PSLE Challenging Math Problems and How to Solve Them

With the PSLE exams drawing near, and revisions happening in full swing, here are three PSLE challenging math problems from previous years, together with their solutions.

I hope this guide helps both parents and children!

PSLE math problem: 2017 PSLE Math Paper 2 Question 10

Jess needs 200 pieces of ribbons, each of length 110 cm, to decorate a room for a party. Ribbon is sold in rolls of 25 m each. What is the least number of rolls of ribbon that Jess needs to buy?

This tough PSLE math problem stumped most of the kids who took the PSLE in 2017.

This is the mistake they did:

110 cm x 200 ribbon pieces = 22 000 cm = 220 m

220 ÷ 25 = 8 rolls, remainder 20 metres

“So, we need one more roll to make 200 ribbons.” 

8 + 1 = 9 rolls

The correct answer, however, is ‘10 rolls’.

Why is it 10 rolls?

Let’s look at the question again.

Step 1: Understand the problem

Let’s draw a ribbon roll of 25 m.

Then, mark out 1.1 m (110 cm) on the ribbon roll. 1.1 m is the length of one ribbon piece.

Let’s divide 25 m by 1.1 m to find out how many ribbons can be made from one roll. 

25 ÷ 1.1 = 22 ribbons rolls, 0.8 metres unused

The 0.8 metres is not enough to make one ribbon. Let’s cast this aside.

“If I have a remainder of 0.8 metres for one roll, then every roll is going to have a remainder of

0.8 metres.”

“I can’t join 0.8 metres to another 0.8 metres. Who makes a ribbon piece out of remainders? It’s going to look ugly.”

Now, for step 2. 

Step 2: Draw a diagram to clearly identify the key information in the problem

Also, 

25 m ÷ 1.1 m = 22 ribbons rolls, 0.8 metres unused.

Let’s mark out 0.8 m to show the unused length on the roll on the same diagram.

We also know that 22 ribbon pieces can be made from one ribbon roll, with 0.8 metres unused.

A-ha! 22 ribbons for one roll. How many rolls do you need for 200 ribbons?

That brings us to Step 3. Solve it.

Step 3: Solve it

Remember that you can make 22 ribbon pieces from one roll. Jess needs to make 200 ribbon pieces.   

Let’s draw a diagram to capture this information clearly.

From the diagram, to solve for the number of ribbon tapes used to make ribbon pieces, we divide 200 ribbon pieces by 20 ribbon pieces to find the number of rolls 

200 ÷ 22 = 9 rolls, remainder 2 ribbon pieces.

9 RIBBON ROLLS! Yes, you…wait.

There is a remainder of 2 ribbons. 9 ribbon rolls are not enough!

So, we add 1 more roll to 9 rolls.

• 9 + 1 = 10
• Jess needs 10 ribbon rolls! Question solved!

So, what is Step 4 after we have solved the question?

Step 4: Check if you have solved it correctly

You have understood the question by this point. You are now checking if your calculations are correct.

So, let’s use 10 rolls to find out if we can make two hundred ribbon pieces.

Remember that 22 ribbons can be made from 1 ribbon roll.

So, 

10 x 22 = 220 ribbon pieces

But, that’s more than 200 ribbon pieces! Is that the right answer?

So, let’s try 9 rolls.

9 x 22 = 198 ribbon pieces.

Wait! That’s less than 200 ribbon pieces!

Summary

Here is the final solution for the ribbons question.

• 25 m ÷ 1.1 m = 22 ribbons, 0.8 m unused
• 200 ÷ 22 = 9 ribbon rolls, remainder 2 ribbon pieces
• 9 + 1 = 10 ribbon rolls

 Answer: 10 ribbon rolls

PSLE challenging math problems: 2016 Paper 2 Question 14

Suyin baked some pies. She gave 1/5 of them to her relatives and 30 of them to her friends. She was left with 2/3 of the pies. Then she packed these into 18 boxes. Some boxes contained 6 pies while the rest contained 12.

•  How many pies were packed into the 18 boxes?
• How many boxes contained 6 pies?

Step 1: Understand the problem

Suyin’s total pies can be put into two groups. One group of pies is given to relatives and friends. The second group is packed into 18 boxes. You should get something like this in Picture 1 below. 

Picture 1

Let’s look at ‘2/3 of total pies’ again from the image above. Where did 1/3 of total pies go to?

Ah! They are given to friends and relatives! So, this leads us to Step 2.

Step 2: Plan a solution (Bar model drawing)

Let’s draw a model to show 1/3 of the total pies given to her relatives and friends and 2/3 packed into 18 boxes. 

