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Part–Whole Model (Finding the Total — One-Step Problems)

Student Worksheet

Guided Example

Jane had 5 packets of beads. Each packet held n beads. Her mother then gave her 20 more beads. How many beads did she have altogether at the end?

Worked Solution

Step 1: Understand the Problem
  • Number of packets of beads → 5
  • Beads in each packet → n
  • Extra beads from her mother → 20
  • What to find → Total number of beads in the end
Step 2: Plan the Solution

Represent the situation using a part–whole bar model:

| n | n | n | n | n |  + 20
Step 3: Solve It

(5 × n) + 20 = 5n + 20
Therefore, Jane had (5n + 20) beads altogether.

Step 4: Check Your Work

(5 × n) + 20 = n + n + n + n + n + 20 = 5n + 20

Practice Questions

  1. Daniel had 20 packs of pens, with q pens in each pack. Later, he bought 35 more pens. How many pens did he have altogether?
    Answer: ________________________________________________
  2. Don, Alice, Gary, and Wayne each had $w. Gary’s father gave him an extra $325. What was the total amount of money the four children had in the end?
    Answer: ________________________________________________
  3. Terry had 8 boxes of marbles, each box containing y marbles. He also bought 38 green marbles and 29 red marbles. How many marbles did he own in total?
    Answer: ________________________________________________
  4. Janet made 4 jugs of apple juice and 6 jugs of orange juice, each jug containing p litres. She then prepared 3 more litres of lemon juice. What was the total amount of fruit juice she made in the end?
    Answer: ________________________________________________

Topic Recap

When several equal groups are given and an additional quantity is added, we can use a part–whole model to find the total.
Total = (Number of Groups × Amount in Each Group) + Extra

Experimental Design Slide Deck

Combined Sciences
(Chemistry)

Topic 1: Experimental & Separation Techniques

1. [cite_start]Measuring Volumes [cite: 995]

Apparatus
Beaker
It is used to contain or collect liquid/solution. [cite_start]Least accurate for measurement. [cite: 997]
Measuring Cylinder
It is used to measure volume of liquid/solution or of small irregular-shaped objects. [cite_start]Accuracy of ~1ml. [cite: 997]
Burette
[cite_start]It is used to measure volume of liquid/solution to an accuracy of 0.1 ml. [cite: 997]
Pipette
[cite_start]It is used to measure accurately a specific volume of liquid (15 cm³, 20 cm³ or 25 cm³). [cite: 997]
Volumetric Flask
[cite_start]It is used to measure accurately a specific volume of liquid (50 cm³ or 100 cm³). [cite: 997]
Gas Syringe
[cite_start]It is used to accurately measure the volume of gases produced in experiments. [cite: 997]

2. [cite_start]Collecting Gases Produced [cite: 999]

Collection Methods
Displacement of Water
Used to collect gases which are not very soluble in water.

[cite_start]Examples: oxygen and hydrogen. [cite: 1000]
Downward Delivery
Used to collect gases which are denser than air.

[cite_start]Examples: carbon dioxide and chlorine gas. [cite: 1000]
Upward Delivery
Used to collect gases which are less dense than air.

[cite_start]Examples: ammonia and hydrogen. [cite: 1000]

3. [cite_start]Separation of Mixtures (Solid-Solid) [cite: 1007]

Solid-Solid
Magnetic Attraction
This technique uses magnets to separate magnetic substances from a mixture. [cite_start]Examples of magnetic substances are iron, steel, nickel and cobalt. [cite: 1008, 1009, 1010]

[cite_start]Example: A magnet can be used to separate iron from sulfur in a mixture of iron and sulfur. [cite: 1012]
Sublimation
It is used when a mixture contains a solid substance which sublimes upon heating. [cite_start]Examples of solid substances which undergo sublimation are iodine crystals, naphthalene (moth balls) and dry ice. [cite: 1015, 1016]

[cite_start]Example: Iodine crystals can be separated from salt crystals by heating the mixture until the iodine crystals sublime. [cite: 1018]
[cite_start]

Separation of Mixtures (Solid-Liquid) [cite: 1019]

Solid-Liquid
Filtration
It is suitable to separate a solid-solid mixture (such as sand with table salt) or solid-liquid mixture (such as sand and water) which contains insoluble solids. The insoluble substance collected on the filter paper is called the residue. [cite_start]The solution collected in the flask is called the filtrate. [cite: 1021, 1022, 1030, 1031]
Evaporation
Used to recover pure soluble solid from a solution which is heat-stable. [cite_start]The solution is heated to dryness. [cite: 1034]

