Insights & Strategies: A Review of The 2023 PSLE Math Paper 2

Com­pared to the 2022 PSLE Math Paper 2, the 2023 PSLE Math Paper 2 was eas­i­er for the most part. Although there was a greater per­cent­age of Area and Perime­ter ques­tion types that test­ed stu­dents on their visu­al spa­tial­i­sa­tion abil­i­ties, stu­dents who are less spa­tial­ly gift­ed can also score well in these sec­tions just by using the tech­niques they have learnt. These tech­niques include rear­rang­ing and form­ing the com­po­nents of a com­plex shape to make more sense of the shapes in the ques­tion, or to solve the ques­tion in few­er steps. Stu­dents may also form equa­tions based on the com­po­nents of a com­plex shape and solve from there. We picked out 6 ques­tions from the Paper 2 and let us take you through an in-depth step-by-step solu­tion to these ques­tions.

Question 9 

TTA stu­dents have been trained to spot cer­tain phras­es that will help them to iden­ti­fy the ques­tion type. This enables them to apply the tech­nique that they have learnt so that they can solve the ques­tions in the most time-effi­cient man­ner. 

As soon as stu­dents see the phrase “40% of the remain­der”, they will recog­nise it as a frac­tion-of-frac­tion ques­tion type and draw the drop-down mod­el. To keep the num­bers in the ques­tion small, stu­dents can change the per­cent­ages in the ques­tion to frac­tions. 

75% will there­fore become 3/4 and 40% will become 2/5 .

Peg­gy starts off by spend­ing 3/4 of her mon­ey on a cal­cu­la­tor. This means that she had 4u of mon­ey and she spent 3u of it on a cal­cu­la­tor and had 1u left. 

She then spent 2/5 of the remain­ing 1u on a book. We drop down the part of the mod­el labelled 1u and call it 5 parts, since the denom­i­na­tor of the frac­tion is 5. 

Of the 5 parts, she spent 2p on a book and had 3p left. 

3p is there­fore equal to $6.90. 

Do you realise that 1u and 5p are essen­tial­ly the same thing? In that case, we make 1u and 5p the same num­ber by mul­ti­ply­ing 1u by 5 to change it to 5p. 

Since we mul­ti­plied 1u by 5, we need to mul­ti­ply all units in the mod­el by 5. 

After going through the process, we can now see that Peg­gy had 20p of mon­ey at first. 

3p -> $6.90

1p -> $2.30 

20p -> $46

The answer is there­fore $46.

Question 11 

This is an inter­est­ing ques­tion that requires stu­dents to find the pat­tern. 

The rods are in between the straws as well as the sides of the grid.

If the fig­ure is 2 rods long, and 3 straws wide, this means that there will be 2 lay­ers (hor­i­zon­tal­ly), and 3 + 1 = 4 lay­ers (ver­ti­cal­ly). There­fore, there will be 2 × 4 = 8 rods. 

As for straws, if the grid is 2 rods high, this means that there will be 2 + 1 = 3 lay­ers of straws.
If this grid is 3 straws wide, there will there­fore be 3 × 3 = 9 straws. 

Let’s apply this pat­tern that we have found to part (a).

In a grid that is 3 rods long and 4 straws wide, there will be 3 lay­ers (hor­i­zon­tal­ly) and 4 + 1 = 5 lay­ers (ver­ti­cal­ly). There­fore, there will be 3 × 5 = 15 rods.

As for straws, if the grid is 3 rods high, this means there will be 3 + 1 = 4 lay­ers of straws.
If this grid is 4 straws wide, there will be 4 × 4 = 16 straws.

There­fore, the answer to part (a) is 15 + 16 = 31

In part (b), the grid is 4 rods long and 52 straws wide. All we need to do is to apply the pat­tern we have found to it. 

If the grid is 4 rods long and 52 straws wide, this means that there will be 4 lay­ers (hor­i­zon­tal­ly) and 52 + 1 = 53 lay­ers (ver­ti­cal­ly). There­fore, there will be 4 × 53 = 212 straws. 

As for straws, if the grid is 4 rods high, this means there will be 4 + 1 = 5 lay­ers of straws.
If this grid is 52 straws wide, there will be 52 × 5 = 260 straws.

The answer to part (b) is there­fore 212 + 260 = 472

Question 12

When­ev­er a ques­tion asks for the perime­ter of the fig­ure, it is a good idea to pick out the lines that form the perime­ter so that no line is left out. 

