Mathematics N3 November 2012 Memo

Mathematics N3 Question Paper

November 2012

Mathematics N3 Question Paper

August 2012

Question 1 Answers only

1.1     C

1.2     B

1.3    B

1.4    A

1.5    D

1.6    C

1.7    B

1.8    D

1.9    D

1.10  there is a printing error: the correct answer is y = - 15cosx   

Question 2

2.1

2.1.1     Answer is 3

2.1.2     Answer is 1

2.1.3     Answer is 5/6

2.2

2.2.1     8x^2 - 2(x + 1)^2

           = 8x^2 - 2 [x^2 + 2x + 1]

          = 8x^2 -2x^2 - 4x -2

          = 6x^2 - 4x - 2

          = 2(3x^2 - 2x - 1)

         = 2(3x + 1) (x - 1)

2.2.2      (a + b)x^2 - 2ax - 2bx + b + a
            = (a + b)x^2 - 2x(a + b) + 1(a + b)
            = (a + b)(x^2 - 2x + 1)
           = (a + b)( x - 1) (x - 1)

2.2.3       m^36 - 15 m ^18  + 54
            = (m^18 - 9)( m^18 - 6)
           = (m^9 - 3) (m^9 + 3)(m^18 - 6)

2.3    x-a will be a factor of f(x) = -x^3 + (a + 3)x^2 -x(3a +3) +3a if f(a) = 0
                                          f(a)  = -a^3 + (a + 3)a^2 - a(3a +3) + 3a
                                                 = -a^3 + a^3 + 3a^2 - 3a^2 - 3a + 3a
                                                 = 0
therefore, x -a is a factor of f(x)

2.4
Answer is 6 sq rt 5/ 5

Question 3

3.1 solve for x:

3.1.1                 3^x + 3^x + 1 = 36

                         3^x  + 3^x . 3^1 = 36

                         3^x ( 1 + 3)           = 36

                         3^x                        = 9 = 3^2

                      therefore      x = 2

3.1.2    algebraic fractions : solve for x

answer : x = 0 or x = 3

3.2

3.3 answer : d = square root( d^2 - 4A/pi)

                   and d = 2,1

3.4    simultaneous equations: if it is given that L = B + 5 and 2 ( L + B) = 30, then through substitution we get that B = 5 and L = 10 

Question 4

4.1

4.1.1 x^2 / 4 + y^2 / 25 = 1

4.1.2    y = x ( x - 1 )( x + 2)

4.1.3 y = ( x - 2 ) ( x - 5 )

4.2     simultaneous equations:    y = - 2x + 14  eq 1

                                                 2y = 7x - 5     eq 2

                           eq 1 x 2:        2y = - 4x + 28

                           eq 2               2y = 7x - 5

subtract                                     0 = - 11x + 33

then x = 3 and y = 8

sketch of two graphs showing point of intersection

August 2012 Engineering Science Memo

August 2012 Engineering Science

Question 1

1.1       Theory: The effects of a force on a physical body.
Answer: Any 2
a force can:

1. bring a moving body to a standstill
2. cause a moving body to slow down ( decelerate )
3. cause a moving body to speed up  ( accelerate )
4. get a stationary body into motion

1.2       Motion  We are given that deceleration takes place. u = 80 km/h and v = 60 km/h We have to convert the speeds into m/s by dividing by 3.6.

therefore u = 80 km/h =     m/s and v = 60 km/h =       m/s

1.2.1     a = (v - u) /t  = - 6,001 m/s^2

1.2.2    t = 0,926 s

1.3

1.3.1    a = 0,667 m/s^2

1.3.2    s = 900 m [total distance in 60 s]

1.4

1.4.1   T2 = 466,667 m/s^2

1.4.2   v = 6,346 m/s

1.4.3   P = 5,923 kW

Question 2

2.1 Theory : Definition of the equilibrant of a system of forces

2.2 Beam

2.2.1      L = 512,5 N

              R = 197,5 N

2.2.2      Shear force diagram (to be scanned in)

Question 3

3.1 Theory: Definition of the RESULTANT

3.2

3.2.1       R = 55 N and L = 45 N

3.2.2      AE = 51,962 N (analytical method)    [to be scanned in]

              AE = 52 N graphical method [to be scanned in]

