Add Maths Marathon!

Additional Mathematics Marathon!

Algebra, sequences and series

*Remainder Theorem

*Factor Theorem

* Quadratics

*Inequalities

*Surds, Indices, Logs

*Sequences and Series

#1 Questions

The polynomial P(x) is given by   $=x^3+x^2-8x-12$  .

(a) Use the remainder theorem to find the remainder when P(x) is divided by   $\left(x-1\right)$  .
(b) Use the factor theorem to show that  $\left(x+2\right)$  is a factor of P(x). 
(c) Hence, express P(x) as the product of linear factors. 

#1 Answers

(a)  $r=P\left(1\right)$  
 $r=\left(1\right)^3+\left(1\right)^2-8\left(1\right)-12$  
 $r=1+1-8-12$  
 $r=-18$  

(b)  $If\ \left(x+2\right)\ is\ a\ factor\ then\ P\left(-2\right)=0$  
 $=\left(-2\right)^3+\left(-2\right)^2-8\left(-2\right)-12$  

 $=-8+4+16-12$  
 $=12-12$  
 $=0$  

(c)

 $\left(x+2\right)\left(x^2-x-6\right)=0$  
 $\left(x+2\right)\left(x-3\right)\left(x+2\right)=0$  

 

#2 Questions

If the roots of the equation  $3x^2+4x-5=0\ are\ \alpha\ and\ \beta.$   Find the values of 

(a)  $\frac{1}{\alpha}+\frac{1}{\beta}\$  
(b)  $\alpha^2+\beta^2$  
(c)  $\alpha^2\beta+\beta^2\alpha$  

#2 Answers

 $\alpha+\beta=\frac{-b}{a}=-\frac{4}{3}$                       (b) $\alpha^2+\beta^2=\left(\alpha+\beta\right)^2-2\alpha\beta$              (c)  $\alpha^2\beta+\beta^2\alpha$  
 $\alpha\beta=\frac{c}{a}=-\frac{5}{3}$                                     $=\left(-\frac{4}{3}\right)^2-2\left(-\frac{5}{3}\right)$                        $=\alpha\beta\left(\alpha+\beta\right)$  
                                                                 $=\frac{16}{9}-\left(-\frac{10}{3}\right)$                                  $=-\frac{5}{3}\left(-\frac{4}{3}\right)$  
(a)  $\frac{1}{\alpha}\ +\frac{1}{\beta}$                                            $=\frac{16}{9}+\frac{10}{3}$                                            $=\frac{20}{9}$  
 $=\frac{\beta+\alpha}{\alpha\beta}\ =\frac{\alpha+\beta}{\alpha\beta}$                             $=\frac{46}{9}$  
 $=-\frac{4}{3}\div-\frac{5}{3}$  
 $=-\frac{4}{3}\times-\frac{3}{5}$  
 $=\frac{4}{5}$  

#3 Questions

Solve the inequalities :

(a)  $x^{2\ }\le3x+4$  
(b)  $\frac{2x+1}{x-1}>0$  

#3 Questions

(a)  $x^2\le3x+4$  
 $x^2-3x-4\le0$  
 $\left(x-4\right)\left(x+1\right)\le0$  
 $x=4\ \ \ x=-1$  
(sketch)

 $\therefore-1\le x\le4$  

(b)

 $\frac{2x+1}{x-1}>0$  

 $\times\left(x-1\right)^2$  
 $\left(2x+1\right)\left(x-1\right)>0$  
 $-\frac{1}{2}\ \ \ \ and\ 1$  
(sketch)
 $\therefore x<-\frac{1}{2}\ or\ x>1$  

#4 Questions

a) Simplify  $3\sqrt{27}\ -5\sqrt{8}-2\sqrt{75}\ +4\sqrt{18}$  
(b) Express  $\frac{\sqrt{3}-1}{\sqrt{3}+1}$  in the form  $a+b\sqrt{3}$  where a and b are integers.

