Q&A - PSLE Math

Academic support for Primary 6 and PSLE
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tianzhu
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Post by tianzhu » Tue Jul 15, 2008 9:27 am

lizawa wrote:Hi Tianzhu,

Your answer is correct. My son has tried this paper before. But I remembered our answer key did have the correct answer.

Lizawa
Hi Lizawa
Thanks, appreciate it if you could share your solution.

lizawa
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Post by lizawa » Tue Jul 15, 2008 10:47 am

Large packet
c:s:t
= 1:2:3

Standard packet
c:s
4:5

Large packet contains 2x as many candies as standard packet.
No. of units (standard packet) = 9
no. of units (large packet) = 9 *2 = 18
therefore, c:s:t = 3:6:9

(a) 1 standard + 1 large, c:s:t = 7:11:9

(b)
before eating : c:s:t = 7:11:9
after eating 21 candies, c:s:t = 2:3:3 = 6:9:9

No toffee candies are eaten
18u - 15u = 21
3u = 21
u =7
candies left = 24*7 = 168.

tianzhu
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Post by tianzhu » Tue Jul 15, 2008 12:25 pm

Hi lizawa

Thank you for your fast response.

tianzhu
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Post by tianzhu » Wed Jul 16, 2008 4:18 pm

Thank You for your time in helping to solve this problem.Trying to figure out without using Algebra.

Image
Last edited by tianzhu on Thu Jul 17, 2008 7:22 am, edited 1 time in total.

lizawa
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Post by lizawa » Wed Jul 16, 2008 4:33 pm

hi tianzhu.

the area refers to the white space labeled A,B and C, or the area of the circle A, B, C ?

lizawa


tianzhu
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Post by tianzhu » Wed Jul 16, 2008 5:37 pm

lizawa wrote:hi tianzhu.

the area refers to the white space labeled A,B and C, or the area of the circle A, B, C ?

lizawa
Hi lizawa
I believe it refers to the areas of circle A,B and C.

lizawa
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Post by lizawa » Thu Jul 17, 2008 9:45 am

Hi tianzhu,

my son said this type of question normally refers to unshaded part of A:B:C

Do you have the answer ?

Is it : 25:18:1 ?

lizawa
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Post by lizawa » Thu Jul 17, 2008 10:25 am

Solution

Total area of cirlces
(B+C) : A
= 4:5
= 24:30

1/6 of A shaded -> 1/6 * 30 = 5 units shaded
unshaded A -> 30 units - 5 units = 25 units
unshaded (B+C) -> 24 - 5 = 19 units

Can draw model here, to see clearer.
B has 5 units, C has 3 units. Size of units in B and C are different.
5 units of B + 3 units of C -> 24

1 unit of B + 1unit of C -> 5
3 units of B + 3 units of C ->15

2 units of B -> 24 - 15 = 9
unshaded B = 4 units of B -> 9 * 2 = 18

2 units of C -> 24 - 18 - 5 = 1
unshaded C = 2 units = 1

unshaded
A:B:C
= 25:18:1

tianzhu
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Post by tianzhu » Thu Jul 17, 2008 10:29 am

Hi lizawa

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

lizawa
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Post by lizawa » Thu Jul 17, 2008 10:41 am

The answer listed in the worksheet is 20:15:1
then, in that case, it's asking for the total area of A : B:C

Extension to the above solution,

unshaded B = 18
total B = 22.5

unshaded C =1
total C = 1.5

A:B:C
30:22.5 : 1.5
= 60:45:3
= 20:15:1

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