<![CDATA[Q&amp;A - P3 Math]]>Moderator's Note: Please use this thread to ask for help on specific questions. Please refrain from out of topic chatter in this thread to make it easier for the community to find the answers they need.


From Ai Tong Pri 3 SA1 Maths:

A teacher gave her students some balloons. If she gave 6 balloons to each student, she will have 2 balloons left. If she gave 8 balloons to each student, she will be short of 2 balloons. What is the smallest possible number of students she has?

Try to do this without using ALGEBRA!

]]>https://forum.kiasuparents.com/topic/149/q-amp-a-p3-mathRSS for NodeTue, 05 May 2026 11:37:52 GMTWed, 16 Apr 2008 14:44:04 GMT60<![CDATA[Reply to Q&amp;A - P3 Math on Thu, 10 Nov 2016 09:50:49 GMT]]>

Bunny27:
Hi I need help on this question,

Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you

Just have to draw a simple diagram

4 units => 300 + 252 = 552
1 unit => 552/4 = 138

sufia only has one unit in the end, so the answer is 138

]]>https://forum.kiasuparents.com/post/1732271https://forum.kiasuparents.com/post/1732271Thu, 10 Nov 2016 09:50:49 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 07 Oct 2016 15:44:43 GMT]]>

love1001:
Hi, can someone help me with this combination type question?


Alice, Ben, Cathy, Daniel and Elle are seated in a row.
Alice can sit either on the extreme left or right.
Cathy must sit in the center.
How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?
Can go by method of listing:
We commonly see similar question with 3 people (X Y and Z) and most P3 are taught to solve by listing.
XYZ XZY YXZ YZX ZXY ZYX (6 ways, \"basic\" list)

So for your question: we know C must be centre. A is either on extreme left or right.
Suppose A is on extreme left like this:
[A ? C ? ?]
Then remaining B, D and E have be like the above \"XYZ\" case, 6 ways to fill the remaining 3 seats.
*Notes (just for understanding):
BDE BED DBE DEB EBD EDB (just need to understand this basic list of 6 choices)
This arrangement is the same as below, since \"_\" are fixed positions taken up by A and C.
_B_DE / _B_ED / _D_BE / _D_EB / _E_BD / _E_DB (same basic list)

But A can also switch to the extreme right like this:
[? ? C ? A]
Then in the same way, there are another 6 ways to for B, D and E to fill the remaining 3 seats.
*Notes (not necessary to list again)
BD_E_ / BE_D_ / DB_E_ / DE_B_ / EB_D_ / ED_B_ (same basic list)

Total is 6+6=12 ways#

]]>https://forum.kiasuparents.com/post/1724603https://forum.kiasuparents.com/post/1724603Fri, 07 Oct 2016 15:44:43 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 07 Oct 2016 15:21:05 GMT]]>

Ender:
love1001:

Hi, can someone help me with this combination type question?


Alice, Ben, Cathy, Daniel and Elle are seated in a row.
Alice can sit either on the extreme left or right.
Cathy must sit in the center.
How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?

The seats, from left to right lets, we label them as S1, S2, S3, S4 and S5

Alice can sit either S1 or S5. That poses two scenario to calculate the combinations.

Senario 1, Alice took S1,
S1 - 1 option, i.e Alice
S2 - 3 option, from Ben, Daniel and Elle
S3 - 1 option, Cathy has to sit here
S4 - 2 option. after S2 is taken, we are left to 2 choices.
S5 - 1 option. After S2 and S4 is taken, we are left with 1 choice.
Total combination for Scenario 1,
1 X 3 X 1 X 2 X 1 = 6

Scenario 2,
We fix Alice at S5
S1 - 3 options, from either Ben, Daniel and Elle
S2 - 2 options. After S1 is chosen, we are left to 2 choices
S3 - 1 option. Only Cathy sits here
S4 - 1 option, After S1 and S2 is taken, we are left with one choice
S5 - 1 option, That is for Alice only
Total combination for scenario 2
3 X 2 X 1 x 1 x1 = 6 combination

Total combination = Combination for Scenario 1 + Combination for Scenario 2
6 + 6 = 12 ######

Let's do it another way. Instead finding the combination via the chairs, we find the how many choices does one individual has.

We start with the one with the most restriction, i.e Cathy and then work ourselves to those with more choices.