#4 Answers

(a)  $3\sqrt{27\ }-5\sqrt{8}-2\sqrt{75}+4\sqrt{18}$  
 $=3\sqrt{9}\sqrt{3}-5\sqrt{4}\sqrt{2}-2\sqrt{25}\sqrt{3}+4\sqrt{9}\sqrt{2}$  
 $=3\left(3\right)\sqrt{3}-5\left(2\right)\sqrt{2}-2\left(5\right)\sqrt{2}-2\left(5\right)\sqrt{3}+4\left(3\right)\sqrt{2}$  
 $=9\sqrt{3}-10\sqrt{2}-10\sqrt{3}+12\sqrt{2}$  
 $=9\sqrt{3}-10\sqrt{3}-10\sqrt{2}+12\sqrt{2}$  
 $=-\sqrt{3}+2\sqrt{2}$  

(b)  $\frac{\sqrt{3}-1}{\sqrt{3}+1}\ \ \ \ \times\ \frac{\sqrt{3}-1}{\sqrt{3}-1}$  

 $=\frac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}$  
  $=\frac{3-2\sqrt{3}+1}{3-1}$  
 $=\frac{4-2\sqrt{3}}{2}$  
 $=\frac{2\left(2-\sqrt{3}\right)}{2}$  = $2-\sqrt{3}$  

#5 Questions

Solve for x for the following equations

 (a)  $5^{2x+1}=\frac{1}{25}$  
(b)  $\sqrt{\left(x+3\right)}\ =x-3$  

#5 Answers

(a)  $5^{2x+1}=\frac{1}{25}$                       (b)  $\sqrt{x+3}=x-3$  

 $5^{2x+1}=\frac{1}{5^2}$                             $x+3=\left(x-3\right)^2$                    
 $5^{2x+1}=5^{-2}$                         $x+3=x^2-6x+9$  
                                                         $x^2-7x+6=0$  
same base                                   $\left(x-6\right)\left(x-1\right)=0$  
 $2x+1=-2$                                          $x=6,x=1$  
 $2x=-3$  
 $x=-\frac{3}{2}$  

#6 Questions

(a) Express  $\log_a16\ -\log_a2\ -\log_ax$  as a single logarithm .

(b) If  $\log_{10}2\ =0.301\ and\ \log_{10}3\ =0.477$  , find the value of  $2\log_{10}21\ -\log_{10}98$  .

#6 Answers

(a)  $\log_a16\ -\ \log_a2\ -\ \log_ax$         (b)  $2\log_{10}21-\log_{10}98$  

 $=\log_a\ \frac{16}{2}\ -\log_ax$                            $=\log_{10}\left(21\right)^2\ -\log_{10}98$  
 $=\log_a\ 8\ -\log_ax\$                              $=\log_{10}441\ -\log_{10}98$            
 $=\log_a\ \frac{8}{x}$                                               $=\log_{10}\frac{441}{98}$ 
                                                                 $=\log_{10}\frac{9}{2}$  
                                                                 $=\log_{10}9\ -\log_{10}2$  
                                                                 $=\log_{10}\left(3\right)^2\ -\log_{10}2$  
                                                                 $=2\log_{10}3-\log_{10}2$  
                                                                 $=2\left(0.477\right)-0.301$  
                                                                 $=0.954-0.301$  
                                                                 $=0.653$  

#7 Questions

(a) If the first term of an Arithmetic Progression is 6 and the tenth term is 24.
(i) Find the common difference
(ii) Write down the first 4 terms

(b) Find the sum of all the terms  in the AP  $2+7+12+...+77$  .

#7Answers

(a)                                                                     (b)  $a=2\$  
(i)  $1st\ term=6\ \ \ \ 10th\ term\ =24$           $d=7-2=5$      Tn = 77

 $T_n=a+\left(n-1\right)d$                               $T_n=a+\left(n-1\right)d$  
 $T_{10}=6+\left(10-1\right)d$                                  $77=2+\left(n-1\right)\left(5\right)$                
 $24=6+9d$                                                 $77=2+5n-5$  
 $18=9d$                                                     $77=-3+5n$                                                 
 $d=\frac{18}{9}$                                                         $80=5n$  
 $d=2$                                                              $n=\frac{80}{5}$                   
                                                                            n = 16 (The Ap has 16 terms)
(ii) 6, 8, 10 ,12

