Mathematics N1







Hi guys
I have added these 2 solutions worked out by hand, as was requested by 2 students via email.

Have a great weekend

Tommy Heugh




Mathematics N1 November 2011 Question Paper

I have received this email:

"I'm not to sure where to ask a specific math question so i hope you guys don't mind me asking here. 

Relating November 2011 Mathematics N1, Question 5, section 4 where the the question states
"A field has a  rectangular shape.The length is 10 m longer than 3 times the breadth.If the length is decreased by 50 m and the breadth is increased by 3 m, the length will be twice the breadth."

How will I go  about answering the question?

My reply:
This is a case of Simultaneous Equations. There are no equations given. You have to draw up two equations from the given facts. They speak of the length and the breadth of a rectangle, and both are unknown. So , we know there are 2 unknowns, and if we want to solve for 2 unknowns simultaneously, we have to have 2 equations.

Let the length be = L and the breadth = B, both in metres

The first fact or statement" the length is 10 m longer than 3 times the breadth.
Therefore, say it was given that "the length is 3 x the breadth", we would have written L = 3B
But now it not the case: so we write L - 10 = 3B [if we subtract 10 m from the length, that will equal 3 x the breadth]

hence, equation 1                                  L - 10 = 3B

Secondly it is stated that : "if the length is decreased by 50 m, and the breadth increased by 3 m, the length would be twice the breadth"

I would say let's start saying                                   L = 2B
Now we bring in the changes L is decreased by 50 m, and  B is increase by 3 m
                                                          now we have L - 50 = 2( B + 3 )
                                                                              or L - 50 = 2B + 6     eq 2
                                                                                  L - 10 = 3B            eq 1
So now we have set up the 2 equations : Subtract equation 1 from equation 2
                                                                                  0 - 40  = - B + 6             { L - L = 0;  - 5- (-10) becomes -50 + 10 = -40; 2B - 3B = -B; 6 - 0 =6}
Hence - B + 6 = - 40  and - B = - 40 -6 = -46
Then B = 46
Breadth = 46 m
and if B = 46, substituting into equation 1 we have that L - 10 = 3(46) =  138
                                                                                       therefore L = 138 + 10 = 148
                                                                                                     Length = 148 m
                                                                                          


Mathematics N1 Question Paper

August 2012

Question 1

1.1          We are given the trinomial 7ax^2b - 4x + 4
1.1.1      x is the variable
1.1.2      2 is the highest component of x
1.1.3      7ab is the coefficient ox x^2
1.1.4      4 is the constant term
1.1.5      the number of terms is 3

1.2   
 1.2.1      270 x 1000/3600 =      75 m / s
1.2.2      340 mm = (340 / 1340) x 100 % = 25,373 %
1.2.3      there is no y-intercept

Question 2

2.1
2.1.1       answer is 16/ x^3  reads 16 divide by x power 3
2.1.2       answer is 2x^2

2.2
2.2.1 answer is 9



****************************************************************

Mathematics N1 Question Paper

November 2005

Question 1

We have have to choose the correct answer only. Answers in brackets

1.1     [4]
1.2     [1]
1.3     [75]
1.4     [1&3]
1.5     [calculator needed]
1.6     [-2; 0; 3]
1.7     [25,373%]
1.8     [6]
1.9     [a^2 + 15a + 36] where a^2 = a squared
1.10   [8/a]

That's it until until tomorrow.

