1) A forklift carries a load of 740kg. The packages in the load are either 20kg or 30kg in weight. If the forklift is carrying 30 packages, how many packages of each weight is it carrying?
Let the number of 20kg packages be \(x\) and the number of 30kg packages be \(y\). Then
\begin{align}x+y & = 30 \quad \enclose{circle}{\color{black}{1}}\\ 20x+30y & =740 \quad \enclose{circle}{\color{black}{2}}\\ multiply \; eqn \; \enclose{circle}{\color{black}{1}} \; by \; 20\\ to \; equate \; x \; coefficients\\ 20x+20y& = 600 \quad \enclose{circle}{\color{black}{3}}\\ subtract \; eqn\; \enclose{circle}{\color{black}{3}}\; from \; eqn \; \enclose{circle}{\color{black}{2}} \; to\; eliminate \; x\\ (20x+30y)-(20x+20y) & =740-600\\ 10y & =140\\ y & =14\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ x+14 & =30\\ x & = 16 \end{align}So there are \(16\) packages weighing 20kg, and \(14\) weighing 30kg.
2) A plane flies at 750km/h with the wind. It flies at 690km/h against the wind. Find (a) the speed of the wind; and (b) the speed of the plane when there is no wind.
Let the speed of the plane be \(x\) and the wind speed be \(y\). Then
\begin{align}x+y & =750 \quad \enclose{circle}{\color{black}{1}}\\ x-y & =690 \quad \enclose{circle}{\color{black}{2}}\\ add \; eqn\; \enclose{circle}{\color{black}{2}}\; to \; eqn \; \enclose{circle}{\color{black}{1}} \; to\; eliminate \; y\\ (x+y)+(x-y) & =750+690\\ 2x & =1440\\ x & =720\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 720+y & =750\\ y & = 30\\ \end{align}So plane speed is 720km/h, and wind speed is 30km/h.
3) The points \((3,7)\) and \((-2,-3)\) lie on the line \(y=mx+c\). Find \(m\) and \(c\), and hence the equation of the line.
Substitute the respective points into the eqaution to obtain a pair of simultaneous equations.
\begin{align} 7 & =m(3)+c \quad \enclose{circle}{\color{black}{1}}\\ -3 & = m(-2)+c \quad \enclose{circle}{\color{black}{2}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{2}} \; from\; eqn\; \enclose{circle}{\color{black}{1}}\\ to \; eliminate\; c\\ 7-(-3) & = 3m-(-2)m\\ 10 & =5m\\ m & = 2\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{2}}\\ 7 & =2(3)+c\\ c & =1 \end{align}So the equation of the line is \(y=2x+1\)
4) Chocolates come in boxes of 20 and boxes of 50. A shop orders 20 boxes, containing a total of 880 chocolates. How many boxes of 50 has the shop ordered?
Let boxes of 20 be \(x\) and boxes of 50 be \(y\). Then
\begin{align} x+y & = 20 \quad \enclose{circle}{\color{black}{1}}\\ 20x+50y & = 880 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{1}}\; by\; 20\\ to \; equate\; the \; coefficients \; of \; x\\ 20x+20y & =400 \quad \enclose{circle}{\color{black}{3}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{3}}\; from\; eqn\; \enclose{circle}{\color{black}{1}}\\ to \; eliminate\; x\\ (20x+50y)-(20x+20y) & =880-400\\ 30y & = 480\\ y & = 16\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 16+y & = 20\\ y & = 4\\ \end{align}So the shop has ordered 4 boxes of 50 chocolates.
5) A carpenter needs a number of 8cm and 12cm lengths of wood to build a frame. He cuts a length of 3.8 metres into 35 smaller lengths of 8cm and 12cm. He needs eleven of the 8cm lengths for the frame. Can he build it?
Let the number of 8cm lengths be \(x\) and the number of 12cm lengths be \(y\). Then
\begin{align} x+y & = 35 \quad \enclose{circle}{\color{black}{1}}\\ 8x+12y & = 380 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{1}}\; by\; 8\\ to \; equate\; the \; coefficients \; of \; x\\ 8x+8y & =280 \quad \enclose{circle}{\color{black}{3}}\\ subtract\; eqn\; \enclose{circle}{\color{black}{3}}\; from\; eqn\; \enclose{circle}{\color{black}{2}}\\ to \; eliminate\; x\\ (8x+12y)-(8x+8y) & =380-280\\ 4y & =100\\ y & = 25\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ x+25 & = 35\\ x & = 10\\ \end{align}So the carpenter does not have sufficient lengths of 8cm to build the frame.
