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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

      chrisu:
      Thanks to all for responding to this post. My daughter came back yesterday evening and told me that the teacher and the Maths HOD agreed to accept both 225 and 450 as answer since the question is ambiguous.


      The answer is 225 if comparison is made after the 1/3 and 1/5 is given to Rahmad. The answer will be 450 if the comparison is made before the 1/3 and 1/5 is given to Rahmad.

      As for proving of Adam = 450;
      If Adam = 450, Rahmad will be 370.
      If Adam gives 1/3 of 450 to Rahmad,
      Rahmad will get : 370 + 150 = 520.
      520 - 450 = 70 (which is 70 more than Adam before the 1/3 is given out).

      If Adam gives 1/5 of 450 to Rahmad,
      Rahmad will get : 370 + 90 = 460.
      460 - 450 = 10 (which is 10 more than Adam).
      The school's HOD ought to be skinned. Nothing in the question (based on what is given here) hinted that the comparison is to be made between \"after\" of Rahmad and \"before\" of Adam.

      This is a common question and the school should know. A typical example of blindly following answer key with no confident to decide that the answer 450 is wrong.

      1 Reply Last reply Reply Quote 0
      • S Offline
        Sun_2010
        last edited by

        atutor2001:
        chrisu:

        Thanks to all for responding to this post. My daughter came back yesterday evening and told me that the teacher and the Maths HOD agreed to accept both 225 and 450 as answer since the question is ambiguous.


        The answer is 225 if comparison is made after the 1/3 and 1/5 is given to Rahmad. The answer will be 450 if the comparison is made before the 1/3 and 1/5 is given to Rahmad.

        As for proving of Adam = 450;
        If Adam = 450, Rahmad will be 370.
        If Adam gives 1/3 of 450 to Rahmad,
        Rahmad will get : 370 + 150 = 520.
        520 - 450 = 70 (which is 70 more than Adam before the 1/3 is given out).

        If Adam gives 1/5 of 450 to Rahmad,
        Rahmad will get : 370 + 90 = 460.
        460 - 450 = 10 (which is 10 more than Adam).

        The school's HOD ought to be skinned. Nothing in the question (based on what is given here) hinted that the comparison is to be made between \"after\" of Rahmad and \"before\" of Adam.

        This is a common question and the school should know. A typical example of blindly following answer key with no confident to decide that the answer 450 is wrong.

        Agreed !
        The question as interpreted by the school is flawed.

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

          chrisu:

          Maybe the school should inform the publisher that there are 2 answers to the question.
          Hi

          Where is your question from?

          You may want to write to the publisher to clarify.

          Best wishes

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

            The question is from MyPal workbook where a number of the parents' kids are using too. Those students from other schools are also given the 450 as answer and not too sure if there is any feedback from their parents.



            tianzhu:
            chrisu:

            Maybe the school should inform the publisher that there are 2 answers to the question.

            Hi

            Where is your question from?

            You may want to write to the publisher to clarify.

            Best wishes

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

              chrisu:
              The question is from MyPal workbook where a number of the parents' kids are using too. Those students from other schools are also given the 450 as answer and not too sure if there is any feedback from their parents.
              Hi

              Questions involving two cases or double if are common in primary maths.

              Let your kid go through these questions, do we compare with the original quantity or remaining?

              1) Wawa had some red and blue stickers. If 72 red stickers were removed, the ratio of the number of red stickers to that of the blue stickers would be 1 : 2. If 180 blue stickers were removed instead, the ratio would become 5 : 1. How many red stickers were there?

              2) There are some red pins and blue pins in a box. If one of the red pin is removed, the number of red pins left will be 1/7 of the total number of pins left. If two blue pins are removed, the number of red pins will be 1/5 of the total number of pins left. How many pins are there in the box?

              3) If Mark gives Lenny 14 stickers, he will have the same number of stickers as Lenny. If Lenny gives 10 stickers to Mark, the ratio of the number of stickers Lenny has to that of Mark will be 2: 5. How many stickers does Mark have?