Question 4

4.1       W perp = Wcos 20 deg = 552,539 N

4.2       W para = Wsin 20 deg =  201,108 N

4.3       Friction force = 201,108 N

4.4       Coefficient of friction = 0,364

4.5       Force up = 201,108 N

Question 5

5.1       Theory :Definition of Heat Value of a fuel

5.2       t2 = 88,689 deg C

5.3       change in length = 10,71 mm

5.4

5.4.1     h wet   =             J

5.4.2     h g      =              J

5.4.3     h =                      J

5.4.4     hss     =               J   

Question 6

6.1 Definition of Density: The density of a body or substance is the ration of its mass to its volume 

6.2

6.2.1   volume per delivery stroke  = area x stroke length

                                                    = pi/4 x (0,16)^2 x 0,09

                                                    = 0,00181 m^3

6.2.2 mass pumped/delivery stroke = density x volume

                                                     = 1000 x 0,00181

                                                     = 1,81 kg

6.2.3   work done/delivery stroke  = density x g x H x vol

                                                     = 1000 x 9,8 x 15 x 0,00181

                                                      266,07 J

6.3

6.3.1                                        W/F = D^2/d^2

                                                    W = 50 x 0,12^2/ 0,03^2

                                                         = 800 N

                                                          = 81,633 kg

6.3.2                                         P = F/a

                                                     = 50 x 4/pi x 0,03^2

                                                     = 70,736 kPa

Question 7

7.1   Definition: The heat generated in a circuit is directly proportional to the square of the current that flows, the resistance and the time the current flows.

7.2 any two: temperature; length; cross sectional area; type of metal

7.3
7.3.1 To calculate the total resistance
                              1 / R par = 1 / 30 + 1 / 45
                                       = 45 + 30/1350
                                       = 75/ 1350
                                  R par = 1350 / 75
                                      = 18 ohm

the total resistance of the circuit is therefore 18 + 12 = 30 ohm

7.3.2 Ammeter reading    I tot  = E / R total
                                                  = 9 / 30
                                                  = 0,3 A

7.3.3  Voltmeter reading       V 12v  = I R
                                                       = 0,3 x 12
                                                       = 3,6 V

7.4
7.4.1         Secondary voltage   Vs / V p = Ns / Np
                                                       V s  = 42 / 420 x 240
                                                              = 24 V

7.4.2         Primary current        I p / I s  = V s / V p
                                                       I p = 24 / 240 x 2
                                                             = 0,2 A

Question 8

8.1   Humid/damp conditions

                   salty air (at coast)

8.2    proton: positive

           neutron : neutral

          electron : negative

8.3  A molecule of a substance or element is a chemical combination of 2 or more atoms of of that substance or element.

8.4      copper; iron; gold; aluminium

Engineering Science N3 November 2012 Memorandum

Engineering Science N3

Engineering Science N3 Question Papers

November 2012

Question 1

1.1

1.1.1      a = - 3 m/s^2

1.1.2      s = 137,5 m/s

1.1.3      Momentum = 70 000 kg.m/s

1.1.4      Change in KE = 825 kJ

1.1.5      Braking force = 6 000 N

1.2

1.2.1     Torque = 300 Nm

1.2.2      Power = 1,8 kW

Question 2

2.1      Theory

2.2    

2.2.1      Reaction C = 63,75 kN

              Reaction B = 16,25 kN

2.2.2      Shear force diagram

Question 3

3.1      Sum of VC = 41,152 kN;     Sum of HC = 62,64 kN

3.2      CA = 25,456 kN   Strut;       AB = 18 kN    Tie

Question 4

4.1      Theory

4.2

4.2.1      Sketch

4.2.2      Fup = 1039,725 N

Question 5

5.1

5.1.1      Heat released by steel = 1 177,6 kJ

5.1.2      Volume of oil = 17,796 litres

5.2         Final area of aluminium plate = 20 023 mm^2

5.3

5.3.1      hf = 962 kJ

5.3.2      hg = 2 801 kJ

5.3.3      hfg = 1 839 kJ

5.3.4      hws = 2 690,66 kJ

5.3.5      h1 - x = 110,34 kJ

Question 6

6.1

6.1.1      Work done by pump = 6 927,212 kJ

6.1.2      Power required = 192,423 W

6.2

6.2.1      Force exerted by ram , W = 3 826,531 kN

6.2.2      Distance moved by ram, H = 0,063 m

6.2.3      Number of pumping strokes = 734

6.2.4      Volume of liquid received = 2 258,02 litres

Question 7

7.1

7.1.1      Total resistance = 6,463 ohms

7.1.2      Total current flow = 2,872 A

7.2

7.2.1      Current through lamp = 0,625 A

7.2.2      Resistance of lamp = 384 ohms

7.2.3      Costs = 2,7 cents

Question 8

8.1     Theory

8.2     Theory

8.3     Theory

Mathematics N2 August 2011 question paper Memo

Mathematics N2

Mathematics N2 Question Paper

August 2011 question paper

Question 1
1.1
1.1.1 answer C
1.1.2             B
1.1.3             B
1.1.4             B
1.1.5             D although the correct answer ia actually 0,096 [34,4 / 360]
1.1.6             A
1.1.7              B
1.1.8             C
1.1.9             B
1.1.10           A

