Q&A - P3 Math

[dazzlego]

Sunflower03:

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?

Hi Big Devil,

May I know how you get 5 8 11 14 17 and 4 9 14 19?

I still can't get it right.

Kindly enlightening me.

Thanks!

Hi Sunflower03,
It is \"multiple of 3 (+ 2)\" and \"multiple of 5 (- 1)\"

Hi Big Devil,
Yes, there is another method. You can use this method if all the cases have the same remainder or have the same shortage.
In this question both cases have the same shortage since \"If she puts them into packets of 3, she will have 2 sweets left over\" has the same meaning as \"if she put them into packets of 3, she will be short of 1\"
Then, find a common multiple of 3 and 5 first and after that subtract 1 (shortage) from it

Cheers :celebrate:

[BigDevil]

:shock: Amazing!


It didn't even occur to me that excess of 2 is equivalent as saying shortage of 1! Certainly makes the solution a lot easier!

Thanks!!

[Sunflower03]

dazzlego:

Sunflower03:

[quote=\"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?

Hi Big Devil,

May I know how you get 5 8 11 14 17 and 4 9 14 19?

I still can't get it right.

Kindly enlightening me.

Thanks!

Hi Sunflower03,
It is \"multiple of 3 (+ 2)\" and \"multiple of 5 (- 1)\"

Hi Big Devil,
Yes, there is another method. You can use this method if all the cases have the same remainder or have the same shortage.
In this question both cases have the same shortage since \"If she puts them into packets of 3, she will have 2 sweets left over\" has the same meaning as \"if she put them into packets of 3, she will be short of 1\"
Then, find a common multiple of 3 and 5 first and after that subtract 1 (shortage) from it

Cheers :celebrate:[/quote]

Thanks Ladies for your explanation. Learn another thing today.

[Sun_2010]

Sunflower03 - BigD is a gentleman


Dazzlego - like your method - really dazzling. Never tot of taking it to the next step and seeing the excess as shortage or the otherway around. I like!

[dazzlego]

Sun_2010:

Sunflower03 - BigD is a gentleman


Dazzlego - like your method - really dazzling. Never tot of taking it to the next step and seeing the excess as shortage or the otherway around. I like!

Dazzlego is also gentleman

Thanks for the compliments. I am blushing now seriously

[Sunflower03]

Opps…Thanks Ladies and gentlemen.

[Applelyn1]

The total length of 3 pcs of ribbon A, B and 210cm. B is 50cm longer than C but 20cm shorter than A. What is the length of ribbon A?


Pls help, thank you.

[dazzlego]

Applelyn1:

The total length of 3 pcs of ribbon A, B and 210cm. B is 50cm longer than C but 20cm shorter than A. What is the length of ribbon A?


Pls help, thank you.

Hi Applelyn1,
Have u tried to draw the model?

http://i50.tinypic.com/2h6hvrp.jpg\">

[Applelyn1]

dazzlego:

Applelyn1:

The total length of 3 pcs of ribbon A, B and 210cm. B is 50cm longer than C but 20cm shorter than A. What is the length of ribbon A?


Pls help, thank you.

Hi Applelyn1,
Have u tried to draw the model?

http://i50.tinypic.com/2h6hvrp.jpg\">

Hi dazzlego

I did and the answer is 100? the answer sheet stated as 80.

Thanks.

[BigDevil]

100cm should be correct.


Check: 100 (A) + 80 (B) + 30 = 210