1/5 of original amount = 1/3 of remainder after buying purse Hence 3/5 of original amount = the remainder after buying purse 2/5 original amount = $120 Original amount = $300

Thanks

Hi Raymond ng, how did you get 135 degrees

AOC = AOD + DOC = 45 + 90

Oh...Okay thank you I understand

Angle WOE = Angle BCE = 45 degrees Angle OAC = 1/2 of angle WOE = 22.5 degrees (Sum of interior angle = exterior angle, OA = OC) X = 22.5 + 45 = 67.5

Thanks Xavier Sng

It is nothing. Thank you.

just wonder of ABC is isosceles, why x not 60? :P

Because BC is shorter than the other 2 lengths

Assign 50u to represent total sum of money at first. 2/5 of the money, or 20u were spent on oranges. 1/10 of the oranges or 2u value worth of oranges were rotten. 18u worth of oranges are the remainder. After removing 1/3 of the remainder or 6u, 12u remain. Take 7/12 of 12u away and one has 5u left. 5u represent $70. 50u represent $700.

Or if the student sees the sequence : $70 × 12/5 × 3/2 × 10/9 × 5/2 = $700

Thanks....Xavier Sng

The pleasure is mine. Regards

May I know, this is a qn for which level?

Serene Ong ....P5

Thanks! Just to let u know that I explained the qn to my p3 using model. First i drew 10 blocks, shade 1/10 which is rotten, then remaining 1/3 (equivalent to 3/10 of the drawing) given to her brother. With that, we are left with 6/10 left. To account for 7/12 used to make juice, we divided the remaining 6 boxes into 2 to make 12 boxes. We must also do the same to the other 4/10 that were shaded earlier. Therefore, 5 units = $70 20 units = $280 2/5 = $280 5/5 = $700 Hope u can understand my explanation. Otherwise let me know... Will redraw my model and send it over

May I know if the answer is (12 x 12) - 2(12/3 x 7) = 88cm sq

Yes it is 88.

How did you get the relationship 12/3 x 7?

12/3 is the width of a rectangle form by A and the triangle besides it. I have problem explaining why it's 12/3. I actually fold a piece of paper to confirm that when you fold the way it's shown you divide the base length into 3 equally. I still cannot explain myself. Shuxx

That's the relationship I have problem with. Why must the bottom triangle be equilateral triangle?

If you reverse fold the 2 triangles, you get a square of 12x12=144. To get the shaded area, you need to subtract 4 right angled triangles (2 after fold + 2 before fold) of base 12÷3=4 vs height 12-5=7. The area will be 4x1/2×7×4=56 Shaded area will be 144-56=88cm sq.

Hi Pang Shing Hsiu may I know the concept behind 12 divide by 3? Thank you.

Hi Pang Shing Hsiu, when we fold the 2 corners to meet at the middle, is the relationship always 1u/1u/1u? What if we fold it at 3cm mark (instead of 5cm), is the relationship still 1u/u/1u ??

I think as long as you fold them and they are able to meet its 1/3 each cos I was using a random paper and it works. Just that I don't know the theory or concept established behind that. Like an origami law or something.

At Pri sch level, I believe there's missing info in this question. I resorted to Secondary trigo and concluded that the isosceles triangle isn't equilateral. Area = 88.56cm^2 to 2 dp.

Raymond Ng is correct. I use Pythagoras thm and similar triangles.

For that triangle to be equilateral (which should be indicated at Pri level), then the breadth of the upper rectangle will have to be 5.07cm but this is redundant info once equilateral triangle condition is stated.

I believe this is the original question - a past exam question. The shaded triangle is a equilateral triangle.

Adrian Ng, thanks for the clarification. The original problem is also erroneous. The 3 angles don't add up to 90deg (87.77deg). It's even worse than the current dimensions

Are you guys saying that technically forming equilateral is not possible? If so the question should not even be valid.

Of course it's possible. But 4 doesn't go with 9, neither does 5 with 12. Fix one & the other can be determined. These 2 lengths are dependent on each other due to geometrical considerations.

If geometrical considerations are ignored at Pri level, then it's a simple subtraction problem -- A Square - 2 Rectangles. If you browse through the history of this & other similar groups, such errors are nothing new. At Pri level, they are not detectable.

It's laying bad foundation and understanding especially math should comes with accuracy and precision. I'm loving this thread!