2.1. Write down the HCF of 72 and 120 . 2.2. Write 1485 as a product of its prime factors. Show your working. 2.3. Determine the sum of all the factors of 100 2.4. Show that 899 is not a prime number. 2.5. Divide 240 in the ratio \( 5: 3: 4 \) 2.6. Ntsako's car uses 1 litre of fuel to travel 12 km . How much fuel will be needed to 2.7. A certain distance is covered in 3 hours at an average speed of \( 120 \mathrm{~km} / \mathrm{h} \). How Long will it take to cover the same distance at an average speed of \( 90 \mathrm{~km} / \mathrm{h} \) ?

2.1. The prime factorization of \(72\) is \[ 72=2^3\times3^2, \] and that of \(120\) is \[ 120=2^3\times3\times5. \] The HCF is found by taking the lowest powers of the common prime factors: \[ \text{HCF}=2^3\times3=8\times3=24. \] 2.2. Factorize \(1485\) as follows: \[ 1485\div3=495, \] \[ 495\div3=165, \] \[ 165\div3=55, \] and \[ 55=5\times11. \] Thus, the prime factorization is \[ 1485=3\times3\times3\times5\times11=3^3\times5\times11. \] 2.3. First, write \(100\) as a product of primes: \[ 100=2^2\times5^2. \] The sum of all factors of a number \(n=p^a\times q^b\) is given by \[ \sigma(n)=\left(\frac{p^{a+1}-1}{p-1}\right)\left(\frac{q^{b+1}-1}{q-1}\right). \] For \(100\): \[ \sigma(100)=\left(\frac{2^{2+1}-1}{2-1}\right)\left(\frac{5^{2+1}-1}{5-1}\right) =\left(\frac{2^3-1}{1}\right)\left(\frac{5^3-1}{4}\right) =\left(8-1\right)\left(\frac{125-1}{4}\right). \] This simplifies to: \[ 7\times\frac{124}{4}=7\times31=217. \] 2.4. To show that \(899\) is not a prime number, we look for factors other than \(1\) and \(899\). Notice that \[ 29\times31=899, \] so \(899\) can be factored as \[ 899=29\times31. \] Thus, \(899\) is not a prime number. 2.5. We need to divide \(240\) in the ratio \(5:3:4\). First, find the total number of parts: \[ 5+3+4=12. \] Then, compute each share: \[ \text{First share}=\frac{5}{12}\times240=100, \] \[ \text{Second share}=\frac{3}{12}\times240=60, \] \[ \text{Third share}=\frac{4}{12}\times240=80. \] 2.6. Ntsako's car uses \(1\) litre of fuel to travel \(12\) km. Therefore, to travel a distance \(d\) km, the fuel needed is given by \[ \text{Fuel}=\frac{d}{12}\text{ litres}. \] (If a specific distance were given, you would substitute it for \(d\). For example, to travel \(60\) km, the required fuel is \[ \frac{60}{12}=5\text{ litres}. \]) 2.7. First, determine the distance covered in \(3\) hours at \(120\) km/h: \[ \text{Distance}=120\times3=360\text{ km}. \] Now, to cover \(360\) km at \(90\) km/h, the time required is \[ \text{Time}=\frac{360}{90}=4\text{ hours}. \]