              Best wishes

              1 Reply Last reply Reply Quote 0
              • J Offline
                James Ang
                last edited by

                When Larry's age was twice Henry's age, Don's age was 27. When Don's age was twice Larry's age, Henry's age was 15. What was Henry's age when Don was 58 years old ?



                Let me show how a P5 student would do to solve by drawing a model. (Remember that P5 students do not know any algebra!)


                L [][]+[31]
                H []+[31]
                D [27]+[31] = 58

                L [15][] + [][16]
                H [15] +[][16]
                D [15][][15][] + [][16] = 58

                from above
                D [15][][15][] + [][16] = 58
                [][][] = 58-15-15-16 = 12
                []=4

                Therefore Henry = [15] +[][16] = 15+4+16 = 35 years old.

                A picture speaks a thousand words! šŸ˜„

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                • B Offline
                  Belle2011
                  last edited by

                  Hello,

                  Can somebody help me with this problem?

                  F, L and W each owned a collection of comics. The total collected owned by L and W was 3/2 as many comics as F owned. L owned 4/5 as many comics as the total collection owned by F and W. If W owned 169 fewer comics than L, how many comics must F and L each give to W in order for the 3 girls to have the same number of comics?

                  Thank-you.

                  Cheers,
                  Belle

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

                    Belle2011:
                    Hello,

                    Can somebody help me with this problem?

                    F, L and W each owned a collection of comics. The total collected owned by L and W was 3/2 as many comics as F owned. L owned 4/5 as many comics as the total collection owned by F and W. If W owned 169 fewer comics than L, how many comics must F and L each give to W in order for the 3 girls to have the same number of comics?

                    Thank-you.

                    Cheers,
                    Belle
                    W = L - 169

                    F = 2/3 ( W + L) = 2/3 (L-169 + L) = 2/3 (2L -169)

                    5L/4 = W+F = (L - 169) + 2/3 (2L -169) = (3L - 3x169 + 4L -2x169)/3 = (7L -5x169)/3

                    15L = 28L -20x169----->13L = 20x169----> L = 20x13 = 260

                    So

                    W = L -169 = 260 -169 = 91

                    F = 2/3 (91+260) = 2/3 (351) = 234

                    (L+W+F)/3 = (260+234+91)/3 = 585/3 = 195

                    So,

                    L must give 260-195 = 65 to W

                    F must give 234 -195 = 39 to W

                    HTH.

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

                      Belle2011:
                      Hello,

                      Can somebody help me with this problem?

                      F, L and W each owned a collection of comics. The total collected owned by L and W was 3/2 as many comics as F owned. L owned 4/5 as many comics as the total collection owned by F and W. If W owned 169 fewer comics than L, how many comics must F and L each give to W in order for the 3 girls to have the same number of comics?

                      Thank-you.

                      Cheers,
                      Belle
                      F : L + W = 2 : 3 = 18 : 27
                      L : F + W = 4 : 5 = 20 : 25

                      L => 20u
                      F => 18u
                      W => 7u

                      20u – 7u = 169
                      13u = 169
                      1u = 13

                      No. of comics that F gives to W = 18u – 15u = 3u = 3 x 13 = 39
                      No. of comics that L gives to W = 20u – 15u = 5u = 5 x 13 = 65

                      1 Reply Last reply Reply Quote 0
                      • B Offline
                        Belle2011
                        last edited by

                        Dear iFruit and Dharma,

                        Thanks for helping.

                        Dharma,
                        F : L + W = 2 : 3 = 18 : 27
                        L : F + W = 4 : 5 = 20 : 25
                        L => 20u
                        F => 18u
                        W => 7u

                        I know you are converting 2:3 = 18:27 as equivalent ratios.
                        But how do you know you have to "stop" at 18:27 so that you can get L is 20u, F is 18u and W is 7u?
                        Thank-you.

                        Cheers,
                        Belle.

                        1 Reply Last reply Reply Quote 0

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