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    Q&A - P5 Math

    Scheduled Pinned Locked Moved Primary 5
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    • M
      mummyv
      last edited by

      Hello


      It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

      1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
      a. How much did Ms Tan spend on teddy bears (Ans$630)
      b. How much did one teddy bear cost (Ans $18)

      2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

      Thanks in advance all the maths gurus πŸ˜†

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      • MathIzzzFunM
        MathIzzzFun
        last edited by

        mummyv:
        Hello


        It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

        1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
        a. How much did Ms Tan spend on teddy bears (Ans$630)
        b. How much did one teddy bear cost (Ans $18)

        2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

        Thanks in advance all the maths gurus πŸ˜†
        Hi
        Q1.
        Amount spent on teddy bears --> $(1750-490)/2 = $630
        Amount spent on toy guns --> $1750 - $630 = $1120
        Ms Tan bought 4 times as many toy guns as teddy bears
        --> 4 groups of toy guns, 1 group of teddy bear
        1 group of toy guns --> $ 1120/4 = $280
        $630-$280= $350 --> 1 group of teddy bears cost $350 more than 1 group of toy guns
        Number of teddy bears/toy guns in 1 group --> $350/$10 = 35
        Cost of 1 teddy bear = $630/35= $18

        Q2.
        You may want to draw a rectangle with 9 beads on each length and 3 beads on each breadth (total 20 beads), when the number of beads on 1 length and 1 breadth are added, we add 2 more beads to 1/2 the total number ie 9+3=12 = 10 + 2.

        Total number of chairs --> 12 x 9 = 108
        Length + breadth --> 4 units --> (108/2) + 2
        1 unit --> 56 / 4 = 14
        Length --> 3 units --> 3 x 14 = 42 chairs.

        cheers.

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        • Z
          Zack7
          last edited by

          mummyv:
          Hello


          It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

          1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
          a. How much did Ms Tan spend on teddy bears (Ans$630)
          b. How much did one teddy bear cost (Ans $18)

          2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

          Thanks in advance all the maths gurus πŸ˜†

          let x be number of teddy bear, no of guns = 4x
          let y be cost of each teddy bear, cost of gun = y-10

          so

          4x(y-10) + xy = 1750 --- eq1
          4x(y-10) - xy = 490 --- eq2

          eq1 - eq2

          2xy = 1260
          xy = 630 (cost of bears)

          cost of guns = 1750 - 630 = 1120 = 4x(y-10)

          so sub x = 630/y into the above

          so

          1120 = 4(630/y) (y-10)
          y = 18 bucks... thats pretty expensive.



          question 2

          this is a perimeter question

          let breathe be x
          length = 3x

          perimeter = 8x

          9 x 12 = 108

          but notice that based on the way we formalise the answer, i.e we let breathe be x, length be 3x,

          if the perimeter is 8x, then we would have double counted the 4 corners. i.e, our 4 corner chairs can act as both the length and the breadth.

          so the total chairs is 9 x 12 + 4 = 112.

          so 8x = 112, x = 14, length = 3x = 42.



          * i have to say these questions are quite tricky, harder than the questions i see at the p6 section

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          • MathIzzzFunM
            MathIzzzFun
            last edited by

            imacsg:
            Can someone help?

            Mark's current age is 1/4 of Ronald's. In 24 years' time, Ronald's age will be 14 years less than twice Mark's age. What is Ronald's age now?

            Thanks,
            Hi
            Now:
            Mark's age --> 1 unit
            Ronald's age --> 4 units

            24 years' time:
            Ronald's age --> 4 units + 24
            Mark's age --> 1 unit + 24
            4 units + 24 = 2 units + 48 - 14
            1 unit --> 5

            Ronald's age now --> 20

            cheers.

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            • M
              mathnoobs
              last edited by

              Zack7:


              * i have to say these questions are quite tricky, harder than the questions i see at the p6 section
              yes, pity our poor children.... so little time for play now....it's just mug, mug, mug...

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              • T
                tianzhu
                last edited by

                mummyv:

                2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)
                Hi

                Good Morning.

                Perhaps, for questions of such genre, it’s useful to have a simple pictorial representation. It might help to organise your thoughts.

                Hope this helps.