Recall that 1/5 of the total pies were given to relatives and 30 to her friends.

So, update the model above with this information.

A-ha! Did you see it? 

1/3 of total pies = 1/5 of total pies + 30 pies

So,

2/3 of total pies

= 2 x 1/3 of total pies

= 2 x (1/5 of total pies + 30 pies)

AH-HA! New discovery!! Let’s update the bar model with this new discovery!

Did you notice? We have 1/5 appearing 3 times. We also have 30 pies appearing 3 times. 

Let’s ‘tidy up’ the model. Let’s collect all the 1/5 parts together and the 30 pies parts as another group. 

This is what the ‘tidied’ model looks like below.

So, there are 3/5 of total pies in the model above. Where did the remaining 2/5 of total pies go to?

A-ha! They represent the 90 pies! Let us update the model with this discovery!

It’s very clear that we can find the total number of pies. Once we find this number, we can solve for the number of pies packed into 18 boxes!

90 ÷ 2/5 = 90 x 5/2 = 225 (total pies)

Now we can solve Part A: the number of pies packed into 18 boxes.

Part A

Step 3: Solve it

Recall that 2/3 of the total pies were packed into 18 boxes. So, 

2/3 x 225 = 150

Let’s check if 150 is the number of pies packed into 18 boxes.

Step 4: Check it

225 x 1/3 = 75 (pies given to relatives and friends)

75 – 30 = 45 (number of pies given to relatives)

Because the pies given to relatives is 1/5 of the total pies,

45 ÷ 1/5 = 225

Answer: 150 pies

Summary 

Here is the final solution for Part A.

90 ÷ 2/5 = 90 x 5/2 = 225 (total pies)

2/3 x 225 = 150

 Answer: 150 pies packed for 18 boxes

Now, for Part B!

Part B

Step 1: Understand it

Recall that there are 18 boxes of pies. Some boxes have 6 pies each. Some boxes have 12 pies each. 

18 boxes. 150 pies. 

Step 2: Plan it

Let’s assume all 18 boxes have 6 pies each. (Assumption method)

With the model, this is how 18 boxes of 6 pies each look like.

So, the total number of pies packed into 18 boxes is

18 x 6 = 108 pies

We have 150 pies, not 108 pies, to be packed into 18 boxes. We missed out some pies to be packed. Let’s find the number of pies we missed out for packing.

150 – 108 = 42 pies missed out for packing

Recall that we can also pack 12 pies into each box.

That means we can add 6 more pies into each box to make 12 pies in a box!

We simply take 6 pies each time from the 42 pies missed out for packing.

How many times do we take 6 pies?

42 ÷ 6 = 7

7 times! Let’s update the model with this new insight! 

This means there are 7 boxes of 12 pies each! Time to solve it!

Step 3: Solve it

18 – 7 = 11

11 boxes of 6 pies each.

Step 4: Check it

7 boxes of 12 pies. 11 boxes of 6 pies. 

Let’s see if they add up to the total of 150 pies that Suyin packed for the 18 boxes.

7 x 12 = 84

11 x 6 = 66

84 + 66 = 150

Bingo! 

Answer: 11 boxes of 6 pies each

Summary

Here is the final solution for Part B.

18 x 6 = 108

150 – 108 = 42

 42 ÷ 6 = 7

18 – 7 = 11

 Answer: 11 boxes of 6 pies each

You must be brimming with confidence now after solving two PSLE challenging math problems.

Let’s take down the last challenging question together!

PSLE challenging math problems: 2015 Paper 2 Question 17  

Three girls, Amy, Beth and Cindy had the same number of coins. Amy and Beth each had a mix of fifty-cent and ten-cent coins. Amy had 9 ten-cent coins while Beth had 15 ten-cent coins. Cindy had only fifty-cent coins.

•  Of the three girls, who had the most money and who had the least?
• What was the difference in the total value of Amy and Beth’s coins?
•  Beth used all her fifty-cent coins to buy some food. She then had $10 less in coins than Cindy. How many fifty-cent coins did Cindy have? 

Step 1: Understand it

Three girls. Two coin types: fifty-cent and ten-cent coins. Let’s draw a simple diagram to understand what these information means. You should get something like the picture below.

That was simple! Now, let’s plan for a way to understand the question deeply.

Step 2: Plan it

Recall that Amy had 9 ten-cent coins and Beth had 15 ten-cent coins. 

Assume they each had 15 coins, 

• Amy would have

 o 15 – 9 = 6 fifty-cent coins,

• Beth would have

 o 15 – 15 = 0 fifty-cent coins

• AND Cindy would have

o 15 fifty-cent coins.

