Q&A - PSLE Math

[CoffeeCat]

atutor2001:

Vanilla Cake:

[quote=\"tianzhu\"]
Hi

I picked up this quote from the thread, Tips for improving in PSLE. It was from a member called livewith_vanilla.

I think the question she posed is a pretty interesting one, it involves two changed quantities, but the number(124) involved in the transaction is lumped into one lot instead of two different portions.

The marbles that A and B has is in the ratio 6:7. Then,124 marbles were added to the total and the ratio became 2:3.How many marbles were there at first?

Have fun.
Best wishes

Hi Uncle Tianzhu,
There seems to be more than one possible correct answer for this question. I have posted a reply to \"livewith_vanilla\" to seek her clarification for this \"One quick(inaccurate) example is this:.....\"

I remember seeing this question in Andrew Er's assessment book.

It is true mathematically that if we have only 1 equation with 2 unknowns, then the number of answers can be infinite. However, this particular question will have only one answer because of the following reasons:
(a) the number of marbles must be \"whole number\"
(b) the numbers in the questions are all prime numbers : 13; 5 and 124=2x2x31

This question was intended to test a student's understanding that if a certain amount is divided into 13 U and it becomes 6 W after adding 124 to 13U, we can convert the 6 W to U by cutting each W into 3 parts. There is no other way of cutting such that 124 can be divided without remainder.[/quote]To assure those who got multiple answers that they are not seeing stars...
I used the old fashioned algebra way ,
i let the marbles at first be 6x and 7x.
and the marbles added to A and B be y and 124-y respectively
using the last ratio, i form the equation
3 (6x+y) = 2( 7x+ 124-y)
i get this diophantine equation 4x + 5y = 248
the smallest solution is form when x=2, and y =48. (26 marbles at first)
the next solution is when x= 2+5 = 7 and y = 48-4 = 44 (91 marbles at first).
I stopped here but i suspect additional solutions can be obtained by adding 5 to x, and subtracting 4 from y.
It's quite late so i will digest what you said tmr morning, so it's possible i might be reading something wrongly.

[Vanilla Cake]

CoffeeCat:

atutor2001:

[quote=\"Vanilla Cake\"]
Hi Uncle Tianzhu,
There seems to be more than one possible correct answer for this question. I have posted a reply to \"livewith_vanilla\" to seek her clarification for this \"One quick(inaccurate) example is this:.....\"

I remember seeing this question in Andrew Er's assessment book.

It is true mathematically that if we have only 1 equation with 2 unknowns, then the number of answers can be infinite. However, this particular question will have only one answer because of the following reasons:
(a) the number of marbles must be \"whole number\"
(b) the numbers in the questions are all prime numbers : 13; 5 and 124=2x2x31

This question was intended to test a student's understanding that if a certain amount is divided into 13 U and it becomes 6 W after adding 124 to 13U, we can convert the 6 W to U by cutting each W into 3 parts. There is no other way of cutting such that 124 can be divided without remainder.

To assure those who got multiple answers that they are not seeing stars...
I used the old fashioned algebra way ,
i let the marbles at first be 6x and 7x.
and the marbles added to A and B be y and 124-y respectively
using the last ratio, i form the equation
3 (6x+y) = 2( 7x+ 124-y)
i get this diophantine equation 4x + 5y = 248
the smallest solution is form when x=2, and y =48. (26 marbles at first)
the next solution is when x= 2+5 = 7 and y = 48-4 = 44 (91 marbles at first).
I stopped here but i suspect additional solutions can be obtained by adding 5 to x, and subtracting 4 from y.
It's quite late so i will digest what you said tmr morning, so it's possible i might be reading something wrongly.[/quote]Hi All,
I am VC's mum and this is the http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=11112&start=20 from thread starter - livewith_vanilla:

Ermmm...sorry but that's not the question i have actually.I actually just thought of some random one.It is not actually a question.Its just an example,something that i just thought of randomly without any thought.Yes,the answers will be vague.I will take an example from my school paper tmr and post it here as an accurate question with only one answer.Sorry!

