Tutor MathsGuru: Ask me for your burning Maths questions!

[Brenda10]

Muffins:

Brenda10:

[quote=\"Muffins\"]Hi Brenda10, your daughter just managed to summarise my workings into a few steps! :shock: How old is your DD???

Thank you Muffins.

She is 11 years old.

:shock: :shock: Very young and VERY BRIGHT!!! :)[/quote]Hi Muffins,

Thank you for thinking highly upon her. :oops:


In fact, DD is just average. She has done more than 50 ratio/percentage questions during the school holiday and may be the numerous practices really help. The questions in this forum really provide her the good opportunity on trial and error. The more important thing is that she could also learn from all of you as well.

[Vanilla Cake]

ruyu:

dont understand :? care to explain?

Thks for pointing out, I have amended the workings.
Average speed should be 3 km/h

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

[Dharma]

Vanilla Cake:

ruyu:

dont understand :? care to explain?

Thks for pointing out, I have amended the workings.
Average speed should be 3 km/h

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h

[YLH88]

[quote]Vanilla Cake wrote:

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Dharma wrote :
Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h[/quote]Hi VC, Dharma,

Why do you change the time as 12/2 and 12/6 ?

For speed questions, when do we use the inverse proportion method ? any standard method we can follow ? Like the question on NanHua P6 SA1 2009 Paper 1 Qn 14, VC uses the rate method. My DS find it very confusing and he is very confused when to use the inverse method and when to use other method for all the speed questions. Can help to advise ? :?

Thank you!
[quote]1) Nan Hua P6 SA1 2009 Paper 1 Qn 14
Adrian and Bob started driving at the same time. Bob took 6 hours to complete his journey while Adrian took 9 hours to complete his. After 3 hours, the two men were left with the same distance to complete. What is the ratio of the distance of Bob's journey to the distance of Adrian's journey ?

Vanilla Cake wrote :
Distance travelled by Bob in 1 hour = 1/6 of the journey
Distance travelled by Adrian in 1 hour = 1/9 of the journey

Distance travelled in 3 hours for Bob and Adrian:
Bob = 1/6x3 = 1/2 of the journey
Adrian = 1/9x3 = 1/3 of the journey

Remaining distance to be completed for Bob and Adrian:
Bob = 1/2 = 2/4 of the journey
Adrian = 2/3 of the journey

You have to make both numerators to be same as \"two men were left with the same distance to complete.\"

Ratio of the distance of Bob's journey to the distance of Adrian's journey = 4 : 3
Answer option is (d)
[/quote]

[Dharma]

YLH88:

[quote]Hi VC, Dharma,


Why do you change the time as 12/2 and 12/6 ?

For speed questions, when do we use the inverse proportion method ? any standard method we can follow ? Like the question on NanHua P6 SA1 2009 Paper 1 Qn 14, VC uses the rate method. My DS find it very confusing and he is very confused when to use the inverse method and when to use other method for all the speed questions. Can help to advise ? :?

Thank you!
[quote]1) Nan Hua P6 SA1 2009 Paper 1 Qn 14
Adrian and Bob started driving at the same time. Bob took 6 hours to complete his journey while Adrian took 9 hours to complete his. After 3 hours, the two men were left with the same distance to complete. What is the ratio of the distance of Bob's journey to the distance of Adrian's journey ?

Vanilla Cake wrote :
Distance travelled by Bob in 1 hour = 1/6 of the journey
Distance travelled by Adrian in 1 hour = 1/9 of the journey

Distance travelled in 3 hours for Bob and Adrian:
Bob = 1/6x3 = 1/2 of the journey
Adrian = 1/9x3 = 1/3 of the journey

Remaining distance to be completed for Bob and Adrian:
Bob = 1/2 = 2/4 of the journey
Adrian = 2/3 of the journey

You have to make both numerators to be same as \"two men were left with the same distance to complete.\"

Ratio of the distance of Bob's journey to the distance of Adrian's journey = 4 : 3
Answer option is (d)

[/quote][/quote]Hi YLH88,

For using ratios, just remember
Distance = Speed x Time
Don’t worry too much about inverse method (there isn’t such a thing) ….just understand the concept (i.e. relationship between distance, speed and time)

For the above qn, we know the distance is the same after Adrian and Bob travelled for 3 hours. So, the speed ratios and time ratios of Bob and Adrian are inversely proportional to each other.