Cathy - 1 choice. She has to take S3
Alice - 2 choices . She can choose either S1 or S5
Ben - 3 choices, after Alice selects either S1 or S5, Ben is left with only 3 choices
Daniel - 2 choices. After all the above have chosen, Daniel has two chairs to choose.
Elle - 1 choice . By now 4 chairs have been chosen. Left one choice.

Total combination = 1 X 2 X 3 X 2 X 1 = 12 combinations ###

]]>https://forum.kiasuparents.com/post/1724601https://forum.kiasuparents.com/post/1724601Fri, 07 Oct 2016 15:21:05 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 07 Oct 2016 01:22:52 GMT]]>

love1001:
Hi, can someone help me with this combination type question?


Alice, Ben, Cathy, Daniel and Elle are seated in a row.
Alice can sit either on the extreme left or right.
Cathy must sit in the center.
How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?
The seats, from left to right lets, we label them as S1, S2, S3, S4 and S5

Alice can sit either S1 or S5. That poses two scenario to calculate the combinations.

Senario 1, Alice took S1,
S1 - 1 option, i.e Alice
S2 - 3 option, from Ben, Daniel and Elle
S3 - 1 option, Cathy has to sit here
S4 - 2 option. after S2 is taken, we are left to 2 choices.
S5 - 1 option. After S2 and S4 is taken, we are left with 1 choice.
Total combination for Scenario 1,
1 X 3 X 1 X 2 X 1 = 6

Scenario 2,
We fix Alice at S5
S1 - 3 options, from either Ben, Daniel and Elle
S2 - 2 options. After S1 is chosen, we are left to 2 choices
S3 - 1 option. Only Cathy sits here
S4 - 1 option, After S1 and S2 is taken, we are left with one choice
S5 - 1 option, That is for Alice only
Total combination for scenario 2
3 X 2 X 1 x 1 x1 = 6 combination

Total combination = Combination for Scenario 1 + Combination for Scenario 2
6 + 6 = 12 ######

]]>https://forum.kiasuparents.com/post/1724405https://forum.kiasuparents.com/post/1724405Fri, 07 Oct 2016 01:22:52 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 16:12:33 GMT]]>I get it now.

Thanks …I can explain to my gal now.

]]>https://forum.kiasuparents.com/post/1720695https://forum.kiasuparents.com/post/1720695Fri, 23 Sep 2016 16:12:33 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 16:07:25 GMT]]>

Bunny27:
nicnac:

[quote=\"Bunny27\"]Hi I need help on this question,





Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you

Draw your models and you should be able to see the answer.
In the end,
Liying has 552 (300 + 252) stamps more than Sufia
Given that in the end Liying has 5 times as many as Sufia, the excess that Liying has must be 4 units.
4u -> 552
U -> 138
Sufia has 138 stamps in the end.

Hi, can you explain why need to plus 252 which Sufia gave away? Thanks[/quote]http://s17.postimg.org/khj7bu6fz/image.jpg\">

]]>https://forum.kiasuparents.com/post/1720694https://forum.kiasuparents.com/post/1720694Fri, 23 Sep 2016 16:07:25 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 16:01:49 GMT]]>

Bunny27:
nicnac:

[quote=\"Bunny27\"]Hi I need help on this question,





Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you

Draw your models and you should be able to see the answer.
In the end,
Liying has 552 (300 + 252) stamps more than Sufia
Given that in the end Liying has 5 times as many as Sufia, the excess that Liying has must be 4 units.
4u -> 552
U -> 138
Sufia has 138 stamps in the end.

Hi, can you explain why need to plus 252 which Sufia gave away? Thanks[/quote]Because when Sufia gives away 252 stamps, this would mean that Liying will have 252 stamps more compared to Sufia.

]]>https://forum.kiasuparents.com/post/1720693https://forum.kiasuparents.com/post/1720693Fri, 23 Sep 2016 16:01:49 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 15:55:57 GMT]]>

nicnac:
Bunny27:

Hi I need help on this question,





Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you

Draw your models and you should be able to see the answer.
In the end,
Liying has 552 (300 + 252) stamps more than Sufia
Given that in the end Liying has 5 times as many as Sufia, the excess that Liying has must be 4 units.
4u -> 552
U -> 138
Sufia has 138 stamps in the end.

Hi, can you explain why need to plus 252 which Sufia gave away? Thanks

]]>https://forum.kiasuparents.com/post/1720692https://forum.kiasuparents.com/post/1720692Fri, 23 Sep 2016 15:55:57 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 15:47:09 GMT]]>

Bunny27:
Hi I need help on this question,





Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you
Draw your models and you should be able to see the answer.
In the end,
Liying has 552 (300 + 252) stamps more than Sufia
Given that in the end Liying has 5 times as many as Sufia, the excess that Liying has must be 4 units.
4u -> 552
U -> 138
Sufia has 138 stamps in the end.

