Tutor MathsGuru: Ask me for your burning Maths questions!

[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? :?

[CoffeeCat]

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? :?

compare
run 5 steps, walk 25 steps --> 99 secs
run 15 steps walk 15 steps --> 99-18 secs
by running 10 more steps instead of walking she will save 18 secs
by running 5 more steps instead of walking she will save 9 secs
so if she run 15 more steps (30 steps in all) and walk 15 steps less (0 steps walked) she will save 9*3=27 secs.
Therefore time taken --> 99-18-27=54 secs

By the way i notice how some tougher psle maths questions in recent years have actually appeared in hci's smops years ago.

[Vanilla Cake]

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?

Hi Mathematical Guru,
Is this speed question from school's homework? To me, it doesn't seem to be.Pls advise the source of this speed question.

5 steps (run) and 25 steps (walk) = 99 seconds
5+10=15 steps (run) and 25-10=15 steps (walk) = 81 seconds (99s-18s)
Increase 1 step (run) and decrease 1 step (walk)=18÷10=1.8s ie decrease by 1.8s
Note : 18 seconds is from the difference between 99 seconds and 81 seconds.
Increase 15 steps (run) and decrease 15 steps (walk)=Decrease by 15x1.8s=27s
15+15=30 steps (run) and 15-15=0 steps (walk) = 81s-27s=54s

Time taken for Wiyathi to run up all the 30 steps if she maintains the same speed = 54 seconds.

[Tang]

CoffeeCat:

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? :?

compare
run 5 steps, walk 25 steps --> 99 secs
run 15 steps walk 15 steps --> 99-18 secs
by running 5 more steps instead of walking she will save 18 secs
so if she run 15 more steps (30 steps in all) and walk 15 steps less (0 steps walked) she will save 18*3=54 secs.
Therefore time taken --> 99-18-54=28 secs

By the way i notice how some tougher psle maths questions in recent years have actually appeared in hci's smops years ago.

Hello,

You may like to check the underlined.


5 R + 25 W --> 99 s

15 R + 15 W --> 99 - 18 = 81 s
1 R + 1 W --> 81/15 = 5.4 s

25 R + 25 W --> 5.4 x 25 = 135 s

20 R --> 135 - 99 = 36 s

30 R --> 30/20 x 36 = 54 s

She will take 54 s to run up all the 30 steps.

[CoffeeCat]

Lol yea what a mistake I had made. I have edited it. Thx for pointing out!

[Vanilla Cake]

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

Hi liketoeat,
Could you pls advise the source of this question as this question requires some thinking? Is this question from school's homework?

Just to share 2 different alternatives for readers.One not using inverse ratio and one using model drawings (posted by the forever helpful Uncle Observer).Another method is algebra which I will not post here.

Alternative method 1
Same height to burn out completely ie equal burnt length
Candle A from 1000 h to 1230 h took 2 hours
Candle B from 1030 h to 1200 h took 1.5 hours

Find lowest common multiple of 1.5 and 2
Multiples of 2 = 2,4,6,8,10.......
Multiples of 1.5 = 1.5,3,4.5, 6, 7.5.......

So, LCM of 1.5 and 2 is 6.

Burning rates
Candle A = 6u/2 = 3u/hour
Candle B = 6u/1.5 = 4u/hour

From start to burn out completely
Candle A: 0630 h to 1230 h ie 6 h -> 6x3u=18u
Candle B: 0800 h to 1200 h ie 4 h -> 4x4u=16u

Candle A was 1.5 cm longer than candle B (Given),
18u-16u=1.5
2u=1.5
u = 0.75

Original length of candle A = 18u = 18x0.75 = 13.5 cm
Original length of candle B = 16u = 16x0.75 = 12 cm

Alternative method 2
By http://psle2010a.blogspot.com/2010/07/challenging-p6.html as posted by the kind and helpful Uncle Observer.

[mathsguru]

Dear all...


Just to let you all know tt I'm back... :peekaboo:

:imsorry: My sincere apologies for not being active for the past month. I've taken a 2-week break during the June hols and when I came back, I was busy conducting lessons. :faint:

Finally can catch a breather now since school re-opened! Many thanks to all kind souls who have been helping out here...so many pages to catch up for me!

Hmm...guess I will start solving questions posted from now on instead of trying to catch up with the backlog. Hehe...Hope to seek your understanding!! :please:

Cheers,
MathsGuru

[ruyu]

mathsguru:

Dear all...


Just to let you all know tt I'm back... :peekaboo:

:imsorry: My sincere apologies for not being active for the past month. I've taken a 2-week break during the June hols and when I came back, I was busy conducting lessons. :faint:

Finally can catch a breather now since school re-opened! Many thanks to all kind souls who have been helping out here...so many pages to catch up for me!

Hmm...guess I will start solving questions posted from now on instead of trying to catch up with the backlog. Hehe...Hope to seek your understanding!! :please:

Cheers,
MathsGuru

Yeah Mathsguru is back!