. Four blue and six red give a total of ten. Both collections have the ratio 2 : 3. Other examples with the same relationship are also valid.
The ratio fixes how the quantities compare. A further amount, such as the total, one colour’s count or their difference, is needed to determine the size of this collection.
Make the equal parts visible
For blue : red = 2 : 3, the following model shows the relationship. Each small box represents the same number of labels within one collection. That number can change when you move to another question.
A “part” is a unit in the model. It is not automatically one label. In RA1a it represents six labels; in RA1c it represents ten. Giving each part the right value connects the ratio to the actual collection.
| If the known amount is… | For blue : red = 2 : 3, it represents… |
|---|---|
| The blue labels | 2 equal parts |
| The red labels | 3 equal parts |
| All the labels | 2 + 3 = 5 equal parts |
| The extra red labels | 3 − 2 = 1 equal part |
Keep the order attached to the names. Blue : red = 2 : 3 also means red : blue = 3 : 2. If you want blue : all labels, the comparison becomes 2 : 5 because the whole contains both colours. These are different descriptions of the same collection.
Use this short sequence: name the quantities → match the known amount to parts → find one part → find the requested quantity → check the original facts. A table or a bar model can help. Keep whichever representation makes your reasoning clearer.
Scaling and adding describe different changes
Two blue and three red labels can be scaled to six blue and nine red labels by multiplying both counts by three. The ratio remains 2 : 3. The quantities have grown together by the same factor.
If you add one label of each colour to the original two and three, the counts become three and four. The ratio is now 3 : 4. After a change, work from the new counts and check the relationship again. Equal additions are not the same instruction as equal multiplication factors.
Check what kind of quantity you are counting
Whole labels, cards and beads must have whole-number counts. For the simplified ratio 2 : 3, possible totals include 5, 10, 15 and other positive whole-number multiples of five. The two groups must fit their equal parts without requiring a fraction of a label.
Lengths are different: part of a centimetre can be a valid measurement. Put compared lengths in the same unit before interpreting their ratio. The ribbon questions here use centimetres throughout. A check should respect what the numbers represent.
RA2: Choose the parts on fresh work
Close the earlier worked answers. Use these separate situations after teaching, recording any model or prompt you still need. Show the parts before calculating.
- RA2a. A stall has only small and large plant pots. Small : large = 3 : 5, and there are 64 pots altogether. How many pots of each size are there?
- RA2b. The lengths of two ribbons are in the ratio shorter : longer = 4 : 7. The shorter ribbon is 28 cm long. Find the length of the longer ribbon.
- RA2c. A collection has wildlife and landscape cards in the ratio wildlife : landscape = 3 : 5. There are 18 more landscape cards than wildlife cards. Find the number of each type.
- RA2d. A tray has 8 blue counters and 12 red counters. Ben adds two counters of each colour. He says the ratio stays 2 : 3. Is he correct? Give the new counts and the new ratio in simplest form.
Open RA2 worked answers after your fresh attempt
RA2a: 24 small pots and 40 large pots
The total is 3 + 5 = 8 parts. One part is 64 ÷ 8 = 8 pots. Small pots: 3 × 8 = 24. Large pots: 5 × 8 = 40. Check that 24 + 40 = 64 and 24 : 40 simplifies to 3 : 5.
RA2b: the longer ribbon is 49 cm
The known 28 cm belongs to the shorter ribbon, which occupies 4 parts. One part is 28 ÷ 4 = 7 cm. The longer ribbon is 7 × 7 = 49 cm. Check that 28 : 49 simplifies to 4 : 7 and the longer ribbon is longer than 28 cm.
RA2c: 27 wildlife cards and 45 landscape cards
The difference is 5 − 3 = 2 parts. One part is 18 ÷ 2 = 9 cards. Wildlife cards: 3 × 9 = 27. Landscape cards: 5 × 9 = 45. Check that 45 − 27 = 18 and 27 : 45 simplifies to 3 : 5.
Here the difference occupies two parts. Carrying over the one-part difference from RA1c would give the wrong scale. Read the new ratio before choosing the number of parts.
RA2d: the new ratio is 5 : 7
There are now 10 blue counters and 14 red counters. The ratio 10 : 14 simplifies to 5 : 7, so Ben is incorrect. The original 8 : 12 simplified to 2 : 3, but adding two to both counts did not preserve that relationship.
You can check without comparing decimal values: if 10 blue counters represented two parts of a 2 : 3 ratio, each part would be five counters and there would have to be 15 red counters. The actual red count is 14.
