Q&A - PSLE Math

[Apphia]

Hi Yu Xuan,


The answer given is a) 0.028cm b) 0.558cm

I think the person who did the answers forgot to convert to cm. :oops:

[tianzhu]

Apphia:

2)Julian had 300 more cards than Faizal. Julian gave 3/5 of his cards to Faizal. Faizal then gave 1/4 of the total number of what he had then to Julian. In the end, Faizal had 300 more cards than Julian. How mny cards did Julian have at first?

You may use MD or Units Method to solve this question.

At First
Julian ------- 5 units + 300
Faizal ------- 5 units

After Julian gave 3/5 of his cards to Faizal.
Julian ------ 2 units + 120
Faizal ----- 5 units + 3 units + 180 ----- 8 units + 180

After Faizal than gave 1/4 of the total number of what he had then to Julian.
Julian ------- 2 units + 120 + 2 units + 45 ------- 4 units + 165
Faizal ------- 6 units + 135

In the end, Faizal had 300 more cards than Julian.
4 units + 165 + 300 ------ 6 units + 135
2units ------- 330
1 unit ------- 165

At First
Julian ------ 5 units + 300 ------ (5*165) + 300 ------1125

Best wishes

[tianzhu]

Apphia:

1) Mr Lee had 36 more guppies than goldfish. He sold 1/4 of the guppies and 1/5 of the goldfish. He found that he had sold 28 more guppies than goldfish. The guppies were sold $0.55 each and the goldfish at $0.85 each. What was the total amount he received?

Hi

It is suggested that MD be used to solve this question.

Please follow these steps.

Guppies ------ 20 units + 36
Goldfish -------20 units

Guppies sold ------- 5 units + 9
Goldfish sold ------- 4 units

He found that he had sold 28 more guppies than goldfish.
Compare the MD

1 unit ----- 19

Guppies ------(5*19)+9 ----- 104
Price of Guppies -------104*0.55 ------- 57.20

Goldfish ------(4*19) ------- 76
Price of Goldfish --------76*0.85 ------- 64.60

(57.20+64.60) ------- 121.80

Best wishes

[tianzhu]

Apphia:

Hi TianZhu,


Thank you for replying so quickly.

Ever so grateful !

Hi

You're welcome.

Best wishes

[Amberz]

iFruit:

Amberz:

The total of 4 numbers P, Q, R,and S was 291.

When P was tripled, Q was halved, R was increased by 20 and S was decreased by 25, the four numbers became equal.
Find the value of R. :?

This is a backwards problem..

Assume at the end, each number became 3u.

Then at the beginning:

P = 3u/3 = u
Q = 3u x 2 = 6u
R = 3u - 20
S = 3u + 25

u+6u+3u-20 + 3u + 25 = 291

13u = 286--->u =22

R = 3u - 20 = 46

HTH.

Got it thanks!

[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!