# 8 Must Know Methods for Solving PSLE Math Problem Sums

While teaching PSLE Math problem sums, I realised that **all my students have problems with problem sums**. I remembered last year, I was teaching a group of unmotivated Primary 6 students. More than half of them have failed their last Math exams. I was going through each problem sum, one by one, on the whiteboard with them. It was a long and painful process and it was not effective at all. Most of the students simply gave up after a while because there was just too many methods and steps to remember.

I wondered if there was a way to group the problem sums into their respective solving methods. Thank god there is!

I have identified **8 methods** which students **must know** to solve PSLE Math problem sums. I started teaching my students each of these methods using examples, and followed through by giving them worksheets with questions that require applications of those methods. Using this systematic teaching process, my students showed great improvements in their results.

Most of them scored ‘A’ or ‘A Star’ for their PSLE Math exams. There is no secret. Here are the 4 steps:

- Understand the question
**Identify the correct method**- Apply the method to solve the question
- Check your answers

What are the 8 methods then?

**1. Model Drawing**

This is the most important and most commonly-used method in problem sums. This is especially useful for problem sums involving fractions where you need to cut the model into different parts and ‘give’ them to another person.

Most students face difficulties in model drawing when they don’t know how to cut models into different parts. Note that when you cut a model into 2 parts, you need to do the same for the other models you have drawn to make it consistent.

*Example: James had 120 more marbles than Dan. After James lost 1/5 of his marbles and Dan lost 3/4 of his marbles, James had 184 more marbles than Dan. How many marbles did Dan have at first?*

**Solutions**

**2. Constant Part, Constant Total, Constant Difference**

For constant part, one side of the ratio changes while the other side remains unchanged.

For constant total, there is an internal transfer from one side to another side. The total amount remains constant or the same.

For constant difference, both sides of the ratio change by the same amount.

*Example: In a class, 30% were boys. After another 9 girls and 9 boys joined the class, there were 3/5 as many boys as girls. How many pupils were there in the end?*

**Solutions**

*Start*

*B : G : Difference*

*= 3 : 7 : 4*

*End*

*B : G : Difference*

*= 3 : 5 : 2*

*= 6 : 10 : 4*

*3 units = 9 pupils*

*16 units = 48 pupils (Answer)*

* *

**3. Grouping**

This method is required when you need to group items into sets. Then divide the total cost by the cost of each set to find the total number of sets.

*Example: John bought 2 more pairs of shorts than shirts and paid a total of $124. Each shirt costs $8 and each pair of shorts costs $12. (a) How many shirts did he buy? (b) How many pair of shorts did he buy?*

**Solutions**

*2 × 12 = 24*

*124 – 24 = 100*

*8 + 12 = 20 (1 shirt and 1 shorts)*

*100 ÷ 20 = 5 (Answer for a)*

*5 + 2 = 7 (Answer for b)*

**4. Excess and Shortage**

This method is for questions which give you an excess when you group items into smaller groups, and a shortage when you group items into larger group.

*Example: Tom packed 5 balls into each bag and found out that he had 7 balls left over. If he packed 6 balls into each bag, he would need another 5 more balls. a) How many bags did he have? b) How many balls did he have altogether?*

**Solutions**

*6 – 5 = 1*

*7 ÷ 1 = 7*

*7 + 5 = 12 bags (Answer for a)*

*12 × 5 + 7 = 67 balls (Answer for b)*

**5. Units and Parts**

This is an algebra way of solving questions which you are given a starting ratio. Then, both sides of the ratio changes by different amounts, and you are given the final ratio.

Common Mistake: Most students treat the starting ratio and final ratio as the same units. This is wrong as the units in the starting and final ratios are different. The correct way is to treat the starting ratio as **units** and final ratio as **parts**.

*Example: Chris had some apples and oranges in the ratio of 9 : 3. He bought 30 more apples and 6 oranges. As a result, the ratio of apples to oranges became 4 : 1. How many apples did he have at first?*

**Solutions**

*A: 9u + 30 —-> 4 parts*

*O: 3u + 6 —-> 1 part*

*12u + 24 —-> 4 parts*

*9u + 30 = 12u + 24*

*12u – 9u = 30 – 24*

*3u = 6*

*1u = 2*

*9u = 18 apples (Answer)*

* *

**6. Guess and Check**

I have found a way to use the Guess and Check method to make it faster and more accurate. It requires you to make ‘smart’ assumptions, followed by making small changes to reach your final goal.

Example: There were a total of 10 birds and rabbits in a pet shop. The shopkeeper counted a total of 26 legs. How many birds and how many rabbits were there?

**Solutions**

*Suppose there are 10 rabbits, 0 bird*

*Number of legs = 10 × 4 = 40*

*Suppose there are 9 rabbits, 1 bird*

*Number of legs = 9 × 4 + 1 × 2 = 38 *

*Notice that when birds increase by 1, the number of legs drop by 2.*

*40 – 26 = 14*

*14 ÷ 2 = 7 birds*

*10 – 7 = 3 rabbits (Answer)*

**7. Number X Value**

This method is applicable when you need to group items into different sets, with a predetermined value in each set (For example, money). It is similar to grouping but you are grouping with unequal items. As the method name says, you need to multiply the number in the ratio, by the value it represents. This will give you the total value in a set. Then, you divide the total value by this answer to get the total number of sets.

*Example: The ratio of the number of 50 cents to 1 dollar coin is 3 : 1. The total value of the coins is $12.50. What is the total number of coins?*

**Solutions**

*3u × $0.5 = $1.5u*

*1u × $1 = $1u*

*$1.5u + $1u = $2.5u*

*$2.5u = $12.50*

*1u = 5*

*4u = 20 coins (Ans)*

**8. Work Backwards**

This method is required when you are given the final conditions, and you need to work backwards to find the original numbers.

*Example: A bus left an interchange carrying some passengers with it. At the bus stop, ¼ of the people alighted and 5 people boarded it. At the 2 ^{nd} stop, ½ of the people alighted and 20 people boarded the bus. When it left the 2^{nd} stop, there were 60 passengers in it. How many passengers were there in the bus when it left the interchange?*

**Solutions**

*60 – 20 = 40*

*40 × 2 = 80*

*80 – 5 = 75*

*75 ÷ 3 × 4 = 100 passengers (Ans)*

*** Now it’s your turn to try. Scroll down and download a free quiz on PSLE Math Problem sums. There are instructions in the quiz on how to submit to me for review. **

Learning the methods to solve PSLE Math problem sums is just the first part. Your child still has to master these common questions that may appear in his or her upcoming test or exam. Read more in this article >>> https://jimmymaths.com/common-types-psle-math-questions/

*By Jimmy Ling*

*Director, Math Tutor*

*Grade Solution Learning Centre*

Contact: 9753 9016

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