Tutor MathsGuru: Ask me for your burning Maths questions!

[newbieparent]

:?: Hi Maths Guru, Need some help with the following questions:


1) ACS 2009 P6 Prelim Paper 2 Q10

2) Henry Park 2009 P6 Prelim Paper 2 Q15

3) Nanyang Pri 2009 P6 Prelim Paper 2 Q14

4) Catholic High 2009 P6 Prelim Paper 1 Q13

Anyone who can help? Will be very grateful!

[Herbie]

Hi Dharma,


Sorry, I have to come back to the speed related issues.

Would u be able to reiterate the inverse formula for speed as indicated below:

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



I can’t locate yr previous explanantion.

So sorry!

many thanks!

[Dharma]

Herbie:

Hi Dharma,


Sorry, I have to come back to the speed related issues.

Would u be able to reiterate the inverse formula for speed as indicated below:

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.

I can't locate yr previous explanantion. :-

So sorry!

many thanks!

Hi Herbie,

For speed, we need to know that distance = speed x time.
We know that if distance is fixed; then speed and time are inversely proportional to each other.

For speed questions, we normally deal with at least 2 moving objects (car, bus,lorry or even 2 or more persons running or walking) either in the same direction or opposite directions.

Therefore, when we have 2 or more moving objects, we try to find out the speed ratio or time ratio if the 2 objects. In this case the. 2 objects must be travelling the same distance, so distance is fixed or constant. This being the case, then the speed ratio of the 2 objects (if given) can be used to find the time ratio of the 2 objects ;because they are inversely proportional to each other. In the same way, if we know the time ratio of the 2 objects we can proceed to find out their speed ratio.

Hope u are able to understand.

[Mathematically Speaking]

MathGuru,


what software do you use to draw all your models? I am truly impressed! Perhaps the teachers in school should start doing this…

[Herbie]

Hi Dharrna,


Thanks for yr reply.

Many thanks!

[liketoeat]

Hi, please help me solve this question. Thank you.


Candle A and candle B were placed on a table. Candle A was 1.5 cm longer than candle B. Candle A and candle B were lighted at 0630 h and 0800 h respectively. They burnt down to the same length at 1030 h. At 1200 h, candle B was burnt out while A only burnt out at 1230 h. Given that the rate of burning of each candle was constant throughout, find the original length of each candle. (Ans: candle A 13.5 cm; candle B 12 cm)

[Dharma]

liketoeat:

Hi, please help me solve this question. Thank you.


Candle A and candle B were placed on a table. Candle A was 1.5 cm longer than candle B. Candle A and candle B were lighted at 0630 h and 0800 h respectively. They burnt down to the same length at 1030 h. At 1200 h, candle B was burnt out while A only burnt out at 1230 h. Given that the rate of burning of each candle was constant throughout, find the original length of each candle. (Ans: candle A 13.5 cm; candle B 12 cm)

Time ratio => A : B = 4 : 3
Burning rate ratio => A : B = 3 : 4

Time taken to burn completely
A : 6 hrs (0630hrs to 1230hrs)
B : 4 hrs (0800hrs to 1200hrs)

Length (A ) = 3u x 6 = 18u
Length (B) = 4u x 4 = 16u

18u – 16u = 2u = 1.5cm
Original Length (A) = 18u = 9 x 1.5cm = 13.5cm
Original Length (B) = 16u = 8 x 1.5cm = 12.0cm

[MOE Hater]

Dharma:

liketoeat:

Hi, please help me solve this question. Thank you.


Candle A and candle B were placed on a table. Candle A was 1.5 cm longer than candle B. Candle A and candle B were lighted at 0630 h and 0800 h respectively. They burnt down to the same length at 1030 h. At 1200 h, candle B was burnt out while A only burnt out at 1230 h. Given that the rate of burning of each candle was constant throughout, find the original length of each candle. (Ans: candle A 13.5 cm; candle B 12 cm)

Time ratio => A : B = 4 : 3
Burning rate ratio => A : B = 3 : 4

Time taken to burn completely
A : 6 hrs (0630hrs to 1230hrs)
B : 4 hrs (0800hrs to 1200hrs)

Length (A ) = 3u x 6 = 18u
Length (B) = 4u x 4 = 16u

18u – 16u = 2u = 1.5cm
Original Length (A) = 18u = 9 x 1.5cm = 13.5cm
Original Length (B) = 16u = 8 x 1.5cm = 12.0cm

My teacher taught us this as well but i could not remeber it just now. :idea: I did manage to solve the question though, but using algebra. :roll: Thanks for sharing the heuristic method. It does save a lot of headache

[liketoeat]

Dharma:

liketoeat:

Hi, please help me solve this question. Thank you.


Candle A and candle B were placed on a table. Candle A was 1.5 cm longer than candle B. Candle A and candle B were lighted at 0630 h and 0800 h respectively. They burnt down to the same length at 1030 h. At 1200 h, candle B was burnt out while A only burnt out at 1230 h. Given that the rate of burning of each candle was constant throughout, find the original length of each candle. (Ans: candle A 13.5 cm; candle B 12 cm)

Time ratio => A : B = 4 : 3
Burning rate ratio => A : B = 3 : 4

Time taken to burn completely
A : 6 hrs (0630hrs to 1230hrs)
B : 4 hrs (0800hrs to 1200hrs)

Length (A ) = 3u x 6 = 18u
Length (B) = 4u x 4 = 16u

18u – 16u = 2u = 1.5cm
Original Length (A) = 18u = 9 x 1.5cm = 13.5cm
Original Length (B) = 16u = 8 x 1.5cm = 12.0cm

Thank you, Dharma.

[Mathematical Guru]

To keep herself fit, Wiyathi climbs up a flight of 30 steps every day. If she runs uo 5 steps and walks up 25 steps, she will take 99 seconds. If she runs up 15 steps and walks up 15 steps, she will take 18 seconds less. How long will she take to run uo all the 30 steps if she maintains the same speed?


How am I to solve this? :?