Simplifying rates – worked solutions
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The document provides worked solutions for simplifying rates and calculating rates of change from given information. It includes examples such as: 1) Calculating the cost per kg for a parcel by dividing the total cost by the total weight. 2) Calculating the rate of temperature change in degrees per hour by dividing the total change by the time period. 3) Calculating a growth rate in cm per year by dividing the change in height by the time period. 4) Calculating a travel rate in km per hour by dividing the total distance by the total time.
Simplifying rates – worked solutions
- 1. Simplifying Rates – Worked Solutions Simplify the following: The 5.5 kg parcel cost $19.25 to post. ∴19.25÷ 5.5 1kg = $3.50 / kg From 6 am to noon, the temperature changed from 10° C to 22° C. From 6am to noon is 6 hours. The temperature changed 12 degrees in 6 hours. ∴In its simplest terms =12 ÷ 6 The temperature therefore changed 2 degrees per hour or 2 C / hr 5f) When Naoum was 10 years old he was 120 cm tall. When he was 18 years old he was 172 cm tall. 8 years gap. 172 −120 = 52cms difference 52 ÷8=6.5 Therefore, Naoum grew at a rate of 6.5cms per year. Written mathematically is: 6.5cms/yr I have divided by 5.5 as there is 5.5 kgs and I wanted to see how much 1kg would cost.
- 2. A cyclist left home at 8.30 am and at 11.00 am had travelled 40 km. 8.30 -11am = 2.5 hours 40 ÷ 2.5 =16kms / hr Which can also be written as 16km/h Be careful with questions like this one above. Best way to work out how many hours is to count from 8.30 to 9 is 30 minutes, and then from 9 to 11 is 2 hours, therefore it is 2.5 hours. If you do these in your head, without working through them logically, you can and will make silly mistakes.