Yes, first topic in P6 Maths text, algebra.However, are children in P6 taught how to do an expansion of (x+y)^2 ? Without that step, it is impossible to solve the equation.

But I think that the solution provided by lantian is must better.

Academic support for Primary 6 and PSLE

- ChiefKiasu
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by **ChiefKiasu** » Sat Sep 13, 2008 9:53 pm

Yes, we will not be able to find the answer if we attempt to first find the values of x and y, the sides of the original rectangle, ie.Priskhipo wrote:...But it is not possible for the area of 4 squares to be 80 cm2 if the perimeter of the rectangle is 20 cm...

x + y = 10

x^2 + y^2 = 40

=> x^2 + (10 - x)^2 = 40

=> x^2 + (100 - 20x + x^2) = 40

=> 2x^2 - 20x + 60 = 0

=> x^2 - 10x + 30 = 0

We cannot solve for x because x is not an integer.

lizawa's method solves this problem because x * y is an integer.

But lantian's method is the best.

Post
by **Priskhipo** » Sun Sep 14, 2008 2:44 pm

Hi,ChiefKiasu wrote:Yes, we will not be able to find the answer if we attempt to first find the values of x and y, the sides of the original rectangle, ie.Priskhipo wrote:...But it is not possible for the area of 4 squares to be 80 cm2 if the perimeter of the rectangle is 20 cm...

x + y = 10

x^2 + y^2 = 40

=> x^2 + (10 - x)^2 = 40

=> x^2 + (100 - 20x + x^2) = 40

=> 2x^2 - 20x + 60 = 0

=> x^2 - 10x + 30 = 0

We cannot solve for x because x is not an integer.

lizawa's method solves this problem because x * y is an integer.

But lantian's method is the best.

I agree lantian's method is best in this case. It illustrates the logic clearly.

(It does not have to check x^2 + y^2 = 40 cm^2 is correct or not.)

I'm only worry that should a student chooses to solve it with Guess and Check strategy, he would not be able to yield an answer because of the error in this question.

A question should yield the same answer no matter which method a student uses.

Below images illustrate why (x^2 + y^2) should not be less than 50 cm^2

The quadratic equation gives a constraint (10 cm) to the relationship of the x and y, therefore it finds x, a sq root of a negative integer (not a real number) if given the total area x^2 & y^2 is 40 cm^2. Hence an impossible case.

Last edited by Priskhipo on Sun Sep 14, 2008 2:48 pm, edited 1 time in total.

- heutistmeintag
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by **heutistmeintag** » Sun Sep 14, 2008 2:47 pm

ok, I need help on this P6 problem. No simultaneous equations pls.

There are several red balls and white balls on the table.

If 1 red ball and 1 white ball are removed together each time until no red balls are left on the table, then the number of remaining white balls is 50.

if 1 red ball and 3 white balls are removed together each time until no white balls are left on the table, then the number of remaining red balls is also 50.

Find the total number of red balls and white balls at first.

Many thanks in advance!

There are several red balls and white balls on the table.

If 1 red ball and 1 white ball are removed together each time until no red balls are left on the table, then the number of remaining white balls is 50.

if 1 red ball and 3 white balls are removed together each time until no white balls are left on the table, then the number of remaining red balls is also 50.

Find the total number of red balls and white balls at first.

Many thanks in advance!

Post
by **Priskhipo** » Sun Sep 14, 2008 3:27 pm

Hi,heutistmeintag wrote:ok, I need help on this P6 problem. No simultaneous equations pls.

There are several red balls and white balls on the table.

If 1 red ball and 1 white ball are removed together each time until no red balls are left on the table, then the number of remaining white balls is 50.

if 1 red ball and 3 white balls are removed together each time until no white balls are left on the table, then the number of remaining red balls is also 50.

Find the total number of red balls and white balls at first.

Many thanks in advance!

Solution:

- heutistmeintag
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**Posts:**385**Joined:**Wed Jun 25,

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by **heutistmeintag** » Sun Sep 14, 2008 5:16 pm

Thanks, Priskhipo. Could you elaborate on the 2nd part pls?

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by **Priskhipo** » Sun Sep 14, 2008 10:38 pm

Hi heutistmeintag,heutistmeintag wrote:Thanks, Priskhipo. Could you elaborate on the 2nd part pls?