So, let’s sum it up in a single diagram what it looks like if each girl had 15 coins each. You should get something like the picture below

From the above picture, 

Amy has 390 cents,

Beth has 150 cents, and Cindy has 750 cents.

So, Cindy had the most money. Beth had the least money.

Let’s solve Part A now based on Step 1 and Step 2.

Part A

Step 3: Solve it

Most money: Cindy

Least money: Beth

Step 4: Check it

What does the ‘+ 50 cents’ mean below each of their totals? 

Recall that Amy had 9 ten-cent coins and Beth had 15 ten-cent coins. So, their number of ten cent coins is fixed. 

So, if they have 16 coins each, it simply means 

• Amy had 7 instead of 6 fifty-cent coins,
• Beth had 1 instead of 0 fifty-cent coins, and
• Cindy had 16 fifty-cent coins instead of 15 fifty-cent coins.

So, they each had 1 more fifty-cent coin if we change their total number of coins from 15 to 16.

And their difference in the amount of money still remains the same.

You can assume 17, 18 or 2000 coins for each of them. You are simply increasing their amount of money by the same number of fifty.

Their difference in value remains the same. 

Cindy will still have the most money. Beth will still have the least money.

Answer:  

Most money: Cindy

Least money: Beth

Summary

Here is the final solution for Part A.

Most money: Cindy

Least money: Beth

Part B will take you only at most 1 minute. Part C will take you at most 3 minutes. 

Let’s solve for Part B.

Part B

390 – 150 = 240 cents

Let’s check for working accuracy.

240 + 150 = 390 cents

Bingo.

Answer: 240 cents or $2.40

Summary

Here is the final solution for Part B.

390 cents – 240 cents = 150 cents

Answer: 150 cents or $1.50

Let’s solve for Part C.

Part C

Recall that Beth used all her fifty-cent coins. 

That means she is left with only ten-cent coins.

Recall that Beth had 15 ten-cent coins. Or $1.50.

Now, she had $10 in coins less than Cindy.

That means…

Cindy had $10 more than Beth.

So, add $10 to $1.50 to find Cindy’s amount of money.

Step 3: Solve it

$10 + $1.50 = $11.50

Cindy has $11.50. 

WAIT!!! We are not done. We have to find the number of fifty-cent coins Cindy had.

So, we divide $11.50 by $0.50 to find the number of fifty-cent coins.

$11.50 ÷ $0.50 = 23

Step 4: Check it

Check if 23 is the correct number of fifty-cent coins for Cindy.

23 x $0.50 = $11.50

$11.50 – $10 = $1.50 (Beth’s 15 ten-cent coins)

Answer: 23 fifty-cent coins

Summary

Here is the final solution for Part C.

$0.10 x 15 = $1.50 

$10 + $1.50 = $11.50

$11.50 ÷ $0.50 = 23 

 Answer: 23 fifty-cent coins

So there, you have just learnt how to solve tricky PSLE Math questions.

 Here are the four steps to solving a tricky PSLE Math question.

Step 1: Understand the problem

Step 2: Draw a diagram to clearly identify the information in the problem

3: Solve it

Step 4: Check if you have solved it correctly 

Discussions over tough PSLE maths questions are undoubtedly an annual affair.

Apart from leaving some, if not many students deflated for not being able to tackle the questions, it has also led to parents taking issue with the difficulty as they take to online platforms to voice their concerns.

As a matter of fact, maths questions in recent years are more than just tackling standard problem sums. More creative questions have emerged that tests students’ ability to approach them with greater thinking and visualisation. 

With this year’s PSLE maths examinations soon to take place on 2 October 2020, we have compiled a list of challenging PSLE Maths Questions from previous years that can better prepare your child in tackling the examinations.

Note: It is still important to note that the level of difficulty of these maths questions may be subjective, which depends on the student’s ability as well.

PSLE Math Prep: More Challenging PSLE Math Questions Over the Years

1. Year 2009

Jim bought some chocolates and gave half of them to Ken. Ken bought some sweets and gave half of them to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim’s sweets to chocolates becomes 1:7 and the ratio of Ken’s sweets to chocolates becomes 1:4. How many sweets did Ken buy?Answer: 68 sweets.

Why it was difficult: This PSLE math question requires more advanced methods to solve, in particular, the simultaneous equations method which is a secondary school maths topic. While pupils can solve it using the ‘Singapore Model Method’ or ‘Guess and Check’, the most effective way would be using the ‘Units and Parts’ method. 

