Q&A - PSLE Math

[tisha]

tianzhu:

Puffer:

Can anyone help?


In planning for a party, Mrs Lim bought enough satay to allow her guests an average of 25 sticks each. If 2 more people joined the party, how many fewer sticks of satay will each of her guests be given?

Thanks in advance.

Hi

Good Afternoon.

I did a quick check; it looks like there are multiple answers.

One is 10 and the other is 5.

It is suggested that you use Tabulated list/Systematic listing to solve this question.

Best wishes

Can no other method that can be used?
How to do tabulation, for this problem I'm not even sure where to start. :?

[atutor2001]

tisha:

[quote]
In planning for a party, Mrs Lim bought enough satay to allow her guests an average of 25 sticks each. If 2 more people joined the party, how many fewer sticks of satay will each of her guests be given?

Can no other method that can be used?
How to do tabulation, for this problem I'm not even sure where to start. :?[/quote]This is question is an example of the superiority of algebra over heuristic (but it is too complicated for Pr. However, I think GEP can still do it) Let me first present the algebraic method then the heuristic method.

Algebraic method :
Let the no. of guests be G
Let the decrease in the average be d

Before :
No. of guests = G
Average no. of satays = 25
Total no. of satay = 25G

After :
No. of guests = (G+2)
Average no. of satays = (25-d)
Total no. of satay = (G+2)(25-d) = 25G + 50 - d(G+2)

The Totals are the same, so we form the algebraic equation : 25G = 25G + 50 - d(G+2)
By simplifying we will get : d(G+2) = 50

The factors of 50 are : 1x50, 2x25, 5x10

By matching d x (G+2) to the above factors:
When d = 1; then (G+2) must be 50 so No. of guests, G = 48
When d = 2; then (G+2) must be 25 so No. of guests, G = 23
When d = 5; then (G+2) must be 10 so No. of guests, G = 8
When d = 10; then (G+2) must be 5 so No. of guests, G = 3
When d = 50; then (G+2) must be 1 so No. of guests, G = -1 (rejected)
When d = 25; then new average becomes (25-55 = 0) (rejected)

So there are 4 possible answers :
New average = 24; Original No. of guest = 48
New average = 23; Original No. of guest = 23
New average = 20; Original No. of guest = 8
New average = 15; Original No. of guest = 3


Now come the interesting part of how to use heuristics to solve this question.
It centers on the ability of the child to see that d(G+2) = 50

The Pr kid must reason that :

1. If the 2 new person is also getting 25 satays, then another 50 satays is needed.

2. Since no new satays are going to be added, these 50 satays must be \"taken back\" by asking everybody to contribute equal amount (including the 2 new guests).

3. This contribution will cause a drop in their original average of 25 i.e. this contribution is the actually the difference in the 2 averages.

4. So the 50 satays = [Contribution x New Total no. of children]

5. The child must then know the mathematical skill of breaking 50 into its factors and match them to [Contribution x New Total no. of guests] to get the different combinations of answers.

My above reasonings actually comes from the algebraic solution. So algebra is a mathematical tool but models and heuristics are not. We can get the heuristic concept from algebra. Good luck!

[pensiveowl]

Could someone pls help us with this problem? It's taken from PROBLEM SOLVING MADE EASY - FRACTIONS (son and I couldn't make the given soln out)


Clement spent 1/5 of his money on some pens, 1/4 of his remaining money on a book and 3/4 of the money he had left on three thumb drives. Each thumb drive cost 18 times as much as a pen he bought. How many pens did he buy?

[qic]

Pen - 1/5

Remain - 1-1/5=4/5
Book - 1/4x4/5 = 1/5
Had left - 1-1/5-1/5 = 3/5
3 thumb Drive - 3/4x3/5 = 9/20
1 thumb Drive = 3/20
Pen - 3/20/18=1/120
number of pens - 1/5 / 1/120 = 24

Pens: 24

[pensiveowl]

qic:

Pen - 1/5

Remain - 1-1/5=4/5
Book - 1/4x4/5 = 1/5
Had left - 1-1/5-1/5 = 3/5
3 thumb Drive - 3/4x3/5 = 9/20
1 thumb Drive = 3/20
Pen - 3/20/18=1/120
number of pens - 1/5 / 1/120 = 24

Pens: 24

Thanks very much! - much simpler than the soln given in the workbook!

[anneshirleygilbert]

Can somebody please help with this question?


Samad bouoght 4 similar pens and 6 similar files. Each pen cost $1.20 more than a file. If the total cost was $6.40 more than the total cost of the pens, how much did Samad spend in all?

[tianzhu]

anneshirleygilbert:

Can somebody please help with this question?


Samad bouoght 4 similar pens and 6 similar files. Each pen cost $1.20 more than a file. If the total cost was $6.40 more than the total cost of the pens, how much did Samad spend in all?

Hi

A quick one, the answers are in odd denominations.

If the total cost was $6.40 more than the total cost of the pens ----- this means that 6 similar files cost $6.40

6F -------6.4
1F ----- 1.067

Total cost ----- 10F +4.8 -------15.47

If it's 8 files instead of 6 files, you'll get nice numbers.

Best wishes

[tisha]

atutor2001:

tisha:

[quote]
In planning for a party, Mrs Lim bought enough satay to allow her guests an average of 25 sticks each. If 2 more people joined the party, how many fewer sticks of satay will each of her guests be given?

Can no other method that can be used?
How to do tabulation, for this problem I'm not even sure where to start. :?

This is question is an example of the superiority of algebra over heuristic (but it is too complicated for Pr. However, I think GEP can still do it) Let me first present the algebraic method then the heuristic method.

Algebraic method :
Let the no. of guests be G
Let the decrease in the average be d

Before :
No. of guests = G
Average no. of satays = 25
Total no. of satay = 25G

After :
No. of guests = (G+2)
Average no. of satays = (25-d)
Total no. of satay = (G+2)(25-d) = 25G + 50 - d(G+2)

.
.
.

My above reasonings actually comes from the algebraic solution. So algebra is a mathematical tool but models and heuristics are not. We can get the heuristic concept from algebra. Good luck![/quote]:thankyou: atutor for this solution. By the time I finished reading the solution, I couldn't even recall what the Q was? :lol:
Will read again and slowly try to chew and digest. Thanks for the detailed explaination.

BTW, if the students draw some neat columns and do a few tabulations will it fetch any marks for the method :?:

[moonee]

:?: Hi can anyone help with this P5 question?


A frog took 5 leaps forward and 6 leaps back. Next it took 7 leaps forward and 8 leaps back. Finally, it took 4 more leaps forward before stopping.

a.Did the frog stop in front of or behind its starting point?
b. How many leaps away was it from its starting point?
c. How many times did it leap over the starting point?
d. How many leaps away was the frog from the starting point when it was furthest away?


~~~~Thank You~~~~

[Cheerfuldad]

Hi all,


Please help!

Q. Tank X and Y are each filled with some water. If water from Tank Y is poured into Tank X until the water in Tank X reaches the brim, there will be 8 litres of water left in Tank Y. If water from Tank X is poured into Tank Y until the water in Tank Y reaches the brim, there will be 26 litres of water left in Tank X. The ratio of the volume of Tank X to the volume of Tank Y is 5:3. How many more litres of water are needed to fill both tanks to their brim?