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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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    • V Offline
      Vanilla Cake
      last edited by

      capricorn28:
      Hi All,
      Actual workings from DD2.


      VC's mum
      capricorn28:
      Paper 2 - Q6 and the ans is 16 degrees
      http://postimage.org/image/s3ym9sis/
      Angle TUV =180⁰-60⁰=120⁰
      Angle XWT=180⁰-170⁰=10⁰
      360⁰-60⁰=300⁰
      Angle XUT =360⁰-300⁰-34⁰-10⁰=16⁰

      Angle XUT = 16⁰.
      capricorn28:
      Paper 2 - Q8 and the ans is 72%
      http://postimage.org/image/240e8e9fo/
      Before
      L: 4u
      F: 6u

      After
      L: 3.2u
      F: 4u

      (3.2u+4u)/10ux100% = 72%

      72% of the stamps was left.
      capricorn28:
      Paper 2- Q13 and the ans is 80cm
      http://postimage.org/image/jk1hymuc/
      5u=8000 cm²
      4u=6400 cm²
      Square root of 6400 cm² = 80 cm

      Length of the square = 80 cm.
      capricorn28:
      Paper 2 - Q14 and the ans is 40.5cm sq
      http://postimage.org/image/jk6gkobo/
      24÷8x5=15 cm
      24-15 = 9 cm
      40-9 = 31 cm
      ½x40x15 = 300 cm²
      ½x31x9 = 139.5 cm²
      ½x24x40 = 480 cm²
      480-139.5-300=40.5 cm²

      Shaded area = 40.5 cm²
      capricorn28:
      Paper 2 - Q15 and the ans is $360
      http://postimage.org/image/2aopcqyys/
      Before
      J: 6u
      K: 5u

      After
      J: 3.6u
      K: 7.4u

      In the end
      J: 7.3u
      K: 3.7u

      7.3u-3.7u=3.6u
      3.6u=$216
      6u = $360

      At first, Jude had $360.
      capricorn28:
      Paper 2 - Q17 and the ans is 120
      http://postimage.org/image/jk4t1btw/
      Hall (2n)
      B : G
      3u/4 : u/4
      3 : 1
      15 : 5

      Canteen (1n)
      B : G
      2p/5 : 3p/5
      2 : 3
      4 : 6

      1u=6
      20u=120

      Number of pupils in the hall = 120.

      1 Reply Last reply Reply Quote 0
      • V Offline
        Vanilla Cake
        last edited by

        wkong:
        The figure is not drawn to scale. It shows a rectangle divided into 6 parts. Each part has a different area.

        a) Find the total area of P and Q.
        b) Find the perimeter of the figure.

        http://www.postimage.org/
        Use guess and check,
        Area of P = 8 cmx 6 cm = 48 cm²
        Area of Q = 5 cm x 4 cm = 20 cm²
        (a) Total area of P and Q = (48+20) cm² = 68 cm²

        Length of rectangle = (5+6+4) cm = 15 cm
        Breadth of rectangle = (8+5) cm = 13 cm
        (15x2)+(13x2) = 30+26 = 56 cm
        (b) Perimeter of the figure = 56 cm

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        • Y Offline
          youngtay
          last edited by

          kancheongmum Thanks for your help.


          Best wishes

          1 Reply Last reply Reply Quote 0
          • W Offline
            wkong
            last edited by

            Thanks again VC for your help 😄

            1 Reply Last reply Reply Quote 0
            • V Offline
              Vanilla Cake
              last edited by

              YLH88:
              Hi all,

              Need help in the below question :
              This is an interesting question but nobody has forward his/her solution yet.I am not please with my algebra method and let's hope that others will forward with better methods for sharing and learning.Could you pls advise the source of this question?

              VC's mum
              YLH88:
              Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
              Both of them were running in the clockwise direction.
              Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
              How long did Sam take to run past Harry for the first time ?
              (ans : 30 minutes)

              http://postimage.org/image/1z70q6hqc/
              LCM of 12 and 15 is 60.
              Work out 60m as the length (circumference) of the circular track even though 60m is not a realistic number.
              Sam's average speed = 60m÷12 = 5 m/min
              Harry's average speed = 60m÷15 = 4 m/min

              Note that both of them started running at the same time from the opposite ends of the diameter of a circular track means the difference in their distance = ½x60m = 30 m.
              For them to meet for the first time, Sam must run more than Harry by 30 m in the same amount of time.

              Assume the time taken to be T,
              5xT= (4xT)+30
              5T=4T+30
              T=30

              Sam took 30 min to run past Harry for the first time.