6) Entry to a park costs $4 for adults and $2 for children. Two hundred entry tickets are sold, for a total of $670. How many adult's tickets and how many children's tickets were sold?
Let the number of adult tickets be \(x\) and the number of child tickets be \(y\). Then
\begin{align} x+y & = 200 \quad \enclose{circle}{\color{black}{1}}\\ 4x+2y & = 670 \quad \enclose{circle}{\color{black}{2}}\\ multiply\; equation \; \enclose{circle}{\color{black}{1}}\; by\; 2\\ to \; equate\; the \; coefficients \; of \; y\\ 2x+2y & =400\\ subtract\; eqn\; \enclose{circle}{\color{black}{2}}\; from\; eqn\; \enclose{circle}{\color{black}{1}}\\ to \; eliminate\; y\\ (4x+2y)-(2x+2y) & =670-400\\ 2x & = 270\\ x & = 135\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ 135+y & =200\\ y & = 65 \end{align}So \(135\) adult tickets and \(65\) children's tickets were sold.
7) The combined age of a dog and its owner is 38 years. In 5 years time the dog will be \(\frac{3}{4}\) as old as the owner was \(8\) years ago. How old is the dog, and the owner?
Let the age of the dog be \(x\), and the age of the owner be \(y\). Then
\begin{align}x+y & = 38 \quad \enclose{circle}{\color{black}{1}}\\ x+5 & = \frac{3}{4}(y-8) \quad \enclose{circle}{\color{black}{2}}\\ subtract \; \enclose{circle}{\color{black}{2}} \; from \; \enclose{circle}{\color{black}{1}} \; to\; eliminate \; x\\ (x+y)-(x+5) & = 38-\frac{3}{4}(y-8)\\ y-5= 38-\frac{3}{4}(y-8) \quad \enclose{circle}{\color{black}{3}}\\ multiply \; eqn \; \enclose{circle}{\color{black}{3}}\: by\; 4\\ to \; eliminate \; the \; fraction\\ 4y-20 & =152-3(y-8)\\ 4y-20 & = 152-3y+24\\ 7y & =196\\ y & = 28\\ then\; by\; substitution\; into\;\enclose{circle}{\color{black}{1}}\\ x+28 & =38\\ x & = 10 \end{align}So the dog is \(10\) and the owner is \( 28\).
8) When both the numerator and denominator of a certain fraction are increased by \(3\) the fraction becomes\( \frac{3}{4}\). When the numerator and the denominator of the same fraction are each decreased by 3, the fraction becomes \(\frac{1}{2}\). Find the fraction.
Let the numerator of the original fraction be \(x\), and the denominator of the original fraction be \(y\).
\begin{align}\frac{x+3}{y+3} & = \frac{3}{4} \quad \enclose{circle}{\color{black}{1}}\\ \frac{x-3}{y-3} & = \frac{1}{2} \quad \enclose{circle}{\color{black}{2}}\\ cross \;multiply \; eqn \; \enclose{circle}{\color{black}{1}}\\ 4(x+3) & = 3(y+3)\\ 4x+12 & = 3y+9\\ 4x-3y & =-3 \quad \quad \enclose{circle}{\color{black}{3}}\\ cross \;multiply \; eqn \; \enclose{circle}{\color{black}{2}}\\ 2(x-3) & = (y-3)\\ 2x-6 & = y-3\\ 2x-y & = 3 \quad \quad \enclose{circle}{\color{black}{4}}\\ multiply \; eqn\; \enclose{circle}{\color{black}{4}} \; by\; 2 \\ to \; equate \; x \; coefficients\\ 4x-2y & = 6 \quad \enclose{circle}{\color{black}{5}}\\ subtract \; eqn \; \enclose{circle}{\color{black}{5}} \; from \; eqn \; \enclose{circle}{\color{black}{3}}\\ (4x-3y)-(4x-2y) & = -3 -6\\ -y & = -9\\ y & =9\\ then\; by\; substitution\; into\; \enclose{circle}{\color{black}{1}}\\ \frac{x+3}{12} & = \frac{3}{4}\\ x+3 & = 9\\ x & = 6\\ \end{align}So the original fraction is \(\frac{6}{9}\)