For constant distance (to complete journey after 3 hours)
Time ratio => Bob : Adrian = (6-3) : (9-3) = 3 : 6 = 1 : 2
Speed ratio => Bob : Adrian = 2 : 1

Now, total distance travelled by
Bob : 2u x 6 = 12u
Adrian : 1u x 9 = 9u
Distance ratio => Bob: Adrian = 4 : 3

[Dharma]

YLH88:

[quote]Vanilla Cake wrote:

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Dharma wrote :
Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h

Hi VC, Dharma,

Why do you change the time as 12/2 and 12/6 ?

[/quote][/quote]

Hi YLH88,

Hope it is clearer now.

For this qn, the distance uphill/downhill is constant.

So, speed ratio=> Uphill : Downhill = 1 : 3
Time ratio => Uphill : Downhill = 3 :1

Total Distance
= Distance Uphill + Distance Downhill (#Distance Uphill = Distance downhill)
= (3u)hrs x 2km/h + (1u)hr x 6km/h
= (12u) km

Total time
= (3u + 1u)hrs
= (4u)hrs

Average Speed = Total Distance  Total time = (12u/4u) km/h = 3 km/h

[Vanilla Cake]

Hi YLH88,

It is not easy to understand the topic on Speed for most people including me. Main reason is that the use of algebra to solve speed problem sums is not encouraged and you have to use other acceptable methods taught by the school.The use of inverse method was not taught by my teacher during my P6 days.Perhaps, this method may not be taught by your DS's school teacher and therefore he is not familiar with this method, right?

In fact, I learn this fast and efficient method from Dharma in KSP forum and Dharma had posted speed concepts on constant distance, constant speed and constant time somewhere in this forum.If I can remember well, these are the key concepts:

Constant distance
Ratio of speed is inversely proportional to the ratio of time

Constant speed
Ratio of distance is directly proportional to the ratio of time

Constant time
Ratio of distance is directly proportional to the ratio of speed

If two objects are travelling to meet each other, required time for them to meet = Total distance apart divided by combined average speeds of both objects.

If one object is ahead of another in the same direction, time needed to catch up = Distance apart divided by difference in the average speeds of both objects.

YLH88:

Mark can climb uphill at an average speed of 2km/h and go downhill at an average speed of 6km/h. Going first uphill and then downhill, without stopping and using the same route, what will his average speed be for the entire trip ?

Vanilla Cake:

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Dharma:

Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h

YLH88:

Hi VC, Dharma,

Why do you change the time as 12/2 and 12/6 ?

This question is almost the same as NMOS 2009 Prelim round Q3 [1 mark]
A boy jogs along a path which has the shape of an equilateral triangle.His speed of jogging on each side of the triangle is 120 metres per minute, 40 metres per minute and 60 metres per minute respectively.What is his average speed to finish jogging along the path?Give your answer to the nearest integer, in metres per minute.

The method used to solve is by algebra.
Let the length of each side of the triangle be L.
Total time taken = L/120+L/40+L/60=L/20
Total distance = 3xL = 3L
Average speed = 3L÷L/20 = 3Lx20/L = 60 metres/minute

Back to question, instead of using algebra, try to use numbers.
Look at their speeds which are 2km/h and 6 km/h and try to look for a whole number which is divisible by both 2 and 6.If you are not comfortable with 12 then use 18 or others that meet the required criteria.

