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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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    • MathIzzzFunM Offline
      MathIzzzFun
      last edited by

      blitz:
      PLs assist.

      Belinda has 20% more stickers than Yi Xizng. Yi Xiang has 25% more stickers than Sabrina.
      (a)What is the ratio of Belinda's stickers to Sabrina stickers?
      (b)Yi Xiang gave some stickers to Sabrina so tha they each had the same number of stickers. Now Belinda has 36 more stickers than Sabrina. How many stickers does Belinda have?
      Hi

      Q1
      Belinda : Yi Xiang = 120 : 100 = 6 :5
      Yi Xiang : Sabrina = 125 : 100 = 5 : 4
      Belinda : Yi Xiang : Sabrina = 6 : 5 : 4

      a) Belinda : Sabrina = 6 : 4 = 3 : 2

      b) Belinda : Yi Xiang : Sabrina = 6u : 5u : 4u
      After Yi Xiang gave some stickers to Sabrina, they each had the same = (5u+4u)/2 = 4.5u

      Belinda had 36 more stickers than Sabrina -> 6u - 4.5u = 1.5u = 36
      1u = 24
      6u = 144

      Belinda had 144 stickers.

      cheers.

      1 Reply Last reply Reply Quote 0
      • MathIzzzFunM Offline
        MathIzzzFun
        last edited by

        blitz:
        One more question...thanks in advance


        60% of people at a performance were adults. 75% of the children at the performance were girls. There were 36 more girls than boys. Halfway thru the performance, some girls left. After that, 2/9 of the remaining people were girls.

        How many girls left the performance?

        I got the answer as 34, is that correct?
        Hi

        I got 18.

        From the first part of question, you will get -
        Adults : Girls : Boys = 108 : 54 : 18
        then 7/9 of remaining people = 108 + 18 = 126, 1/9 = 18, 2/9 = 36 girls remained.

        Number of girls who left the performance = 54 - 36 = 18

        cheers.

        1 Reply Last reply Reply Quote 0
        • B Offline
          blitz
          last edited by

          Thank you MathIzzzFun....will forward this to my child. 😄

          1 Reply Last reply Reply Quote 0
          • 2 Offline
            2DMommy
            last edited by

            Please help me with these questions again :

            1) A stack of bookmarks was shared among Peter, Tom and Jerry. Peter received 3 more than 50% of the bookmarks. Tom received 5 more than half of the remaining bookmarks. After Peter and Tom had taken their share, Jerry received 50% of the remainder and the last 3 bookmarks. What percentage of the stack of bookmarks did Jerry receive ?\t

            2) Kailing and Jim had a total of $63. Kailing gave 0.3 of her share to Jim. Jim then gave 1/3 of the total amount of money he had to his brother. In the end, all the three children had the same amount of money. How much money did Kailing have at first?\t\t

            Thanks !

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            • MathIzzzFunM Offline
              MathIzzzFun
              last edited by

              2DMommy:
              Please help me with these questions again :

              1) A stack of bookmarks was shared among Peter, Tom and Jerry. Peter received 3 more than 50% of the bookmarks. Tom received 5 more than half of the remaining bookmarks. After Peter and Tom had taken their share, Jerry received 50% of the remainder and the last 3 bookmarks. What percentage of the stack of bookmarks did Jerry receive ?\t

              2) Kailing and Jim had a total of $63. Kailing gave 0.3 of her share to Jim. Jim then gave 1/3 of the total amount of money he had to his brother. In the end, all the three children had the same amount of money. How much money did Kailing have at first?\t\t

              Thanks !
              Hi

              you should be able to solve by working backwards:

              1) Jerry received 50% of the remainder (final) and last 3 bookmarks, so the last 3 bookmarks is also = 50% of final remainder, final remainder = 6 .. from here, you can work backwards to solve .. Total bookmarks = 50, Peter 28, Tom 16, Jerry 6.

              2) this was discussed in an earlier thread - http://www.flickr.com/photos/62167097@N02/6020796670/in/photostream

              cheers.

              1 Reply Last reply Reply Quote 0
              • R Offline
                rainbow75
                last edited by

                Mr Tan drives to work everday .The distance from home to office is 40 km.He was sick one day and he drove 20km/h slower and was 10 minutes late for work.what was his usual speed to work?

                1 Reply Last reply Reply Quote 0
                • MathIzzzFunM Offline
                  MathIzzzFun
                  last edited by

                  rainbow75:
                  Mr Tan drives to work everday .The distance from home to office is 40 km.He was sick one day and he drove 20km/h slower and was 10 minutes late for work.what was his usual speed to work?

                  Hi

                  The solution involved solving of quadratic equation. For P6, the usual way is to use Guess & Check to solve.

                  http://i55.tinypic.com/es3slw.jpg\">

                  cheers.

                  1 Reply Last reply Reply Quote 0
                  • R Offline
                    rainbow75
                    last edited by

                    hi mathguru

                    thank you for your help.however I still don’t understand what does ‘n’ represent ? please explain more clearly .Thank you.

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

                      rainbow75:
                      Mr Tan drives to work everday .The distance from home to office is 40 km.He was sick one day and he drove 20km/h slower and was 10 minutes late for work.what was his usual speed to work?

                      Hi rainbow 75

                      This question is actually a typical secondary school question for application of algebra. It should not be in P6 paper.

                      I agree with MathIzzFun that for P6, the only way to solve is to use guess and check but the kid would need very good sense of proportion to get the answer as it involves hours and minutes.

                      Just for information, the secondary solving method using algebra is as follows :

                      Let the normal Time needed = T and the normal Speed be S

                      Time = Distance/Speed

                      T = 40/S (for normal situation)
                      T + 1/6 = 40/(S - 20) (when he is sick) so T = 40/(S - 20) - 1/6

                      Now we have expression of T in 2 ways and can compare them to find S :

                      40/S = 40/(S - 20) - 1/6

                      Solving the above is definitely too difficult for normal pr children

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

                        hi rainbow75,


                        i’ve PMed you the solution with another method. feel free to reply if there is any doubt.

                        1 Reply Last reply Reply Quote 0

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