Xin has been given a 14 day training schedule by her coach - Edexcel - A-Level Maths Pure - Question 11 - 2014 - Paper 2

For Yi, on day 1, she runs for (4 – 13) minutes.

Her running time increases by (2d – 1) minutes each day, thus for day 14:

• Total increase in running time = 13(2d - 1)

• Therefore, on day 14, her running time will be:

  $ext{Running time on day 14, Yi} = (4 - 13) + 13(2d - 1)$

Setting Xin's and Yi's running times equal:

$4 + 13d + 13 = (4 - 13) + 13(2d - 1)$

Simplifying gives:

$4 + 13d + 13 = -9 + 26d - 13$

Which simplifies to:

$4 + 13d + 13 = 26d - 22$

Now merging terms we get:

$36 = 13d + 26d + 22$

Thus,

$36 = 39d + 22$

This results in:

ightarrow d = rac{14}{39}$$.

Given that Xin runs for a total of 784 minutes over 14 days, we can set up the equation:

Using the formula for total running time over 14 days:

ext{Total running time} = 14 imes rac{(A + (A + 13(d + 1)))}{2}

Substituting into the formula:

784 = 14 imes rac{(A + (A + 13(d + 1)))}{2}

This expands to:

$784 = 7(A + A + 13d + 13)$

This results in:

$784 = 7(2A + 13d + 13)$

Now, dividing by 7:

rac{784}{7} = 2A + 13d + 13

This simplifies down:

$112 = 2A + 13d + 13$

Rearranging gives:

$2A = 112 - 13d - 13$

Solving gives:

$2A = 99 - 13d$

Substituting d = 1:

$2A = 99 - 13(1) = 86$

Thus,

$A = 30.$