May i know how you get this 200u (cookies)?

Working backwards, 70% of remainder = 134 + 6 = 140 Remainder = 140 × 10/7 = 200 15% of total = 200 - 5 = 195 Since 15% = 3/20, Total number of cookies = 195 × 20/3 = 1300

Work backwards.. See red ink then black ink.. Hope it helps !

Thank you all. Hopefully my girl can understand.

Candy Chan...u can just assume any number of units..to work from the front But pls do follow Xavier Sng and Pauline Vong Nyit Li methods of working backwards.. I will not be able to do that as my reverse gear is faulty..haahaa. Pauline ...your model methods is quite interesting..

You are most welcome, Candy. Regards

Not to worry, Izam. Your method is perfectly fine. We are all providing different variations that is all, and it is beneficial if Candy's daughter can choose one of the approaches that she is most comfortable with.

You are right meh, Xavier Sng! She has more choice to choose. Thank q.....

Do they give the mass of each coin?

Maggie Lukes, no leh..

How to solve? Make assumption - note the lightest; 50 cents heaviest and 20 cents in the middle. Maybe actually weigh items?

since there are 3 types of coins and there are 9 in total, that will be 3 sets . N 52.47/3 give us 12.47. Hence, implying that there are 3 coins for each set.

If you have the coins, you could weigh them, but that seems a lot for a math question.

I agree with William Chong's answer. I also concur with Maggie Lukes that this question is asking for a lot from a ten-year-old (P4 level).

By the way 52.47 ÷ 3 = 17.49

But in the answer have the weight of the coins. These were done by the child of given in class by teacher?

If you check the workings in the picture, the masses are given as 6.3g, 7.29g, 4.5g respectively. So I believe there were additional information given but not shown in the picture. Based on guess and check, the number of coins are 2, 3, 4, respectively.

If indeed the given masses of the three types of coins were given ab extra, a student strong in mathematics and observation can determine the number of each type of coins without the use of guess-and-check for this particular question. The student can remove the masses of one of each coin initially. 52.47 - 6.3 - 7.29 - 4.5 = 34.38 Since the mass of the 50-cent coin is given to two decimal places, and the hundredths place of that mass is 9 while hundredths place in 34.38 is an 8 suggests that two 50-cent coins can be removed from 34.38g. 34.38 - 7.29 - 7.29 = 19.8 Only the mass of the $1-coin has a 3 in the tenths place. Since there are only 4 more coins to account for, that can only point to the fact that the mass of one single $1-coin can be removed from 19.8g. 19.8 - 6.3 = 13.5 13.5 = 3 × 4.5 accounting for three 20-cent coins. So the numbers of $1, 50-cent and 20-cent coins are 2,3,4 respectively.

Alternatively, the 2nd Decimal place is 7. So there must be 3 50cent coins. So 1st try: 3 3 3 -> exceed 2nd try: 2 3 4 -> bingo ☺

Good point. 52.47 - (3 × 7.29) = 30.6 accounting for three 50-cent coins. 30.6g right? No brainer, without doing a guess-and-check for 3-3-3 there can only be two $1-coins since the tenths place of 30.6 is 6. 2,3,4. No guess-and-check needed. Better.

Thanks everyone!

Glad to help out, Joyce. Regards

15

15..

What techniques did you guys use to solve it?

Depending on the test objective, it may be a poor test item. It is likely used to assess if students can tell parallel sides and students who know parallel sides get this item incorrect. Non-trivial ability in counting is needed to get the correct answer. It is good for class discussion though.

Poorly worded. what do you mean by a "parallel" side? ("parallel" is a relation between at least two objects, not an adjective describing single objects ... unless you define ... )

I think this was from an 11+ (entrance exam to grammar schools and some independent schools) moc test in the UK. A lot of non-verbal reasoning stuff in 11+ too, which I totally hate!

...or I should say I'm very rusty with non-verbal reasoning!

need 3D visualisation. but you can use left-brained logic too: e.g. every long side corresponds to two short sides. But the thing that annoyed me was the word "parallel". it's like the word "equal" in "all men are equal. some are more equal than others".

Lousy question, there are many paired combinations which can easily confuse students. For Shape A, there are 5 lines which make up the longest edges of the figure and that alone has 10 possible pairs of combinations. The shorter edges at the side also have 5 possible pairs of combinations. Therefore, 15 pairs of parallel edges for shape A.

if you define a parallel side as "a side for which there exists at least one other side that is parallel to it", then 15 is the legit answer. Otherwise it is more precise to say that there is one equivalence class of parallel sides consisting of 5 members, and 5 other equivalence classes of parallel sides consisting of 2 members each.

The answer they are looking for is 15.

It is a poor question though. I think most students will write 10 as their answer,

Good day, Mohi. It may not be a good question, but in life most issues we encounter are not good ones too. Haha. Nevertheless it is a fun question to discuss for adults and an eye-opener for youngsters to take a crack at. The question can be a seperator to ferret out youngsters who can visualise a solid from a two-dimensional diagram. Thank you very much for bringing it up.

Definitely good for discussions Xavier!

Can it be 5 pairs since they didn't mention that is a 3D shape?

Jason you got a point there, it just mentions shape so taking the shape to be what we see, there should only be 5 pairs of parallel lines. Good one there.

I am more interested in the thought process of the student in order to solve this.

4+3+2+1 + 5=15

4+2+1+3=15

Sorry 4+2+1+3+5 =15 15 is the answer

15

One part of it is like the subject permutation and combination and the other two ends are probably testing on visual. Some students will like such question.

7 pairs

100% - 30% = 70% 70% -> $245 100% -> $245 / 70 x 100 = $350

The working in the photo shows that the student is not clear about what the figures given mean -- that 245 is after discount and therefore not represented by 100%. Perhaps it's a good idea to point out to him/her that 100% should represent the original value or the whole. So that's why it's not correct to work out 30/100 *245 for the answer. A model can be drawn to show that 70% rep 245 and we are looking for the amount represented by 100%.

Tks to all you😀

Tks Vei Li Soo!!

let take the value to be x x-30%x=245( according to ques.) x-0.3x=245 0.7x=245 x=245/0.7 x=350