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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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    • T Offline
      tianzhu
      last edited by

      kancheongmum:


      thanks very much Dharma and tianzhu your solutions on all question are very helpful.
      Hi kancheongmum

      You're welcome.

      Best wishes

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

        super star:
        HI MATHS GURU

        Please help me with this problem.At present,ronald is 3 times as old as his sister.In 22 years time,ronald's age will be 19 years less than twice his sister's age.how old is ronald?
        Now
        Ron : 3u
        Sister : 1u

        22 years time
        Ron : 3u + 22
        Sister : 1u + 22

        3u + 22 = 2(1u + 22) – 19
        3u + 22 = 2u + 25
        1u = 3

        Ronald age => 3u = 3 x 3 = 9 yrs old

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

          kancheongmum:
          tianzhu:

          Hi clblinym


          You're welcome.

          Best wishes

          Hi tianzhu really like your graphics in your answers. I hope you don't mind explaining to me on the solution of sheep eating grass again. I don't quite get it thank very much.

          Hi clblinym and kancheongmum,

          This is another way to solve the \"sheep eat grass problem\"

          Alternative method
          200 sheep can eat up all the grass in 100 days
          150 sheep can eat up all the grass in 150 days

          Lets say a sheep can eat 1u of grass a day;

          200 sheep in 100 days => 200 x 100 x 1u = 20000u
          150 sheep in 150 days => 150 x 150 x 1u = 22500u

          The difference is due to the fact that the grass in the pastures continues to grow.

          Rate of growth of grass = (22500u – 20000u)/(150 – 100) days = 50u/day

          Original amt of grass = 20000u – 50u(100 days) = 15000u

          The 200 sheep will take 75 days to finish eating the original grass i.e 15000u/200u = 75 days
          The balance 5000u (new grass that grew) will be eaten in 5000u/(200u/day)= 25 days

          Therefore; 200 sheep will take 100 days (75days + 25days) to complete eating all the grass.

          Now lets say that 100 sheep will take N days to eat up all the grass.

          No. of days for the 100 sheep to eat up the original amt of grass = 15000u/(100u/day) = 150 days

          Amt of new grass grown in N days = 50u(N)

          No. of days for the 100 sheep to eat up the original amt of new grass grown
          = 50u(N)/(100u/day) = ½ N days

          100 sheep takes ½ N days to eat up the new grass; so it must have also taken ½ N days to eat up the original grass.

          ½ N = 150 days
          N = 300 days.

          100 sheep will take 300 days to eat up all the grass.

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

            MOE Hater:
            CJM:

            hi all, pls help with this question :


            A large water tank has two inlet pipes ( a large one and a small one ) and one outlet pipe. it takes 3 hours to fill the tank with the large inlet pipe. On the other hand, it takes 4 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 7 hours.

            What fraction of the tank ( initially empty ) will be filled in 2.27 hours if all three pipes are in operation? Give your answer to two decimal places.

            I don't understand. You asked for a fraction then said to give the answer to 2 decimal places 😐

            This is a question given by the teacher.
            We do not have the answer and he wants the answer to be converted to decimal and round off to 2 decimal places

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

              maths6a:
              MOE Hater:

              [quote=\"CJM\"]hi all, pls help with this question :


              A large water tank has two inlet pipes ( a large one and a small one ) and one outlet pipe. it takes 3 hours to fill the tank with the large inlet pipe. On the other hand, it takes 4 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 7 hours.

              What fraction of the tank ( initially empty ) will be filled in 2.27 hours if all three pipes are in operation? Give your answer to two decimal places.

              I don't understand. You asked for a fraction then said to give the answer to 2 decimal places 😐




              Sorry I worked answer to be 0.999 so rounded up will be 1.00 which is like whole tank is filled???[/quote]may i know how did u derive the answer?
              ds need to show the steps in deriving the answer but I have no clue at all.

              thanks

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

                CJM:
                maths6a:

                [quote=\"MOE Hater\"]
                I don't understand. You asked for a fraction then said to give the answer to 2 decimal places 😐






                Sorry I worked answer to be 0.999 so rounded up will be 1.00 which is like whole tank is filled???

                may i know how did u derive the answer?
                ds need to show the steps in deriving the answer but I have no clue at all.

                thanks[/quote]In 1 hr
                Inflow : (1/3 + ¼) = 7/12 of tank
                Outflow : 1/7 of tank

                Net flow : 7/12 – 1/7 = 49/84 – 12/84 = 37/84 of tank filled

                In 2.27 hrs (use calculator)
                Proportion of of tank filled: (37/84 x 2.27 hrs) = 0.9998 = 1.00 ( to 2 decimal places)

                1 Reply Last reply Reply Quote 0
                • K Offline
                  Kiasu Friend
                  last edited by

                  Hi Dharma/ Mathsguru,


                  Please help with this problem (from MGS Prelim).

                  Study the number pattern below:

                  5 x 5 = 25
                  5 x 5 x 5 = 125
                  5 x 5 x 5 x 5 = 625
                  5 x 5 x 5 x 5 x 5 = 3,125

                  (a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., "5 to the power of 6" what is the sum of the digits in the ones and tens place?

