Q&A - PSLE Math

[tianzhu]

Chan09:

Hi Tianzhu,


Sorry still unable to draw this out on a model diagram:
A has $900 more than B. A gave B some money. In the end B had $880 (as not $800 as previously stated more than A. How much did A give B? :?

Can you advise, thank you

Hi

Good Morning.

Hope this helps.

Best wishes

http://farm8.staticflickr.com/7116/7147048583_8af489014b_z.jpg\">

[kklau]

Hi,


Appreciate again to anyone's help to provide solution to this P6 maths question.

The diagram below 2 quadrants and a circle in a square of edge 10cm. Find the area of the shaded region (Take π = 3.14)

http://i46.tinypic.com/hx0y7p.png\">



Thanking you in advance.

Best Regards.

[MathIzzzFun]

Mr OK:

Hi MathIzzzFun, thank you very much for helping me with the 3 questions I posted yesterday. After reading some earlier posts, I understand that there are numerous methods to solve any problem sum. So other than model drawing, is there any other method for these 3 questions that u could share. Thanks once again.

Hi

There are different approaches that can be used.. I have used model approach to benefit the P4/P5s who will not be too accustomed yet to non-visual approach.

For Q1, you can just removed the model, and apply the assumptive method. The model was used to make it easier to see the \"difference\" Many students are not able to comprehend the assumptive method and often get confused with the \"difference\" and most conveniently take the difference as \"3-2=1\" whereas in this case, the difference is \"3+2 =5\"

For the other 2 question, one can use the unit method/cross multiply/unit-part method, or simply \"convert\" the model into words. eg..
Q2-
if 2 blue chips removed,
blue : red --> 2u : 6u

original number blue : red --> 2u + 2 : 6u = 2(u+1) : 6u

if 4 red chips removed,
blue : red --> 2 (u+1) : 8(u+1) -- (ratio 1:4)
red chips --> 6u - 4 = 8u +8 and you can complete the solution.



cheers.

[elenatay]

Michaelia0816:

and also welcome elenatay to here!

Hi Michaelia0816,
Many thanks for the help. I have been searching the pages for a similar posting but gave up after browsing through many pages!

[superhappy]

elenatay:

Michaelia0816:

and also welcome elenatay to here!

Hi Michaelia0816,
Many thanks for the help. I have been searching the pages for a similar posting but gave up after browsing through many pages!

Hi, need help on this questions
1. Varsha had some 20-cent and 50-cent coins.
7/8 of the coins were 20-cent coins and the rest were 50-cent coins.
After Varsha had spent $72.50 worth of 50-cent coins and 5/7 of the 20-cent coins, she had 2/7 of coins left. Find the total amount of money Varsha left.

2. The ratio of the number of beads John had to the number of beads Sally had was 3:7 at first. After Sally had given some beads to John, Sally had 76 beads less than John. The ratio of John's number of beads to Sally's number of beads became 3:2.
a) how many beads did Sally give to John?
b) what percentage of her beads did Sally give to John?

Thank you !

[cimman]

Mr OK:

Hi MathIzzzFun, thank you very much for helping me with the 3 questions I posted yesterday. After reading some earlier posts, I understand that there are numerous methods to solve any problem sum. So other than model drawing, is there any other method for these 3 questions that u could share. Thanks once again.

Hi Mr OK
if you're open to using algebra, I conduct workshops to solve complex problem sums using algebra. The problem with using algebra, is that it is difficult to formulate the equations. I've developed a technique to visualize the equations graphically, so that it is easy to formulate the equations. It's a fusion of modelling and algebra. Modelling meets Algebra. I call it Pictorial Algebra.
The equations that comes out using this technique is also simpler than using traditional methods of translating words to equations. It's a step by step methodical process that leads one to the final solution.
The workshop is free. You can view the details here: http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=43&t=32278

This technique is very suitable for parents as they're already well versed in using algebra and are able to grasp the technique very quickly as compared to modelling. Quite a number of parents have used this method and have found it to be effective.
The 3 problems you've posted can be solved using this technique.

If you're interested, just indicate your interest in the Happening thread and I'll notify you when the next workshop is available. The workshop also gives you the strategies to identify different problem types.

[small]

Need a clarification on this question and TIA.


Question:
Alan is trying to catch up with his sister who is 24m in front of him. For every 60-cm step made by Alan, his sister makes only 20-cm step. How many steps does Alan need to make to overtake his sister?

60 steps or 61-steps?

[kyp71]

Hi, please try to explain n solve this question using model asap (TIA :rahrah: )


Jug X and Jug Y contain different amounts of water at first. 50% of the water in Jug X was poured into Jug Y. Then 40% of the water in Jug Y was poured into Jug X. The final ratio of the amount of water in Jug X to the amount of water in Jug Y was 7:6.

(a) What was the ratio of the amount of water in Jug X to the amount of water in Jug X to the amount of water in Jug Y at first?

(b) If there were 10 litres of water in Jug X at first, how much water was there in Jug X at the end?

[kyp71]

[quote]hi, I did the question n my answer is 60 steps. :imcool: [/quote]

small:

Need a clarification on this question and TIA.


Question:
Alan is trying to catch up with his sister who is 24m in front of him. For every 60-cm step made by Alan, his sister makes only 20-cm step. How many steps does Alan need to make to overtake his sister?

60 steps or 61-steps?

[small]

kyp71:

Hi, please try to explain n solve this question using model asap (TIA :rahrah: )


Jug X and Jug Y contain different amounts of water at first. 50% of the water in Jug X was poured into Jug Y. Then 40% of the water in Jug Y was poured into Jug X. The final ratio of the amount of water in Jug X to the amount of water in Jug Y was 7:6.

(a) What was the ratio of the amount of water in Jug X to the amount of water in Jug X to the amount of water in Jug Y at first?

(b) If there were 10 litres of water in Jug X at first, how much water was there in Jug X at the end?

Hi kyp71,

using ratio method will be more appropriate

(a) In the end, ratio of water in Jug X to Jug Y ----- 7 : 6

Working backwards
X : Y
7 : 6 (at end)
3 : 10
6 : 7 (at first)

The ratio of water in Jug X to Jug Y at first ----- 6 : 7

Explanation:
In the end, ratio of water in Jug X to Jug Y ----- 7 : 6

Before 40% of the water in Jug Y was poured into Jug X, Jug Y will have 10 units of water, so the ratio of water in Jug X to Jug Y now is 3 : 10

This was after 50% of the water in Jug X was poured into Jug Y, so before that Jug X have 6units (3u x 2) of water, thus the amount of water in Jug Y at first is 7 units


(b) I am sure you will be able to solve it.