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    Q&A - PSLE Math

    Scheduled Pinned Locked Moved Primary 6 & PSLE
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    • MathIzzzFunM
      MathIzzzFun
      last edited by

      Beikiasu:
      Hi, can someone help me to solve this problem sum?


      3) A contractor needed to cover an entire hall with tiles. On the first day, he laid 319 tiles. He completed tiling the rest of the hall in 7 days using an equal number of tiles each day. At the end of the 4th day, he was able to tile 8/15 of the hall. How many tiles did the contractor use to cover the entire hall? CHIJ, 2011, Question 15, 4 mark, Ans:1740 tiles.

      Thanks
      using equal fractions/numerators method:

      after laying 319 tiles on 1st day, remaining tiles laid in 7 days

      4/7 remaining tiles = 7/15 total number of tiles

      28/49 remaining tiles = 28/60 total number of tiles

      total number of tiles --> 60 units
      remaining tiles --> 49 units
      11 units --> 319 tiles
      total number of tiles --> 60/11 x 319 = 1740

      cheers.

      1 Reply Last reply
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      • B
        bookwormkids
        last edited by

        Need help on the following questions.


        1)
        Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.

        2)
        Catherine has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in another 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?

        Thanks.

        1 Reply Last reply
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        • T
          tianzhu
          last edited by

          nnchia:
          Hi,


          Need help to solve the following question:-

          A coin box contained only twenty-cent and fifty-cent coins in the ratio of 4:5. When 16 fifty-cent coins were taken out and replaced by some twenty-cent coins, the number of fifty-cent coins left in the box was 7/8 of the twenty-cent coins. The total value of all the coins remained the same. Find the sum of money in the coin box.
          Ai Tong 2011 CA1 Answer - $112.20 (Is this right?)

          Thanks in advance for the help!
          Hi

          One possible approach is to work in term of the number of coins. You may use Units and Parts.

          At first

          twenty-cent coins: fifty-cent coins ----- 4u:5u

          16 number of fifty-cent coins ----- $8 -------- 40 number of twenty-cent coins

          Change

          twenty-cent coins: fifty-cent coins ----- 40:-16

          In the end

          twenty-cent coins: fifty-cent coins ----- 8p:7p

          Make the parts the same

          4u + 40 ------ 8p
          (x7)
          28u + 280 ------ 56p


          5u – 16 ------ 7p
          (x8)
          40u – 128 ------ 56p


          28u + 280 ------- 40u – 128
          12u ------- 408
          1u ------ 34

          Number of twenty-cent coins ------- 4*34 ------136
          Value of twenty-cent coins ------- 27.2

          Number of fifty-cent coins ------- 5*34 ------170
          Value of fifty-cent coins ------- 85

          Total value ------- 85 + 27.2 ------ 112.20

          Best wishes

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          • N
            nnchia
            last edited by

            Hi Tianzhu, thanks a million!

            1 Reply Last reply
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            • T
              tianzhu
              last edited by

              bookwormkids:
              Need help on the following questions.


              1)
              Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.
              Hi

              Please confirm from from your source.

              There should be a difference in length.

              Best wishes

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              • MathIzzzFunM
                MathIzzzFun
                last edited by

                bookwormkids:
                Need help on the following questions.


                1)
                Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.



                Thanks.
                Original question discussed at pg 801


                http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=844766#p844766


                cheers.

                1 Reply Last reply
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                • T
                  tianzhu
                  last edited by

                  bookwormkids:

                  2)
                  Catherine has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in another 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?
                  Hi

                  Scenario 1

                  When she adds in 15 white counters, 65% of the counters in the box are black.

                  White:Black ------ 7:13

                  Scenario 2

                  If she adds in another 40 black counters, 75% of the counters in the box are black.

                  Make the units for white counters for both scenarios the same.

                  White:Black ------ 1:3 -----7:21

                  21 – 13 ----- 8

                  8 units ---- 40

                  1 unit ----- 5

                  7 units ----- 35

                  35 – 15 ----- 20 (white counters@first)

                  Best wishes

                  1 Reply Last reply
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                  • B
                    bookwormkids
                    last edited by

                    tianzhu:
                    bookwormkids:


                    2)
                    Catherine has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in another 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?

                    Hi

                    Scenario 1

                    When she adds in 15 white counters, 65% of the counters in the box are black.

                    White:Black ------ 7:13

                    Scenario 2

                    If she adds in another 40 black counters, 75% of the counters in the box are black.

                    Make the units for white counters for both scenarios the same.

                    White:Black ------ 1:3 -----7:21

                    21 – 13 ----- 8

                    8 units ---- 40

                    1 unit ----- 5

                    7 units ----- 35

                    35 – 15 ----- 20 (white counters@first)

                    Best wishes

                    Hi Tianzhu, my son would like to know:

                    When 40 black counters are added, the number of white counters remain unchanged. So when 15 white counters are added,did the number of black counters remain the same?

                    Thanks for your time.

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                    • B
                      bookwormkids
                      last edited by

                      MathIzzzFun:
                      bookwormkids:

                      Need help on the following questions.


                      1)
                      Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.


                      Thanks.

                      Original question discussed at pg 801


                      http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=844766#p844766


                      cheers.

                      Thank you MathIzzzFun and Tianzhu fpr your help. šŸ˜„

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                      • B
                        Beikiasu
                        last edited by

                        Hi, can someone help me to solve these problem sum?


                        1) A rectangular tank A, 180 cm by 60 cm by 80 cm, is 5/8 filled with water. When some water from tank A is pumped into another rectangular tank B, tank A is left with 15% of its water.
                        (a)How many litres of water is pumped into tank B? Ans:459 litres
                        (b)Given that tank B has a square base of sides 50 cm, what is the height of water in tank B? Ans:183.6 cm.SCG, 2011,SA1, Question 13, 4 mark.

                        2) Tina paid a total of $336 for 2 similar dresses and a skirt after a discount of 20% was given on the price of each item. If the original of each dress was 25% more than that of the original price the skirt, find the price of each dress before the discount. SCG, 2011, SA1, Question 12, 4 mark, Ans: $150.

                        3) Bemie and Ahmad shared the cost of a meal.Bemie paid $15. If Bermie paid $2 less, Ahmad would have had to pay 20% more. What was the cost of the meal? MGS, 2011, SA1, Question 3, 2 mark, Ans: $25.

                        Thanks

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