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    Q&A - P5 Math

    Scheduled Pinned Locked Moved Primary 5
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    • J Offline
      Jamesbond
      last edited by

      PapayaDad:
      Assume 12B and 12G

      So Boys sold 60 tickets ; girls sold 36 tickets
      So boys get $300 ; girls get $180
      Boys get 120 more ; we need to reduce this to 40 only ie need to reduce 80

      If we reduce 1 boy ; we reduce five tickets ; which is reduce $25
      But there will be one more girl ; ie 3 more tickets ; which is $15 more.
      So the difference is reduce by $40

      So we need to reduce 2 boys from our initial assumption.


      Answer 10 boy 14 girls

      Check
      10B = 50 tickets = $250
      14G = 42 tickets = $210
      Correct, boys collected $40 more.
      There are 14 girls.
      Total tickets sold is 92
      :thankyou: PapayaDad.

      1 Reply Last reply Reply Quote 0
      • J Offline
        Jamesbond
        last edited by

        A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

        Pl help.

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

          Jamesbond:
          A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

          Pl help.
          possible approaches - cross-multiply/UP, units, MD.

          Using units:

          10 x $0.50 = 25 x $0.20
          --> 10 fifty-cent coins were replaced by 25 twenty-cent coins

          At first
          number of 50c --> 20 units + 10
          number of 20c (= 3/4 number of 50c) --> 15 units + 7.5

          In the end,
          number of 50c --> 20 units
          number of 20c (=7/5 number of 50c) --> 28 units

          since 25 twenty-cent coins were added, we have:

          15 units + 7.5 + 25 = 28 units
          1 unit --> 2.5

          Sum of money in the box at first
          = sum of money in the box in the end
          = 20 x 2.5 x $0.50 + 28 x 2.5 x $0.20
          = $39

          cheers.

          1 Reply Last reply Reply Quote 0
          • S Offline
            snowball
            last edited by

            4/7 of Ming’s mass is equal to 2/5 of Rob’s mass.

            If Ming puts on 6 kg and Rob loses 12kg, the two men will have the same mass.
            a) Draw a clearly labelled model to solve the problem.
            b) What is Rob’s mass?

            1 Reply Last reply Reply Quote 0
            • S Offline
              snowball
              last edited by

              In January, Mrs Wong spent 2/5 of her salary on rent and $120 on parking fees.

              She spent 1/4 of her remaining salary on food and $480 on shopping.
              She saved the remaining $1896.
              How much was Mrs Wong’s salary for January ?

              1 Reply Last reply Reply Quote 0
              • M Offline
                MathsOlympiadtrainer
                last edited by

                snowball:
                In January, Mrs Wong spent 2/5 of her salary on rent and $120 on parking fees.

                She spent 1/4 of her remaining salary on food and $480 on shopping.
                She saved the remaining $1896.
                How much was Mrs Wong's salary for January ?
                This is a working backwards qn.
                3/4 of her remaining salary-> 480+1896= 2376
                4/4 of her remaining salary-> 2376/3 * 4= 3168

                3/5 of her salary-> 3168+120 = 3288
                5/5 of her salary-> 3288/3 * 5 = 5480

                Hope that helps.

                1 Reply Last reply Reply Quote 0
                • V Offline
                  veraclari
                  last edited by

                  MathIzzzFun:
                  Jamesbond:

                  A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

                  Pl help.

                  possible approaches - cross-multiply/UP, lunits, MD.

                  Using units:

                  10 x $0.50 = 25 x $0.20
                  --> 10 fifty-cent coins were replaced by 25 twenty-cent coins

                  At first
                  number of 50c --> 20 units + 10
                  number of 20c (= 3/4 number of 50c) --> 15 units + 7.5

                  In the end,
                  number of 50c --> 20 units
                  number of 20c (=7/5 number of 50c) --> 28 units

                  since 25 twenty-cent coins were added, we have:

                  15 units + 7.5 + 25 = 28 units
                  1 unit --> 2.5

                  Sum of money in the box at first
                  = sum of money in the box in the end
                  = 20 x 2.5 x $0.50 + 28 x 2.5 x $0.20
                  = $39

                  cheers.

                  Hi Mathizzfun, can I ask.. how to get the 20units in the first place? 😉
                  And why cross multiply? TIA!!!

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

                    veraclari:
                    MathIzzzFun:

                    [quote=\"Jamesbond\"]A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

                    Pl help.

                    possible approaches - cross-multiply/UP, lunits, MD.

                    Using units:

                    10 x $0.50 = 25 x $0.20
                    --> 10 fifty-cent coins were replaced by 25 twenty-cent coins

                    At first
                    number of 50c --> 20 units + 10
                    number of 20c (= 3/4 number of 50c) --> 15 units + 7.5

                    In the end,
                    number of 50c --> 20 units
                    number of 20c (=7/5 number of 50c) --> 28 units

                    since 25 twenty-cent coins were added, we have:

                    15 units + 7.5 + 25 = 28 units
                    1 unit --> 2.5

                    Sum of money in the box at first
                    = sum of money in the box in the end
                    = 20 x 2.5 x $0.50 + 28 x 2.5 x $0.20
                    = $39

                    cheers.

                    Hi Mathizzfun, can I ask.. how to get the 20units in the first place? 😉
                    And why cross multiply? TIA!!![/quote]
                    At first, number of 20c = 3/4 number of 50c,
                    In the end, number of 20c = = 7/5 number of 50c

                    4 x 5 = 20

                    cross multiply method is another way that can be used get the solution.

                    here are some questions that were solved using the cross multiply method:

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=951428&hilit=cross+multiply#p951428

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=960600&hilit=cross+multiply#p960600

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=961764&hilit=cross+multiply#p961764

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=915223&hilit=cross+multiply#p915223

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=974082&hilit=cross+multiply#p974082

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=974032&hilit=cross+multiply#p974032

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=878744&hilit=cross+multiply#p878744

                    http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=866815&hilit=cross+multiply#p866815

                    cheers.

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

                      snowball:
                      4/7 of Ming's mass is equal to 2/5 of Rob's mass.

                      If Ming puts on 6 kg and Rob loses 12kg, the two men will have the same mass.
                      a) Draw a clearly labelled model to solve the problem.
                      b) What is Rob's mass?

                      http://i45.tinypic.com/2nret7p.png\">


                      cheers.

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

                        there were as many men as women in a bus. 2/7 of the men and 1/3 of the women alighted the bus. what fraction of the people did not alight the bus?



                        two ribbon X and Y were cut into 5 equal pieces. each piece cut from ribbon y was 3/5m longer than each piece of ribbon x. if ribbon x was 4m long, what is the lenght of ribbon y?

                        what i did was:

                        3/5m x 100 = 60cm (each piece of ribbon y is 60cm longer than ribbon x)
                        ribbon y = 5u + (60x5)
                        = 5u + 300cm
                        5u = 4m
                        so ribbon y is = 4m + 300cm = 7m

                        correct?

                        1 Reply Last reply Reply Quote 0

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