Q&A - P5 Math

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Suz855
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Q&A - P5 Math

Post by Suz855 » Mon Oct 11, 2010 11:45 am

Moderator's Note: Please use this thread to ask for help on specific questions. Please refrain from out of topic chatter in this thread to make it easier for the community to find the answers they need.

Pls help, thanks


Image

:?

Vanilla Cake
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CHIJ St Nic 2007 P5 Maths SA2 Q45

Post by Vanilla Cake » Mon Oct 11, 2010 12:28 pm

Suz855 wrote:Pls help, thanks

Image

:?

There is missing information as you don't see the letters R and S of the triangle in the picture. Most likely, it was erased away with the letters R and S unintentionally and the important information that base of the triangle PRS has the measurement of 9 cm ie RS = 9 cm.

With this important information that RS = 9 cm, here's the solution:

Area of ΔPRS = ½x(12+4)x9= 72 cm²
(a) Area of shaded part = 72/8x7 = 63 cm²

Area of rectangle = (129+63) cm² = 192 cm²
Length of rectangle = 192÷12=16 cm
12/16x100% = 75%
(b) Breadth of the rectangle as a percentage of the length of the rectangle = 75%

Mods: Pls merge this post with Q&A - PSLE - Maths.
Thks.

Suz855
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Post by Suz855 » Mon Oct 11, 2010 2:01 pm

Thanks VC,

I was thinking along using the ratio given.

Since the shaded part is 7 and given that area of the triangle is 8, the unshaded triangle at the tip will be 1 vs the complete triange as 8.

Taking the height of whole triangle as 16 and the height of the unshaded triangle as 4, the base of the triangle should have the ratio of 1 : 2 which will bring us to area of triange as 32 or 64cm2 and the shaded portion as 28 or 56cm2.....

However probably you are right that the base of 9cm might already been given thus making this a simple area of triangle question instead of a more challenging one :D

ck123
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P5 SA2 SCGS Q18 , Ai Tong Q17

Post by ck123 » Fri Oct 22, 2010 3:49 pm

Hi anyone

Please help with the following questions:

1) Zara had 70 more dresses than skirts. She sold 3/4 of the dresses and 3/5 of the skirts. She sold 126 more dresses than skirts. What fraction of the remaining clothes that Zara had were skirts?

2) A seamstress had some white and some black buttons. 60% of the buttons were white. He bought another 120 buttons. Now, the percentage of white buttons became only 40% of the total. How many white buttons did he have at first?

Thanks a lot... :?:

[/img][/list]

Brenda10
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Re: P5 SA2 SCGS Q18 , Ai Tong Q17

Post by Brenda10 » Fri Oct 22, 2010 4:14 pm

ck123 wrote:Hi anyone

Please help with the following questions:

2) A seamstress had some white and some black buttons. 60% of the buttons were white. He bought another 120 buttons. Now, the percentage of white buttons became only 40% of the total. How many white buttons did he have at first?

Thanks a lot... :?:

[/img][/list]


Hi, dd working for Q2)

W:B
3:2 (Before)
Plus 120
4:6 (After)

4-3=1
6-2=4
1+4=5
120/5=24
24*3= 72

There are 72 white marbles at frist.

Thank you.


Brenda10
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Re: P5 SA2 SCGS Q18 , Ai Tong Q17

Post by Brenda10 » Fri Oct 22, 2010 4:39 pm

ck123 wrote:Hi anyone

Please help with the following questions:

1) Zara had 70 more dresses than skirts. She sold 3/4 of the dresses and 3/5 of the skirts. She sold 126 more dresses than skirts. What fraction of the remaining clothes that Zara had were skirts?Thanks a lot... :?:

[/img][/list]


Hi, you may refer the following link for solution.

http://www.kiasuparents.com/kiasu/forum ... start=2880

iFruit
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Re: P5 SA2 SCGS Q18 , Ai Tong Q17

Post by iFruit » Fri Oct 22, 2010 10:58 pm

ck123 wrote:Hi anyone

Please help with the following questions:

2) A seamstress had some white and some black buttons. 60% of the buttons were white. He bought another 120 buttons. Now, the percentage of white buttons became only 40% of the total. How many white buttons did he have at first?

Thanks a lot... :?:

[/img][/list]


I think this is question is incomplete. There can be many solutions as it is.

ck123
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Post by ck123 » Sat Oct 23, 2010 1:50 am

A big Thank You to all. :celebrate:

How can i insert a picture for a question to ask?

Thanks[/img]

greenfinger
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Area of Circle & revolutions

Post by greenfinger » Fri Apr 08, 2011 11:27 pm

Hi ,anyone can help me, pls


question:

A wheel made 72 revolutions.It has a radius of 23 cm.How far did it travel? :?

jieheng
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Re: Area of Circle & revolutions

Post by jieheng » Fri Apr 08, 2011 11:53 pm

greenfinger wrote:Hi ,anyone can help me, pls


question:

A wheel made 72 revolutions.It has a radius of 23 cm.How far did it travel? :?


The distance it travelled = No. of revolutions * Circumference = No. of revolutions * Pi * Diameter = 72 * 3.14 * ( 2 * 23 ) = 10400 cm (Ans)

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