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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • V Offline
      Vanilla Cake
      last edited by

      kancheongmum:
      Harry and Sally were at a party. Harry shook hands with 4 times as many boys as girls while Sally shook hands with 5 times as many boys as girls. How many boys and how many girls were at the party? (ans 25 boys, 6 girls)

      Harry
      Boys: [unit][1][unit][1][unit][1][unit][1][1]
      Girls: [unit][1]

      Sally
      Boys: [unit][unit][unit][unit][unit]
      Girls: [unit][1]

      Comparing the number of boys for both models,
      4 units+5=5 units
      1 unit = 5

      Boys: 5 units = 5x5=25
      Girls: unit+1 = 5+1 = 6

      There were 25 boys and 6 girls at the party.

      OR

      The trick of such question is that the number of girls is always the number that follows the 2 numbers given in the question itself.Since the 2 given numbers are 4 and 5, the number of girls will be 6

      The number of boys (From Harry's perspective) will be 4 times as many boys as girls-> 4x6+1 (1 is Harry himself) = 24+1 = 25

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      • V Offline
        Vanilla Cake
        last edited by

        Just to share an interesting 5-mark speed question from 3rd Annual Mathlympics - 2010 Prelim round Q28 (Not in exact words). It's not a typical speed question (in my humble opinion) and do you have any alternative solutions to share?


        Downhill: 72 km/h
        Ground level: 63 km/h
        Uphill: 56 km/h
        Car travels from Town A to Town B in 4 hours. It returns to Town A using the same route in 4 h 40 min. What is the distance between Town A and Town B?

        D1 - Uphill, D2 - Downhill, D3 - Ground level
        Time from Town A to Town B --> D1/56 + D2/72 + D3/63

        D3 - Ground level, D2 - Uphill, D1 - Downhill
        Time from Town B to Town A --> D3/63 + D2/56 + D1/72

        Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

        Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/63 + 1/63)
        = 2/63 D1 + 2/63 D2 + 2/63 D3
        = 2/63 D (Distance D = D1+D2+D3)

        2/63 D -> 26/3
        D -> 26/3x63/2 = 273 km

        Distance between Town A and Town B = 273 km

        A similar http://psle2010a.blogspot.com/2010/08/speed-p6.html posted on 27 Aug 2010 can be found in Uncle Observer's blog.

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        • T Offline
          tianzhu
          last edited by

          Hi kancheongmum


          Good Morning

          Vanilla Cake has provided the MD solution to your question.
          Another way is to do it by Units Method. The point to note is the total number of people remains unchanged.

          Please let me know if you need more information.

          Best wishes

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          • M Offline
            Maths Monster
            last edited by

            kancheongmum:
            Hi tianzhu

            I remember you have posted the solution for this question before but I can't find it,pls help tks
            Harry and Sally were at a party. Harry shook hands with 4 times as many boys as girls while Sally shook hands with 5 times as many boys as girls. How many boys and how many girls were at the party? (ans 25 boys, 6 girls)
            Let u be the number of girls Harry shook hand with.
            Harry
            Girls [u]
            Boys [4u]

            Sally
            Girls [u-1]
            Boys [5 (u-1)] or [5u-5]

            The number of boys that Harry shook hand with is 1 less than the boys that Sally shook hand with. We can write:

            4u+1=5u-5
            u=6

            Therefore, there are 6 girls and 4u+1 =25 boys at the party.

            1 Reply Last reply Reply Quote 0
            • M Offline
              Maths Monster
              last edited by

              Vanilla Cake:
              Just to share an interesting 5-mark speed question from 3rd Annual Mathlympics - 2010 Prelim round Q28 (Not in exact words). It's not a typical speed question (in my humble opinion) and do you have any alternative solutions to share?


              Downhill: 72 km/h
              Ground level: 63 km/h
              Uphill: 56 km/h
              Car travels from Town A to Town B in 4 hours. It returns to Town A using the same route in 4 h 40 min. What is the distance between Town A and Town B?

              D1 - Uphill, D2 - Downhill, D3 - Ground level
              Time from Town A to Town B --> D1/56 + D2/72 + D3/63

              D3 - Ground level, D2 - Uphill, D1 - Downhill
              Time from Town B to Town A --> D3/63 + D2/56 + D1/72

              Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

              Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/63 + 1/63)
              = 2/63 D1 + 2/63 D2 + 2/63 D3
              = 2/63 D (Distance D = D1+D2+D3)

              2/63 D -> 26/3
              D -> 26/3x63/2 = 273 km

              Distance between Town A and Town B = 273 km

              A similar http://psle2010a.blogspot.com/2010/08/speed-p6.html posted on 27 Aug 2010 can be found in Uncle Observer's blog.
              To share my thoughts on this.... this question is only applicable if the average speed of the uphill and downhill is equal to the ground level. In this case:

              (1/56 + 1/72) / 2 = 1/63

              Or in the other question:

              (1/12 + 1/4) / 2 = 1/6

              Otherwise, there will be multiply answers.

