Q&A - P5 Math

[dazzlego]

whywhy:

hi dazzlego, how did u get 65 degree? the question I have show angle QRT is 45 degree. tks

Maybe the teacher ask them to change to 65. There was an amendment in my paper.
If it is 45, the answer would be 165 (which does not match with the figure given since the angle is obviously more than 180)

Cheers

[dazzlego]

lapapillion:

whywhy:

Roysth SA2 P2 Q17 - PLS HELP

in May, the average no of participants in one pottery class was 24. In June, 5 new pottery classes were added. The ave no of participants in all the pottery classes became 26. if there were 32 participants in each of the new pottery classes, find the total no of participants in all the pottery classes in June. thanks

I took some time to do it in algebra, but obviously that can't be taught to pri 2 kid right?

So I think the next best approach is by trial and error. First we have to come up with a algorithm to solve the problem.

Average no. of participants is equal to the sum of all the students in May plus all the new students in June, divided by the total number of classes. And that number is equal to 26. But the sum of all the number of students in May is an unknown figure, since we do not have the number of classes in May. So we'll have to trial and error a number for that:

Say no. of classes in May is 1. Then we'll have 1x24 + 5(32) students in all and 1+5 classes. The average works out to be 30+, which is not 26, hence our assumption is wrong.

Let's go on to 2 classes in May. Then we'll have 2x24 + 5(32) students in all and 2+5 classes. The average works out to be 29+, which is not 26 again.

Now we don't have to do 1 by 1 until we hit the answers, so we do in multiples of 5. I'll try 5, then 10, then 15, then 20.

You'll find that the right number of classes in may is 15. You know you're right when the average turns out to be 26 exactly. So the number of participants is 15x24+5x32 = 520

Hi whywhy,
Is this p2 question? If it is then you should post it on p2 thread, not p5

Hi lapapillion,
I got an alternative method for this type of question by using the concept of Area
I am not sure whether this method can be applied for p2 or not

Anyway, I draw this diagram at first. (by assuming the no. of class was 1 unit at first)
The blue color is for the individual average and the red color is the combined average.

http://i49.tinypic.com/35cemc1.jpg\">


The shaded area in this diagram is actually the number of students.
Therefore, Blue shaded = Red shaded
Blue shaded = 24 x 1 unit + 32 x 5 = 24 units + 160
Red shaded = 26 x 1 unit + 26 x 5 = 26 units + 130
2 units = 30
1 unit = 15
total no. of class now: 15 + 5 = 20
average no. of students now = 26
total no. of students = 20 x 26 = 520

Cheers :celebrate:

[whywhy]

Oic. noted with thanks

[whywhy]

Oic. noted with thanks

[whywhy]

Oic. noted with thanks

[cimman]

dazzlego:

lapapillion:

[quote=\"whywhy\"]Roysth SA2 P2 Q17 - PLS HELP

in May, the average no of participants in one pottery class was 24. In June, 5 new pottery classes were added. The ave no of participants in all the pottery classes became 26. if there were 32 participants in each of the new pottery classes, find the total no of participants in all the pottery classes in June. thanks

I took some time to do it in algebra, but obviously that can't be taught to pri 2 kid right?

So I think the next best approach is by trial and error. First we have to come up with a algorithm to solve the problem.

Average no. of participants is equal to the sum of all the students in May plus all the new students in June, divided by the total number of classes. And that number is equal to 26. But the sum of all the number of students in May is an unknown figure, since we do not have the number of classes in May. So we'll have to trial and error a number for that:

Say no. of classes in May is 1. Then we'll have 1x24 + 5(32) students in all and 1+5 classes. The average works out to be 30+, which is not 26, hence our assumption is wrong.

Let's go on to 2 classes in May. Then we'll have 2x24 + 5(32) students in all and 2+5 classes. The average works out to be 29+, which is not 26 again.

Now we don't have to do 1 by 1 until we hit the answers, so we do in multiples of 5. I'll try 5, then 10, then 15, then 20.

You'll find that the right number of classes in may is 15. You know you're right when the average turns out to be 26 exactly. So the number of participants is 15x24+5x32 = 520

Hi whywhy,
Is this p2 question? If it is then you should post it on p2 thread, not p5

Hi lapapillion,
I got an alternative method for this type of question by using the concept of Area
I am not sure whether this method can be applied for p2 or not

Anyway, I draw this diagram at first. (by assuming the no. of class was 1 unit at first)
The blue color is for the individual average and the red color is the combined average.

http://i49.tinypic.com/35cemc1.jpg\">


The shaded area in this diagram is actually the number of students.
Therefore, Blue shaded = Red shaded
Blue shaded = 24 x 1 unit + 32 x 5 = 24 units + 160
Red shaded = 26 x 1 unit + 26 x 5 = 26 units + 130
2 units = 30
1 unit = 15
total no. of class now: 15 + 5 = 20
average no. of students now = 26
total no. of students = 20 x 26 = 520

Cheers :celebrate:[/quote] this cannot be a P2 question. The concept of Average is not taught at P2. This question was raised earlier by pineapple_tarts in the P5 forum: http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&start=1260

[zeemimi]

this is a P5 question. My DS just did this yesterday.

[zinc6539]

http://img.photobucket.com/albums/v187/veron/Capture2.jpg\">


P5 Maths 2011 SA2 CHIJ Qn 24

Pls help to solve, thks. Ans given is 330 deg.

[zeemimi]

zinc6539:

http://img.photobucket.com/albums/v187/veron/Capture2.jpg\">


P5 Maths 2011 SA2 CHIJ Qn 24

Pls help to solve, thks. Ans given is 330 deg.

s = 60

t + u + v = 360 - 90 = 270

total = 270 + 60 = 330

[zinc6539]

Thanks! Now then I remember that the interior angles of a 4-sided figure always add up to 360 degrees. =P