O-Level Elementary Mathematics Revision Exercise Solution

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					O-Level Elementary Mathematics Revision Exercise 1 Solution
  1. The temperature in the house is 20°C. The temperature outside the house is
     –17°C. Find the difference between the two temperatures.

     20°C – (–17°C) = 37°C

  2. Find the HCF of 112 and 280.




      2  2  2  7  56

  3. The temperature inside a greenhouse is m°C, and outside it is –n°C, where m
     and n are positive integers.
     Write down an expression for
     (a) the difference between the two temperatures,
     (b) the mean of the two temperatures.

     (a) (m – [-n])°C = (m + n)°C

            m  - n       m n
     (b)               C       C
               2             2

  4. Express 2.45 hours in minutes.

           45
      2        60  147 mins
          100

  5. Rearrange the following numbers in an ascending order.
                     34
       45   6.65            6.78
                      5
                            34
     6.65    45     6.78
                             5

  6. A train take 4 hours 30 minutes to reach Town B from Town A.
     (a) One train left Town A at 0840. When did it arrive at Town B?
     (b) On the return journey it reached Town A at 2010. When did it leave
         Town B?
     (c) The distance between Town A and Town B is 432 kilometres.
         Calculate the average speed of the train.

     (a) 0840 + 4h 30 mins = 1310

     (b) 2010 – 4h 30 mins = 1540
   (c) 4h 30 minutes = 4.5 h
               Distance
       Speed =
                 Time
               432
             =
                4.5
             = 96 km/h

              27
7. Express       as a decimal.
              60
    27 9
      = = 0.45
    60 20

8. (a) Express, correct to two significant figures,
       (i) 452.923.
       (ii) 0.01073
   (b) Hence, estimate, correct to one significant figure, the value of
                             452.923 0.01073.
   (a) 450; 0.011
   (b) 452.923 0.01073  450  0.011
                            = 4.95

9. Evaluate the following.
             2
       1
   (a)  
       5
   (b) 2 2  4 2  8
                         2
                     
              1        3
   (c) 15 0   
              8

             2
       1
   (a)           5
                             2

       5
                  = 25

   (b) 2 2  4 2  8 = 2 2  2 4  2 3
                     = 2 2  4 3
                     = 23
                     =8

                         2
                     
              1
                             = 1 8 3
                         3            2
   (c) 15 0   
              8
                                   
                             = 1 3 8
                                          2


                             =4
10. Expand the following algebraic expressions.
    (a) 2a  b 
                 2


                 
   (b) x 2  2 y 3  y 

   (a) 2a  b  = 2a   22a b   b 2
                 2        2


                   = 4 a 2  4 ab b 2
   (b) x 2  2 y 3  y   3 x 2  x 2 y  6 y  2 y 2

11. Factorise completely
    (a) 9a – 12a²
    (b) x 2  2 x  15
    (c) 5ac + 10ad – 2bc – 4bd

   (a) 9a – 12a² = 3a(3 – 4a)
   (b) x 2  2 x  15




       (x – 3)(x + 5)
   (c) 5ac + 10ad – 2bc – 4bd
       = 5a(c + 2d) – 2b(c + 2d)
       = (5a – 2b)(c + 2d)
                                                              4    3
12. Express as a single fraction in its simplest form                .
                                                            y 3 y 5
      4    3    4 y  5  3 y  3
             =
    y 3 y 5       y  3 y  5
                4 y  20  3 y  9
              =
                   y  3 y  5
                      y  29
              =
                 y  3 y  5
            5a 2 25 a 2b
13. Simplify            .
             8b    12
    5a 2 25 a 2b 5a 2      12
               =     
    8b    12      8b 25 a 2b

                      =

                         1 3
                      =    
                        2b 5b
                          3
                      =
                        10b 2
                       0.4325
14. Evaluate 3               , giving your answer correct to 3 significant figures.
             16 .23  2.013
   Using a calculator, the answer is 0.0952.

15. Solve the equation 3x² + 5x – 7 = 0, giving your answer correct to 3 significant
    figures.

   3x² + 5x – 7 = 0
        5   52  43 7
    
                23
       5  109       5  109
               or 
           6              6
     0.907     or  –2.57

16. There are a total of 2.37  108 living in a country. 9.3  107 of them were under
    18 years old. Find the number of people who were 18 or over. Give your
    answer in standard form.

    2.37  10   9.3  10 
               8                7


     2.37  10   0.93  10 
                   8                8


     2.37  0.93   10 8
     1.44 108

17. Solve the following exponential equations.
    (a) 3x  1
    (b) 4a  22  32
    (c) x 125  25 2

   (a) 3x  1
       3 x  30
      x=0
   (b) 4a  22  32
        22a  22  25
       2a – 2 = 5
            2a = 7
                 7
             a
                 2
   (c) x 125  25 2
        5 
             1
           3 x
                  54

        5 
             1
           3 x
                  54
          3
        5  54
          x

        3
          4
        x
           3
        x
           4

18. Simplify 2n 102n1  25 . Leave your answer in index form.

    2n 102n1  25
     2 n  2 2n1  5 2n1  5 2
     2n 2n-1  52 n12
     23 n -1  52 n 3

19. Given that 3h + 2x = 2f – gx, express x in terms of f, g and h.

   3h + 2x = 2f – gx
   2x + gx = 2f – 3h
   x(2 + g) = 2f – 3h
        2 f  3h
    x
         2 g

20. The plan of a house is drawn to a scale of 1 : 50.
    (a) Find the actual length in metres, represented by 16 cm on the plan.

