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[Maths Hub]

pinky88:

Hi,


I need help for the following questions:

2.\tDiana is to go for a business trip. She needs a collection of wigs and sunglasses, 8 items altogether. At least 3 of these have to be wigs and at least 4 sunglasses. She has 10 different wigs and 12 different sunglasses to choose from. In how many ways can she organize the collection of wigs and sunglasses she needs for the trip.

TIA.

2. There are 2 possible cases.
Case 1: 3 wigs, 5 sunglasses
10C3 X 12C5 = 120 X 792 = 95040
Case 2: 4 wigs, 4 sunglasses
10C4 X 12C4 = 210 X 495 = 103950

Total: 95040 + 103950 = 198990

Cheers.

[pinky88]

Many thx, Maths Hub

[Maths Hub]

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

[BeContented]

Maths Hub:

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

Thank you.....been waiting until neck long long liao.
DD will be delighted

[TanJ]

Maths Hub:

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

Does gettng a gold qualify for invitation round? How will my kid know?
But then.... second round heavyweights.... chances as good as nil....

[MathsOlympiadtrainer]

TanJ:

Maths Hub:

Congrats to all students who did well in the SMOPS 2013! :rahrah: :rahrah: :rahrah:


http://www.hci.sg/aphelion/apmops/smops2013results.htm

Does gettng a gold qualify for invitation round? How will my kid know?
But then.... second round heavyweights.... chances as good as nil....

Both Platinum and Gold medalist will definitely qualify for invitation round since there are less than 150 people for P and G altogether. They might include some \"better\" Silvers too.

Cheers.

[TanJ]

MathsOlympiadtrainer:

TanJ:

Does gettng a gold qualify for invitation round? How will my kid know?
But then.... second round heavyweights.... chances as good as nil....

Both Platinum and Gold medalist will definitely qualify for invitation round since there are less than 150 people for P and G altogether. They might include some \"better\" Silvers too.

Cheers.

Thanks.

[pinky88]

Hi, need help again on the flwg qns, thx...


1. In the diagram below, ABCD is a square. The points A, B & G are collinear. The line segments AC & DG intersect at E, and the line segments DG & BC intersect at F. Suppose that DE = 15 cm, EF = 9 cm & FG = x cm, find the value of x.

http://i42.tinypic.com/21l7uwz.png\">

2. Find the ratio of the area of the larger equilateral triangle PQR to that of the smaller equilateral triangle LMN.

http://i42.tinypic.com/21mc76b.jpg\">

http://i44.tinypic.com/24g0g36.png\">

[MathsOlympiadtrainer]

pinky88:

Hi, need help again on the flwg qns, thx...


1. In the diagram below, ABCD is a square. The points A, B & G are collinear. The line segments AC & DG intersect at E, and the line segments DG & BC intersect at F. Suppose that DE = 15 cm, EF = 9 cm & FG = x cm, find the value of x.

http://i42.tinypic.com/21l7uwz.png\">

Let the length of the square be 1 unit. We try to make use of similar triangle. (shall skip the proving of similar triangles here)
Triangle ADE~Triangle CFE
So AD/CF=DE/FE
CF--> 3/5 unit
FB--> 2/5 unit

Next,
Triangle GDA~Triangle GFB
So DA/FB=DG/FG=5/2
5 units--> x + 15 + 9
2 units--> x
Then we have 3 units--> 24 and thus, 2 units-->16

[MathsOlympiadtrainer]

pinky88:

Hi, need help again on the flwg qns, thx...



2. Find the ratio of the area of the larger equilateral triangle PQR to that of the smaller equilateral triangle LMN.

http://i42.tinypic.com/21mc76b.jpg\">

First of all, we need to apply fraction of figure knowledge to this question. For example, if we cut a triangle A into ratio of 1:2, the 2 smaller triangles B and C will be 1/3 and 2/3 of A respectively. Now if we cut the smallest triangle B into the ratio of 1:3, the new smallest triangle D will be 1/4 of B, which is equivalent to 1/4 X 1/3 = 1/12 of A.

Now let's look at the question,
Triangle PML-->1/6 X 5/6 = 5/36 of Triangle PQR
Triangle MRN-->1/6 X 5/6 = 5/36 of Triangle PQR
Triangle NQL-->1/6 X 5/6 = 5/36 of Triangle PQR
Triangle LMN--> 1 - 5/36 X 3 = 21/36 = 7/12 of Triangle PQR

PQR:LMN=12:7