The average of two numbers is half way between them

Label the shortest distance $n$

$11n=$ distance from $40$ to $150$

$11n=110\Rightarrow n=10$

$a=150+n+4n=200$

$b$ is made from $a$ and $40$, weighted equally

$c$ is made from $a$ and ($a$ and $40$), but $a$ should have weight equal to the weight of ($a$ and $40$). That is the same as using $a$ twice and ($a$ and $40$) once.

$d$ is made from ($a$ and $40$) and (three $a$s and $40$), but again ($a$ and $40$) must be stretched to give equal weights:

And the same for $150$:

$150$ is the average of eleven $a$s and five $40$s

$150=\dfrac {11a+200}{16}$

$\Rightarrow 16\times150 = 11a+200$

$\Rightarrow 2400 = 11a+200$

$\Rightarrow 2200 = 11a$ so $a=200$

$\dfrac{40+a}2=b$ and $\dfrac{a+b}2 = c$

So $\dfrac{a+\frac{40+a}2}2=c \Rightarrow \dfrac{40+3a}4=c$

$d=\dfrac{b+c}2 = \dfrac{\frac{40+a}2+\frac{40+3a}4}2 = \dfrac{2(40+a)+40+3a}8 = \dfrac {120 + 5a} 8$

$150=\dfrac{c+d}2 = \dfrac{\frac{40+3a}4+\frac {120 + 5a} 8}2 = \dfrac{80+6a + 120+5a}{16} =\dfrac{200+11a}{16}$

$\Rightarrow 16\times150 = 200 + 11a$

$\Rightarrow 2400 = 200 + 11a$

$\Rightarrow 2200 = 11a$ so $a=200$

$150$ is the mean of $c$ and $d$ so $c +d =300\Rightarrow d = 300 - c$ (make $d$ the subject to get rid of $d$, then get rid of $c$ etc until only $a$ remains)

$d$ is the mean of $b$ and $c$ so $b+c=2d=2(300-c)$

$\Rightarrow b+c=600-2c \Rightarrow 3c=600-b \Rightarrow c = \frac13(600-b)$

$c$ is the mean of $a$ and $b$ so $a+b=2c=2\times \frac13(600-b)$

$\Rightarrow 3a+3b=1200-2b \Rightarrow 5b = 1200-3a \Rightarrow b = \frac15(1200-3a)$

$b$ is the mean of $40$ and $a$ so $40+a=2b=2\times\frac15(1200-3a)$

$\Rightarrow 200+5a = 2400-6a \Rightarrow 11a = 2200 \Rightarrow a=200$

You can find more short problems, arranged by curriculum topic, in our short problems collection.