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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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    • A
      acehkr3009
      last edited by

      Hi,


      Here is another question:

      The ratio of Apples to Pears in a fruit stall is 5 : 8. If 60% of the pears are sold, what percentage of the apples must be sold for the number of pears and apples in the stall to be the same?

      Appreciate with Thanks.

      1 Reply Last reply
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      • A
        acehkr3009
        last edited by

        tianzhu:
        acehkr3009:

        Here is another interesting questions...


        The average of the 5 numbers in a list is 54. The average of the first 2 numbers is 48. What is the average of the last 3 numbers?

        Really tough, need your help on this....

        Thanks in advance.

        Hi

        5*54 ----- 270
        2*48 ------96
        270- 96 -- ----174
        174/3 ------58

        Best wishes

        Many thanks.

        Solution looks simple, but when I am trying it at first, brain just got stuck and freeze!

        You are really fantastic!

        1 Reply Last reply
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        • T
          tianzhu
          last edited by

          acehkr3009:
          The ratio of Apples to Pears in a fruit stall is 5 : 8. If 60% of the pears are sold, what percentage of the apples must be sold for the number of pears and apples in the stall to be the same?

          Hi

          You may use MD or Units Method to solve this question.

          Apples: Pears ----- 5:8 -----25:40

          60% of pear% sold ----- 60/100*40 ----- 24
          40% of pear left ---- (40-24) ------16

          Number of apples to be sold ----- (25-16) -----9
          9/25*100% ---- 36 %( percentage of apples to be sold)

          Best wishes

          1 Reply Last reply
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          • T
            tianzhu
            last edited by

            acehkr3009:

            Solution looks simple, but when I am trying it at first, brain just got stuck and freeze!
            Hi

            The KS way.

            Keep (in) Syllabus
            Keep (it) Simple
            Keep (it) Short

            Keep Supporting your kids

            Best wishes

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            • Y
              Yu Xuan
              last edited by

              Dharma:
              clblinym:

              [quote=\"Dharma\"]
              Ratio of Area of AEB : Area of AEC : Area of ADE = 1 : 4 : 2

              Area of ABC = 1/2 X AB x Perpendicular Height
              Area of ADC = 1/2 x DC x Perpendicular Height

              AB : DC = Area of ABC : Area of ADC = (1+4 ) : 2 = 5 : 2



              Dear Dharma,
              Thank you so much for your great help.
              My son and I fully understand your explanation.
              We really appreciate your helpfullness.

              I just want to check with you if it is a typo for the sentence.
              Ratio of Area of AEB : Area of AEC : Area of ADC = 1 : 4 : 2

              Sorry for the typo error. Yes , it is ADC[/quote]Hi Dharma... how did you derive the ration of AEC : ADC 4 :2 pls? Thank you for your time.

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              • D
                Dharma
                last edited by

                Yu Xuan:
                Dharma:

                [quote=\"clblinym\"]





                Dear Dharma,
                Thank you so much for your great help.
                My son and I fully understand your explanation.
                We really appreciate your helpfullness.

                I just want to check with you if it is a typo for the sentence.
                Ratio of Area of AEB : Area of AEC : Area of ADC = 1 : 4 : 2

                Sorry for the typo error. Yes , it is ADC

                Hi Dharma... how did you derive the ration of AEC : ADC 4 :2 pls? Thank you for your time.[/quote]Area of ABE : Area of AECD = 1: 6
                Area of ABE : Area of AEC = 1: 4

                So, Area of ABE : Area of AEC : Area of ACD = 1 : 4 : (6 -4 ) = 1 : 4 : 2

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                • Y
                  Yu Xuan
                  last edited by

                  Hi Dharma… now I see it. Thank you!!

                  1 Reply Last reply
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                  • D
                    Dharma
                    last edited by

                    Yu Xuan:
                    Hi Dharma... now I see it. Thank you!!

                    You are welcome.

                    1 Reply Last reply
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                    • D
                      Dharma
                      last edited by

                      acehkr3009:
                      Hi,


                      Need some help on this question.

                      Not sure why question appear so small.

                      Here is the retype of the question:

                      If W & Y is the mid-point of AB and CD respectively, and Z & X are any points along AD and BC respectively. What is the fraction of the shaded portion ?

                      Thanks.

                      http://www.postimage.org/image.php?

                      Your question looks incomplete. Can u check whether triangle AWZ is also shaded?

                      1 Reply Last reply
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                      • K
                        Kiasu Friend
                        last edited by

                        tianzhu:
                        trytry:

                        Hi, need help.

                        This question was given as prelim practice:

                        Tom was driving from his office back to his house. As he was sick, he drove 20km/h slower. As a result, he reached home 10 minutes later.The distance between his office and house is 40km. What is his usual speed?

                        Tia.

                        Hi

                        Try Guess and Check

                        40/60 ---2/3h ----- 40min
                        40/80 --- 1/2h -----30min

                        Difference -----10min -----OK

                        Therefore speed -----80km/h

                        Best wishes


                        Hi tianzhu,

                        Thank you for the elegantly simple solution based on 'guess & check' method. But in order to understand the concept involved, can you provide a method to 'derive' the answer systematically rather than 'guessing'?

                        Because in the pressure of the exam-hall, 'guess & check' may be difficult to come up and quite a few wrong guesses may be tried using up precious time.

                        I am sorry for taking your time. Thank you very much for your tireless contribution to this forum.

                        1 Reply Last reply
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