 $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\ or\ \frac{n}{2}\left(a_1+a_n\right)$  
 $S_{16}=\frac{16}{2}\left(2\left(2\right)+\left(16-1\right)\left(5\right)\right)$    or            $=\frac{16}{2}\left(2+77\right)$  
 $=8\left(4+15\left(5\right)\right)$                                                $=8\left(79\right)$  
 $=8\left(4+75\right)$                                                      $=632$          
 $=8\left(79\right)$         
 $=632$  

#8 Questions

The third term of a GP is 10 and the sixth term is 80. Find :

(a) The common ratio

(b) The sum of the first 6 terms.

(c) The sum to infinity .

#8 answers ( $ar^{n-1}$  )

(a)  $3rd\ term=10\rightarrow\ ar^2=10\left(1\right)$  

 $6th\ term\ =80\rightarrow ar^5=80\left(2\right)$  
(2) divided by (1)
 $\frac{ar^5}{ar^2}=\frac{80}{10}$                                  (b) $S_n=\frac{a\left(1-r^n\right)}{1-r}$  
 $r^3=8$                                              $S_6=\frac{\frac{5}{2}\left(1-\left(2\right)^6\right)}{1-2}$            $=\frac{\frac{5}{2}\left(1-64\right)}{-1}$  
 $r=^3\sqrt{8}$                                               $=\frac{\frac{5}{2}\left(-63\right)}{-1}$       $=\frac{315}{2}\ OR\ 157\ \frac{1}{2}$  
 $r=2$  
sub r = 2 into (1)                                         (c)      $S_{\infty}=\frac{a}{1-r}=\frac{\frac{5}{2}}{1-2}$  
 $a\left(2\right)^2=10$                                                                 $=\frac{5}{2}\div-1$  
 $4a=10$                                                                       $=\frac{5}{2}\times-\frac{1}{1}$                                               
 $a=\frac{10}{4}$  $=\frac{5}{2}$                                                              $=-\frac{5}{2}$  

Coordinate geometry , Vectors , Trigonometry

*Coordinate Geometry

* Vectors including unit vectors

*Trigs including use of double/compound angle formulae

#1 Questions

A circle C  has equation  $x^2+y^2-4x+6y-3=0$  .

(a) Determine the coordinates of the center of C.
(b) Find the radius of C.
(c) Given that the point (p,1) lies on C, find the value of p.

#1 Answers

(a)                                                       (b) $r=\sqrt{f^2+g^2-C}$  
 $2f=-4\ \ \ \ 2g=6$                                   $=\sqrt{\left(-2\right)^2+\left(3\right)^2-\left(-3\right)}$  
 $\ \ f=-2\ \ \ \ \ \ g=3$                                   $=\sqrt{4+9+3}$  
 $C=\left(-f,-g\right)=\left(2,-3\right)$                       $=\sqrt{16}=4$  
(c) Using  $\ \ r=\ \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\ ,\ \left(2,-3\right)\ and\ \left(p,1\right)\ to\ find\ p$  
 $4=\sqrt{\left(2-p\right)^2+\left(-3-1\right)^2}$  
 $4=\sqrt{\left(2-p\right)^2+16}$  (square both sides)
 $16=\left(2-p\right)^2+16$  
 $\left(2-p\right)^2=0$  
 $2-p=0$  
 $p=2$  

#2 Questions (2018 c)

The position vectors of 2 points , A and B , relative to a fixed origin , O , are given by  $\overrightarrow{OA}\ =2i+j\ and\ \overrightarrow{OB}\ =3i-5j$  , where i and j represent unit vectors in the x and y directions respectively. Calculate 

(a) the magnitude of  $\overrightarrow{AB}$  
(b) the angle  $AOB$  , giving your answer to the nearest whole number 
(c) the unit vector of  $\overrightarrow{BA}$  

#2 Answers

(a)  $\overrightarrow{AB}\ =\overrightarrow{OB}\ -\overrightarrow{OA}$                   (c)Unit vector =  $\frac{vector}{magnitude}\$    

