Q&A - P3 Math

[Keroppi30]

I got this question from a P3 maths assessment book:-


The mass of 50 twenty-cent and one-dollar coins is 600g. The mass of 2 one-dollar coins is as heavy as 3 twenty-cent coins. How much heavier is a one-dollar coin than a twenty-cent coin? The answer is 5g per the answer booklet.

[atutor2001]

Keroppi30:

I got this question from a P3 maths assessment book:-


The mass of 50 twenty-cent and one-dollar coins is 600g. The mass of 2 one-dollar coins is as heavy as 3 twenty-cent coins. How much heavier is a one-dollar coin than a twenty-cent coin? The answer is 5g per the answer booklet.

I think there is something missing in this question. Based on the question itself, there can be 26 (sorry should be 50) possible answers.

When mass of 2 $1-coin = mass of 3 20-cent coins it means that :

Ratio of the mass of 1 $1-coin : mass of 1 20-cent coin = 3 : 2 (flip!)

Let mass of 1 $1-coin = 3U
Let mass of 1 20-cent coin = 2U

Possible Answer 1 : If no. of $1-coin = no. of 20-cent coin

mass of 25 no. $1-coin + mass of 25 no. 20-cent coin = 25x3U + 25x2U = 600
Therefore 125U = 600g
Difference in weight between 1 $1-coin & 1 20-cent coin = 3U - 2U = 1U = 600/125 = 4.8g (possible answer 1)

Possible Answer 2 : If no. of $1-coin = 26 and no. of 20-cent coin = 24

mass of 26 no. $1-coin + 24 no. 20-cent coin = 26x3U + 24x2U = 600
Therefore 126U = 600g
Difference in weight between 1 $1-coin & 1 20-cent coin = 3U - 2U = 1U = 600/126=4.76... (possible answer 2 - this answer is acceptable because weight can be in fraction)

We can following the say way to work out the remaining 48 possible combinations

Based on the given answer of 5g we can calculate backwards to find the missing information in the question :

Difference in weight = 1U = 5g

600/5 = 120U

If all 50 nos. of coins are $1, TOTAL no. of units = 50 x 3 = 150U
When we change 1 no. of $1 to 1 no. of 20-cent, the TOTAL no. of units = 150n - 3U + 2U = 149U
It means that each time we change a $1 to a 20-cent there will be drop of 1U (150U - 149U = 1U)

The actual units is 120U, that is we need to reduce by 30U (i.e. 150U-120U = 30U)

So we need to change 30 nos. of $1-coin to 30 nos. of 20-cent coins

It means that, in the actual question, there are 20 nos. of $1-coins and 30 nos. of 20-cent coins

The missing statement could be : \"The ratio of the number of $1 coins to the number of 20-cent coins is 2 : 3\"

[Keroppi30]

:thankyou: atutor2001.


When I first saw the question, I thought there should be info on at least the ratio of the one-dollar coins to twenty-cent coins in the question too. But as this question was found under the challenging section thus I wasn't so sure. I was thinking there might be some techniques that requires higher-order thinking skills beyond me.

[tianzhu]

Hi

Hope this helps.

For P3 Maths, you could also use GC, but this method is more tedious.

Best wishes.

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[atutor2001]

Keroppi30:

I got this question from a P3 maths assessment book:-


The mass of 50 twenty-cent and one-dollar coins is 600g. The mass of 2 one-dollar coins is as heavy as 3 twenty-cent coins. How much heavier is a one-dollar coin than a twenty-cent coin? The answer is 5g per the answer booklet.

If the question is modified to :
[quote]There are 50 pupils sharing a total of 600 cards. Every boy is given same number of cards. Every girl is also given same number of cards but different from each boy. The sum of the number of cards from 3 boys is equal to the sum of the number of cards from 2 girls. What is the difference in the number of cards between 1 boy and 1 girl?[/quote]In this modified question, although the ratio of the number of boys and girls is also not given, however, there is only 1 possible answer because the number of cards of each boy and each girl cannot be a fraction and must be a whole number.

Have been wondering whether beside using \"Guess & Check\", is there any other \"mathematical method\" to find the answer.

Thanks in advance.

[Keroppi30]

Hi tianzhu


using your method are we assuming the ratio of one-dollar coins and twenty-cents coins is 2:3

Hi atutor2001

The answer to the original question has to be whole numbers too cos P3 hasn’t touch on decimals yet.

[tianzhu]

Keroppi30:

using your method are we assuming the ratio of one-dollar coins and twenty-cents coins is 2:3

Hi

There is no assumption; the solution is based on the information provided in the question.

When dealing with primary Maths and Science, it’s important to see what they’ve studied in their syllabuses, do not try to go beyond that. See them through the eyes of our young ones.

In this particular question, we are given that the mass of 2 one-dollar coins is as heavy as 3 twenty-cent coins. What does this mean? Imagine 2 one-dollar coins and as 3 twenty-cent coins put into a box (A group).So, you’ll have a total of 5 coins in a box.

Next, we are told that there was 50 coins altogether. Therefore, you’ll have 10 such boxes to make up a total of 50 coins.

The total mass of 50 coins is 600g; therefore the mass per box is 600/10 which is 60g.

Hope this helps.

[atutor2001]

Keroppi30:

Hi atutor2001

The answer to the original question has to be whole numbers too cos P3 hasn't touch on decimals yet.

I agree that the book implicitly expected the answer in whole number.

However, it is important for the book to state clearly by including a statement, for example : \"The difference in the weight is a whole number\".

Otherwise, many students develop the wrong impression that all answers must be in \"whole number\". This is a common problem many average students in upper Pr are facing. Whenever they get an answer in fraction (which is actually correct), they thought that it is wrong because they have been getting whole number answers in the past and nobody highlight to them that answers can also be in fraction. Some students even manipulated the working just to get a whole number and end up with a wrong answer.

It is important for assessment books for lower Pr to state clearly if it only wants answer in \"whole number\" when there are other possible answers. By doing so, it will make the student realise that there are other possible answers that may not be in whole number. In later years, they will have confident to accept answers in \"fraction.\"

[Keroppi30]

:thankyou: tianzhu & atutor2001


I really appreciate your help. I’m not trying to challenge your answers but I’m trying to address the questions raised by my dd. I’m not very good at explaining maths concepts to my dd in fact what I usually do when I encounter such problem sums is I’ll try to obtain the answers via simultaneous equations.

[atutor2001]

Keroppi30:

:... I’m not trying to challenge your answers but I’m trying to address the questions raised by my dd. ....

Hi Keroppi30

Don't be too concerned, questioning our proposed answers. In fact, your query sets me thinking. Which is why I modified the question - hoping to find a solution to a more tricky type of math condition where only the whole number solution is allowed but is not explained explicitly in Pr Math. Lets just continue to question when there is ambiguity. It makes math more interesting and \"alive\".

Regards