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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • P Offline
      pensiveowl
      last edited by

      Nice! - thanks!

      wanderlust-alm:
      After resignation of members , % of man is 10% ( of the original number ) as 1/4 of 40% is 10%.
      After reignation of members, % of woman is 48% ( of original number ) as 80% x 60% ( original % ) is 48%.
      hence members who quit is 100-10-48 = 42% ==> 210 members.

      Number of woman in the club at first = 60% x 210 / 42 = 300 women
      Number of men resigned from club = 30% x 210 / 42 = 150 men.

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      • R Offline
        raghupavithra
        last edited by

        round off 378.71 to the nearest whole number


        1. 376
        2. 378
        3. 379
        4. 380

        pls help

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        • C Offline
          CloudeeDaz
          last edited by

          Hi

          need help withthis question

          Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal no of beads in each small box and equal no of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box?

          thanks

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          • MathIzzzFunM Offline
            MathIzzzFun
            last edited by

            CloudeeDaz:
            Hi

            need help withthis question

            Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal no of beads in each small box and equal no of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box?

            thanks
            Hi

            this was discussed in an earlier post..

            http://www.flickr.com/photos/62167097@N02/5664739745/in/photostream

            cheers.

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            • C Offline
              CloudeeDaz
              last edited by

              thanks for the link!

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              • W Offline
                wanderlust-alm
                last edited by

                CloudeeDaz:
                Hi

                need help withthis question

                Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal no of beads in each small box and equal no of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box?

                thanks
                Each big box has 5 more beads ==> extra beads over the small box ==> 75 beads.

                14 small box + 15 small box + 75 beads = total amount of beads ( Y beads )

                each small box contain ( 3/8 y ) / 14 beads ==> 3/112 y
                Beads in big box = 3/8 y + 3/112 y + 75 beads

                hence total number of beads = 3/8 y + ( 3/8 y + 3/112 y + 75 )

                Work out to common denominator

                42/112 + 42/112 + 3/112 + 75 beads = total number of beads

                87/112 + 75 beads = total number of beads

                ==> 25 /112 = 75 beads
                hence 3 / 8 = 126 beads

                hence each small box has 126/14 = 9 beads

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                • W Offline
                  wanderlust-alm
                  last edited by

                  wanderlust-alm:
                  CloudeeDaz:

                  Hi

                  need help withthis question

                  Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal no of beads in each small box and equal no of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box?

                  thanks

                  Each big box has 5 more beads ==> extra beads over the small box ==> 75 beads.

                  14 small box + 15 small box + 75 beads = total amount of beads ( Y beads )

                  each small box contain ( 3/8 y ) / 14 beads ==> 3/112 y
                  Beads in big box = 3/8 y + 3/112 y + 75 beads

                  hence total number of beads = 3/8 y + ( 3/8 y + 3/112 y + 75 )

                  Work out to common denominator

                  42/112 + 42/112 + 3/112 + 75 beads = total number of beads

                  87/112 + 75 beads = total number of beads

                  ==> 25 /112 = 75 beads
                  hence 3 / 8 = 126 beads

                  hence each small box has 126/14 = 9 beads

                  Thanks for link too. The illustration by units is much clearer šŸ˜„ . My method is more like the old style šŸ˜‚ although eventually answer still the same but definately easier to view through units . Thanks Too !!

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                  • P Offline
                    pixiedust
                    last edited by

                    Need some help, thanks in advance :


                    A box contained 50-cent and 20-cent coins in the ratio 2:3.
                    Some 50-cent coins were taken out and exchanged for 20-cent coins.
                    The ratio of the number of 50-cent coins to the number of 20-cent coins became 2:7.
                    Find (a) the total value of money in the box (b) the number of 50-cent coins which were taken to exchange for 20-cent coins.

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                    • A Offline
                      acehkr3009
                      last edited by

                      Let the number of coins taken out (50 cents) and exchange (20 cents) be U numbers.


                      Before exchange,

                      50 cents : 20 cents
                      2 : 3
                      After the exchange of U numbers of 50 cents with 20 cents,

                      50 cents = 2 - U
                      20 cents = 3 + U

                      New ratio becomes,
                      2 : 7

                      So 7 times number of 50 cents = 2 times number of 20 cents

                      2 X (3 + U) = 7 X (2 - U)
                      6 + 2U = 14 - 7U
                      9U = 8
                      U = 8/9 (fraction)

                      Now we know each of ratio unit of 50 cents & 20 cents is subdivided into 9 smaller units. Of which, 8 units (or 8 numbers of coins) of 50 cents coins is used to exchange for 8 units of 20 cents coins.

                      Before the exchange,
                      Based on the ratio (50 cents: 20 cents) 2:3,
                      There will be 2 x 9 = 18 number of 50 cents coins
                      Value of 50 cents coins = 18 x 50 cents = $9
                      There will be 3 x 9 = 27 number of 20 cents coins
                      Value of 20 cents coins = 27 x 20 cents = $5.40
                      Total value at first will be $14.40

                      After exchanging 8 number of 50 cents with 20 cents,
                      There will be 18 - 8 = 10 number of 50 cents &
                      10 number of 50 cents coins: 10 x 50 = $5
                      There will be 27 + 8 = 35 number of 20 cents.
                      35 number of 20 cents coins: 35 x 20 = $7
                      Total value after the exchange = $12.

                      Hope the above lenghtly solution is correct & acceptable.
                      Pls advise if there is simpler solutions.
                      Thanks in advance.

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                      • T Offline
                        tianzhu
                        last edited by

                        pixiedust:
                        Need some help, thanks in advance :


                        A box contained 50-cent and 20-cent coins in the ratio 2:3.
                        Some 50-cent coins were taken out and exchanged for 20-cent coins.
                        The ratio of the number of 50-cent coins to the number of 20-cent coins became 2:7.
                        Find (a) the total value of money in the box (b) the number of 50-cent coins which were taken to exchange for 20-cent coins.
                        Hi

                        You may use a heuristics called \"Systematic Listing\".

                        We assume that some 50 cents were taken out and exchanged by a number of 20 cents of the same value so that the total sum of money in the box stays the same.

                        Due to time constraints, I shall not present the answer in slide but give some pointers for you to work it out.

                        1)Make a table of equivalent ratios of the number of 50 cents coins and 20 cents coins before and after the coins are replaced.

                        2)Calculate the total value of coins in the box before and after the replacement.

                        Before: (2*50)+(3*20) ----- 160
                        After: (2*50)+(7*20) ----- 240

                        The total value of coins remains the same before and after the replacement of coins.

                        From the table, the total sum of money ------ 480 cents.

                        Before:(6*50)+(9*20) ----- 480
                        After:(4*50)+(14*20) ----- 480

                        The number of 50 cents being exchanged ------ 2

                        Best wishes

                        PS ----- In this question, we are calculating the least possible amount of money in the box.Usually, for such question, the range of money in the box is usually specified, otherwise, there'll be multiple answers.

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