                  (b) What is the sum of the last THREE digits in "5 to the power of 15" ?

                  © What is the sum of the last FOUR digits in "5 to the power of 210"?

                  Thanks a lot.

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

                    Kiasu Friend:
                    Hi Dharma/ Mathsguru,


                    Please help with this problem (from MGS Prelim).

                    Study the number pattern below:

                    5 x 5 = 25
                    5 x 5 x 5 = 125
                    5 x 5 x 5 x 5 = 625
                    5 x 5 x 5 x 5 x 5 = 3,125

                    (a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., \"5 to the power of 6\" what is the sum of the digits in the ones and tens place?

                    (b) What is the sum of the last THREE digits in \"5 to the power of 15\" ?

                    (c) What is the sum of the last FOUR digits in \"5 to the power of 210\"?

                    Thanks a lot.
                    a)\tSum of digits in ones and tens place = 2 + 5 = 7

                    b)\t5 to the power of odd number => last 3 digits = 125
                    5 to the power of even number => last 3 digits = 625
                    5 to the power of 15 => 15 is odd number ; so last 3 digits = 125
                    Sum of last 3 digits = 1 + 2 + 5 = 8

                    c)\t5 to power of 5 = 3125 => Sum of last 4 digits =11
                    5 to power of 6 = 15,625 => Sum of last 4 digits =18
                    5 to power of 7 = 78,125 => Sum of last 4 digits = 16
                    5 to power of 8 = 390,625 => Sum of last 4 digits = 13
                    5 to power of 9 = 1,953,125=> Sum of last 4 digits = 11

                    210 divide by 4 =52R2
                    => Last 4 digits of 5 to the power of 210 = 5625
                    5 to power of 210 => Sum of last 4 digits = 18

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

                      Dharma:
                      CJM:

                      [quote=\"maths6a\"]





                      Sorry I worked answer to be 0.999 so rounded up will be 1.00 which is like whole tank is filled???

                      may i know how did u derive the answer?
                      ds need to show the steps in deriving the answer but I have no clue at all.

                      thanks

                      In 1 hr
                      Inflow : (1/3 + ¼) = 7/12 of tank
                      Outflow : 1/7 of tank

                      Net flow : 7/12 – 1/7 = 49/84 – 12/84 = 37/84 of tank filled

                      In 2.27 hrs (use calculator)
                      Proportion of of tank filled: (37/84 x 2.27 hrs) = 0.9998 = 1.00 ( to 2 decimal places)[/quote]thank U so much for the solution !

                      1 Reply Last reply Reply Quote 0
                      • K Offline
                        Kiasu Friend
                        last edited by

                        Dharma:
                        Kiasu Friend:

                        Hi Dharma/ Mathsguru,


                        Please help with this problem (from MGS Prelim).

                        Study the number pattern below:

                        5 x 5 = 25
                        5 x 5 x 5 = 125
                        5 x 5 x 5 x 5 = 625
                        5 x 5 x 5 x 5 x 5 = 3,125

                        (a) In (5 x 5 x 5 x 5 x 5 x 5), i.e., \"5 to the power of 6\" what is the sum of the digits in the ones and tens place?

                        (b) What is the sum of the last THREE digits in \"5 to the power of 15\" ?

                        (c) What is the sum of the last FOUR digits in \"5 to the power of 210\"?

                        Thanks a lot.

                        a)\tSum of digits in ones and tens place = 2 + 5 = 7

                        b)\t5 to the power of odd number => last 3 digits = 125
                        5 to the power of even number => last 3 digits = 625
                        5 to the power of 15 => 15 is odd number ; so last 3 digits = 125
                        Sum of last 3 digits = 1 + 2 + 5 = 8

                        c)\t5 to power of 5 = 3125 => Sum of last 4 digits =11
                        5 to power of 6 = 15,625 => Sum of last 4 digits =18
                        5 to power of 7 = 78,125 => Sum of last 4 digits = 16
                        5 to power of 8 = 390,625 => Sum of last 4 digits = 13
                        5 to power of 9 = 1,953,125=> Sum of last 4 digits = 11

                        210 divide by 4 =52R2
                        => Last 4 digits of 5 to the power of 210 = 5625

                        5 to power of 210 => Sum of last 4 digits = 18

                        Dear Dharma,

                        Thanks a lot for giving solution so fast. Very very fast indeed.
                        But I do not understand the part marked in red. My questions are:

                        (1) Why divide 210 by 4? Is it because we want the sum of last 4 digits? If we want the sum of last 3 digits, will we divide by 3?

                        (2) When we divide 210 by 4, we get 52R2. From that, how do we conclude that the last 4 digits of \"5 to the power of 210\" are 5625?

                        (3) Is there any concept or formula I should learn to understand the above?

                        Kindly explain these. I am most grateful for sparing your time.
                        Thank you so much.

                        1 Reply Last reply Reply Quote 0

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