{"id":1490,"date":"2023-09-29T08:31:15","date_gmt":"2023-09-29T08:31:15","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=1490"},"modified":"2023-09-29T08:34:41","modified_gmt":"2023-09-29T08:34:41","slug":"week-5-and-6-jss-3-first-term-mathematics-lesson-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-and-6-jss-3-first-term-mathematics-lesson-notes\/","title":{"rendered":"Week 5 and 6 – Jss 3 First Term Mathematics Lesson Notes"},"content":{"rendered":"

WEEK 5
\n<\/strong>WORD PROBLEMS
\n<\/strong>Sum and Difference<\/p>\n

    \n
  1. \n
    When 30 is added to a number the result is -18. Find the number.\n<\/div>\n<\/li>\n<\/ol>\n

    SOLUTION
    \n<\/strong>The number be
    \n + 30 = -18
    \n = – 48<\/p>\n

      \n
    1. \n
      The difference between a number and 5 is 20 what are the two positive numbers\n<\/div>\n<\/li>\n<\/ol>\n

      SOLUTION
      \n<\/strong>Let the number be
      \n \u2013 (-5) = 20
      \n + 5 = 20
      \n \u2013 15
      \nIf is less than -5
      \n-5 \u2013 = 20
      \n = -5 \u2013 20
      \n = -25<\/p>\n

        \n
      1. \n
        The sum of three consecutive integers is 138. Find the numbers.\n<\/div>\n

        SOLUTION
        \n<\/strong><\/li>\n<\/ol>\n

        Let the numbers be n, n + r, and n + 2
        \nn + (n + 1) + (n + 2) = 138
        \n3n + 3 + 138
        \n3n = 138
        \nn = 45
        \nn + 1 = 46
        \nn + 2 = 47
        \nThe numbers are 45, 46, and 47
        \nExercise: Ex 2.1 No 6, 7, 8<\/p>\n

        \u00a0
        \n\u00a0
        \n\u00a0Product
        \n<\/strong>Examples: (1) Find the product of 1, -0.8 and – 2 (2) what number must be multiplied by 25 to make ? (3) The product of 3 numbers is 3600. If two of the numbers are equal and the third number are 25. Find the two equal numbers.
        \nSOLUTION
        \n<\/strong>1.\u00a0\u00a0\u00a0\u00a0 () x \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2) Let the number be \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(3) 25 x x = 3600
        \n\u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a025 = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0252<\/sup> = 3600
        \n\u00a0\u00a0\u00a0\u00a0= 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 100x = 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02<\/sup> =
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02<\/sup> = 144
        \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 0.03\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 12\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
        \nSolving combine products with sum and difference
        \n<\/strong>Examples: (1) Add the sum of the 20 and 30.5 to the position difference between 25 and 45. (2) From the sum of 8 ad 7 subtract the negative difference between 15 and 24 (3) find the product of 3 and
        \nSOLUTION
        \n<\/strong>1.\u00a0\u00a0\u00a0\u00a0(20 + 30.5) + (45 \u2013 25)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2) -8 + 7 \u2013 115 \u2013 25\u00a0\u00a0\u00a0\u00a0(3) Sum = +
        \n\u00a0\u00a0\u00a0\u00a0= 50.5 + 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = -8 + 7 + 9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
        \n\u00a0\u00a0\u00a0\u00a0= 70.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =
        \nProduct = x
        \n= = 1
        \nExercise:<\/strong> Ex. 2.3 No 1, 2, 3, page 17.
        \nWord problems with fractions
        \n<\/strong>Examples: <\/p>\n

          \n
        1. \n
          Find the three-fifth of the sum of 45 and -60. (45 \u2013 60) = (-15) = -9\n<\/div>\n<\/li>\n
        2. \n
          Divide the difference between 25 and 10 by the product of 6 and 5. = = 0.5\n<\/div>\n<\/li>\n
        3. \n
          Find one-quarter of the sum of the product of 2 and 3 and the product of 1 and\n<\/div>\n<\/li>\n<\/ol>\n

