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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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    • M Offline
      maths6a
      last edited by

      Dear Gurus out there, need help with 2 questions:-


      1. If Andy runs up 7 steps of the escalator, he will take 36 sec to reach the top of the escalator. If he runs up 13 steps, he will take him only 27 sec to reach the top. How many seconds would it take him to reach the top if he did not run up any steps of the escalator at all?

      2. When asked how many gold coins a miser had, his reply was, " If I divide the coins into 2 unequal numbers, then 49 times the difference between the two numbers equals the difference between the squares of the two numbers."
      How many coins did he have?

      Thank you in advance fof the solution.

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

        Hi maths6a


        Just curious, where are your two questions from? They don’t look like questions from past years papers.
        [quote]If Andy runs up 7 steps of the escalator, he will take 36 sec to reach the top of the escalator. If he runs up 13 steps, he will take him only 27 sec to reach the top. How many seconds would it take him to reach the top if he did not run up any steps of the escalator at all? [/quote]If Andy runs up 7 steps, he needs 36 seconds to reach the top of the escalator.

        If he runs up 13 steps, he needs 27 seconds to reach the top the escalator .

        36-27 ----- 9
        Therefore each additional step takes 9/6 -------1.5 seconds

        Total steps in escalator = 7 + 36 /1.5 ----- 31

        or 13 + 27/1.5 ------ 31

        If he did not run up any steps, he will take 31*1.5 ------ 46.5 seconds.
        [quote]When asked how many gold coins a miser had, his reply was, \" If I divide the coins into 2 unequal numbers, then 49 times the difference between the two numbers equals the difference between the squares of the two numbers.\"
        How many coins did he have? [/quote]The second question involves difference of squares formula.

        a2 - b2 = (a + b) (a - b) (a2 means a to the power of 2)

        Given a2 - b2 = 49(a-b)

        Therefore (a+b) (a-b) = 49(a-b)
        Hence (a+b) =49

        Use listing or GC
        Try a=25, b=24
        625-576 = 49(25-24)

        There are 49 coins

        1 Reply Last reply Reply Quote 0
        • M Offline
          maths6a
          last edited by

          tianzhu:
          Hi maths6a


          Just curious, where are your two questions from? They don’t look like questions from past years papers.
          [quote]If Andy runs up 7 steps of the escalator, he will take 36 sec to reach the top of the escalator. If he runs up 13 steps, he will take him only 27 sec to reach the top. How many seconds would it take him to reach the top if he did not run up any steps of the escalator at all?
          If Andy runs up 7 steps, he needs 36 seconds to reach the top of the escalator.

          If he runs up 13 steps, he needs 27 seconds to reach the top the escalator .

          36-27 ----- 9
          Therefore each additional step takes 9/6 -------1.5 seconds

          Total steps in escalator = 7 + 36 /1.5 ----- 31

          or 13 + 27/1.5 ------ 31

          If he did not run up any steps, he will take 31*1.5 ------ 46.5 seconds.
          [quote]When asked how many gold coins a miser had, his reply was, \" If I divide the coins into 2 unequal numbers, then 49 times the difference between the two numbers equals the difference between the squares of the two numbers.\"
          How many coins did he have? [/quote]The second question involves difference of squares formula.

          a2 - b2 = (a + b) (a - b) (a2 means a to the power of 2)

          Given a2 - b2 = 49(a-b)

          Therefore (a+b) (a-b) = 49(a-b)
          Hence (a+b) =49

          Use listing or GC
          Try a=25, b=24
          625-576 = 49(25-24)

          There are 49 coins[/quote]Wow, what kind of questions are these? They are just 2 of the questions in my P4 neighbour's worksheet over the september holidays! Managed to help her with the rest but have no clue how to help her with these 2 so better check with the experts here. Thanks!

          1 Reply Last reply Reply Quote 0
          • H Offline
            Hiroii-kun
            last edited by

            Dear Gurus,

            need help with this question.

            At first, warehouse X and Y had a total of 10400 sacks of rice. After warehouse X has cleared 3/4 of its stock and warehouse Y has cleared 3/5 of its stock, warehouse Y now has 520 more sacks of rice than warehouse X. How many sacks of rice did each warehouse have at first?

            Ans is 5600 & 4800 but cant seem to get it.

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

              Hiroii-kun:
              Dear Gurus,

              need help with this question.

              At first, warehouse X and Y had a total of 10400 sacks of rice. After warehouse X has cleared 3/4 of its stock and warehouse Y has cleared 3/5 of its stock, warehouse Y now has 520 more sacks of rice than warehouse X. How many sacks of rice did each warehouse have at first?

              Ans is 5600 & 4800 but cant seem to get it.
              4u + 5p = 10400 ........ (1)
              2p - 1u = 520 ..............(2)
              8p - 4u = 2080 .............(2) x 4 = (3)

              13p = 12480 ........(1) + (3)
              1p = 960
              1u = 1400
              X => 4u = 4 x 1400 = 5600
              Y => 5p = 5 x 960 = 4800

              1 Reply Last reply Reply Quote 0
              • H Offline
                Hiroii-kun
                last edited by

                Dharma:
                Hiroii-kun:

                Dear Gurus,

                need help with this question.

