Q&A - PSLE Math

[tianzhu]

pixiedust:

My solution is using listing multiples method.
Is there a faster way?
Such questions, analysing the gap (like the oranges question) doesn't work?

Hi

Good Afternoon.

Is there a faster way? Is there a more efficient method to make life a little easier for our young ones?

I gave your question some thoughts last night.

This question is usually solved with “Systematic Listing” .It takes a bit of time but it also ensures that students gain some marks as long as they don’t make careless mistakes.

I am not too sure whether working it this way is more efficient than SL, do take a look and decide if it works for you.

First the thought process, we are given a number of cadets between 20 and 70.

So we have a big group of NCC cadets in a hall.

For case 1, they are grouped into 7 cadets in a team, this means multiples of 7.As for groups of 9, it also means multiples of 9.

For comparison, we have some groups of 7 plus 4 in case 1 and some groups of 9 plus 6 in case 2.Remember the total number of cadets in the hall is the same in both cases.

Hence, it’s likely we would have more groups of 7 than groups of 9.The difference between 6 and 4 is 2.Therefore, we need to find the multiples of 7 which is greater than the multiples of 9 by 2.

The closest multiples of 7 and 9 to 70 is 63.The difference between them is zero.

Next move down, (8*7) ------ 56 and (6*9) -------54.This gives a difference of 2.

The answer is (56+4) ------60 or (54+6) -------60

For my solution, I’ve drawn more MD than necessary to show the thought process. Once you become more familiar, you may do away with some of them.

First, find the difference between the remaining cadets after the groupings. Next proceed to find the multiples which give the required difference and you are on the way to the answer.

Hope you’ll find it useful.

Best wishes

http://farm6.static.flickr.com/5171/5511424062_c0bfacc8c2_b.jpg\">

[pixiedust]

:goodpost: Thank you VERY much tianzhu !

It is useful. It is like using listing method + intelligence.

Much appreciate your time and help.

[atutor2001]

pixiedust:

Appreciate help with the following question.


My solution is using listing multiples method.
Is there a faster way?
Such questions, analysing the gap (like the oranges question) doesn't work?

Some NCC cadets are gathered in a hall.
If they are grouped into 7 cadets in a team, there will be 4 cadets left out.
If they are grouped into 9 cadets in a team, there will be 6 cadets left out.
How many cadets are there in the hall if the number of cadets is between 20 and 70 ?

Mathematically, there is no faster way because the number of unknowns (i.e. The total no. of cadets; The no. of groups of 4 in case 1; The no. of groups of 6 in case 2 - 3 unknowns) is more than the number of relationship (equations) given by the question.

[ozora]

tianzhu:

Emelyn:

Hi,


Brain jammed....cannot solve the problem. Sign of old age too....
Asking for help here .......


http://postimage.org/image/2kyiwxjj8/

In the figure above, which is not drawn to scale, 1/4 of the circle is shaded. The ratio of the area of the square to the area of the rectangle and circle is 5:7. 40% of the square is shaded. 1/3 of the rectangle is shaed. What is the ratio of the area of the circle to the square to the rectangle ?

MTIA !!

Hi

In this question, the units are in pretty nice numbers.

You may use Units Method or MD.

Circle
Shaded ----- 1 unit
Unshaded ------ 3 units
Total -------4 units

Rectangle
Shaded ----- 1 unit
Unshaded ------ 2 units
Total ------- 3 units

Square
Shaded ----- 2 unit
Unshaded ------ 3 units
Total --------5 units

C:S:R ------4:5:3

Now you may wish to try this one.

Best wishes

http://farm3.static.flickr.com/2407/2673911274_a92537168a_o.jpg\">

hi Tianzhu
is the answer: 6:5:3?

[MathIzzzFun]

atutor2001:

pixiedust:

Appreciate help with the following question.


My solution is using listing multiples method.
Is there a faster way?
Such questions, analysing the gap (like the oranges question) doesn't work?

Some NCC cadets are gathered in a hall.
If they are grouped into 7 cadets in a team, there will be 4 cadets left out.
If they are grouped into 9 cadets in a team, there will be 6 cadets left out.
How many cadets are there in the hall if the number of cadets is between 20 and 70 ?

Mathematically, there is no faster way because the number of unknowns (i.e. The total no. of cadets; The no. of groups of 4 in case 1; The no. of groups of 6 in case 2 - 3 unknowns) is more than the number of relationship (equations) given by the question.

Hi

newbie here. Just to share that there is a quick method for this type of problem.

Look at the numbers carefully and you will see that by adding 3 to the total will make the total divisible by both 7 and 9. This means that the new total is a multiple of both 7 and 9. So the number is question is 7 x 9 - 3 = 63 - 3 = 60.

cheers.

[tianzhu]

pixiedust:

:goodpost: Thank you VERY much tianzhu !

It is useful. It is like using listing method + intelligence.

Much appreciate your time and help.

Hi pixiedust

Good Morning.

You're welcome.

Best wishes

[tianzhu]

ozora:

hi Tianzhu
is the answer: 6:5:3?

Hi ozora

Good Morning

The answer listed in the worksheet is 20:15:1

Best wishes

[atutor2001]

MathIzzzFun:

atutor2001:

[quote=\"pixiedust\"]Appreciate help with the following question.


My solution is using listing multiples method.
Is there a faster way?
Such questions, analysing the gap (like the oranges question) doesn't work?

Some NCC cadets are gathered in a hall.
If they are grouped into 7 cadets in a team, there will be 4 cadets left out.
If they are grouped into 9 cadets in a team, there will be 6 cadets left out.
How many cadets are there in the hall if the number of cadets is between 20 and 70 ?

Mathematically, there is no faster way because the number of unknowns (i.e. The total no. of cadets; The no. of groups of 4 in case 1; The no. of groups of 6 in case 2 - 3 unknowns) is more than the number of relationship (equations) given by the question.

Hi

newbie here. Just to share that there is a quick method for this type of problem.

Look at the numbers carefully and you will see that by adding 3 to the total will make the total divisible by both 7 and 9. This means that the new total is a multiple of both 7 and 9. So the number is question is 7 x 9 - 3 = 63 - 3 = 60.

cheers.[/quote]Hi MathIzzzFun

Thank you for sharing this technique. The addition of '3' to make the total divisible by both 7 and 9 because their remainders are 4 (which needs 3 more to form 7) and 6 (also needs 3 more to form 9) is a short cut that students who a trained in math olympiad may know.

Thanks for highlighting, I have forgotten about this technique - a very powerful tool for solving questions whereby the remainders need the same amount to form the exact group. If the above condition does not exist, we still need to list to get the answer.
Cheers

[tisha]

http://postimage.org/image/j5ukeupw/


Need help with this Q. It is from RGPS 2009 SA1 paper.

The given ans in the answer key is
------------------------------------------
1/2 X 12 X 18 = 108
12 X 18 = 216
216-40=176
108-40=68cm^2

------------------------------------------

Is the solution correct? Can anyone explain?

[tianzhu]

Hi tisha


Good Evening.

WXYZ is counted twice when we are calculating the areas of the two triangles.

So we need to subtract 40 from the total areas of the two triangles.

There are some unnecessary steps in the WS’s solution.

You may work it this way.

1/2 X 12 X 18 = 108 ----- this gives the areas of the two triangles.
108-40=68cm^2

Best wishes