1.2
1.2.1                True
1.2.2                 True
1.2.3                False
1.2.4                True
1.2.5                 False
1.2.6                True
1.2.7                True
1.2.8                True
1.2.9                 False
1.2.10               False

Question 2

Factorisation, LCM, algebraic fractions and simplification.
Since I have dificulty in expressing powers, I will only post the answers here.
2.1
2.1.1 Answer         sin @/4 (x - y)(x + y)
2.1.2 Answer         (1 - p)(3x +b)
2.1.3 Answer         3a(a + 5)(a - 2)
2.1.4 Answer         3x(1 + 4y^2)

2.2
         Answer                  LCM = 2x(x - y)^2(x + y)(2x + 3y)

2.3 
2.3.1 Answer          (3p + 4y)/xyp                 
2.3.2 Answer             1

Question 3

3.1 We have to solve for x and y: if y = x^2 - 4 and y = - x - 2
The easiest method is to equate the right hand side of each equation, since the left hand sides are equal, therefore                                     x^2 - 4 = - x - 2
which we must rearrange as a quadratic equation of x,  all to the left hand side
then we have to factorise , then  x = -2, or x = 1
then we have to calculate the corresponding y-value for each x-value
(-2;0) and (1;-3) are the final answers

3.2          answer : d = eT/mv by applying the rules of manipulation

3.3 
3.3.1       answer: a^(x + 7)  [read a raised to the power(x + 7)
3.3.2      answer: 1
3.3.3      answer: the answer is difficult to express here
                             1 / 6th root of (ab)^5 [reads: THE 6th root of (ab) raised to the power 5
3.4
3.4.1     using LCM of 3 and 4 which is 12 to help us, and rules
               we get that x = 4
3.4.2 we have to use the rule that (anything)^0 = 1 [reads anything raised to the power of 0 = 1]
and also the rule (root x)^2 = x [reads: the square root of x, squared = x]
then x =1

3.4.3 here we have to rewire square root of x, as well as 32, into power form first.
then x = 2^10 = 1024

3.5 the answer, applying the rules of logs = 3,071

3.6 
3.6.1
 log 100 base 5 - log 4 base 5 = log 100/4 base5
                                                        = log 25 base 5
                                                        = 2 log 5 base 5
                                                        = 2

3.6.2 
4 raised to the power log 16 base 4
we can first simplify the power part, which is log 16 base 4 = log 4^2 base 4 = 2 log 4 base 4
                                                                                                = 2 x 1 = 2
then finally we have 4 raised to the power log 16 base 4 = 4 ^2
                                                                                            16

Question 4

4.1 convert degrees, minutes and seconds to radians. First we have to convert everything to degrees. Convert 12 seconds into minutes first (by dividing by 60), then add the answer to 36 minutes. Now convert the total minutes into degrees (by dividing by 60), and add this answer to 58. now we have a value expressed in degrees only. To convert the degrees into radians we multiply by pi, and divide by 180.

4.2 angular motion
4.2.1   w = 2 x pi x n where n is the rotational frequency in revs per sec.
4.2.2    angular displacement = angular velocity x time
4.2.3   no of revs in 1 minute [ here we have to divide our previous answer of 4.2.2 by 2 pi to convert the radians into revs.

4.3 here we apply the applicable formulae from the formula sheet

Answers
4.1     answer = 1,023 radians

4.2
4.2.1     n = 47, 746 revs/sec
4.2.2     theta = 18000 radians
4.2.3     theta = 2864,789 revs

4.3
4.3.1          s = 418,879 mm
4.3.2        A  = 83 775,804 mm^2

4.4
4.4.1                   Volume = 3 078 760,801 mm^3
4.4.2           surface area = 118 752,202 mm^2

Question 5

5.1      A graph has to be drawn, and answers need to be determined off of the graph:

The answers are as follows:

5.1.1    x = 30 degrees, and x = 150 degrees

5.1.2    x = 15 degrees and x = 165 degrees

5.1.3 amplitude = 2

5.2 graph of y = x^2 - 2x - 8

y-intercept (0; - 8)

roots           (-2;0) and (4;0)

axis of symmetry   x = 1

turning point  (-1; -9)

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