Question 2

2.1.1       (7y^0 - 3x^0)^2
            = (7.1 - 3.1)^2
            = 4^2
           = 16

2.1.2      Answer [-9x^15]

2.2       log 256 base 2 + log 100 base 10 - log e base e
           = log 2^8 base 2 + log 10^2 base 10 - 1
           = 8 log 2 base 2 + 2 log 10 base 10
           = 8.1 + 2.1
           = 10                      NB 8.1 means 8 x 1. For fractions we use a comma eg. 8,1

2.3     Calculation

Question 3


3.1              Answer: 2x^2 - 3x + 7 remainder: 28

3.2          
3.2.1           16ab -18ac + 4ad = 2a (8b - 9c + 2d)

3.2               3a + 6b - at -2bt = 3(a + 2b) -t(a + 2b)
                                               = (a + 2b)(3 - t)

3.3               Final answer is [  6x / 5 ]       6x divided by 5

3.4              Final answer : LCM  = 2.3.7^2.x^2.y^2 = 294x^2y^2 which means 294 x-squared y-squared

Question 4

4.1   Solve for x

3 (2x - 3)/ 7 - 3 = 0
3 (2x - 3 ) / 7    = 3
x 7 both sides: 3 (2x - 3) = 21
divide by 3 both sides: 2x - 3 = 7
2x = 10
x = 5

4.2 Solve for ages

Let mother = M and son = S
it is given: M = 2 S              eq 1
also: M - 10 = 3 (S -10)     eq 2

Solution:

Substitute M = 2 S into eq 2
then 2S - 10 = 3 ( S - 10)
2S - 10 = 3 S - 30
2S - 3S  = - 30 + 10
       - S  = - 20
         S  = 20
then M = 40 meaning the mother is 40 years old, and the son 20 years old.

4.3 Answer  g = 4pi^2l / T^2 meaning g = (4 x pi Squared x l) divided by T squared

4.4 through substitution and calculation, g = 1,974

4.5   the baloon rises at 50 m / 80 s or at 50/80 m/s or 0,625 m/s

therefore after 4 minutes or 4 x 60 seconds, it will reach a height of 0,625 x 240

                                                                                                = 150 m

Question 5 Graphs

5.1

5.1.1 if y = x + 2

5.1.1      gradient = 1
5.1.2      y-intercept = 2
5.1.3     straight line

5.2 

 xy = 9      The graph represents a hyperbola in the 1st and 3rd quadrants

That's it for now.


Question 6

6.1 Trigonometry

6.1.1      If angle BAD = 30 deg, then angle DAC = 30 degrees since BD = DC and AB = AC
              Therefore angle A = 60 deg.
               Hence angle A = 60 deg since angle B = angle C and angle B + angle C = 120 deg

6.1.2      Angle C = 60 degrees. B + C = 180 - A = 120
                and angle C = angle B = 120 / 2
               also because AB = AC and also in triangle ADC , angle C = 180 - angle ADC + angle DAC
                                                                                                 = 180 - (30 + 90)
                                                                                                 = 60 degrees

6.2       AC = Rt (40 ^ 2 + 30 ^ 2)
                  = 50 mm

6.3  
6.3.1    Sq Rt3 tan 60 - Sq Rt 2 cos 45 + Sq Rt  2 sin 45
           = Sq Rt 3 x Sq Rt 3/1 - Sq Rt 2 x 1 / Sq Rt 2 + Sq Rt 2 x 1 / sq rt 2
            =    3 / 1 - 1 + 1
            = 3

6.3.2       3 tan 45 x 2 sin 30 / 3 cos 60
            = 3 (1)  x 2 (1 / 2) /3 (1/ 2)
           = 3 x 1/ 3/2
          = 3 x 2/3
          = 2
6.4
6.4.1

if cos A = 11 / 18 then A = Cos ^ -1 (11/18) meaning inverse cos of 11/18
therefore A = 52.33 degrees

6.4.2     sin 67 deg 18 minutes = A 
             18 minutes = 18/60 degrees = 0.3 degrees
              therefore sin 67 degrees 18 minutes = sin 67,3 degrees
                                                                           = A
                                                                    A = sin 67,3 degrees
                                                                        = 0,923


Question 7

7.1 Radius = Sq Rt 180 ^ 2 - 120 ^ 2
                 = 134,164 mm
Therefore the volume of the cone  =  1/3 Pi 134,164 ^ 2 x 120
                                                    = 2261944,059 mm ^ 3 meaning 2261944,059 millimetre cubed

