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    Tutor MathsGuru: Ask me for your burning Maths questions!

    Scheduled Pinned Locked Moved Primary Schools - Academic Support
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    • L Offline
      liketoeat
      last edited by

      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)

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      • D Offline
        Dharma
        last edited by

        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

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        • M Offline
          MOE Hater
          last edited by

          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 πŸ˜„

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          • L Offline
            liketoeat
            last edited by

            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.

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            • M Offline
              Mathematical Guru
              last edited by

              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? :?

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              • CoffeeCatC Offline
                CoffeeCat
                last edited by

                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.

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                • V Offline
                  Vanilla Cake
                  last edited by

                  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)=18Γ·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.

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                  • T Offline
                    Tang
                    last edited by

                    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.

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                    • CoffeeCatC Offline
                      CoffeeCat
                      last edited by

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

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                      • V Offline
                        Vanilla Cake
                        last edited by

                        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.

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