]]>https://forum.kiasuparents.com/post/1720691https://forum.kiasuparents.com/post/1720691Fri, 23 Sep 2016 15:47:09 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 15:26:52 GMT]]>Hi I need help on this question,





Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you

]]>https://forum.kiasuparents.com/post/1720689https://forum.kiasuparents.com/post/1720689Fri, 23 Sep 2016 15:26:52 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 23 Sep 2016 15:26:25 GMT]]>Hi I need help on this question,

Liying and Sufia has an equal number of stamps. Liying bought another 300 stamps and Sufia gave away 252 stamps. After that, Liying had five times as many stamps as Sufia. How many stamps did Sufia have in the end?
Thank you

]]>https://forum.kiasuparents.com/post/1720688https://forum.kiasuparents.com/post/1720688Fri, 23 Sep 2016 15:26:25 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Sun, 31 Jul 2016 00:50:57 GMT]]>Yes, it is from MO. I can only solve by doing a matching but I do not think it is the right method because when number of people get larger, I can’t possibly match.

]]>https://forum.kiasuparents.com/post/1705510https://forum.kiasuparents.com/post/1705510Sun, 31 Jul 2016 00:50:57 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Sat, 30 Jul 2016 16:57:40 GMT]]>

love1001:
Hi, can someone help me with this combination type question?


Alice, Ben, Cathy, Daniel and Elle are seated in a row.
Alice can sit either on the extreme left or right.
Cathy must sit in the center.
How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?
Not a common p3 question. From MO?

]]>https://forum.kiasuparents.com/post/1705488https://forum.kiasuparents.com/post/1705488Sat, 30 Jul 2016 16:57:40 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Sat, 30 Jul 2016 14:29:09 GMT]]>Hi, can someone help me with this combination type question?


Alice, Ben, Cathy, Daniel and Elle are seated in a row.
Alice can sit either on the extreme left or right.
Cathy must sit in the center.
How many different sitting combinations can there be?

What is the easiest method to solve within the need to draw it out?

]]>https://forum.kiasuparents.com/post/1705451https://forum.kiasuparents.com/post/1705451Sat, 30 Jul 2016 14:29:09 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Fri, 10 Jun 2016 14:50:00 GMT]]>

s277400:
There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.

http://i65.tinypic.com/vgstfr.jpg\">

By drawing the above diagrams, the question becomes easy to understand. Hope your kids can understand this too.

]]>https://forum.kiasuparents.com/post/1688598https://forum.kiasuparents.com/post/1688598Fri, 10 Jun 2016 14:50:00 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Tue, 07 Jun 2016 02:34:42 GMT]]>Yes I believe so...now 😆

]]>https://forum.kiasuparents.com/post/1687505https://forum.kiasuparents.com/post/1687505Tue, 07 Jun 2016 02:34:42 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Tue, 07 Jun 2016 02:08:04 GMT]]>I believe 1820 isn’t supposed to be the final total. If it is, then just divide by 2 (equal no. of fish) and add 520 and you can get Y (orig)…all other info redundant

]]>https://forum.kiasuparents.com/post/1687498https://forum.kiasuparents.com/post/1687498Tue, 07 Jun 2016 02:08:04 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 06 Jun 2016 15:14:47 GMT]]>

PiggyLalala:
MrsKiasu:

[quote=\"s277400\"]There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.


Dd tried...

...X1...+340 :X

...Y1...+340+520 :Y

Total X+Y=1820

X1+Y1+340+340+520=1820
X1+Y1=620 ...Y1=310 ...Y=310+340+520 =1170 (Ans) pls check 😂

base on what i can understand fr yr dd's solution...
before:pond Y =1170 fishes and X = 310+340=650
after: pond Y = 1170-520=650 fish
pond X= 650+340=990.
they are not equal.

could i have misread some information .... i hope not.[/quote]Based on her model drawn...but if total changed then her ans will be wrong..

Prior:X= 310, After: X = 310+340=650

Prior Y=310+340+520=1170, After Y= 310+340=650

Edit: the Prior doesn't add up to 1820..so our answer is wrong.