Time taken to climb uphill @ 2 km/h = 18/2 = 9h
Time taken to go downhill @ 6 km/h = 18/6 = 3h
Total time taken = 12h
Total distance covered = 18x2 = 36 km
Average speed for the entire trip = 36km/12h = 3 km/h

Otherwise use algebra where L is the route of going uphill which is the same route(distance) of going downhill.
Time taken to climb uphill @ 2 km/h = L/2 = L/2 h
Time taken to go downhill @ 6 km/h = L/6 = L/6 h
Total time taken = L/2+L/6 = 3L/6+L/6 = 4L/6 = 2L/3
Total distance covered = Lx2 = 2L km (x2 due to going uphill and coming downhill)
Average speed for the entire trip = 2L÷2L/3 = 2Lx3/2L = 3 km/h

OR

Using Dharma's inverse method which is fast and efficient

Constant distance
Ratio of speed
Climb Uphill : Go Downhill
2 : 6
Reduce to simplest form
1 : 3

Ratio of time
Climb Uphill : Go Downhill
3 : 1

Total time taken =(3+1) = 4u hours
Distance travelled to climb uphill @ average speed of 2 km/h =2x3u=6u km
Distance travelled to go downhill @ average speed of 6 km/h=6x1u=6u km
Note that both distances of climbing uphill and going downhill are the same ie 6u km.
Total distance = 6u+6u = 12u
Average speed = 12u km÷4u hours = 3 km/h

Sorry for the lengthy explanation and hope that you are able to understand my explanation in one way or another while waiting for Mathsguru and other members' solutions.

[YLH88]

Hi Dharma, VC,


Thank you so much for the detailed explanation!! I will need to digest them first before explaining to my ds. May still need your help if I'm still blur

VC,
Yes you are right. The inverse method is not taught in school. But Dharma has provided a very good and clear explanation in this forum earlier and I always use it as a guide to explain to my ds. But sometimes due to the complexity of the question, the mummy also blur and confuse... :?

[CoffeeCat]

YLH88:

Hi Dharma, VC,


Thank you so much for the detailed explanation!! I will need to digest them first before explaining to my ds. May still need your help if I'm still blur

VC,
Yes you are right. The inverse method is not taught in school. But Dharma has provided a very good and clear explanation in this forum earlier and I always use it as a guide to explain to my ds. But sometimes due to the complexity of the question, the mummy also blur and confuse... :?

hey just my comments. Earlier ylh88 is correct to describe this method as the inverse proportion method. Dharma could have meant that such methods weren't given that official name or he could have misread as inverse method, which doesn't exist as far as I know.
Such questions are common in nmos, not so common for psle but this inverse proportion in speed concept is described on onsponge. For many p6 students who happened to be comfortable with algebra they will tend to use the algebra method over the inverse proportion method because it is more natural. In a way the inverse method is invented because use of algebra is discouraged. Anyway I think you won't be penalized if u use the algebra method.

[Vanilla Cake]

CoffeeCat:

hey just my comments. Earlier ylh88 is correct to describe this method as the inverse proportion method. Dharma could have meant that such methods weren't given that official name or he could have misread as inverse method, which doesn't exist as far as I know.

Such questions are common in nmos, not so common for psle but this inverse proportion in speed concept is described on onsponge. For many p6 students who happened to be comfortable with algebra they will tend to use the algebra method over the inverse proportion method because it is more natural. In a way the inverse method is invented because use of algebra is discouraged. Anyway I think you won't be penalized if u use the algebra method.

Just to add comments. I think the inverse proportion method to solve P6 speed questions may not be taught in all Primary schools in Singapore.My time when I was in P6 was an example as I didn't pick up this method until Dharma introduces it in KSP forum. For whatever reasons, use of algebra to solve P6 speed questions is not encouraged and don't ask me why. You may read that in the these examples from Onsponge forum that the thread starters didn't like the use of algebra to solve P6 speed questions and the admin staff will chip in to provide non-algebra methods if algebra methods are presented by online helpers.

http://www.onsponge.com/forum/35-thinkingmathonsponge/3290-speed-question-p6.html
http://www.onsponge.com/forum/35-thinkingmathonsponge/3258-speed.html
http://www.onsponge.com/forum/35-thinkingmathonsponge/3175-p6-speed.html