              Note to calculate average speed. Example a car travels at 10 km/h from town A to B and takes 20 km/h to travel back from town B to A. What is the average speed?

              Average speed = 1 / ((1/10 + 1/20) / 2) = 13 1/3 km/h

              It is not = (10 + 20) / 2 = 15 km/h........this is a common mistake student make!

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              • T Offline
                tianzhu
                last edited by

                Maths Monster:

                Otherwise, there will be multiply answers.
                Hi

                For the benefits of members,please help to provide more details.

                Best wishes

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

                  Hi all thank you for the solutions on the shook hand question. I have come across something interesting while teaching my ds. I felt that I have to share but I hope it will not cause any confusion. Tianzhu, Dharam and all the maths guru please contribute.


                  Tony had 15%more 50cts coins than 20cts coins. If there were 6 more 50cts coins than 20cts coins, how much did Tony have?

                  My solution:
                  100% — 20cts coins
                  115% — 50cts coins
                  15% ---- 6 coins
                  1% — 6/15
                  100% ---- 6/15 x 100 — 40 coins (no. of 20cts coins)
                  total money Tony have — (40 x 20cts) + (46 x 50cts) — $31.00

                  My ds did the above and was marked wrong. This is my understanding of the question.

                  The school teacher solution which I think is the answer key. This question is from My pals Maths Test book:

                  100% - 15% — 85%
                  85%/2 — 42.5% (20cts coins)
                  42.5% + 15% — 57.5% (50cts coins)
                  15% — 6 coins
                  42.5% — 6/15 x 42.5 — 17 20cts coins
                  57.5% — 6/15 x 57.5 — 23 50cts coins
                  total money Tony have — (17 x 20cts) + (23 x 50cts) — $14.90

                  Very confuse now please help. Thanks

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                  • M Offline
                    Maths Monster
                    last edited by

                    tianzhu:
                    Maths Monster:


                    Otherwise, there will be multiply answers.

                    Hi

                    For the benefits of members,please help to provide more details.

                    Best wishes

                    Sure, if we use the same question but now we change the ground speed to say 64km/h instead of 63km/h.

                    The equation will be:

                    Total time -> 4 h + 4 h 40 min = 8 h 40 min = 8⅔ h = 26/3 h

                    Total time --> D1 (1/56 + 1/72) + D2 (1/72 + 1/56) + D3 (1/64 + 1/64)
                    = 2/63 D1 + 2/63 D2 + 2/64 D3
                    = 2/63 (D1 + D2) + 2/64 D3 = 26/3

                    This is one equation with 3 variables that means there are infinitely many solutions to this equation.

                    1 Reply Last reply Reply Quote 0
                    • M Offline
                      Maths Monster
                      last edited by

                      kancheongmum:
                      Hi all thank you for the solutions on the shook hand question. I have come across something interesting while teaching my ds. I felt that I have to share but I hope it will not cause any confusion. Tianzhu, Dharam and all the maths guru please contribute.


                      Tony had 15%more 50cts coins than 20cts coins. If there were 6 more 50cts coins than 20cts coins, how much did Tony have?

                      My solution:
                      100% --- 20cts coins
                      115% --- 50cts coins
                      15% ---- 6 coins
                      1% --- 6/15
                      100% ---- 6/15 x 100 --- 40 coins (no. of 20cts coins)
                      total money Tony have --- (40 x 20cts) + (46 x 50cts) --- $31.00

                      My ds did the above and was marked wrong. This is my understanding of the question.

                      The school teacher solution which I think is the answer key. This question is from My pals Maths Test book:

                      100% - 15% --- 85%
                      85%/2 --- 42.5% (20cts coins)
                      42.5% + 15% --- 57.5% (50cts coins)
                      15% --- 6 coins
                      42.5% --- 6/15 x 42.5 --- 17 20cts coins
                      57.5% --- 6/15 x 57.5 --- 23 50cts coins
                      total money Tony have --- (17 x 20cts) + (23 x 50cts) --- $14.90

                      Very confuse now please help. Thanks
                      I would agree with the school teacher solution if the question is phased as \"Tony had more 50cts coins than 20cts coins and the different between 50cts coins and 20cts coins is 15% of the total number of coins.\", or something similar.

                      Perhaps others can comment.

                      1 Reply Last reply Reply Quote 0
                      • T Offline
                        teachingmum
                        last edited by

                        The teacher’s solution is incorrect. Since the no of 50 cents is compared to the no of 20 cents. the no of 20 cents is taken as 100%.

                        1 Reply Last reply Reply Quote 0

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