The cash price of a new motorcycle is $25 120. Jack paid a down payment of 20% of the price of the motorcycle and the balance with a fixed simple interest rate of 4% per annum for a period of 3 years.

  1. ((a))

    Calculate his monthly instalment.

    Answer $...

  2. ((b))

    The technical specification of his new motorbike, as stated by the manufacturer, is shown in Table 1.

    An example of the dimensions for a motorbike tyre 195 / 55 R16 is explained below.

    195 is the section width of the tyre in millimetres.

    55% aspect ratio represents the height of the tyre sidewall as a percentage of the section width.

    Diameter of the tyre's inner rim is 16 inches.

    The cross-section of a wheel can be presented in Diagram II, where d is the diameter of inner rim, and h the height of the sidewall.

    1. ((i))

      Find the fuel capacity of his motorbike, correct to the nearest litre.

      Answer ... litres

    2. ((ii))

      Jack starts riding with a full tank of petrol. After riding for a total distance of 190 km, he filled up the petrol tank to the brim again, by pumping 13 litres of petrol.

      Determine if the fuel efficiency stated by the manufacturer in Table 1 is true.

  3. ((c))

    Jack replaces his tyres at the recommended distance travelled.

    What would be the number of complete revolutions made, when Jack has to first replace a front tyre?

    Use the information in Table 1 and show your calculations clearly.

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Worked solution

12((a))

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

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

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

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

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

=$2411.52=\char36{}2411.52

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

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

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

Monthly instalment =22507.52÷36=625.208…=22507.52\div 36=625.208\ldots

Answer

$586.13

12((b))((i))

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

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

Correct to the nearest litre, this is 1414 litres.

Answer

1414 litres

12((b))((ii))

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 66 litres per 100100 km.

For 190190 km, expected petrol used is
6×190100=11.4 litres.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=13 litres.

Since
13>11.4,13>11.4,
Jack used more petrol than stated by the manufacturer for 190190 km.

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

Answer

No. The stated fuel efficiency is not true, since Jack used 1313 litres for 190190 km, which is about 6.846.84 litres per 100100 km, more than 66 litres per 100100 km.

12((c))

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

For the front tyre 120/70 R17120/70\,R17, the section width is 120120 mm and the aspect ratio is 70%70\%.

h=70%×120=84 mmh=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 1717 inches.

Since 1 inch=2.54 cm1\text{ inch}=2.54\text{ cm},

17 inches=17×2.54=43.18 cm=431.8 mm17\text{ inches}=17\times 2.54=43.18\text{ cm}=431.8\text{ mm}

Outer diameter of tyre:

431.8+2(84)=599.8 mm431.8+2(84)=599.8\text{ mm}

3. Find the circumference of the front wheel.

Convert the outer diameter to metres:

599.8 mm=0.5998 m599.8\text{ mm}=0.5998\text{ m}

Circumference of wheel:

C=πd=π(0.5998)=1.884327… mC=\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 60006000 km.

6000 km=6 000 000 m6000\text{ km}=6\,000\,000\text{ m}

Number of revolutions:

6 000 0001.884327…=3 184 160.2…\frac{6\,000\,000}{1.884327\ldots}=3\,184\,160.2\ldots

Hence the number of complete revolutions is

3 184 1603\,184\,160

Answer

3 184 1603\,184\,160 complete revolutions

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