                At first, warehouse X and Y had a total of 10400 sacks of rice. After warehouse X has cleared 3/4 of its stock and warehouse Y has cleared 3/5 of its stock, warehouse Y now has 520 more sacks of rice than warehouse X. How many sacks of rice did each warehouse have at first?

                Ans is 5600 & 4800 but cant seem to get it.

                4u + 5p = 10400 ........ (1)
                2p - 1u = 520 ..............(2)
                8p - 4u = 2080 .............(2) x 4 = (3)

                13p = 12480 ........(1) + (3)
                1p = 960
                1u = 1400
                X => 4u = 4 x 1400 = 5600
                Y => 5p = 5 x 960 = 4800

                ooh thanks!

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

                  maths6a:
                  Dear Gurus out there, need help with 2 questions:-


                  1. If Andy runs up 7 steps of the escalator, he will take 36 sec to reach the top of the escalator. If he runs up 13 steps, he will take him only 27 sec to reach the top. How many seconds would it take him to reach the top if he did not run up any steps of the escalator at all?

                  2. When asked how many gold coins a miser had, his reply was, \" If I divide the coins into 2 unequal numbers, then 49 times the difference between the two numbers equals the difference between the squares of the two numbers.\"
                  How many coins did he have?

                  Thank you in advance fof the solution.

                  1.\t

                  If Andy walks up (instead of running up) 6 more steps (13 – 7) he will take 9 secs (36s – 27s) more..
                  So, every 1 step that Andy walk up instead of running he will take 1.5 seconds more
                  Total time taken if Andy walks up all the steps = 36s + 7(1.5s) = 46.5s

                  2.

                  If the 2 numbers are A and B

                  49(A – B) = A^2 – B^2 (Given in Question)
                  (Imagine a small square of side B within a larger square of side A)
                  A^2 – B^2 is the difference in area between the 2 squares. (See below)

                  A^2 – B^2 = (A + B)(A – B)
                  So, A + B = 49 (Total number of coins)



                  http://www.postimage.org/image.php?v=TshfZPi

                  1 Reply Last reply Reply Quote 0
                  • starlight1968sgS Offline
                    starlight1968sg
                    last edited by

                    Hi,

                    I need some help here:
                    Jane has 70 dresses than skirts. She sold 3/4 of the dresses and 3/5 of the skirts. She sold 126 more dresses than skirts. What fraction of the remaining clothes that Jane had were skirts?
                    Many thanks.

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

                      starlight1968sg:
                      Hi,

                      I need some help here:
                      Jane has 70 dresses than skirts. She sold 3/4 of the dresses and 3/5 of the skirts. She sold 126 more dresses than skirts. What fraction of the remaining clothes that Jane had were skirts?
                      Many thanks.
                      Hi starlight1968sg,
                      Just curious, your question is from http://www.orlesson.org/orp/09Ma/2009-Math-SA2-SCGS.pdf.It's a 5-mark question and you changed the original name - Zara to Jane. Why?

                      Before
                      Dresses : 4u
                      Skirts: 4u-70

                      Sold
                      Dresses: 3u
                      Skirts: 3/5x(4u-70)= 2.4u-42

                      Sold 126 more dresses than skirts-> 3u=2.4u-42+126
                      0.6u=84
                      u = 140

                      After
                      Dresses: 1u = 140
                      Skirts: 4u-70-(2.4u-42) = 4u-70-2.4u+42 = 1.6u-28 = 1.6(140)-28=224-28=196

                      196/(140+196)=196/336 = 7/12

                      Fraction of the remaining clothes that Jane had were skirts = 7/12.

                      Hope that other members will come with model drawing solutions and other alternatives.

                      1 Reply Last reply Reply Quote 0
                      • F Offline
                        firebird
                        last edited by

                        Dear maths guru


                        Good evening.

                        Please help me on the following maths which I am assisting my child:

                        1) Mdm Ang bought some highlighters, pens and mechanical pencils. 1/4 of them were highlighters. The number of pens she bought was 6 more than 1/2 of the total number of all items and the remaining were mechanical pencils. Each of the highlighters, pens and mechanical pencils cost $2.10, $4.05 and $1.60 respectively. She spent a total of $227.10 on all items. How many pens did she buy altogether?

                        This is Nanyang Primary SA 2 2009 question 18.

                        2) Simple sum but I want to check with you

                        What is the sum of all the common factors of 12, 18 and 24?

                        12 - 1,2,3,4,6,12 = sum of all = 28
                        18 - 1,2,3,6,9,18 = sum of all = 39
                        24 - 1,2,3,4,6,12,24 = sum of all = 60

                        So sum all three above = 28+39+60= 127

                        Kindly correct me.

                        Thank you.
                        With best regards
                        Firebird

                        1 Reply Last reply Reply Quote 0

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