7.2    Circumference = 77,285 cm. Since 2 x  pi x  Radius = Circumference
                                                                             Radius = 77, 285/ (2 x pi)
                                                                                        = 12,3 cm

7.3    Area = 1 /2 x base x perpendicular height
          perpendicular height = Sq Rt ( 60 ^ 2 - 40 ^ 2)
                                         = 44,721 mm

Area in cm ^ 2 = 1/2 x base x perpendicular height
                       = 1/2 x 8 x 4,4721
                      = 17,888 cm ^2

That's it. That brought us to the end of November 2005

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@


Mathematics N1

April 2006 Question Paper 

Question 4



4.1 Manipulation of formulae
If s =1/2gt^2 what is the value of t if s = 1500 and g = 9,8.
Firstly we need to manipulate (rearrange the formula.
1/2gt^2 = s
x 2: gt^2 = 2s
divide by g both sides: t^2 = 2s/g
take square root both sides:  t = square root of (2s/g)
Substituting for s and g. we have that t = 17,496

4.2 Convert minutes into hours:
45 minutes = 45/60 hours
                 = 0,75 hours

4.3 Convert km/h into m/s
14 km/h = (14 x 1000) /3600
             = 3,89 m/s

4.4 Simultaneous equations
If screws cost "x" rands/kilogram and nails cost "y" rands/kilogram:
5x + 6y = 28,60     eq 1
3x + 2y = 13,50     eq 2

                               5x + 6y = 28,60
multiply eq 2 by -3: -9x - 6y = -40,5
Add:                         -4x     = -11,9
therefore                      x     =    -11,9/-4
                                    x      =      2,98
and therefore if we substitute x = 2,98 into eq 2:
we have that 3x + 2y = 13,5
then              3(2,98) + 2y = 13,5
                                    2y = 13,5 - 8,94
                                      y = 2,28
Therefore the screws costs R 2,98 per kilogram
and nails cost R 2,28 per kilogram

4.5 Angular motion
Given: gear spin at 90 r/min , therefore N = 90 rpm
Wanted:
4.5.1 rotational frequency in rps
Rotational frequency = N/60
                              = 90/60
                              =1,5 rps

4.5.2 Angular velocity
                               w = 2#n where # stands for pi (and pi = 22/7)
therefore                  w = 2 x 22/7 x 1,5
                                   = 9,42 rad/s


Question 5 - GRAPHS


5.1
5.1.1     xy = 2
5.1.2     1/3
5.1.3     negative
5.1.4     3
5.1.5     hyperbola
5.1.6     quadrants 1 and 3

5.2
we are give that x = {-2; -1; 0; 1; 2}
and we are required to draw the graph:        8x - 4y = 4
we need to simplify the equation and rewrite into standard form
therefore,                                              if 8x - 4y = 4 then -4y = -8x + 4
                                                             divide by -4                             y = 2x - 1
Draw up a table
                      x     -2     -1     0    1     2
                     2x   -4     -2     0    2     4
                     -1    -1    -1     -1   -1   -1
y = 2x - 1             -5     -3     -1   1    3 (add the coloured rows)

Draw the straight line graph as shown below:




Question 6 - TRIGONOMETRY


6.1
6.1.1     180 degrees
6.1.2      lll
6.13       duplicate
6.1.4      180 degrees
6.1.5       equal
6.1.5       ^M = 90 deg. KL = 8, and LM = 3. what is the length of KM?
According to Pythagoras       KM^2 = KL^2 - LM^2
                                                        = 8^2    - 3^2
                                                KM  =  sq root of 55
                                                         =  7,146   cm




@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@




April 2006

We are busy reviewing this question paper. I have discussed the first 3 questions already.
I also uploaded the solutions for Questions 1 and 2.

Tomorrow we will discuss Question 4, whilst I upload the solutions of Question 3. That is so that you can have time to try each question on your own.

Cheers

No comments:

Post a Comment

Note: Only a member of this blog may post a comment.