Thx :)

Hint: Area of A and C = Area of B and D = half area of rectangle

Raymond Ng should be 1/2 x 10 x 5

Tks

Edited. Tks 😂

We know SU is 10 cm and SV:SU = 2:5. So 5u=10, u=2, so SV = 4. The area of SQV = 8 cm^2, so you can find the area of the shaded region by subtracting 8 from the area of the whole triangle.

In an isoceles triangle with right angle p, the 2 sidesPR and PQ wil be the same length so that the angle R = Q. Since PQR and STU are identical isoceles triangles, ST=SU=PQ=PR. So the shaded area would be the area of triangle STU (1/2 10 *10= 50) minus area of triangle SQV (1/2 4*4=8) = 42cm2. No need to find length of PS, but in case you are wondering, it would then be 10 - 4=6cm

Tks all

Tks. But how to find the length PS?

You don't need PS. PQR is isosceles, so PQ = 10.

Hi. The number of $5-notes must be even and the number of $2-notes must be a multiple of 5 so that we can add up nicely to $110.

Just found a shorter and easier solution

https://www.facebook.com/media/set/?set=oa.538423516314309&type=1

Grand Strategy: Systematic Search we work with the bigger notes $10 and $5 first. First, assume # $10 notes = 0 starting off with 22 × $5 + 0 × $2 = $110 20 × $5 + 5 × $2 = $110 we can get 18 × $5 + 10 × $2 = $110 *** Every time we take away two $5 notes, we add five $2 notes, so the number of notes increases by 3 in this process. Next, assume # $10 notes = 1 we can have 20 × $5 + 0 × $2 = $100 and try to change the $5 to $2 notes as above total # notes = 21, 24, 27, 30, ... you will never get 28 Next, assume # $10 notes = 2 we can have 18 × $5 + 0 × $2 = $90 and try to change the $5 to $2 notes as above total # notes = 20, 23, 26, 29, ... you will never get 28 Next, assume # $10 notes = 3 Start with 3 × $10 + 16 × $5, and 19 notes total By exchanging $5 with $2 notes, it is possible to get total # notes = 19, 22, 25, 28 We eventually get another solution 3 × $10 + 18 × $5 + 10 × $2 = $110 *** Every time we add one $10 note, we take away two $5 notes. effectively reducing the number of notes by 1 *** Observe the 3-cycle pattern for the number of $10 notes If the number of $10 notes is not a multiple of 3, you will never get to 28 notes. So the next thing to try is 6 × $10, 6 × $10 + 2 × $5 + 20 × $2 = $110 So the next thing to try is 9 × $10, but this does not work from here onwards. So there are only three solutions.

Christina Cheong Beng Na is this comprehensible?

I would use guess&check using Teo Kai Meng tips...

My latest solution incorporates Teo Kai Meng tips, plus some more insights to narrow down the search. Otherwise, you still have a lot of random guessing.

Guess & check + assumption

Using trial and error method looks more presentable to a Pr 4 child. 1) By telling them that since total notes used is 28 pieces, likely there will be more $5 and $2 notes than the $10 notes. 2) Also total amount from $5 and total amount from $2 added up must be an even number examples $100, or $90 or $80. 3) $10x 1 + $5 x 14 + $2×15 = $110, but there are 30 pieces of notes, so this ans is wrong. 4) $10× 2 + $5×12 + $2×15 = $110, but there are 29 pieces of notes, so this ans not accepted. 5) $10× 3 + $5×10 + $2×15 = $110, and there are 28 pieces of notes, so this ans is correct and accepted. I hope this method, ppl find more user friendly and easy to understand. All we need to note is sum total amount is $110, and all the pieces of notes added up total 28, then this is 100% correct answer. The catch is each time u reduced two $5 bill, u have to increase one $10 bill, like money exchange.

Key Insights A) Every time you exchange one $10 note for two $5 notes, the total number of notes increases by 1 (decreases by 1 if it's the other way round) B) Every time you exchange two $5 notes for five $2 notes, the total number of notes increases by 3 (decreases by 3 if it's the other way round) Starting with only 22 $5 notes, we can do 'B' procedure to get 25 notes, and do 'B' again to get 28 notes. that's one solution. 10 × $2 + 18 × $5 + 0 × $10 = $110. Once you get one solution, you can get the other two solutions by maintaining the same number of notes as follows. Key Insight C) To maintain the same number of notes, every time you do a procedure B (exchange two $5 notes for five $2 notes), you can do procedure A in reverse 3 times (i.e. exchange six $5 notes for three $10 notes). Altogether you exchange eight $5 notes for three $10 notes and five $2 notes. 15 × $2 + 10 × $5 + 3 × $10 = $110. Do combo C again. Get the third solution. 20 × $2 + 2 × $5 + 6 × $10 = $110. You cannot do it anymore. So stop.

Iterate by reducing 10s, and increasing 2, then 5. 5 iterations but very straight forward. Cheers.

Shortest iteration but you need to know you are reducing your total notes to just below 28.

If you start with all S$5 notes. Can be done too. Same concept, reduce to just under 28 notes first.

Primary kid student can't really understand. I tried algebra with my kids.. theybsay so confusing

somehow we need to reduce the "search space" and look for patterns instead of random guessing. i uploaded another solution below. Hope it is acceptable for young kids.

Wow.. let me digest it.. 😊

Actually this is the idea I used. For those who claim there is a better or easier method, I am not sure if they even tried to work it out in full. If they did, they would realise that there is still a lot of trial and error to do. You need one more key observation to cut down the amount of work.

Calculated guess is easier than algebra for such and several other types of questions. Students do not demonstrate understanding in solving such problems with algebra, but with calculated guess and check, if they understand grouping well enough, it does not take a lot of time to get the answer. Algebra isn't always the best method and may not be able to solve all problems and it takes longer than calculated guess and check method to solve the problem.