Milana has had six tests in her math class so far. She knows the average (mean) of some of her marks.
The average of her first and second test marks is \(85\%\).
The average of her second and third test marks is \(82\%\).
The average of her third and fourth test marks is \(74\%\).
The average of her fourth and fifth test marks is \(79\%\).
The average of her fifth and sixth test marks is \(80\%\).
Determine the average of her first and sixth test marks.
Let \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) represent Milana’s six test marks, in order. We are looking for the average of \(a\) and \(f\). We know the following.
The average of her first and second test marks is \(85\), so \(\dfrac{a+b}{2}=85\), which gives \(a+b=170\).
The average of her second and third test marks is \(82\), so \(\dfrac{b+c}{2}=82\), which gives \(b+c=164\).
The average of her third and fourth test marks is \(74\), so \(\dfrac{c+d}{2}=74\), which gives \(c+d=148\).
The average of her fourth and fifth test marks is \(79\), so \(\dfrac{d+e}{2}=79\), which gives \(d+e=158\).
The average of her fifth and sixth test marks is \(80\), so \(\dfrac{e+f}{2}=80\), which gives \(e+f=160\).
Adding these six equations gives the following: \[\begin{aligned} (a+b) + (b+c) + (c+d) + (d+e) + (e+f) &= 170 + 164 + 148 + 158 + 160\\ a + 2b + 2c + 2d + 2e + f &= 800\\ a + 2(b+c) + 2(d+e) + f &= 800\\ a + 2(164) + 2(158) + f &= 800\\ a + 328 + 316 + f &= 800\\ a + f &= 156\\ \frac{a+f}{2} &= 78 \end{aligned}\] Therefore, the average of Milana’s first and sixth test marks is \(78\%\).