一辆新摩托车的现金价格为 $25 120。 杰克（Jack）付了摩托车价格 20% 的首付款，余款在 3 年内以每年 4% 的固定单利计算。

1. Find the down payment and balance to be paid by instalments.

Down payment $=20\%\times 25120=\char36{}5024$

Balance $=25120-5024=\char36{}20096$

2. Calculate the simple interest on the balance for 3 years.

Simple interest $=4\%\times 20096\times 3$

$=\char36{}2411.52$

3. Find the total amount to be repaid and divide by the number of months.

Total repayment $=20096+2411.52=\char36{}22507.52$

Number of months $=3\times 12=36$

Monthly instalment $=22507.52\div 36=625.208\ldots$

1. Convert the fuel capacity from gallons to litres and round to the nearest litre.

Fuel capacity $=3.7\times 3.785=14.0045$ litres

Correct to the nearest litre, this is $14$ litres.

1. Find the amount of petrol Jack would be expected to use according to the manufacturer's stated fuel efficiency.

Manufacturer's stated fuel efficiency is $6$ litres per $100$ km.

For $190$ km, expected petrol used is
$6 \times \frac{190}{100}=11.4\text{ litres}.$

2. Compare the expected petrol used with the actual petrol pumped and make a conclusion.

Actual petrol used $=13$ litres.

Since
$13>11.4,$
Jack used more petrol than stated by the manufacturer for $190$ km.

So the manufacturer's stated fuel efficiency is not true.

1. Find the height of one sidewall of the front tyre.

For the front tyre $120/70\,R17$, the section width is $120$ mm and the aspect ratio is $70\%$.

$h=70\%\times 120=84\text{ mm}$

2. Convert the inner rim diameter to millimetres and find the outer tyre diameter.

The inner rim diameter is $17$ inches.

Since $1\text{ inch}=2.54\text{ cm}$,

$17\text{ inches}=17\times 2.54=43.18\text{ cm}=431.8\text{ mm}$

Outer diameter of tyre:

$431.8+2(84)=599.8\text{ mm}$

3. Find the circumference of the front wheel.

Convert the outer diameter to metres:

$599.8\text{ mm}=0.5998\text{ m}$

Circumference of wheel:

$C=\pi d=\pi(0.5998)=1.884327\ldots\text{ m}$

4. Divide the replacement distance by the circumference to get the number of complete revolutions.

The front tyre is replaced every $6000$ km.

$6000\text{ km}=6\,000\,000\text{ m}$

Number of revolutions:

$\frac{6\,000\,000}{1.884327\ldots}=3\,184\,160.2\ldots$

Hence the number of complete revolutions is

$3\,184\,160$