The PSLE Math paper has always been a topic of discussion after its completion. While the paper is designed to be accessible to most students, according to the Ministry of Education (MOE), challenging questions are capped at 15% of the paper. Most students have commented that the 2024 PSLE Math paper was relatively manageable but we also noted that the way the questions were phrased can be tricky.
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2024 PSLE Math Questions: A Step-By-Step Guide
Let’s look at 2024 7 toughest PSLE Math questions and how you can solve these questions.
1. Corn, Sardine and Tuna Question
(a) Average mass of 3 cans: 420g
Total mass: Average mass x number of items
= 420 x 3
= 1260g
Total mass of sardine can and tuna can: Total mass of 3 cans – mass of corn
= 1260 – 430
= 830g
(b) To find the largest difference between sardine can and tuna can, there are 4 conditions to fulfill.
- Total mass of the sardine can and tuna can must be 830g
- Mass of the sardine can and tuna can must be 3 digits
- Mass of sardine can needs to start with 4
- Mass of tuna can needs to end with 9
Since the total mass of the sardine can and tuna can must be equal to 830g and the mass of the tuna can needs to end with 9, the last digit of the mass of the sardine must be 1.
In order to achieve the largest difference in mass of the two cans, the mass of the sardine needs to be as large as possible while the mass of the tuna needs to be as small as possible. As a result, you can derive that the second digit for the sardine can is 9. Then, find the mass of the tuna.
Mass of tuna: 830 – 491 = 339g
The difference: 491-339 = 152g
2. Adult and Child tickets
Since they are asking for the least amount needed to purchase 4 adults and 3 children tickets, we will need to consider the various combinations that a person can make to buy 4 adult tickets and 3 children tickets.
Case 1:
1 A + 1 C = $36
3A + 3 C = $108
$108 + 1 adult ticket price ($24) = $132
However, one of the key pitfalls of this question is not considering other combinations that may cost less.
Case 2:
2A + 1 C = $56
4A + 2 C= $112
$112 + $18 = $130
But are there other ways you can make up 4 adult tickets and 3 children tickets?
Case 3:
1 A + 1 C = $36 (x 2)
2A + 1 C = $56
By multiplying 2 to the 1 adult and 1 child ticket set and adding it to the 2 adults and 1 child ticket set, the total cost will be $128 which is the least amount needed to purchase 4 adults tickets and 3 children tickets.
3. Common factors
Listing out the conditions, the two numbers must:
- Be less than 40
- Add up to 60
- Have common factors of 1, 2, 4
- Exactly 6 factors each
To solve this question, let’s take a look at their factors. Since they have common factors of 1, 2 and 4, we can conclude that they must be a multiple of 4 (and 2). We can eliminate looking at the multiples of 2 as all multiples of 4 must be a multiple of 2 but not the other way around.
Now let’s list out all the multiples of 4 that are less than 40 and add up to 60.
0, 4, 8, 12, 16, 20, 24, 28, 32 and 36
There are only 2 sets of numbers that add up to 60. They are 24 & 36 and 28 & 32. Now let’s list out all of their factors. Only 28 and 32 fulfill all of the conditions. So here’s your answer!
3. Fractions Question
Since A and B spent the same amount of money and the fraction spent by A and B are 1/4 and 2/7 respectively, you can make the numerator the same to give you the ratio of the total amount A and B has.
Additionally, you are given that C spent 3 times as much as B. By multiplying the fraction C spent (2/5) by 3 (top and bottom), you will be able to get the ratio of the total amount of A, B and C. You will also arrive at the following final answer.
Least (at first): B
Most (at first): C
Total amount spent: $1560
8 u + 7 u + 15 u = $1560
1 u = $1560 ÷ 30 = $52
C spent: 6 u = $52 x 6 – $312
5. 2024 PSLE Math Paper 2 Q17 – Area and Perimeter
The concept behind this question is frequently tested in PSLE and is one that you need to master. A folded portion of paper has the same dimensions.
Based on the question, BC : BD = 1 : 4. BC is 1 u while CD is equal to 3 u.
Perimeter of Y: AB + AB + 2 u = 2AB + 2 u
Perimeter of X: AB + 4 u + AB + 1 u + 3 u = 2AB + 8 u
Difference in perimeter of X and Y: (2AB + 8 u) – (2AB + 2 u) = 60
6 u = 60
1 u = 60 ÷ 6 = 10cm
Length of AE: 4u = 10 x 4 = 40cm
Area of X: Area of the rectangle – (2 x area of the triangle)
Area of rectangle: 50 x 40 = 2000
Area of triangle x 2 : (1/2 x 50 x 10) x 2 = 500
Area of X: 2000 – 500 = 1500 cm²
Common pitfall: Students need to be very careful in finding the difference between the perimeter of X and Y. They tend to miss out on adding AB and 1 u (folded inwards) into the perimeter of X, resulting in the wrong final answer.
6. Hearts and stars
A key strategy to tackle creative questions is to identify the problem-solving tools you have previously learned and apply them to such questions. A key technique taught in excess-shortage questions is to find the big difference and small difference.
In this case, you are required to find the number of students (assuming only 1 turn each) to make an equal number of hearts and stars,
If every 2 students pin a set of hearts and stars, the difference of stars will increase by 3
Small difference: 8-5 = 3
However, currently, the difference in hearts and stars is 20, which is not divisible by 3. The way to solve this question is to get 1 student to add 1 set of stars such that the big difference can be divisible by 3.
Big difference: 22 – 10 = 12
Sets: 12 ÷ 3 = 4 (4 sets of 2 students)
Total number of students: 4 x 2 + 1 = 9
7. Magnets
The easiest way to solve this question is to work backwards.
(a) Tina gave away 4 small magnets, total magnets: 97 – 4= 93
She then bought some large magnets and had 114 magnets left. Find the number of large magnets she bought by finding the increase in large magnets.
Increase in large magnets (50%): 114-93= 21
The amount of large magnets increased by 50%. This means that the percentage of large magnets she has is 150%.
50% – 21
150% – 21 x 3 = 63
(b) The number of large magnets at first (100%) – 21 x 2 = 42
Number of small magnets at first – 97 – 42 = 55
Difference in small and large magnets (at first): 55-42 = 13
There are 13 more smaller magnets than large magnets.
Watch our full 2024 PSLE Math Post-Mortem
Now you know how to solve some of the toughest 2024 PSLE Math questions. You might also be interested in our 2024 PSLE Math Same-Day Post Mortem where we solved the PSLE Math questions based on students’ recollection. If video tutorial is not your thing, you can also download our worked solutions for the 2024 PSLE Math questions that we have collected.
Students and parents who are preparing for the PSLE can also join our WhatsApp community for support.
Especially for 2025 Primary 6 students, the upcoming year will be crucial in terms of mastering knowledge from Primary 4 and Primary 5 as well as learning new topics such as circles. Gain a basic understanding of the topic of the circle through our $9 trial class!
It’s not enough to simply master the concepts tested at PSLE. Here are some strategies that your child can use as they prepare for the upcoming PSLE Math paper:
- Understand the basic definitions
- Memorise formula
- Strengthen mental calculation (for Paper 1)
- Master concepts and problem application
- Practice (both topical practices and timed practices)
Contact our education consultant to find out more about how you can help your child to achieve AL1 in the PSLE.
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