Logo
    • Education
      • Pre-School
      • Primary Schools Directory
      • Primary Schools Articles
      • P1 Registration
      • DSA
      • PSLE
      • Secondary
      • Tertiary
      • Special Needs
    • Lifestyle
      • Well-being
    • Activities
      • Events
    • Enrichment & Services
      • Find A Service Provider
      • Enrichment Articles
      • Enrichment Services
      • Tuition Centre/Private Tutor
      • Infant Care/ Childcare / Student Care Centre
      • Kindergarten/Preschool
      • Private Institutions and International Schools
      • Special Needs
      • Indoor & Outdoor Playgrounds
      • Paediatrics
      • Neonatal Care
    • Forum
    • ASKQ
    • Register
    • Login

    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
    9.3k Posts 673 Posters 4.2m Views 1 Watching
    Loading More Posts
    • Oldest to Newest
    • Newest to Oldest
    • Most Votes
    Reply
    • Reply as topic
    Log in to reply
    This topic has been deleted. Only users with topic management privileges can see it.
    • MathIzzzFunM
      MathIzzzFun
      last edited by

      Essential:
      Hi please help my DS for question 14 . John shifted the decimal point of a number twice to the left to obtain a new number . The difference between the new number and the original number was 136.62. A) how many times of the new number is the original number ? B) what is the sum of the 2 numbers ?

      Hi

      decimal point shifted twice to the left = dividing by 100
      so, new number is 1/100 of original number.
      Difference = 99/100 of original number
      original number = 136.62/99 x 100 = 138
      .. with this, part b should be a breeze šŸ˜„

      cheers.

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

        Essential:
        Hi please help my DS for question 14 . John shifted the decimal point of a number twice to the left to obtain a new number . The difference between the new number and the original number was 136.62. A) how many times of the new number is the original number ? B) what is the sum of the 2 numbers ?

        Hi

        Original number ----- 100 units

        New number ------ 1 unit

        100 – 1 ----- 99

        99 units ------- 136.62

        1 unit ----- 1.38

        Original number ----- 100 units ------- 138

        New number ------ 1 unit ------ 1.38

        The original number is 100 times the new number.

        Sum of 2 numbers ------ 101 units ------- 139.38 (138 + 1.38)

        Best wishes

        1 Reply Last reply
        Reply Quote 0
        • W
          Winx5015
          last edited by

          Hi, need help for the following questions.


          (1) If Ahmad and Ben work together, they can complete a job in 8 days. If Ahmad were to work alone for 2 days, he can finish 3/20 of the job in 2 days. How many days will Ben take to complete the job if he were to work alone?

          (2) Steve and Huiling took part in a walkathon which started at 6.30am. Steve's average speed was 15 m/min faster than Huiling. Both of them did not change their speed throughout the walkathon. When Steve reached the finishing line after 40 minutes, Huiling had only walked 8/9 of the distance.
          (a) What time was it when Huiling completed the walkathon?
          (b) Find Steve's average speed for the walkathon in m/min.

          (3) At a concert, tickets were sold at $12, $30 and $50. The sales of $12 tickets made up 25% of the total amount of money collected. The sales of $30 tickets made up 60% of the total amount. given that 5 more $12 tickets were sold than the $30 tickets, how many $50 tickets were sold?

          (4) At a funfair, the children can choose to ride on the Merry-Go-Round of the Train or both with their coupons. 7/10 of the children chose the Merry-Go-round, 45% chose to ride on the Train, and 10% of the children did not take any of the two rides. A total of 75 children went for both rides.
          (a) How many children took ONE ride at the Merry-Go-Round only?
          (b) How many children did not take any rides?

          Thanks:)

          1 Reply Last reply
          Reply Quote 0
          • MathIzzzFunM
            MathIzzzFun
            last edited by

            Winx5015:
            Hi, need help for the following questions.


            (1) If Ahmad and Ben work together, they can complete a job in 8 days. If Ahmad were to work alone for 2 days, he can finish 3/20 of the job in 2 days. How many days will Ben take to complete the job if he were to work alone?

            Thanks:)
            Q1.

            Ahmad & Ben --> 8 days ---> 20 units of work
            Ahmad --> 2 days --> 3 units of work
            Ahmad --> 8 days --> 12 units of work
            Ben --> 8 days --> (20-12) = 8 units of work
            Ben --> 20 units of work --> 20 days
            Ben will take 20 days to complete the work on his own.

            cheers.

            1 Reply Last reply
            Reply Quote 0
            • MathIzzzFunM
              MathIzzzFun
              last edited by

              Winx5015:
              Hi, need help for the following questions.


              (2) Steve and Huiling took part in a walkathon which started at 6.30am. Steve's average speed was 15 m/min faster than Huiling. Both of them did not change their speed throughout the walkathon. When Steve reached the finishing line after 40 minutes, Huiling had only walked 8/9 of the distance.
              (a) What time was it when Huiling completed the walkathon?
              (b) Find Steve's average speed for the walkathon in m/min.

              Thanks:)
              Q2.

              when Steve reached finishing line,
              distance covered by Steve : Huiling --> 9 : 8
              time ratio, Steve : Huiling --> 8:9
              Time taken by Huiling to reach finish line = 9/8 x 40 min = 45 min
              6.30 am + 45 min --> 7.15 am
              Huiling reached the finish line at 7.15 am

              Steve's speed: Huiling's speed = 9 : 8
              Steve's speed = 9 x 15 m/min = 135 m/min

              cheers.

              1 Reply Last reply
              Reply Quote 0
              • MathIzzzFunM
                MathIzzzFun
                last edited by

                Winx5015:
                Hi, need help for the following questions.