 $=\left(_{-5}^3\right)-\left(_1^2\right)$                                                           $\overrightarrow{BA}\ =\overrightarrow{OA}\ -\overrightarrow{OB}$  
 $=\left(_{-6}^1\right)$                                                                               $=\ \left(_1^2\right)-\left(_{-5}^3\right)$  
 $\left|\overrightarrow{AB}\right|\ =\sqrt{\left(1\right)^2+\left(-6\right)^2}$                                               $=\left(_6^{-1}\right)$  
 $=\sqrt{1+36}=\sqrt{37}$                               $\left|\overrightarrow{BA}\right|=\sqrt{\left(-1\right)^2+\left(6\right)^2}=\sqrt{37}$  
                                                                                                                        $U.V=\frac{-i+6j}{\sqrt{37}}$  
(b) $\overrightarrow{OA}\cdot\overrightarrow{OB}\ =\left|\overrightarrow{OA}\right|\left|\overrightarrow{OB}\right|\cos\ AOB$  
 $\left(_1^2\right)\cdot\left(_{-5}^3\right)\ =\sqrt{\left(2\right)^2+\left(1\right)^2}\cdot\sqrt{\left(3\right)^2+\left(-5\right)^2}Cos\ AOB$  
 $6-5=\sqrt{5\ }\cdot\sqrt{34}\cos\ AOB$  
 $1=\sqrt{170}\cos\ AOB$  
 $\cos\ AOB=\frac{1}{\sqrt{170}}$  
 $AOB=\cos^{-1}\left(\frac{1}{\sqrt{170}}\right)=86^{\degree}$  

#3 Questions

Prove the identities
(a)  $\frac{1}{1+\cos x}+\frac{1}{1-\cos x}=\frac{2}{\sin^2x}$  
(b)  $\sin\left(\alpha+\beta\right)-\sin\left(\alpha-\beta\right)=2\cos\alpha\sin\beta$  

#3 Answers

(a)  $\frac{1}{1+\cos x}+\frac{1}{1-\cos x}=\frac{2}{\sin^2x}$  
 $=\frac{\left(1-\cos x\right)+\left(1+\cos x\right)}{\left(1+\cos x\right)\left(1-\cos x\right)}$  (combined into 1 fraction)
 $=\frac{1-\cos x+1+\cos x}{1-\cos x+\cos x-\cos^2x}$  
 $=\frac{2}{1-\cos^2x}$  
 $=\frac{2}{\sin^2x}$  
(b)  $\sin\left(\alpha+\beta\right)-\sin\left(\alpha-\beta\right)=2\cos\alpha\sin\beta$  

 $=\sin\alpha\cos\beta+\cos\propto\sin\beta-\left(\sin\alpha\cos\beta-\cos\alpha\sin\beta\right)$  
 $=\sin\alpha\cos\beta+\cos\alpha\sin\beta-\sin\alpha\cos\beta+\cos\alpha\sin\beta$ 
 $=\cos\alpha\sin\beta+\cos\alpha\sin\beta$   
 $=2\cos\alpha\sin\beta$  

#4 Questions

A sector OPQ of a circle of radius r cm has an Area of  $100cm^2$  
(a) Show that the perimeter of the sector is  $2r+\frac{200}{r}cm$  .
(b) Determine the value of r if the sector angle equals 60 degrees (leave answer is surd form).

#4 Answers

(a)Area of sector =  $\frac{1}{2}r^2\theta$          (b)  $60\degree=60\times\frac{\pi}{180}$  