As colour cod­ed in pur­ple, blue, and yel­low, the pur­ple lines form the cir­cum­fer­ence of a cir­cle, and so do the blue lines and yel­low lines respec­tive­ly. 

To get the perime­ter of the shad­ed part, we find the perime­ter of the 3 cir­cles which each has a radius of 7cm. 

22/7 × 14cm = 44cm 

44cm × 3 = 132 cm

Anoth­er trick to solv­ing ques­tions quick­ly would be to apply the rearrange and form method. The spaces in between the cir­cles are actu­al­ly made up of 4 L‑brackets. When rearranged around a cir­cle, these 4 L‑brackets and the cir­cle form a square. Through some sim­ple shift­ing, you will see that the shad­ed parts can be rearranged to form 2 squares and 1 cir­cle. 

Area of 1 square -> 14cm × 14cm = 196cm²

Area of 1 cir­cle -> 22/7 × 7cm × 7cm = 154cm²

Shad­ed part -> 196cm² + 196cm² + 154cm² = 546cm²

Question 13

This has been a pop­u­lar vol­ume ques­tion type in recent years and TTA stu­dents have been taught the fol­low­ing tech­nique when it involves an unknown height of the water lev­el. 

From the ques­tion, we know that the total vol­ume of water in both tanks is 21 400 cm². We also have the base areas of both tanks. The height of the water lev­el is unknown but what we do know is that the height of the water lev­el in tank Y is 5cm high­er. There­fore, let the height of the water lev­el in Tank X be rep­re­sent­ed by 1u, and the height of the water lev­el in Tank Y be rep­re­sent­ed by 1u + 5. 

The vol­ume of water in Tank X is -> 20 × 10 × 1u = 200u

The vol­ume of water in Tank Y is -> (30 × 20 × 1u) + (30 × 20 × 5) 

                                                                    = 600u + 3000

Total vol­ume in both tanks is -> 200u + 600u + 3000

The total vol­ume of water in both tanks is equal to 21 400cm³. 

Let’s form an equa­tion and solve for 1u, which rep­re­sents the height of the water lev­el in Tank X. 

200u + 600u + 3000 = 21 400

800u + 3000 = 21 400

800u = 18 400

1u = 23

The height of the water lev­el in tank X is there­fore 23cm. 

For part (b), let’s focus on Tank Y. 

If the height of the water lev­el in tank X is 23cm, then the height of the water lev­el in tank Y is 23cm + 5cm = 28cm. 

Vol­ume of water in Y -> 30cm × 20cm × 28cm = 16800 cm³

Half of tank Y -> 16800 cm³ – 3300 cm³ = 13500 cm³ 

Capac­i­ty of tank Y -> 13500cm³ × 2 = 27000 cm³

Height of tank Y -> 27500 cm³ ÷ 600 cm² = 45 cm

The height of tank Y can also be found by tak­ing 3300cm³ ÷ 600cm² = 5.5cm to find out the decrease in height. The height of the water lev­el when tank Y is half-filled is 28cm – 5.5cm = 22.5cm. Hence, the full height of Tank Y is 22.5cm × 2 = 45cm. 

Question 16

This is a gap and sub­ject ques­tion type. Stu­dents have 1 rule to remem­ber: The num­ber of gaps is always 1 less than the num­ber of sub­jects. 

This means that if there are 5 lamp­posts, there will be 4 gaps. If there are 10 lamp­posts, there will be 9 gaps. 

24 pins were placed at an equidis­tance apart along the rec­tan­gu­lar board, and we want to find out how many pins there are along the length. 

First of all, we need to elim­i­nate the four cor­ners so that we do not dou­ble count them.
24 – 4 = 20

We are left with 20 pins now that we have placed a pin at each cor­ner.

Accord­ing to the fig­ure, there are 4 pins along the breadth, with 2 pins in between the cor­ner pins. Along 2 breadths, we will be able to place anoth­er 2 + 2 = 4 pins. 

This leaves us with 20 – 4 = 16 pins

These 16 pins will then be dis­trib­uted equal­ly between the 2 lengths.

16 ÷ 2 = 8. 

We will end up plac­ing 8 pins in between the cor­ner pins along the length. 

This means that we will have a total of 8 + 2 = 10 pins along the length of the rec­tan­gu­lar board. 