RA3: Check the requested quantity and the data
Return on another day and write the actual date. Keep the earlier models closed at first. If you have already studied these answers, use an unfamiliar teacher-selected task for the later check.
- RA3a. A collection contains only notebooks and pencils. Notebooks : pencils = 5 : 3. There are 72 items altogether. How many pencils are there, and how many notebooks?
- RA3b. The lengths of a green ribbon and a gold ribbon are in the ratio green : gold = 7 : 4. The gold ribbon is 28 cm long. How long is the green ribbon?
- RA3c. The numbers of tickets for morning and afternoon sessions are in the ratio morning : afternoon = 4 : 9. There are 35 more afternoon tickets than morning tickets. How many tickets are there altogether?
- RA3d. A tray contains only red and white beads. A record gives the exact ratio red : white = 3 : 4 and the total as 25 whole beads. Can both records be correct? Explain without rounding bead counts.
Open RA3 worked answers after your later check
RA3a: 27 pencils and 45 notebooks
The total represents 5 + 3 = 8 parts. One part is 72 ÷ 8 = 9 items. Pencils occupy three parts, giving 27 pencils. Notebooks occupy five parts, giving 45 notebooks.
The question asks for pencils first, while the ratio names notebooks first. Keep each result attached to its label. Check that 45 + 27 = 72 and notebooks : pencils = 45 : 27 = 5 : 3.
RA3b: the green ribbon is 49 cm
The gold ribbon’s 28 cm represents 4 parts. One part is 7 cm, so the green ribbon is 7 × 7 = 49 cm. Check the given order: green : gold = 49 : 28 = 7 : 4.
RA3c: 91 tickets altogether
The difference is 9 − 4 = 5 parts. One part is 35 ÷ 5 = 7 tickets. The total occupies 4 + 9 = 13 parts, so there are 13 × 7 = 91 tickets altogether.
A fuller check gives 28 morning tickets and 63 afternoon tickets. Their difference is 35, their total is 91, and 28 : 63 simplifies to 4 : 9. Stopping at 28 or 63 would leave the requested total unanswered.
RA3d: the two exact records cannot both be correct
The simplified ratio 3 : 4 requires seven equal parts in all. Whole-bead collections with this ratio can have totals such as 7, 14, 21 or 28. A total of 25 does not fit.
Dividing 25 by seven and rounding the resulting bead counts would change the exact relationship. At least one record needs checking. We cannot decide which record is wrong from the information given, and should not replace 25 with a nearby total without recounting or checking the original source.
Use the attempt to choose the next step
In a fictional Punggol family review, Ben reaches 28 morning tickets in RA3c and stops. Jo asks what quantity the question requests. Ben finds the afternoon count and adds the two. They keep his first stopping point and record Jo’s prompt, so the next task can check whether he finishes at the requested quantity himself.
Mira’s difficulty appears earlier. She treats the 35 extra tickets as the total. Adrian asks her to point to the gap between the two bars. They discuss what “35 more” describes before asking her to calculate again. The next useful example should test that reading decision.
| What the work shows | A useful next action |
|---|---|
| The amounts are attached to the wrong names. | Write the two quantity names above their ratio numbers and label each calculated answer. |
| The total is used where one group or a difference is known. | Mark exactly which part of the model the known amount describes before dividing. |
| The learner finds one part and stops. | Return to the requested quantity and count how many parts it occupies. |
| A ratio is carried forward after the collection changes. | Write the new quantities first, then simplify their ratio and check the claim. |
| A fractional count of whole objects is accepted. | Check the arithmetic, the stated conditions and whether the original records are consistent. |
Record the task, date, original model, calculation and help used. Note one decision that changed and one point that remains uncertain. The sets use different quantities and include different demands, so compare the reasoning and support as well as the number of completed answers.
Use the weekly practice record to choose a realistic return date. If the concern came from a marked school question, use the returned-paper review to keep the original response beside later work.
Connect the ratio to the next piece of learning
For a collection with blue : red = 2 : 3, blue labels occupy two of the five total parts. That is two-fifths of the collection, or 40%. Naming the whole is what makes the connection work. Continue with Find the Percentage Whole when you are ready to practise the percentage version.
If the uncertain step is what division or an answer’s unit means, use Relationships Before Operations. For teaching across related topics, the existing Primary 5 Mathematics practice guide connects fractions, percentages, ratios, models and checking. The Mathematics learning pathway provides the wider subject routes.
Choose the next example from the decision that needs attention. A clear explanation on a new question gives you something useful to review together.