Hi atutor2001,
I believe that this question is invalid and will not likely to appear in any upcoming PSLE Maths exams. If such example does appear in Andrew Er's assessment books, pls let us know which assessment book? Pls elaborate more on your 2 reasons that this particular question has only one answer instead of multiple answers.
Thanks.

[Brenda10]

Hi tianzhu


Thank you for your encouragement.

There is still a long way to go and hopefully we can keep up our stamina.

We do hope that the effort will be pay off.

Have a nice day.

[trytry]

2008

http://www.jamesangtutors.com/forum/viewtopic.php?p=5520&sid=dd197883788c1e3bdaea959288b33365

7. Mrs Smith bought some mangoes and mangosteens for $16.50. She bought 5 fewer mangoes than mangosteens. Each mango cost $1.80 more than the mangosteen. How many mangosteens did she buy?

Anyone has solution to the above question?
Crack my head can't solve.

[Brenda10]

trytry:

2008

http://www.jamesangtutors.com/forum/viewtopic.php?p=5520&sid=dd197883788c1e3bdaea959288b33365

7. Mrs Smith bought some mangoes and mangosteens for $16.50. She bought 5 fewer mangoes than mangosteens. Each mango cost $1.80 more than the mangosteen. How many mangosteens did she buy?

Anyone has solution to the above question?
Crack my head can't solve.

HI,

I would like to give a try:

Mangoesteen 6 7 8 9 10

Mango 1 2 3 4 5

Additional $1.80 $3.6 $5.4 $7.2 $9


$16.5-$9= $7.5
$7.5/15 = $0.5
(10*$0.5) + (5*$2.3) = $16.50

Therefore total she bought 10 mangoesteen.

Hopefully the answer is correct.

Thanks

[atutor2001]

CoffeeCat:

To assure those who got multiple answers that they are not seeing stars...
I used the old fashioned algebra way ,
i let the marbles at first be 6x and 7x.
and the marbles added to A and B be y and 124-y respectively
using the last ratio, i form the equation
3 (6x+y) = 2( 7x+ 124-y)
i get this diophantine equation 4x + 5y = 248
the smallest solution is form when x=2, and y =48. (26 marbles at first)
the next solution is when x= 2+5 = 7 and y = 48-4 = 44 (91 marbles at first).
I stopped here but i suspect additional solutions can be obtained by adding 5 to x, and subtracting 4 from y.
It's quite late so i will digest what you said tmr morning, so it's possible i might be reading something wrongly.

Hi Coffeecat

Thanks for bring out my mistake. It is correct that there are 2 possible answers because the additional marble given was 124 (equal to 2x2x31). However, if the additional marble is 62 (equal to 2x31) then there should be only 1 answer because 1 unit will become 31 and cannot be split into any other way as 31 is a prime number.

Hi Vanilla Cake

Sorry, I cannot remember which book but it was one of those on P5 challenging math question (it is 5 years ago when my youngest kid was taking PSLE). I remember this question because I strongly disagree with it then, thinking it will generate infinite answers - until a math expert explained to me the rationale behind it.

There will be more than one answer if we use a \"composite number\" (numbers with more than 2 factors) for the additional number of marbles added to the total. This is because the marbles can then be distributed between A and B in many different ways using the different combination of factors of that \"composite number\" For the given example, the additional marble is 124 and so 1 unit is 62 = 2 x 31 or 1 x 61 allowing 2 possible answers.

As for the fact that marbles cannot be added in fraction, this limit the answer to \"whole numbers\". If instead of using marbles but let say the weight of something in kg or volume in litres, then the answers can be in fractions and there will be infinite answers as any number can be broken up into the product of 2 fractions in infinite number of ways. Hope I am right.

[trytry]

Brenda10:

trytry:

2008

http://www.jamesangtutors.com/forum/viewtopic.php?p=5520&sid=dd197883788c1e3bdaea959288b33365

7. Mrs Smith bought some mangoes and mangosteens for $16.50. She bought 5 fewer mangoes than mangosteens. Each mango cost $1.80 more than the mangosteen. How many mangosteens did she buy?