Part (a) is ask­ing for the length of the rec­tan­gu­lar board. 

If there are 10 pins, it means that there are 10 – 1 = 9 gaps.

In the ques­tion, it was stat­ed that PQ = QR. This means that the tri­an­gu­lar cards are isosce­les tri­an­gles. If you look at the gap in between the shad­ed tri­an­gles, the white tri­an­gles have the same 2 equal sides as PQ and QR, this means that they will form the same angle and the longest side of the white tri­an­gle will also be 40cm.

In that case, the length of AB will there­fore be 9 × 40cm = 360cm.

For part (b), it is not dif­fi­cult to find the area of 1 card, but it may be tricky to fig­ure out how many cards were placed along the lengths. 

If you noticed, the cards placed along the breadths have the long side of the card sit­ting along the breadth. How­ev­er, the cards along the lengths of the board have only the tip of the card sit­ting along the length. 

You can think of the white tri­an­gles as the “sub­jects”, and the black tri­an­gles in between as the “gaps”. Since you already know that there are 9 white tri­an­gles along the length, this means that there will be 1 less black tri­an­gle than white tri­an­gles. There are 9 – 1 = 8 tri­an­gles along the length of the board. 

In total, we have 3 + 3 = 6 tri­an­gles along the breadths, and 8 + 8 = 16 tri­an­gles along the lengths, which gives us 22 tri­an­gles.

Area of 1 tri­an­gle -> 1/2 × 40cm × 20cm = 400cm²

Total area of the cards -> 400 cm² × 22 = 8800 cm²

Question 17

This is not a tricky ques­tion if stu­dents know the trick to it.

Let’s say that the three trapez­i­ums in fig­ure 2 are not touch­ing. The total perime­ter of these trapez­i­ums would have been 96cm × 3 = 288cm.

When you com­bine the trapez­i­ums, the cir­cled parts were sub­tract­ed from the perime­ter of the fig­ure. Do you see how many PQ were removed in Fig­ure A below?

The cir­cled parts con­sist of 4 PQ.
There­fore, 4 PQ is equal to 288cm – 204cm = 84cm. 

Line PQ is there­fore 84cm ÷ 4 = 21cm. 

For 17(b), there are a few ways to solv­ing it but we will focus on the con­cept that 2 trapez­i­ums join togeth­er to form a rec­tan­gle. 

In Fig­ure 3, the 4 trapez­i­ums have an area of 1932cm².

Area of 1 trapez­i­um -> 1932cm² ÷ 4 = 483cm² 

Area of 1 rec­tan­gle -> 483cm² x 2 = 966cm²

The breadth of the rec­tan­gle is line PQ, which you already know the length of 21cm. 

There­fore, the length of the rec­tan­gle is 966cm² ÷ 21cm = 46cm

Line RS is 46cm – 21cm – 12cm = 13cm.

Conclusion

Over the years, we see shifts in trends and the evo­lu­tion of ques­tion types, and our cur­ricu­lum is updat­ed annu­al­ly to reflect the changes in the MOE syl­labus. We want to place our eggs in the right bas­kets, and have our stu­dents reap as much fruit from their labour. Our role as teach­ers is to help stu­dents con­nect the dots in the ques­tion and teach them how to make sense of infor­ma­tion in the ques­tion and solve it in the most time-effi­cient man­ner.

At Think Teach, Math lessons are struc­tured so that learn­ing is a breeze and not a tor­ture. Your child will be chal­lenged by tricky ques­tions and trained to solve com­mon­ly test­ed ones with ease. Con­sis­ten­cy is the key here to suc­cess. With a struc­tured approach to prob­lem-solv­ing, every Math ques­tion would seem famil­iar. Your child will no longer per­ceive every Math ques­tion as a daunt­ing moun­tain that he/she has not con­quered. As I like to say, “Prac­tice makes per­fect”. If some­thing at first seems unfa­mil­iar to you, keep on work­ing at it until you know it like the back of your hand!

This arti­cle was proud­ly writ­ten for you by TTA’s Math Team. Our strat­e­gy in help­ing stu­dents achieve exam excel­lence is sim­ply to expose them to every type of ques­tion that could come out for PSLE and arm them with our easy-to-apply tech­niques and would help them approach these ques­tions sys­tem­at­i­cal­ly

For 2023 PSLE Math Paper 1

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