If there is change to the final total of fishes in pond..I get your answer of $1340

But if final total of fishes remain 1820...then I get 1430 (which was dd's first answer before she redo, told me she missed out something). P3 qn, my side flattened liao :rotflmao:

]]>https://forum.kiasuparents.com/post/1687401https://forum.kiasuparents.com/post/1687401Mon, 06 Jun 2016 15:14:47 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 06 Jun 2016 14:43:58 GMT]]>

MrsKiasu:
s277400:

There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.


Dd tried...

...X1...+340 :X

...Y1...+340+520 :Y

Total X+Y=1820

X1+Y1+340+340+520=1820
X1+Y1=620 ...Y1=310 ...Y=310+340+520 =1170 (Ans) pls check 😂

base on what i can understand fr yr dd's solution...
before:pond Y =1170 fishes and X = 310+340=650
after: pond Y = 1170-520=650 fish
pond X= 650+340=990.
they are not equal.

could i have misread some information .... i hope not.

]]>https://forum.kiasuparents.com/post/1687390https://forum.kiasuparents.com/post/1687390Mon, 06 Jun 2016 14:43:58 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 06 Jun 2016 14:26:04 GMT]]>

PiggyLalala:
s277400:

There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.


the key words are at the end of adding 340 fishes to pond x and removing 520 fishes fr pond Y, both ponds had EQUAL no of fish.

total no of fishes at the end of the process =1820+340-520=1640.
therefore at the end... pond x and pond y has 1640/2=820 each
hence... before.. pond X: 820-340=480
pond Y:820+520=1340

I thought of this..but later agreed with dd's method ie fixed the total...what is the answer on the answer sheet?

]]>https://forum.kiasuparents.com/post/1687382https://forum.kiasuparents.com/post/1687382Mon, 06 Jun 2016 14:26:04 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 06 Jun 2016 14:09:43 GMT]]>

s277400:
There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.

the key words are at the end of adding 340 fishes to pond x and removing 520 fishes fr pond Y, both ponds had EQUAL no of fish.

total no of fishes at the end of the process =1820+340-520=1640.
therefore at the end... pond x and pond y has 1640/2=820 each
hence... before.. pond X: 820-340=480
pond Y:820+520=1340

]]>https://forum.kiasuparents.com/post/1687379https://forum.kiasuparents.com/post/1687379Mon, 06 Jun 2016 14:09:43 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 06 Jun 2016 13:29:41 GMT]]>

s277400:
There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.

Dd tried...

...X1...+340 :X

...Y1...+340+520 :Y

Total X+Y=1820

X1+Y1+340+340+520=1820
X1+Y1=620 ...Y1=310 ...Y=310+340+520 =1170 (Ans) pls check 😂

Edit: Please note that the ans is incorrect!

]]>https://forum.kiasuparents.com/post/1687364https://forum.kiasuparents.com/post/1687364Mon, 06 Jun 2016 13:29:41 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 06 Jun 2016 05:30:12 GMT]]>There are 1820 fish in two ponds X and Y. After 340 fish were released into pond X and 520 fish were removed from the pond Y , both ponds had the same number of fish. How many fish were there in pond Y? Need help. Thks.

]]>https://forum.kiasuparents.com/post/1687248https://forum.kiasuparents.com/post/1687248Mon, 06 Jun 2016 05:30:12 GMT<![CDATA[Reply to Q&amp;A - P3 Math on Mon, 23 May 2016 01:04:02 GMT]]>

JingIes:
Hi, can anybody explain how to go about solving guess-and-check questions? I understand that there is another method that does not require one to guess and check his/her answers (the one where you have to draw up a simple table and literally ticking or crossing off your guess). I do not understand how that method works, and hopefully someone can enlighten me over here.


I have a sample question here:

There are 20 chickens and goats in a farm altogether.
There are 60 legs altogether. How many chickens and how many goats are there in the farm?


:thankyou:
Not sure what is taught in schools but you may wanna consider this.

Chicken = 2 legs, Goat = 4 legs.
Start with the one which has fewer coz adding is easier.

If ALL chicken, 20 x 2 = 40, short of 20. (ie, 60-40)
For every chicken => goat, you get 2 more legs.
To get from 40 to 60 (20 more legs), you need to change 10 chicken to goat,
Chicken 20 becomes 10
Goat 0 becomes 10.

Check answer,

Chicken = 2 x 10 = 20
Goat = 4 x 10 = 40
TOTAL = 60

]]>https://forum.kiasuparents.com/post/1682672https://forum.kiasuparents.com/post/1682672Mon, 23 May 2016 01:04:02 GMT