O-Level Additional Math

[Muffins]

Getting right on that now.... http://www.postimage.org/

[Muffins]

Hi VC, I have found out that ACS(I) has sent out questions from the preliminary round of Mathlympics 2009 to the participating schools, so I hope that helps. Gonna find more now…

[cchew]

Vanilla Cake:

BTW,my younger sister is taking part in http://www.acs.sch.edu.sg/acs_indep/announcements.php?id=93 for All Singapore Primary Schools 2010.Do you know whether ACSI sell the worked solutions for the past years' questions of Mathlympics?Do let us know your findings.
Thanks.

You can get from HotHouse Educational Services Pte Ltd
http://www.econnections.com.sg/index1.htm

[Vanilla Cake]

cchew:

Vanilla Cake:

BTW,my younger sister is taking part in http://www.acs.sch.edu.sg/acs_indep/announcements.php?id=93 for All Singapore Primary Schools 2010.Do you know whether ACSI sell the worked solutions for the past years' questions of Mathlympics?Do let us know your findings.

Thanks.

You can get from HotHouse Educational Services Pte Ltd
http://www.econnections.com.sg/index1.htm

Hi cchew,
Thanks for the link. Will place order for them.

[midnightspark]

Hi,


Just asking some sec 2 maths questions, please help


1)
A rectangle of sides x cm and y cm has an area of 72 cm2. Another rectangle of sides (x+1.5) and (y-4)cm has the same area. Find the values of x and y.

2)
When 5 is added to both the numerator and denominator of a fraction, the result becomes 1/2. When 1 is subtracted from both the numerator and the denominator, the fraction becomes 1/5. Find the fraction.

3)
The scale of map X is 1: x and the scale of map Y is 1:y. If the same distance is represented as 5cm on map X and 7.5cm on map Y, calculate the ratio x:y.

Thanks.

[CoffeeCat]

midnightspark:

Hi,


Just asking some sec 2 maths questions, please help


1)
A rectangle of sides x cm and y cm has an area of 72 cm2. Another rectangle of sides (x+1.5) and (y-4)cm has the same area. Find the values of x and y.

2)
When 5 is added to both the numerator and denominator of a fraction, the result becomes 1/2. When 1 is subtracted from both the numerator and the denominator, the fraction becomes 1/5. Find the fraction.

3)
The scale of map X is 1: x and the scale of map Y is 1:y. If the same distance is represented as 5cm on map X and 7.5cm on map Y, calculate the ratio x:y.

Thanks.

For qns 2, use algebra.
let the original fraction be x/y.
(x+5)/(y+5) = 1/2
cross multiplying, 2x + 10 = y+ 5
2x + 5 = y
(x-1)/ (y-1) = 1/5
5x - 5 = y - 1
......

For qns 1
xy = 72
(x+1.5)(y-4) = 72
let y= (72/x) and substitute . You will get a quadratic equation after multiplying the whole equation by x.

For qns 3,
5x = 7.5y
x/y = 7.5/5 = ...

hmmm too busy to give you full solutions. hope this helps.

[midnightspark]

CoffeeCat:

midnightspark:

Hi,


Just asking some sec 2 maths questions, please help


1)
A rectangle of sides x cm and y cm has an area of 72 cm2. Another rectangle of sides (x+1.5) and (y-4)cm has the same area. Find the values of x and y.

2)
When 5 is added to both the numerator and denominator of a fraction, the result becomes 1/2. When 1 is subtracted from both the numerator and the denominator, the fraction becomes 1/5. Find the fraction.

3)
The scale of map X is 1: x and the scale of map Y is 1:y. If the same distance is represented as 5cm on map X and 7.5cm on map Y, calculate the ratio x:y.

Thanks.

For qns 2, use algebra.
let the original fraction be x/y.
(x+5)/(y+5) = 1/2
cross multiplying, 2x + 10 = y+ 5
2x + 5 = y
(x-1)/ (y-1) = 1/5
5x - 5 = y - 1
......

For qns 1
xy = 72
(x+1.5)(y-4) = 72
let y= (72/x) and substitute . You will get a quadratic equation after multiplying the whole equation by x.

For qns 3,
5x = 7.5y
x/y = 7.5/5 = ...

hmmm too busy to give you full solutions. hope this helps.

Thanks for the solution!

[SKT]

Hi,


LCM of A&B is 4x², B&C is 12x³, A&C is 6x³. Find all possible expressions C.

[CoffeeCat]

SKT:

Hi,


LCM of A&B is 4x², B&C is 12x³, A&C is 6x³. Find all possible expressions C.

lol this is like a deductive puzzle.
Remember the concept of lcm is about choosing the highest index of prime factor.
lcm (A & B) = 4 x ^2 --- [1]
lcm (B & C) = 12 x^3 --- [2]
lcm (A & C) = 6 x^3 --- [3]

From [1] and any of [2] or [3] you can tell C must have x^3.
From [3] & [1] you can tell A has only factor 2 and not 4. B is the one with 4.
From [3] & [2] you can tell C must have factor 3 and possibly 2 but not 4.
Therefore possible expressions of C are
x^3, 3x^3, 6x^3.

[SKT]

CoffeeCat:

SKT:

Hi,


LCM of A&B is 4x², B&C is 12x³, A&C is 6x³. Find all possible expressions C.

lol this is like a deductive puzzle.
Remember the concept of lcm is about choosing the highest index of prime factor.
lcm (A & B) = 4 x ^2 --- [1]
lcm (B & C) = 12 x^3 --- [2]
lcm (A & C) = 6 x^3 --- [3]

From [1] and any of [2] or [3] you can tell C must have x^3.
From [3] & [1] you can tell A has only factor 2 and not 4. B is the one with 4.
From [3] & [2] you can tell C must have factor 3 and possibly 2 but not 4.
Therefore possible expressions of C are
x^3, 3x^3, 6x^3.

Thank you CoffeCat.