                (3) At a concert, tickets were sold at $12, $30 and $50. The sales of $12 tickets made up 25% of the total amount of money collected. The sales of $30 tickets made up 60% of the total amount. given that 5 more $12 tickets were sold than the $30 tickets, how many $50 tickets were sold?

                Thanks:)
                Q3 - this was discussed on pg 697 --> http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=6960

                cheers.

                1 Reply Last reply
                Reply Quote 0
                • MathIzzzFunM
                  MathIzzzFun
                  last edited by

                  Winx5015:
                  Hi, need help for the following questions.


                  (4) At a funfair, the children can choose to ride on the Merry-Go-Round or the Train or both with their coupons. 7/10 of the children chose the Merry-Go-round, 45% chose to ride on the Train, and 10% of the children did not take any of the two rides. A total of 75 children went for both rides.
                  (a) How many children took ONE ride at the Merry-Go-Round only?
                  (b) How many children did not take any rides?

                  Thanks:)
                  Q4
                  None of the rides --> 10%
                  Merry-Go-Round --> 70%
                  Train ONLY --> 100% - 10%- 70% = 20%
                  Train --> 45%
                  Merry-Go-Round & Train --> 45% - 20% = 25%
                  a) Merry-Go-Round ONLY --> 70% - 25% = 45%

                  b) 25% --> 75
                  10% --> 30
                  30 children did not take any rides

                  cheers.

                  1 Reply Last reply
                  Reply Quote 0
                  • W
                    whitety
                    last edited by

                    MathIzzzFun:
                    peggy:

                    Please help :

                    2.an equal number of men and women were at a party. The ratio of the number of woman who wore watches to the number of men who wore watches was 3:11. The ratio of the number of women to the number of men who did not wear watches was 5:3.

                    A. Find the ratio of the number of women who wore watches to the number of women who did not wear watches.

                    Many thanks.

                    Q2.
                    a)
                    wearing watches, women : men = 3 : 11 (difference of šŸ˜Ž
                    not wearing watches, women : men = 5:3 (difference of 2)
                    --> difference must be the same since equal number of men and women

                    wearing watches, women : men = 3u : 11u
                    not wearing watches, women : men = 20u : 12u
                    women wearing watches : women not wearing watches = 3:20



                    Hi MathIzzzFun,

                    wearing watches, women : men = 3u : 11u need to x4? = 12u : 44u
                    not wearing watches, women : men = 20u : 12u
                    women wearing watches : women not wearing watches = 3:20
                    -> then answer will be 3:5, is my understanding wrong ?

                    Thanks!

                    1 Reply Last reply
                    Reply Quote 0
                    • MathIzzzFunM
                      MathIzzzFun
                      last edited by

                      whitety:
                      MathIzzzFun:

                      [quote=\"peggy\"]Please help :

                      2.an equal number of men and women were at a party. The ratio of the number of woman who wore watches to the number of men who wore watches was 3:11. The ratio of the number of women to the number of men who did not wear watches was 5:3.

                      A. Find the ratio of the number of women who wore watches to the number of women who did not wear watches.

                      Many thanks.

                      Q2.
                      a)
                      wearing watches, women : men = 3 : 11 (difference of šŸ˜Ž
                      not wearing watches, women : men = 5:3 (difference of 2)
                      --> difference must be the same since equal number of men and women

                      wearing watches, women : men = 3u : 11u
                      not wearing watches, women : men = 20u : 12u
                      women wearing watches : women not wearing watches = 3:20



                      Hi MathIzzzFun,

                      wearing watches, women : men = 3u : 11u need to x4? = 12u : 44u
                      not wearing watches, women : men = 20u : 12u
                      women wearing watches : women not wearing watches = 3:20
                      -> then answer will be 3:5, is my understanding wrong ?

                      Thanks![/quote]hope this MD helps...

                      http://i48.tinypic.com/2qledjr.png\">

                      cheers.

                      1 Reply Last reply
                      Reply Quote 0
                      • W
                        whitety
                        last edited by

                        Hi MathIzzzFun,


                        Thank you very much!

                        1 Reply Last reply
                        Reply Quote 0

                        Hello! It looks like you're interested in this conversation, but you don't have an account yet.

                        Getting fed up of having to scroll through the same posts each visit? When you register for an account, you'll always come back to exactly where you were before, and choose to be notified of new replies (either via email, or push notification). You'll also be able to save bookmarks and upvote posts to show your appreciation to other community members.

                        With your input, this post could be even better šŸ’—

                        Register Login
                        • 1
                        • 2
                        • 699
                        • 700
                        • 701
                        • 702
                        • 703
                        • 931
                        • 932
                        • 701 / 932
                        • First post
                          Last post



                        Online Users
                        thebottomsupblogT
                        thebottomsupblog

                        Statistics

                        8

                        Online

                        211.3k

                        Users

                        34.6k

                        Topics

                        1.8m

                        Posts
                        Apps Shared on Our Forum

                        Maths

                        Ruiya.ai
                        math support
                        →
                        Tallytide
                        P1-P6 maths app
                        →
                        MugCorner
                        Sec 1-4 maths practice
                        →

                        English

                        tutor.agenticolab.com
                        P1-6 English portal
                        →
                        wordwizards.sg
                        English practice
                        →

                        Chinese

                        chinese.getbuzzi.com
                        P1-P6 Chinese
                        →
                        Wonder Chinese
                        P1-P6 Chinese writing
                        →
                        Xiaozhu Free trial
                        Mandarin chat buddy
                        →

                        Multiple subjects

                        Pickytic.ai
                        AI study buddy for pri and sec
                        →

                          About Us Contact Us global:forum Terms of Service Privacy Policy