Megan goes on walking holiday for five days.
The table shows how far she walked on the first four days.
Monday-14km, Tuesday-23km, Wednesday-13km, Thursday-13km
Megan says,'My average for the first four days is more than 15km.'
Explain why Megan is correct.
Friday is her last day.
She wants to increase her average to 17 km.
How many kilometres must she walk on Friday?

  a) Megan's total so far is 3 km more than necessary for an average of 15 km

  b) 22 km

Step-by-step explanation:

a) The total of differences from 15 is ...

  -1 + 8 + (-2) +(-2) = 3

Since this is more than 0, Megan's average is more than 15.

__

b) The total of differences from 17 is ...

  -3 +6 -4 -4 = -5

To increase her average to 17 km, Megan must walk 17+5 = 22 km on Friday.

Comment on the approach

In general, I find it easier mentally to deal with small numbers than with larger ones. So adding small differences can be easier than computing larger sums.

a) By computing the difference from the supposed average, we have effectively found the difference of the sum of values from the total necessary to give the required average. That is, Megan's average would be 15 if her total distance were 4·15 = 60. Her total distance is (4·15 +3) = 63, so her average is more than 15.

b) By finding out how much the total is below that necessary for the desired average, we determine the amount above that average the final number needs to be.

  (14 -17) +(23 -17) +(13 -17) +(13 -17) +(Friday -17) = 0 . . . . the Friday value to make an average of 17

  14 +23 +13 +13 +Friday = 5·17 . . . . . . . add 5·17

  (14 +23 +13 +13 +Friday)/5 = 17 . . . . . . shows the Friday amount gives the required average

  Friday = 17 -((14 -17) +(23 -17) +(13 -17) +(13 -17)) . . . . first equation rearranged to show how we computed Friday

  Friday = 17 -(-5) = 17 +5 = 22

hope this .

r3t40