              1 Reply Last reply Reply Quote 0
              • Y Offline
                YLH88
                last edited by

                Vanilla Cake:
                YLH88:

                Hi all,

                Need help in the below question :

                This is an interesting question but nobody has forward his/her solution yet.I am not please with my algebra method and let's hope that others will forward with better methods for sharing and learning.Could you pls advise the source of this question?

                VC's mum
                YLH88:
                Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
                Both of them were running in the clockwise direction.
                Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
                How long did Sam take to run past Harry for the first time ?
                (ans : 30 minutes)

                http://postimage.org/image/1z70q6hqc/
                LCM of 12 and 15 is 60.
                Work out 60m as the length (circumference) of the circular track even though 60m is not a realistic number.
                Sam's average speed = 60m÷12 = 5 m/min
                Harry's average speed = 60m÷15 = 4 m/min

                Note that both of them started running at the same time from the opposite ends of the diameter of a circular track means the difference in their distance = ½x60m = 30 m.
                For them to meet for the first time, Sam must run more than Harry by 30 m in the same amount of time.

                Assume the time taken to be T,
                5xT= (4xT)+30
                5T=4T+30
                T=30

                Sam took 30 min to run past Harry for the first time.

                Hi VC's mum,

                Thank you for the solution. But I find it a bit difficult to explain to my DS on this.

                This question is taken from the A star booklet on Rate and Speed.

                1 Reply Last reply Reply Quote 0
                • K Offline
                  kancheongmum
                  last edited by

                  YLH88:
                  Hi all,


                  Need help in the below question :

                  Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
                  Both of them were running in the clockwise direction.
                  Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
                  How long did Sam take to run past Harry for the first time ?
                  (ans : 30 minutes)

                  http://postimage.org/image/1z70q6hqc/

                  Thank you!
                  Hi YLH88

                  This is catching up question without the distance given. I too had a hard time explaining to my DS. Let me try and see it helps.

                  Sam took 12 min to complete one round
                  Harry took 15 min to complete one round
                  So Sam is for sure faster than Harry
                  We name Sam starting point S and Harry starting point H
                  We imagine both Sam and Harry start from point S. Harry start running first and when he reached point H, then Sam start running. So Sam has half the circular track to catch up.

                  Time ratio --- Sam: Harry ---- 12 : 15 ----4:5
                  Same distance, constant speed, time ratio is inversely proportional to distance ratio and speed ratio
                  distance ratio --- 5:4
                  5u - 4u = 1u ---- 1/2 of the track
                  When Sam past Harry the 1st time Sam would have run

                  1/2 x 5 = 5/2 or 2 1/2 of the track
                  He would take --- 5/2 x 12 min = 30 mins

                  You can also use Harry to get the answer
                  When Sam past Harry the lst time Harry would have run

                  1/2 x 4 = 2 round
                  He would take --- 2 x 15 min = 30 mins

                  hope this help

                  1 Reply Last reply Reply Quote 0
                  • I Offline
                    iFruit
                    last edited by

                    YLH88:
                    Hi all,


                    Need help in the below question :

                    Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
                    Both of them were running in the clockwise direction.
                    Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
                    How long did Sam take to run past Harry for the first time ?
                    (ans : 30 minutes)

                    http://postimage.org/image/1z70q6hqc/

                    Thank you!
                    Sam took 12 mins to run round the track.

                    So distance run by Sam in 1 min = 1/12 of track.

                    distance run by Harry in 1min = 1/15 of track.

                    Distance caught up by Sam in 1 min = (1/12 - 1/15) of track = 1/60 of track

                    Time taken by sam to cover the distance between them, which is 1/2 the track or 30/60 of track = 30mins.

                    1 Reply Last reply Reply Quote 0
                    • Z Offline
                      zyberk
                      last edited by

                      Hi all,


                      Appreciate if anyone can help with the steps to achieve the answer to this math number problem. Thanks

                      | Y |
                      | 16 | 5 |
                      | 28 | 10 | 3 |
                      | 50 | 20 | 8 | 3 |

                      The answer given is Y=23

                      Thanks again.

                      1 Reply Last reply Reply Quote 0
                      • C Offline
                        CJM
                        last edited by

                        Dave had 59 toy cars and toy soldiers. After giving away 1/3 of his toy cars and 4 of his toy soldiers, he had an equal number of toy cars and toy soldiers. How many toy soldiers did he have at first ?

                        1 Reply Last reply Reply Quote 0

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