        1 : 50
        16 : 50  16
        16 : 800
        800 cm = 8 m
        The actual length is 8 m.

   (b) A wall has a length of 12.2 m.
       Find the length, in centimetres, of this wall on the plan.

        12.2 m = 1220 cm
        1 : 50
        1220  50 : 1220
        24.4 : 1220
        The length of the wall on the plan is 24.4 cm.
   (c) A room has a floor area of 20 m².
       Find the floor area, in square centimetres, of this room on the plan.

       1 : 50
       1 cm : 0.5 m
       1 cm² : 0.25 m²
       20  0.25 cm² : 20 m²
       80 cm² : 20 m²
       The floor area of the room on the plan is 80 cm².

21. The average speed of a car was x km/h. The average speed of a van was
    30 km/h lower than that of the car.
    (a) Write down, in terms of x, an expression for
        (i) the speed of the van,
           (x – 30) km/h

       (ii) the time taken by the car to travel 45 km,
              45 
              h
              x 
       (iii) the time taken by the van to travel 45 km.
              45 
                     h
              x  30 
   (b) The car took 15 minutes less than the van to cover 45 km.
       (i) Write down an equation in x and show that it reduces to
           x² – 30x – 5400 = 0.

           45      45 15
                      
         x  30 x 60
         45x  45x  30 15
                           
             xx  30       60
         45x  45x  1350 15
                            
              x 2  30 x      60
                        
         601350  15 x  30 x
                         2
                                   
         15x 2  450x  81000  0
          x 2  30 x  5400  0
         (Hence shown)
       (ii) Hence, solve the equation and find the speed of the van.
            x 2  30 x  5400  0
                  30      30 2  41 5400 
            
                                  21
             30  22500               30  22500
           =                       or    =
                  2                        2
           = 90 km/h         or = –60 km/h
                                     (Rejected)
            The speed of the van is (90 – 30) = 60 km/h.

22. The total cost of 3 chairs and 2 tables are $375. The total cost of 2 chairs and
    3 tables are $450. Find the cost of 1 chair and 1 table.

   Let the cost of one chair be x and the cost of one table be y.
   3x + 2y = 375 -------- (1)
   2x + 3y = 450 -------- (2)
   From (1), 9x + 6y = 1125 (  3) --------- (3)
   From (2), 4x + 6y = 900 (  2) ---------- (4)
             (3) – (4)
   5x = 225
    x = 45
   Substitute x = 45 into (1)
   3(45) + 2y = 375
           2y = 240
             y = 120
   The cost of 1 chair and 1 table are $45 and $120 respectively.

23. The speed of light is 3 108 m/s. The distance between the Sun and the Earth
    is 1.44  1011 m. Calculate the time taken for the light to reach from the Sun to
    the Earth. Leave your answer in minutes.
     1.44  10 11   3  10 8   1.44  3  10 118
                               = 0.48  103
                               = 480 s
                               = 8 mins

24. Convert 72 km/h into m/s.
    72  1000
              = 20 m/s
     60  60

25. Convert 15 m/s into km/h.
    15  1000
              = 54 km/h
        1
     60  60
26. Three paths, AB, BC and CA, run along the edges of a horizontal triangular
    field ABC. BC = 51 m, AC = 72 m and angle ACB = 81°.
    (a) Calculate the length of AB.
    (b) Calculate the area of the field ABC.
    (c) Calculate the shortest distance from
        C to AB.
    (d) A vertical tree, CT, has its base at C.
        The angle of elevation of the top of
        the tree from A is 15°.
        Calculate the height of the tree.
    (e) John measured the largest angle of
        elevation of the top of the tree as seen
        from the path AB.
        Calculate this angle of elevation.

   (a) AB² = 72² + 51² – 2(71)(51) cos 81°
       AB  81.5 m (3 sig. fig.)

         1
   (b)     (72)(51) sin 81° = 1810 m² (3 sig. fig.)
         2

           72        81.5
   (c)            =
        sin∠ABC sin81°
       72 sin 81° = 81.5 sin ∠ABC
                      72 sin 81
        ABC  sin 1
                         81.5
               = 60.8°
                 CT
   (d) tan 15° =
                  72
       CT = 72 tan 15°
           = 19.3 m

   (e) The base length of the triangle is the shortest distance from C to AB.
                19.3
        tan  
                44.5
                      19.3
             tan 1
                      44.5
              = 23.4°

				
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