O-Level Additional Math

[Xiao Hu]

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.

Hi CoffeeCat,
Can you help double check your answer for question 3?
Since for the same distance, it's represented shorter at 5cm than 7.5cm on X vs Y map. So isn't the ration the other way round 5/7.5=2/3?
Hope you wouldn't mind to help clarify and to point out where my mistake in reasoning is.

Thanks,
Xiao Hu.

[FrekiWang]

Xiao Hu:

CoffeeCat:

[quote=\"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.

Hi CoffeeCat,
Can you help double check your answer for question 3?
Since for the same distance, it's represented shorter at 5cm than 7.5cm on X vs Y map. So isn't the ration the other way round 5/7.5=2/3?
Hope you wouldn't mind to help clarify and to point out where my mistake in reasoning is.

Thanks,
Xiao Hu.[/quote]This is the ration:

Map X has a scale of 1:x, this implies 5cm on map X represents an actual distance of 5x cm.
Similarly, 7.5cm on map Y represents an actual distance of 7.5y cm.

Now you are given they represent the same distance, which implies 5x = 7.5y. Do a reverse of cross-mutiplication we have x/y=7.5/5=3/2

Comment on your ration:

For the same actual distance, if you have a map with a very large scale 1:10000000, the distance measured on the map will be very short (i.e. 10km becomes 1mm); Similarly, if you have a map with a very small scale, 1:10, the distance measured on the map will be very long (i.e. 10km becomes 1km). Therefore, for the same actual distance. the longer the distance shown on the map, the smaller the scale of the map is. So x>y since 5cm<7.5cm in this question.

[Xiao Hu]

Hi FrekiWang,

Yes, you are right and the solution was correct. I was not clear about the map scale myself, how the scale was used, that it’s 1 on the map scaled to longer length on actual land.

Thanks for pointing out.
Xiao Hu

[listener]

Hi, i need help in a probability question.


Gabriel takes a bus or a cab to go to school. On a rainy day, the probability that Gabriel takes a bus to school is 1/6 and on a non-rainy day the probability that he takes a cab is 1/12.
Assuming Gabriel’s choice of transport is independent of the weather, find the probability that
a) it will rain when Gabriel goes to school and hence, draw a probability tree for it.
b) Gabriel will take a bus to school on a non-rainy day.

[FrekiWang]

listener:

Hi, i need help in a probability question.


Gabriel takes a bus or a cab to go to school. On a rainy day, the probability that Gabriel takes a bus to school is 1/6 and on a non-rainy day the probability that he takes a cab is 1/12.
Assuming Gabriel's choice of transport is independent of the weather, find the probability that
a) it will rain when Gabriel goes to school and hence, draw a probability tree for it.
b) Gabriel will take a bus to school on a non-rainy day.

Draw a tree and solve it please, I dont want to give any detail as this is somewhat even easier than the textbook example.

[Herbie]

if 600= 22235*5. Find the smallest value of k such tat 600k is a perfect square. Can help to solve it?

[listener]

Hi, i need help in a probability question.


Gabriel takes a bus or a cab to go to school. On a rainy day, the probability that Gabriel takes a bus to school is 1/6 and on a non-rainy day the probability that he takes a cab is 1/12.
Assuming Gabriel's choice of transport is independent of the weather, find the probability that
a) it will rain when Gabriel goes to school and hence, draw a probability tree for it.
b) Gabriel will take a bus to school on a non-rainy day.

Anyone knows how to even complete the tree diagram since Gabriel's choice of transport is independent of the weather?

http://i56.tinypic.com/rig3rb.jpg\">

[liketoeat]

Hi, I need help with the following Qn…


A motorist travelled the first part of his journey at an average speed of 56 km/h. He the increases his speed to 64 km/h for the rest of his journey. If he travels 130 km in 2 hours and 15 minutes, find the distance he travelled for the first part of the journey.

Thanks.

[FrekiWang]

Herbie:

if 600= 2*2*2*3*5*5. Find the smallest value of k such tat 600k is a perfect square. Can help to solve it?

600=2^3 * 3^1 * 5^2

For a perfect square, all the prime factors must have an EVEN power.

Therefore we need to mutiply 600 by another 2 and 3.

so k=2 * 3 =6

This type of question is standard and the approach is standard too.

[FrekiWang]

liketoeat:

Hi, I need help with the following Qn..


A motorist travelled the first part of his journey at an average speed of 56 km/h. He the increases his speed to 64 km/h for the rest of his journey. If he travels 130 km in 2 hours and 15 minutes, find the distance he travelled for the first part of the journey.

Thanks.

primary? secondary?

Let assume it is a secondary question first, then we can solve it by using an equation.
2hour15mins=2.25hour(15mins=1/4hour)
Let the time of the first part be t hours, then the time of the second part will be (2.25-t) hours.
Here we have
56t+64(2.25-t)=130
Solve, we get t=1.75
Therefore the distance travelled in the first part = 56 x 1.75 = 98km.

If it is a primary question,
Assumed he travelled 2.25hours at a speed of 56km/h
Distance he would have travelled = 56 x 2.25 = 126km
He actually travels 4km more than 126km because he has increased the speed somewhere. In each hour, he could travel 64-56=8km more, which means he need to travel 4/8=0.5hour at the new speed to increase his distance travelled to 130km.

So time taken for the first part of the journey is 2.25-0.5=1.75
56 x 1.75=98km.