          \u00a0Solution: <\/strong> = = 1
          \nExercise: <\/strong>Ex 2.4 No 1, 2, 3, page 18
          \nProblems leading to equations
          \n<\/strong>Examples<\/p>\n

            \n
          1. \n
            When of a number is added to 30. The result is 20 added to the number find the number.\n<\/div>\n<\/li>\n
          2. \n
            When the sum of 28 and a certain number is divided by 5. The result is equal to treble the original number. What is the number?\n<\/div>\n<\/li>\n<\/ol>\n

            SOLUTION
            \n<\/strong>(1) Let the number be (2) Let the number be
            \nHence + 30 = + 30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 3
            \nMultiply each term by 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a015 = 28 +
            \n3 + 150 = 5 + 100\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =
            \n5 – 3 = 150 -100\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 2
            \n2 = 50
            \n = 25
            \nASSIGNMENT
            \n<\/strong>EXERCISE 2.6; NO 1, 2, AND 3 PAGE 20.
            \n<\/strong>
            \n\u00a0WEEK 6
            \n<\/strong>
            \n\u00a0CHANGE OF SUBJECT OF A FORMULA
            \n<\/strong>Subject of formula: <\/strong>A formula is always written in terms of the subject e.g. V<\/strong> = 2<\/sup>h is a formula for volume of a cone, r is the radius, height (h) =.
            \nExamples 1: Make r the subject of the formula in A = 2<\/sup> (2) Make T the subject of the formula in = K (3) Make x the subject of the formula in y =km<\/sup>\/x
            \nSOLUTION
            \n<\/strong>a)\u00a0\u00a0\u00a0\u00a0A = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup>b) = K \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0c)\u00a0\u00a0\u00a0\u00a0y =
            \n\u00a0\u00a0\u00a0\u00a02<\/sup> =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0PT = KV\u00a0\u00a0\u00a0\u00a0 xy = km
            \n\u00a0\u00a0\u00a0\u00a02 <\/sup>= \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup> x =
            \nExercise<\/p>\n

              \n
            1. \n
              Make b the subject of the formula in A = (a + b) x\n<\/div>\n<\/li>\n
            2. \n
              Make x the subject of the formula in a = b ( 1 \u2013 x)\n<\/div>\n<\/li>\n
            3. \n
              Make w the subject of the formula in L =\n<\/div>\n<\/li>\n<\/ol>\n

              Substitution in formula
              \n<\/strong>Examples: The total surface area of a closed cylinder of base radius rcm and height hcm is given by A = 2(r+h)<\/p>\n

                \n
              1. \n
                Write in terms of A and r\n<\/div>\n<\/li>\n
              2. \n
                Find the height of the cylinder of base radius 7cm and the total surface area of 396cm2<\/sup>. = .\n<\/div>\n<\/li>\n<\/ol>\n

                SOLUTION
                \n<\/strong><\/p>\n

                  \n
                1. \n
                  A = 2(r+h)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0b) h = – 7\n<\/div>\n

                  r + h = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 h = 2m
                  \nh = – r\n<\/li>\n<\/ol>\n

                  Exercise: The mass of water in a rectangular tank 1m long. 6m wide and him deep is Mkg where M = 1000 lbh (a) What is the mass of water in a tank 5m long, 4m wide and 5m deep? (b) What is the depth of the water in a tank 5m long and 3m wide of its mass is 24000kg?
                  \nAssignment
                  \n<\/strong><\/p>\n

                    \n
                  1. \n
                    if T = (a) Express m in terms of T an K (b) Find m when T = 20 and K = 50, =\n<\/div>\n<\/li>\n
                  2. \n
                    make h the subject of the formula V<\/strong> = 2<\/sup>h\n<\/div>\n<\/li>\n
                  3. \n
                    hence find the value of h when V = 256, = , r = 21\n<\/div>\n<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                    WEEK 5 WORD PROBLEMS Sum and Difference When 30 is added to a number the…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,138],"tags":[],"class_list":["post-1490","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-jss-3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1490","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=1490"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1490\/revisions"}],"predecessor-version":[{"id":1491,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1490\/revisions\/1491"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=1490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=1490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=1490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}