Answer:
a) Qs = 50 + 20p - 7ps
= 50 + 20p - 7×(2)
= 50 + 20p - 14
= 36 + 20p
At equilibrium, [tex]Q_{d}[/tex] = [tex]Q_{s}[/tex]
So, 150 - 10p + 5[tex]p_{b}[/tex] = 36 + 20p
So, 20p + 10p = 30p
= 150 - 36 + 5[tex]p_{b}[/tex]
= 114 + 5[tex]p_{b}[/tex]
So, p = (114/30) + (5/30)[tex]p_{b}[/tex]
= 3.8 + 0.17[tex]p_{b}[/tex]
Thus, [tex]p_{e}[/tex] = 3.8 + 0.17[tex]p_{b}[/tex]
Q = 36 + 20p
= 36 + 20(3.8 + 0.17[tex]p_{b}[/tex])
= 36 + 76 + 3.4[tex]p_{b}[/tex]
= 112 + 3.4[tex]p_{b}[/tex]
Thus, [tex]Q_{e}[/tex] = 112 + 3.4[tex]p_{b}[/tex]
b) [tex]p_{e}[/tex] = 3.8 + 0.17[tex]p_{b}[/tex]
= 3.8 + 0.17×(5)
= 3.8 + .85
= 4.65
[tex]Q_{e}[/tex] = 112 + 3.4[tex]_{b}[/tex]
= 112 + 3.4(5)
= 112 + 17
= 129
c) Qd = 150 - 10p + 5pb = 150 - 10(2.5) + 5(5) = 150 - 25 + 25 = 150
Qs = 36 + 20p = 36 + 20(2.5) = 36 + 50 = 86
Thus, there is excess demand as [tex]Q_{d}[/tex] > [tex]Q_{s}[/tex]
d) New [tex]Q_{d}[/tex]= 180 - 10p + 5[tex]p_{b}[/tex]
= 180 - 10p + 5×(5)
= 180 - 10p + 25
= 205 - 10p
Now, new [tex]Q_{d}[/tex] = [tex]Q_{s}[/tex] gives,
205 - 10p = 36 + 20p
So, 20p + 10p = 205 - 36
So, 30p = 169
So, p = 169÷30
So, [tex]p_{e}[/tex] = 5.63
Q = 205 - 10p = 205 - 10×(5.63) = 205 - 56.3 = 148.7
So, [tex]Q_{e}[/tex] = 148.7