KSP2013777:

Pls help me with this question.


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Thank you

Hi there! You are IP Yr4 I believe.
This pair of simultaneous equations can't be solved simply with the knowledge you have at the moment. You should be using GC.
Pt A can be found easily of cos.

As for the area between these two curves, simply integrate from the (x) limits you have found. This is can done easi
Eg integrate wrt x; from 0 to 3.50; exponential minus hyperbola.
Hope this is useful.

Verysuperkiasu:

Thank you for your clear explanation, ADoc!

Initially, the confusion my child has is this - If dry ice goes straight to gaseous (which should be invisible) then why we still see the mist which suggests liquid state. Correct me if I'm wrong - that means to say the mist (liquid) we see is from the condensed water vapour from the air and is not the gaseous state of dry ice. Am I right?

hi! That's right, just as we can't observe CO2 in the air mixture around us.

Just to elaborate further. Some kids may have played with dry ice and they observe that \"white bubbles\" are formed when water is added to dry ice. And so wrongfully conclude that carbon dioxide is \"white\".

cheers!

Yes, the bubbles are the CO2 (gaseous) that is released. But the \"white outer surface\" is actually condensed water vapour. Think of it as an invisible ball with condensation on its surface. The ball itself is invisible or transparent but due to the condensation, it's as though the ball is \"white\".

Verysuperkiasu:

I know that dry ice sublimes (from solid straight to gaseous without going thru liquid state). But the appearance of the mist - doesn't it suggest that it is in the liquid state? because anything that u can see like clouds, mist, fog are not in gaseous state but they are actually water droplets (liquid state) right?

Hi there! A quick explanation would be:
The white mist/smoke/fog is in fact condensed water vapour from the surrounding air.

Perhaps a useful way to make yout kid understand without having to explain sublimation process would be:

(1) dry ice is solidified carbon dioxide

(2) it is very cold

(3) when dry ice sublimes, the carbon dioxide vapour (gaseous) is also very cold

(4) the surrounding water vapour in the air (which is at a higher temperature) loses heat to the carbon dioxide vapour since heat flows from regions of higher temperature to regions of lower temperature

(5) water vapour cools and condenses to form the white mist/cloud that we see when dry ice sublimes

Of cos the mechanics is a little more complicated than above explanation, but these should be sufficient to \"cure\" our kids' curiosity and \"why why why\"...

hope this is useful. cheers!

Michaelia0816:

Ok now I need help! Pls help!

Question:
A) Mr Chan had a total of 160 pieces of $2 notes and $5 notes. He gave 50% of the $2 notes and 25% of the $5 notes to Susan and had 92 notes left.
a) Find the percentage of the number of $2 notes at first.
b) How much money did Mr Chan have at first?

Hi!
Let's look at the remainder and attempt to \"work backwards\":

gave away 50% of $2 notes --> left 50%
gave away 25% of $5 notes --> left 75%

=> 50% of $2 notes + 75% of $5 notes --> 92
that means 100% of $2 notes + 150% of $5 notes --> 92 x 2 = 184 } multiply by 2 throughout

qs says 100% $2 notes + 100% $5 notes --> 160
therefore if we compare (subtract) the two, we have 50% of $5 notes --> 184 - 160 = 24

=> 100% of $5 notes --> 24 x 2 = 48
=> 100% of $2 notes --> 160 - 48 = 112

(a) percentage of $2 notes of total number of notes = (112/160) x 100% = 70%

(b) amount of $5 notes = 48 x $5 = $240
amount of $2 notes = 112 x $2 = $224

total = $240 + $224 = $464

cheers
eugene

[again this is pseudo algebra (solving simultaneous by elimination method) in disguise but our primary kids are taught to solve such qs in this manner though not formally introduced to the concept until Sec1]

MathIzzzFun:

hi Zack7,

if you were to go through the solution and understand the logic, you should see that it was just a typo omission by ADoc, the method is correct.

2/3 ali = 3/5 raja.. \"equalize\" the numerator ie 2/3 x 3/3, 3/5 x 2/2
we will get 6/9 Ali = 6/10 Raja --> So, Ali --> 9 units & Raja --> 10 units

.. this is a common method taught and it is easy to understand because one can easily draw a model to see that indeed Ali has 9 blocks and Raja has 10 blocks.

there will always be different approaches to solving a problem - no one method is better than the other.. it is a question of preference. One person may prefer the algebra approach, another may like the unit approach, while another will simply stick to model drawing. So, be open to various approaches because for some questions, the algebra approach is not the most efficient one.

cheers.

Hi MathIzzzFun! Tks for clarifying my mess-up!

cheers
eugene

Zack7:

ADoc:

[quote=\"peggy\"]Hi, my mind just isn't working. This should be an easy one but probably I might have solved too difficult questions lately till I can't solve the easy one. Please help.


2/3 of Ali's story books was the same as 3/5 of Raju's story books. If Ali had 100 fewer books, how many books would Ali have ?

Hi! Other than algebra, which some primary students may not be too comfortable with, we explain to them that before we can compare ratio or fraction units of the SAME qty, we must change them to the same number of units. This is an important concept in order to score those ratio qs, that I'm sure parents and students here are already very familiar with.

(2/3) x 3/3 = 6/9 } ali
(3/5) x 2/2 = 6/10 } raju

We see that Ali has 1 unit less than raju ---> 100 books
therefore Ali has 9 x 100 = 900 books.

