Intermediate June 2018 Unit 2 Q7
7. Calculate the mean of these four time periods.
You must give your answer in hours and minutes. [4]
5 hours 20 minutes 2 hours 44 minutes 6 hours 18 minutes 4 hours 34 minutes
Mean time = ………… hours ………… minutes
| Scheme | Marks | Comments |
|---|---|---|
| An attempt to find the total of the four time periods. | M1 | Allow any convincing attempt. A total has to be found. e.g. sight of 18·16 or 18h 16min etc. Not enough to simply list e.g.5 20 + 2 44 + 6 18 + 4 34 |
| (Sum of time periods =) 18(hr) 56(min) OR 1136(min) | A1 | C.A.O. Allow 18·93(…)(hr) but mark final answer. |
| ÷ 4 | m1 | FT for m1, only if ‘their sum of time periods’ is between 17h (1020min) and 21h (1260min) inclusive. |
| = 4 hours 44 minutes | A1 | Allow FT A1 only if the sum of their time period is x hrs y min where x is not a multiple of 4 and y ≠ 0. OR the sum of their time period is t minutes, where t is not a multiple of 60. Sight of 284 (min) implies M1A1m1. Note 1: If time is incorrectly added as ‘decimals’. 18·16 or 18h 16min is M1A0. Further work of 18·16 ÷ 4 = 4·54 (or 4h54m) is m1A0 BUT 18h 16m ÷ 4 = 4h 34m is m1A1 (FT) Note 2: Incorrect use of calculator. e.g. M1A0m1A0 for sight of 930·5(min) (From 320 + 164 + 378 + 274 ÷ 4) |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| Attempt to add time periods as ‘hours + min’ | M1 | |
| 17hours (+) 116 minutes | A1 | |
| ÷ 4 | m1 | FT for m1, only if ‘their sum of time periods’ is between 17h (1020min) and 21h (1260min) inclusive. |
| 4 hours 44 minutes | A1 |