 $\therefore100cm^2=\frac{1}{2}r^2\theta$                                  $=\frac{1}{3}\pi^c$  
 $100\div\frac{1}{2}=r^2\theta$   
 $200=r^2\theta$                                       $100cm=\frac{1}{2}r^2\theta$                                  $r^2=\frac{600}{\pi}$  
 $200=r\ r\theta$                                           $100=\frac{1}{2}r^2\left(\frac{1}{3}\pi\right)$              $\therefore r=\sqrt{\frac{600}{\pi}}cm$           
 $\therefore r\theta=\frac{200}{r}$  (arc length)                   $100\div\frac{1}{2}=r^2\left(\frac{1}{3}\pi\right)$                       
 $PQ=r\theta=\frac{200}{r}$                                 $200=r^2\left(\frac{1}{3}\pi\right)$  
 $\therefore Perimeter\ =2r+\frac{200}{r}cm$                 $\frac{200}{\frac{\pi}{3}}=r^2$  

#5 Question ( 2015 4(b))

Solve the following equation , giving our answer correct to 1 decimal place.

 $8\sin^2\theta=5-10\cos\theta$  , where  $0\degree\le\theta\le360\degree$  

#5 Answers

 $8\sin^2\theta=5-10\cos\theta$                                  $\cos\ is\ -ve\ in\ the\ 2nd\ and\ 3rd\ quad.$  
 $8\left(1-\cos^2\theta\right)=5-10\cos\theta$                       $\theta_2=180-75.5=104.5^{\degree}$  
 $8-8\cos^2\theta=5-10\cos\theta$                           $\theta_3=180+75.5=255.5\degree$  
 $8\cos^2\theta-10\cos\theta+5-8=0$                 $\therefore\theta=104.5\degree,\ 255.5\degree$  
 $8\cos^2\theta-10\cos\theta-3=0$  
 $let\ x\ =\ \cos\theta$  
 $8x^2-10x-3=0$  (factorize)
 $\left(2x-3\right)\left(4x+1\right)=0$  
 $x=\frac{3}{2}\ \ \ \ x=-\frac{1}{4}$  
 $\cos\theta=\frac{3}{2}\ \ \ \ \ \ \ \ \ \ \cos\theta=-\frac{1}{4}$  
Invalid                         $=\cos^{-1}\left(-\frac{1}{4}\right)$  
                                    $=104.5^{\degree}$  (ignore sign)
                                       $\propto orPV=180-104.5=75.5\degree$  

Introductory Calculus

*Differentiation to include stationary values and kinematics

*Integration to include area under a curve and Kinematics

#1 Question

The equation of a curve is y  $=x^3-3x^2+4.$  
(a) Find the values of the x and y at the stationary points on the curve.
(b) Determine the nature of these two stationary points.
(c) Calculate the area of the region bounded between the curve , the X axis and the lines x = 2 and x = 4.

#1 Answers

(a)  $y=x^3-3x^2+4$                (b)  $\frac{d^2y}{dx^2}=6x-6$  
 $\frac{dy}{dx}=3x^2-6x$                            $at\ \left(0,4\right),\ \frac{d^2y}{dx^2}=6\left(0\right)-6\ =\ -6$  
 $At\ a\ S.p.\ \frac{dy}{dx}=0$                    $-6<0\therefore It\ is\ a\ \max\ pt$  
 $3x^2-6x=0$                                 $At\ \left(2,0\right),\ \frac{d^2y}{dx^2}=6\left(2\right)-6=6$  
 $3x\left(x-2\right)=0$                                       $6>0\therefore It\ is\ a\ \min\ pt$  
 $3x=0\ \ \ x-2=0$  
 $x=0\ \ \ x=2$  
 $when\ x=0,\ y=4$  
 $when\ x=2,\ y=\left(2\right)^3-3\left(2\right)^2+4=0$  
 $\therefore Sps\ are\ \left(0,4\right)\ and\ \left(2,0\right)$  
 

(c)  $A=\int_{x_2}^{x_1}y\ dx$  

 $=\int_2^4\left(x^3-3x^2+4\right)$  dx
 $=\left[\frac{x^4}{4}-x^3+4x\right]\$  
 $=\left(\frac{\left(4\right)^4}{4}-\left(4\right)^3+4\left(4\right)\right)-\left(\frac{\left(2\right)^4}{4}-3\left(2\right)+4\left(2\right)\right)$  
 $=16-4=12squ\ units$  

#2 Question

 $y=10\sqrt{x}-\frac{4}{x^4}+\sin x$  : 

Differentiate the function above .

