hC+10/0/01110 (a) The cash price of a mobile set is 10% lessthan the hire purchase price.instalments of M150.H.P=300+12(150)Calculate=M2100(i) the cash priceCash Price=H.P_1xH.P=M2100-10xM2100=M1890Answer (a)(i) M.1.8.9.O[3](ii) the rate of interest.=Interestx100%cash Price=M210/0x100%=11.1%Answer(a)(ii)11.1%[3](b) A bus travellingat an average speed of 45 km/h left the bus stop at 8.15 am.A taxi left the bus stop at 9.00 am and overtook the bus at 10.45 am.How much time, in hours, did the taxi take before overtaking the bus.Time taken by:Answer(b)(i)1.75hours[2](ii) Find the average speed of the taxi./taxci =■=112.5km=64.3km/h64.3Answer (b)(ii)km/h[3]Distance covered is common for both bus ■ Taoci::Distance=Sxt=45km/x2.5h=112.5kmECoL 20200178/03/O/N/20[Turn over

Let's break down each part of this question and provide clear explanations for the steps involved.

### Part (a)
#### (i) The cash price of a mobile set:
- The hire purchase (H.P.) price is given as:
  \[
  \text{H.P.} = 300 + 12 \times 150 = 300 + 1800 = 2100
  \]
- The cash price is 10% less than the hire purchase price, so:
  \[
  \text{Cash Price} = \text{H.P.} - \left(0.10 \times \text{H.P.}\right)
  \]
  \[
  \text{Cash Price} = 2100 - \left(0.10 \times 2100\right) = 2100 - 210 = 1890
  \]
  Therefore, the cash price is **M1890**.

#### (ii) The rate of interest:
- The rate of interest is calculated as:
  \[
  \text{Interest} = \text{H.P.} - \text{Cash Price} = 2100 - 1890 = 210
  \]
- The interest rate as a percentage of the cash price is:
  \[
  \text{Interest Rate} = \frac{\text{Interest}}{\text{Cash Price}} \times 100 = \frac{210}{1890} \times 100 \approx 11.11\%
  \]
  Therefore, the interest rate is **11.1%**.

### Part (b)
#### (i) Time taken by the taxi to overtake the bus:
- The bus left the bus stop at 8:15 AM and the taxi left at 9:00 AM.
- The taxi overtook the bus at 10:45 AM. So, the time difference is:
  \[
  10:45 \text{ AM} - 9:00 \text{ AM} = 1.75 \text{ hours}
  \]
  Therefore, the time taken by the taxi to overtake the bus is **1.75 hours**.

#### (ii) The average speed of the taxi:
- The distance covered by both the bus and taxi is the same.
- The bus traveled for 2.5 hours (from 8:15 AM to 10:45 AM) at a speed of 45 km/h:
  \[
  \text{Distance} = 45 \, \text{km/h} \times 2.5 \, \text{hours} = 112.5 \, \text{km}
  \]
- The taxi traveled the same distance in 1.75 hours. So, the speed of the taxi is:
  \[
  \text{Speed of taxi} = \frac{\text{Distance}}{\text{Time}} = \frac{112.5}{1.75} \approx 64.29 \, \text{km/h}
  \]
  Therefore, the average speed of the taxi is **64.3 km/h**.

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Let me know if you need any clarifications or further details. Here are 5 related questions to expand your understanding:

1. How is the cash price of an item related to its hire purchase price?
2. What factors determine the rate of interest in a hire purchase agreement?
3. How would the speed of the bus affect the time it takes for the taxi to overtake it?
4. If the interest rate increased, how would that affect the cash price and hire purchase price?
5. How could we calculate the time for the bus to reach a different distance, given its speed?

**Tip**: When calculating interest rates or percentages, always ensure that you're using consistent units and values.

The cash price of a mobile set is 10% less than the hire purchase price. The hire purchase price is M2100 (12 instalments of M150 plus a deposit of M300). Calculate (i) the cash price and (ii) the rate of interest. Additionally, a bus traveling at 45 km/h left at 8:15 am and was overtaken by a taxi that left at 9:00 am and overtook the bus at 10:45 am. Calculate (i) how much time the taxi took to overtake the bus and (ii) the average speed of the taxi.

Learn how to calculate the cash price, interest rate, and average speed through real-world examples. This detailed explanation covers percentage calculations, simple interest, and speed-distance-time relationships. Suitable for students in Grades 9-10.

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