Q&A - P3 Math

[BigDevil]

Alicia has fewer than 20 sweets.

She puts the sweets into packets such that each packet will have the same number of sweets.
If she puts them into packets of 3, she will have 2 sweets left over.
If she puts them into packets of 5, she will be short of 1 sweet.
How many sweets does Alicia have?

I know this can be solved by listing...

n x 3 + 2: 5 8 11 14 17
n x 5 - 1: 4 9 14 19
Ans: 14 sweets

Are there any other methods?

[qms]

BigDevil:

Alicia has fewer than 20 sweets.

She puts the sweets into packets such that each packet will have the same number of sweets.
If she puts them into packets of 3, she will have 2 sweets left over.
If she puts them into packets of 5, she will be short of 1 sweet.
How many sweets does Alicia have?

I know this can be solved by listing...

n x 3 + 2: 5 8 11 14 17
n x 5 - 1: 4 9 14 19
Ans: 14 sweets

Are there any other methods?

BigD,

At P3 level, this is the only method to solve the question. The other method is taught at...university level.

The answer is not unique if you increase the number of sweets. If you were to continue the list, you would realise that 29 could be an answer, too.

[BigDevil]

qms:

BigD,

At P3 level, this is the only method to solve the question. The other method is taught at...university level.

Thanks, qms. So DD just have to learn this method of solving this kind of problem.
[quote]The answer is not unique if you increase the number of sweets. If you were to continue the list, you would realise that 29 could be an answer, too.[/quote]I realise that, and had told DD that every increase of 15 (common multiple of 3 and 5) is a possible answer to the question. That's why the question had to state that Alicia has fewer than 20 sweets to make the answer unique.

[qms]

BigDevil:

Thanks, qms. So DD just have to learn this method of solving this kind of problem.

In a way, if such question appears in SA2, the number (as in the total number of sweets) will not be big. Hence, it won't take too long before you arrive at the answer.

If you review your list again, you will notice that the number of bags is not constant. If you put 3 sweets in a bag, there will be 4 bags. If you put 5 sweets in a bag, there will be 3 bags. Since the numbers of bags are different in the two scenarios, the standard formula cannot be applied.

[Daddy]

Hi,


Need help on this question:

Bobby places 12 cards on a table. Each card is either red or black in colour. If the fraction of black cards is 7/12, what is the smallest number of additional cards that Bobby needs to put on the table so that the function of black cards becomes 5/8?

[qms]

Daddy:

Bobby places 12 c


ards on a table. Each card is either red or black in colour. If the fraction of black cards is 7/12, what is the smallest number of additional cards that Bobby needs to put on the table so that the function of black cards becomes 5/8?

There are 12 cards, out of which 7 are black.

10/16, in its simplest form, is equivalent to 5/8.

Hence, Bobby needs to put at least another 4 cards on the table, hoping to get 3 black cards and 1 red card so that the function of black cards becomes 5/8, since he currently has 7 black cards out of the 12 cards.

Correct? http://i827.photobucket.com/albums/zz200/pepper2010_bucket/ththbnbnb.gif\">

[BigDevil]

Not sure if I understand the question correctly.


Currently, black cards are 7/12 = 14/24

Target 5/8 = 15/24.

So smallest number of additional cards to be placed = 1

[qms]

BigDevil:

Not sure if I understand the question correctly.


Currently, black cards are 7/12 = 14/24

Target 5/8 = 15/24.

So smallest number of additional cards to be placed = 1

BigD, if you pick one more card, there will be a total of 13 cards. So, even if you pick a black card, the function of black card becomes 8/13.

The number of cards to be picked must be such that we should be able to eventually achieve 5/8 of black card.

Right? http://i114.photobucket.com/albums/n264/pcreasy/questionmarkbiggersmiley.gif\">

[plum-cake]

BigDevil:

Not sure if I understand the question correctly.


Currently, black cards are 7/12 = 14/24

Target 5/8 = 15/24.

So smallest number of additional cards to be placed = 1

12 to 24 means 12 more cards not 1 right?
:?

[qms]

plum-cake:

12 to 24 means 12 more cards not 1 right?

:?

What BigD meant was that the question asked for function of black cards to be 5/8. If you consider multiples of 8 and 12, the common denominator is 24. Hence, converting 7/12 gives 14/24, while converting 5/8 gives 15/24.

Since the target is 5/8, which is the same as 15/24, you just need to place one more card to make it 15/24.

However, my point is, placing one more card will mean the final fraction is either 7/13 or 8/13. It won't give you 5/8, regardless of the multiples you consider.