#2 Answer

 $y=10\sqrt{x}-\frac{4}{x^4}+\sin x$  
 $=10x^{\frac{1}{2}}-4x^{-4}+\sin x$  
 $\frac{dy}{dx}=5x^{-\frac{1}{2}}+16x^{-5}+\cos x$  
 $=\frac{5}{\sqrt{x}}+\frac{16}{x^5}+\cos\ x$  

#3 Question (2018 8a)

A particle movies in a straight line so that its distance , s metres after t seconds, measured from a fixed point O, is given by the function   $s=t^3-2t^2+1-1$  .

Dtermine 
(a) its velcoity when t = 2
(b) the values of t when the particle is at rest

#3 Answers

(a)  $v=\frac{ds}{dt}$  

 $s=t^3-2t^2+t-1$  
 $\frac{ds}{dt}=3t^2-4t+1$  
 $\therefore v=3t^2-4t+1$  
 $when\ t=2$  
 $v=3\left(2\right)^2-4\left(2\right)+1$  
 $v=5ms^{-1}$  

(b)  $at\ rest\ \frac{ds}{dt}=0$  

 $3t^2-4t+1=0$  
 $\left(t-1\right)\left(3t-1\right)=0$  
 $t=1\ or\ t=\frac{1}{3}$  

#4 Question

(a) Integrate  $\int\sqrt{\left(2x-3\right)^3}$  dx.

(b) Find the volume generated when the region between the curve  $y=\sqrt{4-x^2}$  , x = 1  and  x = 2 is rotated 360 degrees about the x axis. 

#4 Answers

(a)  $^2\sqrt{\left(2x-7\right)^3}dx$           (b) $V=\pi\int_{x_2}^{x_1}y^2dx$          $V=\pi\int_1^2\left(4-x^2\right)dx$  
 $=\int\left(2x-7\right)^{\frac{3}{2}}$                       $y=\left(4-x^2\right)^{\frac{1}{2}}$                $=\pi\left[4x-\frac{x^3}{3}\right]\$  
 $=\frac{\left(2x-7\right)^{\frac{3}{2}+1}}{\frac{3}{2}+1}\times\frac{1}{2}+c$       $y^2=\left(\left(4-x\right)^{\frac{1}{2}^{ }}\right)^2$         $=\pi\left[\left(4\left(2\right)-\frac{\left(2\right)^3}{3}\right)-\left(4\left(1\right)-\frac{\left(1\right)^3}{3}\right)\right]$  
 $=\frac{\left(2x-7\right)^{\frac{5}{2}}}{\frac{5}{2}\left(2\right)}+c$                    $=4-x^2$                         $=\pi\left(\frac{16}{3}-\frac{11}{3}\right)$  
 $=\frac{\sqrt{\left(2x-7\right)^5}}{5}+c$                                                          $=\frac{5}{3}\pi cubic\ units$  

# 5 Questions (2015 8b)

A particle moves in a straight line so that t seconds after passing through a fixed point O, its acceleration, a, is given by   $a=\left(3t-1\right)m\ s^{-2}$   . The particle has a velocity, v, of  $4ms^{-1}$  when   t = 2 and its displacement , s, from O is 6 metres when   t = 2. Find :
(i) the velocity when t = 4
(ii) the displacement of the particle from O when t = 3.

#5 Answers

 $\left(a\right)\ \frac{dv}{dt}=a\$                    (b) $v=\frac{ds}{dt}$                                        $when\ t=3$  
 $v=\int a\ dt$                            $s=\int v$                           $s=\frac{\left(3\right)^3}{2}-\frac{\left(3\right)^2}{2}+4$  
 $v=\int\ \left(3t-1\right)dt$             $s=\int\left(\frac{3}{2}t^2-t\right)dt$               $s=\frac{27}{2}-\frac{9}{2}+4$                     
 $v=\frac{3t^2}{2}-t+c$                     $s=\frac{3t^3}{6}-\frac{t^2}{2}+c$                    $=13m$  
 $when\ t=2\ ,\ v=4$           $s=\frac{1}{2}t^3-\frac{t^2}{2}+c$  
 $4=\frac{3\left(2\right)^2}{2}-2+C$               $when\ s=6,t=2$  
 $4=4+C$                               $6=\frac{\left(2\right)^3}{2}-\frac{\left(2\right)^2}{2}+c$  
 $C=0$                                                 $6=2+c$  
 $v=\frac{3}{2}t^2-t$     ,When t = 4 :                $c=4$  
 $v=\frac{3\left(4\right)^2}{2}-4=20ms^{-1}$           $s=\frac{t^3}{2}-\frac{t^2}{2}+4$  

