Question

Rob is a temporary excavator operator. He is just learning how to use the machine, so on the first day he is only able to move 3 tons of dirt. Each day he gets a little bit better and is able to move 6% more than the previous day. On his last day of work he is able to move approximately 6.8 tons.a) How long did Rob work? (round to the nearest day)b) How much dirt in total did Rob move from his first day to his last? Round to one decimal.

Answer 1

The question given requires the knowledge of geometric progression

If he moves 6% more, this is equivalent to:

$(100+6)\text{ \%=}\frac{106}{100}=1.06$

On the first day, he moved 3 tons of dirt

On the second day, he will move :

$3(100+\frac{6}{100})=3(1.06)\text{ tons}$

On the third day, he will move :

$\begin{gathered} 3(1.06)\text{ x(1.06) tons} \\ \Rightarrow\text{ 3}(1.06)^2 \end{gathered}$

So we can see a pattern here.

This pattern is the model of the function which is given by

$T_n=3(1.06)^{n-1}$

where n is the number of days.

For part A.

To get how long he took to move 6.8 tons

$6.8=3(1.06)^{n-1}$

we will then solve for n

$\begin{gathered} (1.06)^{n-1}=\frac{6.8}{3}=2.667 \\ (1.06)^{n-1}=2.667 \end{gathered}$

find the logarithms of both sides

$\begin{gathered} (n-1)\log 1.06=\log 2.667 \\ n-1=\frac{\log 2.667}{\log 1.06}=16.835 \\ \end{gathered}$

Make n the subject of the formula

Approximately

$\begin{gathered} n=16.835+1 \\ n=17.835 \end{gathered}$

Approximately, this is about 18 days

For part B

To get the total dirt moved from his first day to the last day, we will apply the formula

$\begin{gathered} S_n=\frac{a(r^n-1)}{r-1} \\ \text{where a= first term, n is the number of days, r is the common ratio} \end{gathered}$$S_n=\frac{3(1.06^{18}-1)}{1.06-1}=\frac{5.563}{0.06}=92.7tons$