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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • A Offline
      atutor2001
      last edited by

      youngmum:
      My boy got this problem. Can someone help me to solve this?


      Monkey King had 2000 pears. He gave all the pears to some monkeys. On the first day, the monkeys ate half of the total number of pears. On the second day, they ate 1/3 of the remaining pears. On the third day, they ate 1/4 of the rest of the pears. They went on to eat the pears as 1/5,1/6,1/7,.....,1/2000 of the remainder of every previous day's pears. Find the number of pears left at the end.

      Thanks. By the way, please explain clearly because i m a not so smart mum.
      The answer is 4

      1st Remainder\t4/5 x 2000
      2nd Remainder\t4/5 x 5/6 x 2000
      3rd Remainder\t4/5 x 5/6 x 6/7 x 2000


      2nd last Remainder\t 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 2000

      Last Remainder 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 1999/2000 x 2000 = 4

      1 Reply Last reply Reply Quote 0
      • F Offline
        Full.010894Cream
        last edited by

        atutor2001:
        youngmum:

        My boy got this problem. Can someone help me to solve this?


        Monkey King had 2000 pears. He gave all the pears to some monkeys. On the first day, the monkeys ate half of the total number of pears. On the second day, they ate 1/3 of the remaining pears. On the third day, they ate 1/4 of the rest of the pears. They went on to eat the pears as 1/5,1/6,1/7,.....,1/2000 of the remainder of every previous day's pears. Find the number of pears left at the end.

        Thanks. By the way, please explain clearly because i m a not so smart mum.

        The answer is 4

        1st Remainder\t4/5 x 2000
        2nd Remainder\t4/5 x 5/6 x 2000
        3rd Remainder\t4/5 x 5/6 x 6/7 x 2000


        2nd last Remainder\t 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 2000

        Last Remainder 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 1999/2000 x 2000 = 4

        think you missed the 1st 3 days of the pear feast! šŸ˜„

        1 Reply Last reply Reply Quote 0
        • D Offline
          Dharma
          last edited by

          Full.Cream:
          atutor2001:

          [quote=\"youngmum\"]My boy got this problem. Can someone help me to solve this?


          Monkey King had 2000 pears. He gave all the pears to some monkeys. On the first day, the monkeys ate half of the total number of pears. On the second day, they ate 1/3 of the remaining pears. On the third day, they ate 1/4 of the rest of the pears. They went on to eat the pears as 1/5,1/6,1/7,.....,1/2000 of the remainder of every previous day's pears. Find the number of pears left at the end.

          Thanks. By the way, please explain clearly because i m a not so smart mum.

          The answer is 4


          1st Remainder\t4/5 x 2000
          2nd Remainder\t4/5 x 5/6 x 2000
          3rd Remainder\t4/5 x 5/6 x 6/7 x 2000


          2nd last Remainder\t 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 2000

          Last Remainder 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 1999/2000 x 2000 = 4

          think you missed the 1st 3 days of the pear feast! :D[/quote]http://www.postimage.org/

          Poor monkeys feasting on rotten pears. Most may have suffered from food poisoning.

          1 Reply Last reply Reply Quote 0
          • A Offline
            atutor2001
            last edited by

            Full.Cream:


            think you missed the 1st 3 days of the pear feast! šŸ˜„
            Thanks the answer is 1, was dreaming.

            1 Reply Last reply Reply Quote 0
            • youngmumY Offline
              youngmum
              last edited by

              Thanks everyone!

              1 Reply Last reply Reply Quote 0
              • N Offline
                N3SKiasu
                last edited by

                Hi all,

                Need your help on this Rosyth Prelim 2008 Q48 (I have searched thru past pages and could not find any discussion on this question, hope my eyes are working fine by browsing thru the 50+ pages and it is not repeated):

                Ning and Zongwei jogged to and fro repeatedly along a straight path in a park between 2 points A and B.
                Ning jogged at a uniform speed of 4m/s.
                Zongwei jogged at uniform speed of 6m/s.
                They started jogging from opposite directions at the same time as shown below.
                Ning ------> <----------Zongwei
                A---------------------X---------------------Y-------------------B

                They first met one another at point X. The second time they met was at point Y.
                a) Given that the distance between X and Y is 160m. Find the distance between A and B. (Answer given : 400m)
                b) If they started jogging at 8am, how long did they take to meet again for the third time? (Express your answer in minutes and seconds). (Answer given : 3m20s)

                Do you consider this is as a challenging question?
                A bit worried for DD now as she could not solve this sort of question at this point...and so do I, helpless mom...counting down... šŸ™
                TIA.

                1 Reply Last reply Reply Quote 0
                • N Offline
                  N3SKiasu
                  last edited by

                  Would like to check on the answer for SCGS Prelim 2008 Q23:


                  A 200m long MRT train took 20 min to pass through a tunnel completely.
                  If the MRT train travelled at an average speed of 120 m/min, find the length of the tunnel.

                  Solution:
                  120m/min x 20 min = 2400 m
                  2400m - 200 m = 2200m

                  But the answer given is 2300 m. Just wondering if the answer given is correct?

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

                    N3SKiasu:
                    Hi all,

                    Need your help on this Rosyth Prelim 2008 Q48 (I have searched thru past pages and could not find any discussion on this question, hope my eyes are working fine by browsing thru the 50+ pages and it is not repeated):
                    Hi
                    Dharma had answered this Q.Please go to pg 54.
                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&postdays=0&postorder=asc&start=795

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

                      N3SKiasu:

                      120m/min x 20 min = 2400 m
                      2400m - 200 m = 2200m
                      But the answer given is 2300 m. Just wondering if the answer given is correct?
                      Hi
                      You are right.
                      The answer is 2200m.

                      1 Reply Last reply Reply Quote 0
                      • N Offline
                        N3SKiasu
                        last edited by

                        tianzhu:
                        N3SKiasu:

                        Hi all,

                        Need your help on this Rosyth Prelim 2008 Q48 (I have searched thru past pages and could not find any discussion on this question, hope my eyes are working fine by browsing thru the 50+ pages and it is not repeated):

                        Hi
                        Dharma had answered this Q.Please go to pg 54.
                        http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&postdays=0&postorder=asc&start=795

                        Thank you so much Tianzhu, oh it was just 3 pages before this, I must have browsed thru too quickly. Millions thanks!

                        1 Reply Last reply Reply Quote 0

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