The idea in brief. The unchanged holding has the same value before and after the other holding changes. Put the multiple on the correct stage, before or after the change. In the lesson. In an earlier lesson, we drew bars of equal units and found one unit. Today, we will find whose shells stay the same, and use them as 1 unit before and after the change. Find whose shells stay the same. Their bar is 1 unit, before and after the change.
Lesson video, with the working
One quantity unchanged
5 min 18 s
The film teaches 3 questions in turn. Each question shows first; its working is written as the film explains it.
Mei had 3 times as many shells as Kavin. Mei gave away 14 shells. Then Mei had the same number of shells as Kavin. How many shells did Mei have at first?
3 units − 1 unit = 2 units
2 units = 14
1 unit = 14 ÷ 2 = 7
Mei's shells at first = 3 × 7 = 21
Ans: 21 shells
Example 1 of 2
Siti had 4 times as many shells as Bala. Siti gave away 18 shells. Then Siti had 2 times as many shells as Bala. How many shells did Siti have at first?
4 units − 2 units = 2 units
2 units = 18
1 unit = 18 ÷ 2 = 9
Siti's shells at first = 4 × 9 = 36
Ans: 36 shells
Example 2 of 2
Ravi had 2 times as many shells as Wei Jie. Then Ravi found 24 more shells. Now Ravi has 5 times as many shells as Wei Jie. How many shells does Wei Jie have?
5 units − 2 units = 3 units
3 units = 24
1 unit = 24 ÷ 3 = 8
Ans: 8 shells
How to solve it
Find whose shells stay the same. That child's bar is 1 unit, before and after the change.
Draw the other child's bar before the change: 3 times as many is 3 units.
Draw it after the change too: the same number as Kavin is 1 unit.
The units that changed are the shells given away. 3 units minus 1 unit is 2 units, and 2 units are 14 shells.
Divide to find 1 unit: 14 divided by 2 is 7. Then use it to answer the question asked.
Find whose shells stay the same. Their bar is 1 unit, before and after the change.
Practice questions with worked solutions
Question 1 is free to try. The other 10 open in the free trial, each with its solution film.
Question 1 of 11 · Basic · free to tryTwo boxes held 80 tiles altogether. After 11 tiles were removed from Box A…Two boxes held 80 tiles…
Two boxes held 80 tiles altogether. After 11 tiles were removed from Box A, A held 2 times as many tiles as B. No tiles were added to or removed from B. How many tiles were in B?
Show the working and the answer
Working
Total after removal = 80 − 11 = 69
Total equal units = 2 + 1 = 3
3 units = 69
1 unit (tiles in b) = 69 ÷ 3 = 23
Answer: 23 tiles
For parents
How do you solve P3 one quantity unchanged word problems?
In these before-and-after word problems only one person's amount changes.
What your child learns here
In these before-and-after word problems only one person's amount changes. The unchanged amount keeps the same size throughout, so it can serve as the unit in both stages and the change can be counted in units. The page applies Primary 3 multiplication and division within the tables to a two-stage bar model.
Spot whose amount does not change and draw it as 1 unit in both stages.
Draw the other bar before and after, using the multiple given for each stage.
Count how many units the change covers, then find one unit.
Use one unit to find the amount at the stage the question names.
Does the answer make sense?
B's amount never changes, so when A ends up several times as much as B, A's amount after the change must be an exact multiple of B's. Your child can also check the direction: when A loses some, A at first is more than A afterwards; when A gains some, it is less.
How to practise these questions
The Basic questions practise drawing the after picture and working back to the start. The Intermediate questions add stories where only one quantity changes and the other is used as the unit. In the Challenge questions, the quantity that stays the same must be found from the story, then used to compare before and after. If an unchanged-quantity question stalls, watch its solution film and attempt it again later from Revision.
Questions parents ask
What is a before-and-after word problem in P3? It describes two moments: the amounts before a change and after it. At P3 the comparisons use 'times as many', and the change is a number added or given away. The lesson on this page draws the bars for both moments, one under the other.
Why is the unchanged amount kept as 1 unit? It is the same size before and after, so it can be the measuring stick for both moments. The other person's change can then be counted in units, and one unit found by division.
Does my child need to draw two sets of bars? Drawing both stages keeps the 'before' multiple and the 'after' multiple apart, and shows the change as extra units. The MOE syllabus does not require a particular method, but the bars make the reasoning easy to follow and check.