Another variation of this sort of 2-mark qs is to ask for the fraction or ratio of say, Ali's to raju's.

cheers
eugene

your method is quite unorthodox i must say...

firstly, (2/3) x 3 = 2, not 6/9. same with 3/5

secondly, 6/9 is not 1 unit less than 6/10... in fact, 6/9 > 6/10. it might work as a shortcut in this question, but i don't think it is clear to the kids or even the right concept.

a clearer way would be

2/3 ali = 3/5 raja (as per the question)
ali = 9/10 raja (cross multiply)

now here comes the crucial step namely, how to interpret the above equation.
the equation tells you ali has only 9/10 as many books as raja.
always ask yourself when you are forming your conclusion, who has more books who has less. does it agree with the question? in this case, ali has less.

the equation might look like raja has fewer since there is a 9/10 factor, but this is not true because of the equal sign. it is important to note the equality sign.

so in this case, one should interpret it as : the WHOLE of ali is only EQUAL to 9/10 of raja. that means everything that ali has is not even equal to the whole of raja, but rather only equal to 9/10 of raja's books.

so this means 1/10 of raja's books is the amount that is more than ali. so ali has 100 books less than raja, which means raja has 100 more than ali.

9/10 of raja, which is = ali, is then 100x9 = 900.[/quote]
Hi! My bad! tks for pointing out Zack. What i meant was 3/3 and 2/2 (as corrected above) to change to equivalent fractions. The concept is logically sound and in practice and can be construed as algebra in disguise in fact. It's just that most primary students prefer working in this manner to the ill-perceived horrors of algebra, especially cross multiplying and simultaneous linear eqn (which are not taught \"formally\" in most schools except by some primary teachers in certain schools that I know of).

cheers
eugene

jewelbox:

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Hi! Again, without using formal algebra as our kids may be uncomfortable with, we can choose to work simply with ratios that they are well-trained at.

A recurring concept for most, if not all, ratio problems (including fractions), is to change to equivalent ratios. Second is to IDENTIFY the qty that remains constant.

In this case we can choose either (1) the shaded area that is cut is the SAME for both Rect and Sq, or (2) the difference between the areas of the Rect and Sq remains the SAME if we cut the same amount of area from each of them.

Before cutting
R : S = 5 : 2

After cutting
R : S = 3 : 1

using (2) we see that the difference in areas of R and S before and after cutting must remain the SAME as the same qty was cut.

Now we check:

before cutting: R - S = 5 - 2 = 3 units
after cutting: R - S = 3 - 1 = 2 parts [not the same]

the all-too-familiar step is to cross multiply each other:

so we have:

before cutting R : S = 5 : 2 = 10 : 4 (MULTIPLY throughout by \"2\")
after cutting R : S = 3 : 1 = 9 : 3 (MULTIPLY throughout by \"3\")

indeed now the difference in units are the SAME = 6 units

area of S before cutting = 4 units --> 36 cm square
after cutting = 3 units => shaded area = 1 unit
1 unit -- > 9 cm square

again sorry for the lengthy reply. These are meant to guide our kids in their understanding. The workings are much much shorter than these certainly.

For experienced primary students, this should be at most a 4-5min qs as the steps involved are simple.

hope this is useful

cheers
eugene

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Hi! A couple of good practices to enhance our kids' understanding for this type of qs:

(1) explain to the kid the concept of finding the area of a fraction of a circle

Our primary students are familiar with area of semi-circles ( x 1/2) and quadrants ( x 1/4). To reinforce their understanding, we explain to them the \"whys\": half a circle is 180 deg --> what fraction is 180 deg of a full circle? --> 180/360 = 1/2 --> that's why we multiply 1/2 for area of semi-circle. Similar explanation for quadrants. Now we can stretch this understanding and explain to them that for any part of a circle, so long as we know the angle that it subtends, say 45 deg in this example, it is 45/360 = 1/8 of a circle.

(2) do not be afraid to add in extra alphabets (or points) on the figure if it helps you to understand this sort of plus and minus areas of various shapes problem.

In this example: adding a point F, where the arc BE and Line DC intersect, may be useful.

So we have:
area of part of a circle ABE = area of shaded part ADFE + area of BDF
area of triangle BCD = area of shaded part BCF + area of BDF

Now perhaps our kids can see the solution clearer. Qs asks for difference between the shaded areas, hence it's area of part of a circle ABE minus area of triangle BCD, since the area of BDF cancels out each other.

Sorry for the wordy explanation. These are meant to help parents to explain to their kids should they face difficulty.

Hope this is useful to some.

cheers
eugene

peggy:

Hi, my mind just isn't working. This should be an easy one but probably I might have solved too difficult questions lately till I can't solve the easy one. Please help.


2/3 of Ali's story books was the same as 3/5 of Raju's story books. If Ali had 100 fewer books, how many books would Ali have ?

Hi! Other than algebra, which some primary students may not be too comfortable with, we explain to them that before we can compare ratio or fraction units of the SAME qty, we must change them to the same number of units. This is an important concept in order to score those ratio qs, that I'm sure parents and students here are already very familiar with.

(2/3) x 3 = 6/9 } ali
(3/5) x 2 = 6/10 } raju

We see that Ali has 1 unit less than raju ---> 100 books
therefore Ali has 9 x 100 = 900 books.

Another variation of this sort of 2-mark qs is to ask for the fraction or ratio of say, Ali's to raju's.

cheers
eugene

JadeDry:

Hello,


I would appreciate assistance with the following question:

3 divided by (x-3) + (1 divided by (x+1)) divided by (2x divided by (x-3))

Is the answer:

(7(x squared) +9)/((2x)(x+1)(x-3))

or

2x/(x+1) ?

Thanks in advance.

hihi! First answer too, assuming the correct order of operations from what you typed above.

cheers
eugene