[cite_start]Example: Sodium chloride crystals are heat-stable which can be obtained by evaporation to dryness. [cite: 1037]
Crystallisation
Used to recover pure soluble solid from a solution which is not heat-stable. The solution is heated until it becomes saturated. [cite_start]It is allowed to cool slowly so that crystals can be formed easily. [cite: 1043, 1044, 1045]

Example: Sugar crystals are not heat-stable. They can only be obtained by crystallisation. [cite_start]Evaporation to dryness will decompose the sugar crystals to carbon. [cite: 1047, 1048]
[cite_start]

Distillation & Chromatography [cite: 1055, 1071]

Advanced Methods
Simple Distillation
Distillation is used to separate a liquid from a solid-liquid mixture.

Applications: Obtain pure water from seawater by desalination. [cite_start]Produce pure distilled water from normal water. [cite: 1056, 1057, 1058, 1060]
Fractional Distillation
Used to separate a liquid-liquid mixture which is miscible (liquids that dissolve well in each other). [cite_start]It can only be used if the liquids have different boiling points. [cite: 1119, 1120, 1122]

[cite_start]Applications: Separation of liquid air, crude oil, and ethanol from sugar solution. [cite: 1136, 1137, 1138, 1139]
Chromatography
It is used to separate a mixture which is based on the extent of solubility in a given solvent. [cite_start]The more soluble the component in the solvent, the further the distance it travels in the chromatography paper. [cite: 1072, 1097]

[cite_start]Advantages: Gives quick/accurate analysis, requires only small amount of sample, able to separate complex mixtures like food dyes. [cite: 1100, 1101, 1102]
Interactive MCQ Quiz

Loading question…

1
1
1
A
1
B
.model{ max-width:520px; font-family:system-ui; } .bar{ display:flex; border:2px solid #333; border-radius:10px; overflow:hidden; height:44px; margin:10px 0 4px; } .unit{ flex:1; border-right:2px solid #333; display:flex; align-items:center; justify-content:center; font-weight:700; } .unit:last-child{ border-right:none; } .label{ font-weight:800; margin-bottom:8px; }
{ rows: [ { name:”A”, units: 3, highlight: [0,1,2] }, { name:”B”, units: 1 } ]}
Which bar model correctly shows A : B = 3 : 1 (in equal units)?
Model A
A B 1 1 1 1
Model B
A B 1 1 1
Model C
A B 1 1 1 1 1
Model D
A B 1 1 1 1 1
Explanation
A : B = 3 : 1 means A has 3 equal units for every 1 equal unit of B. So the correct model shows A split into 3 same parts and B as 1 same part.

Question 1: Ratio Problem

Topic: Ratio & Proportion | Difficulty: ★★★☆☆

John and Mary have some money in the ratio 3:5. After John spends $24, the ratio of John’s money to Mary’s money becomes 1:3. How much money does Mary have?

Model Method (Using LCM):

💡 Why use 3u per block? Since Mary’s amount stays unchanged, we find LCM(5,3) = 15.
Before (ratio 3:5): John = 9u, Mary = 15u
After (ratio 1:3): John = 5u, Mary = 15u (unchanged!)

Before spending: John: 3u 3u 3u 9u Mary: 3u 3u 3u 3u 3u 15u Ratio 9:15 = 3:5 After John spends $24: John: 5u 4u 9u total (same as before) 4u = $24 Mary: 3u 3u 3u 3u 3u 15u (unchanged!) Same length New ratio 5:15 = 1:3
📝 Click to view solution

Step 1: Use LCM to find a common unit

Before ratio = 3:5, After ratio = 1:3
Since Mary’s amount doesn’t change, we need to make her “5 parts” equal to her “3 parts”
LCM(5, 3) = 15
So Mary has 15u throughout

Step 2: Find John’s units before and after

Before: Ratio 3:5 → John:Mary = 9u:15u
After: Ratio 1:3 → John:Mary = 5u:15u

Step 3: Calculate the value of 1 unit

John went from 9u to 5u
Difference = 9u – 5u = 4u
4u = $24
1u = $24 ÷ 4 = $6

Step 4: Calculate Mary’s money

Mary has 15u
15u = 15 × $6 = $90

Answer: Mary has $90

💡 Tip: In ratio “before and after” problems, use the LCM method to keep the unchanged person’s bar the same length. This helps you see clearly what changed! Find LCM of the two ratios’ parts for the person who didn’t change, then work backwards to find everyone’s values.

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