1) The sequence \(4, \; 7,\;12, \; 19...\) contains 172 as a term. Form an expression for the sequence and use this in an equation to find what number term this is.
The 'first-differences' of the original sequence are \(3, \;5,\;7...\)
Therefore the 'second-difference' is \(2\).
Half the 'second-difference' gives us the coefficient of \(n^2\) (i.e. the squared part of our expression).
So our expression is \(n^2+x\) where we need to find \(x\).
Substitute \(n=1\) and equate to the first term of the original sequence.
\begin{align} 1^2+x & =4 \\ So \quad x & = 3\\ and \; the \; expression \; for \; our \; sequence \; must \; be \\ n^2+3\\ forming \; an \; equation \; with \; our \; term \\ n^2+3 & =172\\ n^2 & = 169 \\ n & = \pm 13\\ we \; reject \; the \; negative \; solution \\ So \quad n & = 13 \\ \end{align}Therefore 172 is the thirteenth term of our sequence.
2) A certain positive number exceeds 8 times its reciprocal by 2. Find the number.
Our answer must be positive, therefore the number is \(4\).
3) A boy increases his walking speed by \(2 \;km/h\) and thereby reduces the time for his \(12 \;km\) journey by \(1 \; hour\). Find his original speed.
Let \(v\) be speed. First equate the two times, then solve.
\begin{align} \frac{12}{v+2} & = \frac{12}{v}-1\\ 12 & = 12+ \frac{24}{v}-v-2\\ 0 & = \frac{24}{v}-v-2\\ 0 & =-v^2-2v-24\\ 0 & = v^2+2v-24\\ 0 & = (v+6)(v-4) \end{align}Therefore his original speed was \( 4 \; km/h\).
4) A standard light-weight rod being made in a factory weighs \(60 \; g \). The factory increases the volume of the rod by \( 2 \; cm^3\) and its density is thereby reduced by \(1 \; g/cm^3\). What was the volume of the original standard rod?
Let \( v \) be volume. First equate the densities, then solve.
\begin{align} \frac{60}{v+2} & = \frac{60}{v}-1\\ 60 & = 60 +\frac{120}{v}-v-2\\ 0 & =v^2+2v-120\\ 0 & = (v+12)(v-10)\\ \end{align}Therefore the original volume was \(10 \; cm ^3\).
5) The square of the largest of three consecutive positive integers is \(32\) less than the sum of the individual squares of the two preceding numbers. Find the three numbers.
Let the three numbers be \(x, \; x+1, \; and \; x+2\)
\begin{align} (x+2)^2 & = x^2 +(x+1)^2-32\\ x^2+4x+4 & = x^2+x^2+2x+1-32\\ x^2+4x+4 & = 2x^2+2x-31\\ 0 & = x^2-2x-35\\ 0 & = (x-7)(x+5)\\ x = 7 \; or \; x & = -5 \end{align}We need positive integers, therefore the three consecutive numbers are 7, 8 and 9 .
6) The surface area of a cube with side-length \( (x-2) cm\) exceeds the area of a square with side-length \((x-2) cm \) by \(405 \; cm^2\). Find the side-length in centimeters.
First form an equation, then solve.
\begin{align} \; 6(x-2)^2 & = (x-2)^2+405\\ 6(x^2-4x+4) & = x^2-4x+4+405\\ 6x^2-24x+24 & = x^2-4x+409\\ 0 & = 5x^2-20x-385\\ 0 & = x^2-4x-77\\ (x-11)(x+7) & = 0 \\ So \; x=11 \; or \; x & =-7\\ we \; reject \; the \; negative \; solution\\ \therefore \; x & =11 \end{align}Therefore the side-length is \( 9 \; cm\).
7) The denominator of an improper fraction is 8 less than the numerator. If we deduct 3 from both the numerator and the denominator, we create a number that is one fifth of the original numerator. Find the two possible original fractions.
First form an equation, then solve.
\begin{align}\frac{x-3}{(x-8)-3} & = \frac{1}{5}x\\ \frac{x-3}{x-11} & = \frac{1}{5}x\\ x-3 & = \frac{1}{5}x(x-11)\\ x-3 & = \frac{x^2-11x}{5} \\ 5x-15 & = x^2-11x\\ 0 & =x^2-16x+15\\ 0 & = (x-15)(x-1)\\ \therefore \; x= 15 \; or \;x & =1 \end{align}Therefore the two possible original fractions are \(\frac{15}{7}\) or \(\frac{1}{-7}\)
8) A gold prospector walks a certain distance on a bearing of 225°. He finds nothing there, and then turns and walks 3 times the distance he has already walked, this time on a bearing of 315°. He stops at a place 100 metres from his original starting point. What is the exact value for the length of the first section of his walk?
Between the two bearings, the prospector has turned through an angle of 90°. Therefore we can use Pythagoras' Theorem to form our equation.
\begin{align}(3x)^2+x^2 & =100\\ 9x^2+x^2 & = 100\\ 10x^2 & =100\\ x^2 & =10\\ x & = \sqrt{10}\\ \end{align}Therefore the first section of the prospector's walk was \(\sqrt{10} \; metres\)