Anyone has solution to the above question?
Crack my head can't solve.

HI,

I would like to give a try:

Mangoesteen 6 7 8 9 10

Mango 1 2 3 4 5

Additional $1.80 $3.6 $5.4 $7.2 $9


$16.5-$9= $7.5
$7.5/15 = $0.5
(10*$0.5) + (5*$2.3) = $16.50

Therefore total she bought 10 mangoesteen.

Hopefully the answer is correct.

Thanks

Thanks for the solution.
You used \"guess and check\" method?
Why do you stop at 10 mangoesteen and 5 mangoes? :?

[CoffeeCat]

atutor2001:

CoffeeCat:

To assure those who got multiple answers that they are not seeing stars...
I used the old fashioned algebra way ,
i let the marbles at first be 6x and 7x.
and the marbles added to A and B be y and 124-y respectively
using the last ratio, i form the equation
3 (6x+y) = 2( 7x+ 124-y)
i get this diophantine equation 4x + 5y = 248
the smallest solution is form when x=2, and y =48. (26 marbles at first)
the next solution is when x= 2+5 = 7 and y = 48-4 = 44 (91 marbles at first).
I stopped here but i suspect additional solutions can be obtained by adding 5 to x, and subtracting 4 from y.
It's quite late so i will digest what you said tmr morning, so it's possible i might be reading something wrongly.

Hi Coffeecat

Thanks for bring out my mistake. It is correct that there are 2 possible answers because the additional marble given was 124 (equal to 2x2x31). However, if the additional marble is 62 (equal to 2x31) then there should be only 1 answer because 1 unit will become 31 and cannot be split into any other way as 31 is a prime number.

I believe you are talking about the diophantine (we call such equations diophantine because we are only interested in whole number solutions) equation
13x -5y = 62.
I understand your concern about the marbles added not being whole number, as such i formed my equation differently by defining the marbles being added as a variable.
Again let the marbles at first be 6x and 7x.
and the marbles added to A and B be y and 62-y respectively
By changing the 124 to 62 like this ... 3 (6x+y) = 2( 7x+ 124-y)
you will get 4x + 5y = 124
the smallest solution is x= 1, y=24, (13 marbles at first)
the next solution will be x= 1+5=6, y=24-4=20 (78 marbles at first).
I suspect one can get additional solutions by adding 5 to x and subtracting 4 from y.

Perhaps this is not very similar to the example you seen in the Andrew Er assessment book. It will be interesting to see that question though.
By the way no offence, it's not like I am targeting you or anything, it's a professional sickness =).

[atutor2001]

CoffeeCat:

I believe you are talking about the diophantine (we call such equations diophantine because we are only interested in whole number solutions) equation
13x -5y = 62.
I understand your concern about the marbles added not being whole number, as such i formed my equation differently by defining the marbles being added as a variable.
Again let the marbles at first be 6x and 7x.
and the marbles added to A and B be y and 62-y respectively
By changing the 124 to 62 like this ... 3 (6x+y) = 2( 7x+ 124-y)
you will get 4x + 5y = 124
the smallest solution is x= 1, y=24, (13 marbles at first)
the next solution will be x= 1+5=6, y=24-4=20 (78 marbles at first).
I suspect one can get additional solutions by adding 5 to x and subtracting 4 from y.

Perhaps this is not very similar to the example you seen in the Andrew Er assessment book. It will be interesting to see that question though.

Thank you coffeecat for confirming that I am wrong. Don't think I can find that book again or to hunt down my kid's teacher to clarify. Maybe I will go bookshop to see if I can find the assessment book.

Once again thank you for the correction.

Regards

[atutor2001]

Hi Coffeecat


Since 4x+5y can be set to be equal any number, if I change the additional marbles to 9, i.e. 4x+5y=9, then there will be only 1 solution, where x = 1 and y = 1 because negative number is not allowed.

Similarly I can change the additional marbles to 14, i.e. x=1 and y=2 and so on…

Maybe the question I saw was crafted in this way which generates only 1 solution but my explanation on "prime numbers" was definitely wrong.

Regards.