2. Year 2012

A bakery and a library are 120m apart. They are located between Hong’s house and Jeya’s house, as shown below. The bakery is exactly half-way between the two houses.

One day, Hong and Jeya started cycling from their houses at the same time and they arrived at the library together. Jeya cycled at 70m per min while Hong cycled at a speed 15m per min faster than Jeya.

a) How much further did Hong cycle than Jeya? Answer: 240m

b) How far is Jeya’s house from the library? Answer: 1,120m 

Why it was difficult: While the question appeared straightforward, there is more than meets the eye. In the event of using the straightforward approach, students might derive that Hong cycled 120m further than Jeya. 

However, upon closer look will students notice that apart from Hong who went 120 metres extra, Jeya went 120 metres less than half the distance between two houses. Therefore, the conclusion is Hong cycled 240 m more than Jeya. 

3. Year 2013

One machine took 70 minutes while another took 100 minutes to print the same number of copies of a newsletter. The faster machine printed six more copies of the newsletter per minute that the slower one.

a) The slower machine completed the job at 1pm. At what time was the printing started? Answer: 11.20am

b) What was the total number of copies printed by the two machines? Answer: 2,800

Why it was difficult: Like the maths question that perplexed students in 2012, this is another concept—rate—that isn’t taught until secondary school. The question requires students to apply the concept of speed in which they have learnt, to the concept of rate in their own terms. 

4. Noteworthy: Year 2015

While not exactly meant for PSLE-level maths examinations, one of the most puzzling questions for students and parents alike is this question that has even made its rounds on U.S. site BuzzFeed.

Question: Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates. 

May 15, May 16, May 19

June 17, June 18

July 14, July 16

August 14, August 15, August 17 

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.

Albert: Then I also know when Cheryl’s birthday is.

So when is Cheryl’s birthday?Answer: July 16

Why it was difficult: There is no direct logic to follow in this question that also involves a tricky word play. Tackling the question would require creative thinking and using the process of elimination.

5. Year 2017

Jess needs 200 pieces of ribbon, each of length 110cm, to decorate a room for a party. Ribbon is sold in rolls of 25m each. What is the least number of rolls of ribbon that Jess needs to buy? Answer: 10 rolls

Why it was difficult: This question requires some creative problem solving on students’ part. The key is to first recognise that the unused portion of each roll is discarded. Following that, students can use division to solve the problem. 

6. Year 2019

The year in which many found the maths questions to be particularly difficult, such as this pattern question:

a) Fill in the graph above. Answer: Total number of triangles in Figure 1: 1, Figure 2: 4, Figure 3: 9, Figure 4: 16, Figure 5: 25; white triangles in Figure 5: 15; grey triangles in Figure 5: 10.

b) Find the total number of grey and white triangles for Figure 250. Answer: 62,500 triangles

c) Find the percentage of grey triangles in Figure 250. Answer: 50.2 per cent

Why it was difficult: This is unlike what students have been exposed to in previous years’ PSLE Maths papers and papers from other primary schools. For such pattern questions, if a particular row’s value is higher than the other, it should be consistent when moving from one figure to another, in ascending order. However, in this question, their values alternate for the white and grey triangles. 

Another complex question from the 2019 PSLE maths paper came into spotlight as well, involving five identical semi-circles. 

Why it was difficult: The question requires a level of analytical and creative thinking, not expected of a Primary 6 student. Students can ultilise a variety of ways to solve this question, from using multiple equations, drawing a model and labelling overlapping lengths—or even redrawing the figure to complete the circles. 

One Singaporean mother was so upset at the difficulty of the 2019 PSLE math paper that she took to Facebook to write an open letter to Education Minister Ong Ye Kung. 

“Make it challenging. Make it doable, I agree. But what I don’t understand is the cruel decision to make it so unreasonably tough that children came out crying, deflated, demoralised and crushed.”

While it’s understandable that parents would be concerned about their children’s well-being, PSLE is just one of the many checkpoints in a child’s education journey. Besides, there are varying difficulty of questions in these papers to test different students’ abilities—which is essentially the point of these examinations.

Perhaps it would offer a sense of relief for parents to know that having the young ones face such challenges now will give them the opportunity to build resilience and rise through tougher times in the future. 

This article was contributed by Cai Shaoyang (or Brandon) who is a former primary school teacher with over 12 years of teaching experience in Singapore Maths. 

He is now a full-time tutor who helps 7 -12-year-old children score an A in Maths exams and tests with minimal stress and maximal confidence.

To help your child learn math with confidence, sign up for his weekly newsletter here.

With input from Jia Ling

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PSLE Maths Prep: Tough Examination Questions From Over the Years To Challenge Your Child