                Best wishes

                http://farm6.staticflickr.com/5272/6897367566_8d8cd99d93_z.jpg\">

                1 Reply Last reply
                Reply Quote 0
                • M
                  mummyv
                  last edited by

                  MathIzzzFun:
                  mummyv:

                  Hello


                  It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

                  1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
                  a. How much did Ms Tan spend on teddy bears (Ans$630)
                  b. How much did one teddy bear cost (Ans $18)

                  2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

                  Thanks in advance all the maths gurus πŸ˜†

                  Hi
                  Q1.
                  Amount spent on teddy bears --> $(1750-490)/2 = $630
                  Amount spent on toy guns --> $1750 - $630 = $1120
                  Ms Tan bought 4 times as many toy guns as teddy bears
                  --> 4 groups of toy guns, 1 group of teddy bear
                  1 group of toy guns --> $ 1120/4 = $280
                  $630-$280= $350 --> 1 group of teddy bears cost $350 more than 1 group of toy guns
                  Number of teddy bears/toy guns in 1 group --> $350/$10 = 35
                  Cost of 1 teddy bear = $630/35= $18

                  Q2.
                  You may want to draw a rectangle with 9 beads on each length and 3 beads on each breadth (total 20 beads), when the number of beads on 1 length and 1 breadth are added, we add 2 more beads to 1/2 the total number ie 9+3=12 = 10 + 2.

                  Total number of chairs --> 12 x 9 = 108
                  Length + breadth --> 4 units --> (108/2) + 2
                  1 unit --> 56 / 4 = 14
                  Length --> 3 units --> 3 x 14 = 42 chairs.

                  cheers.

                  Thanks MathIzzzFun :salute: you are really good at maths

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                  • M
                    mummyv
                    last edited by

                    Zack7:
                    mummyv:

                    Hello


                    It's time from SA2 prep & Would appreciate some help solving the following questions 1b and 2

                    1. Ms Tan bought 4 times as many toy guns as teddy bears. She spent $1750 altogether. A toy gun cost $10 less than a teddy bear. Total cost of toy guns was $490 more than total cost of teddy bears.
                    a. How much did Ms Tan spend on teddy bears (Ans$630)
                    b. How much did one teddy bear cost (Ans $18)

                    2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

                    Thanks in advance all the maths gurus πŸ˜†


                    let x be number of teddy bear, no of guns = 4x
                    let y be cost of each teddy bear, cost of gun = y-10

                    so

                    4x(y-10) + xy = 1750 --- eq1
                    4x(y-10) - xy = 490 --- eq2

                    eq1 - eq2

                    2xy = 1260
                    xy = 630 (cost of bears)

                    cost of guns = 1750 - 630 = 1120 = 4x(y-10)

                    so sub x = 630/y into the above

                    so

                    1120 = 4(630/y) (y-10)
                    y = 18 bucks... thats pretty expensive.



                    question 2

                    this is a perimeter question

                    let breathe be x
                    length = 3x

                    perimeter = 8x

                    9 x 12 = 108

                    but notice that based on the way we formalise the answer, i.e we let breathe be x, length be 3x,

                    if the perimeter is 8x, then we would have double counted the 4 corners. i.e, our 4 corner chairs can act as both the length and the breadth.

                    so the total chairs is 9 x 12 + 4 = 112.

                    so 8x = 112, x = 14, length = 3x = 42.



                    * i have to say these questions are quite tricky, harder than the questions i see at the p6 section

                    Thanks Zack7. I myself got a shock at the questions. Every year I have to work harder to solve questions. Not easy being a kid and not easy trying to be a responsible, supportive mother. πŸ˜†

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                    • M
                      mummyv
                      last edited by

                      tianzhu:
                      mummyv:


                      2. There are 9 stack of chairs. There are 12 chairs in each stack. Mohan places each chair equally spaced apart to for a perimeter of a rectangle. If the number of chairs places along the length is thrice the number of chairs placed along the breadth, how many chairs are placed along one of the lenghts of the rectangle. (Ans 42)

                      Hi

                      Good Morning.

                      Perhaps, for questions of such genre, it’s useful to have a simple pictorial representation. It might help to organise your thoughts.

                      Hope this helps.

                      Best wishes

                      http://farm6.staticflickr.com/5272/6897367566_8d8cd99d93_z.jpg\">

                      Thanks tianzhu. I hope I can reach your level sometime this year πŸ˜“

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                      • Z
                        Zack7
                        last edited by

                        mathnoobs:
                        Zack7:



                        * i have to say these questions are quite tricky, harder than the questions i see at the p6 section

                        yes, pity our poor children.... so little time for play now....it's just mug, mug, mug...

                        problems of a country that strives too hard for excellence

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