# 3 PSLE Challenging Math Problems and How to Solve Them

Here are three PSLE challenging math problems from previous years, together with their solutions.

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 challenging math problems: 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 is the question that 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

## 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

### 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.&nbsp;

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.

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

### 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

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 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.

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.

### Summary

Here is the final solution for Part B.

390 cents – 240 cents = 150 cents

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)

### 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

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

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.  