Probability and Statistics

*Probability theory, including solving probability problems using sample / possibility space diagram

*Data representation and Analysis

#1 Question (2014 7b)

Two ordinary six-sided dice are thrown together. The random variable S represents the sum of the two score of their face landing uppermost.

(a) Copy and complete the sample space diagram below.

(b) Find

(i) P ( S > 9)

(ii) P ( S ≥ 4)

(c) Let D be the difference between the scores of the faces landing uppermost. The table below gives the probability of each possible value of d.

Find the values of a,b,c.

#1 Answers

b. (i) $P\left(S>9\right)=\frac{6}{36}=\frac{1}{6}$  
 $\left(ii\right)P\left(S\le4\right)=\frac{6}{36}=\frac{1}{6}$  

(c)
 $a=\frac{10}{36}=\frac{5}{18}$  
 $\left\{\left(2,1\right),\left(3,2\right)\left(4,3\right)\left(5,4\right)\left(6,5\right)\right\}$  
 $\left\{\left(1,2\right)\left(2,3\right)\left(3,4\right)\left(5,4\right)\left(5,6\right)\right\}$  

 $b=\frac{6}{36}=\frac{1}{6}$  
 $\left\{\left(4,1\right)\left(5,2\right)\left(6,3\right)\right\}\left\{\left(1,4\right)\left(2,5\right)\left(3,6\right)\right\}$  

 $c=\frac{2}{36}=\frac{1}{18}$  
 $\left\{\left(6,1\right)\left(1,6\right)\right\}$  

#2 Question

Blood Samples were taken from 40 blood donors and the lead concentration (in mg per 100) of each sample was determined. The results are given below.

(a) Construct a stem and leaf diagram to represent the data.

(b) For this data, write down the values of the range and median.

#2 Answers

(a) Key : 1 | 7means 17 mg PER 100ML

(b) range = highest value - lowest value

 $=65-17=48$  
 $median\ =\ \frac{20^{th}+21^{st}}{2}=\frac{30+30}{2}=30$  

#3 Questions

The ages (in years) of a group of people visiting a public toilet are given below.

(a) Compute the quartiles for the data.

(b) Determine the Interquartile range (IQR)

(c) Illustrate the data using a box and whisker plot.

# 3 Answers

Data in ascending order  $14,\ 20,\ 23,\ 25,\ 26,\ 26,\ 27,\ 27,\ 27$  

 $28,\ 28,\ 29,\ 29,\ 30,\ 31,\ 31,\ 32,\ 34\$  
 $34,\ 36,\ 37,\ 40,\ 40,\ 40,\ 42,\ 52,\ 76$  
(a) $Q2=\left(\frac{n+1}{2}\right)^{th}=\frac{27+1}{2}=14th\ =\ 30$   $Q1=\left(\frac{n+1}{2}\right)^{th}=\frac{13+1}{2}=7th\left(1st\ set\right)\ =27$  

 $Q3=\left(\frac{n+1}{2}\right)^{th}=\frac{13+1}{2}=7th\left(2nd\ set\right)=37$  
(b)   $IQR=Q3-Q1$   
  $=37-27$  
  $=10$  

Min Value = 14, Max value = 76

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CONGRATS YOU ARE AT THE END !

GOOD JOB KEEP STRIVING AND GOOD LUCK IN YOUR CSEC ADDITIONAL MATHEMATICS EXAMINATION! YOU GOT THIS !

~KASSIA~