Referring to Model representing White Balls:

1 Part + 50 in 1st scenario = 3 units in 2nd scenario

because it says for every 3 white balls & 1 red ball are removed each time till no white ball left. Therefore all of the white balls are divided into 3 units.

From the 2 models, 1 Part = 2 units and 50 = 1 unit

Referring to Model representing Red Balls:

(1 unit = 50) is confirmed in 2nd scenario.

1 unit of Red Balls is removed and left with 50.

(Remember 1 Part = 2 units?)

Hope this can help you.

Post
by **suiyuan** » Mon Sep 15, 2008 3:54 pm

Thank you for your time

At a party, Sarah distributed 18 sweets to each of the boys and 20 sweets to each of the girls and had 48 sweets left. If she had distributed 20 sweets to each of the boys and 18 sweets to each of the girls instead, she would have 32 sweets left. If Sarah had 40 pupils in her class,

a) How many more boys than girls were there in her class?

b) How many sweets did she have at first?

At a party, Sarah distributed 18 sweets to each of the boys and 20 sweets to each of the girls and had 48 sweets left. If she had distributed 20 sweets to each of the boys and 18 sweets to each of the girls instead, she would have 32 sweets left. If Sarah had 40 pupils in her class,

a) How many more boys than girls were there in her class?

b) How many sweets did she have at first?

Post
by **lizawa** » Mon Sep 15, 2008 8:25 pm

Scenario 1:

Boys : 18 sweets each

Girls : 18 sweets + 2 sweets each

Left : 48 sweets

Scenario 2 :

Boys : 18 sweets + 2 sweets each

Girls : 18 sweets each

Left : 32

So, no. of sets of 18 sweets will be the same in both scenarios.

The only difference is the sets of "2 sweets".

Difference in sets of 2 sweets = 48 - 32 = 16

16 = 8 sets of 2 sweets

(a) There are 8 more boys than girls.

(b) Total no. of pupils = 40

no. of girls = 16

no. of boys = 24

no. of sweets = (18 * 24) + (20*16) + 48 = 800

Boys : 18 sweets each

Girls : 18 sweets + 2 sweets each

Left : 48 sweets

Scenario 2 :

Boys : 18 sweets + 2 sweets each

Girls : 18 sweets each

Left : 32

So, no. of sets of 18 sweets will be the same in both scenarios.

The only difference is the sets of "2 sweets".

Difference in sets of 2 sweets = 48 - 32 = 16

16 = 8 sets of 2 sweets

(a) There are 8 more boys than girls.

(b) Total no. of pupils = 40

no. of girls = 16

no. of boys = 24

no. of sweets = (18 * 24) + (20*16) + 48 = 800

Post
by **youngmum** » Tue Sep 16, 2008 2:41 pm

How to solve these 3 questions pretty please?

(1) Mrs Tan has some pencils. If she packs them into bundles of 5 pencils in each bundles, she will have 4 extra pencils. If she packs them into bundles of 9 pencils in each bundles, she will need 7 mor epencils. How many pencils are there?

(2) At a bk fair, Tom spent $6 less than of his money on a storybook and $2 more than of his remaining money on reference books. After buying thebooks, he still had $10 left.

(a) How much money was left after he bought his story books and before hte bought the reference bk?

(b) How much money did Tom have at first?

(3) There were 2 identica flights of steps. For the first flights of steps, Siti walked up some steps adn ran 4 steps and took a total of 75 seconds. For the second flight of steps, she walked up some steps and ran 11 steps and took a total of 40 seconds. how long will siti take if she had walked up both flight of steps? Leave your answer in seconds.

TIA.

(1) Mrs Tan has some pencils. If she packs them into bundles of 5 pencils in each bundles, she will have 4 extra pencils. If she packs them into bundles of 9 pencils in each bundles, she will need 7 mor epencils. How many pencils are there?

(2) At a bk fair, Tom spent $6 less than of his money on a storybook and $2 more than of his remaining money on reference books. After buying thebooks, he still had $10 left.

(a) How much money was left after he bought his story books and before hte bought the reference bk?

(b) How much money did Tom have at first?

(3) There were 2 identica flights of steps. For the first flights of steps, Siti walked up some steps adn ran 4 steps and took a total of 75 seconds. For the second flight of steps, she walked up some steps and ran 11 steps and took a total of 40 seconds. how long will siti take if she had walked up both flight of steps? Leave your answer in seconds.

TIA.

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