a. Let [tex]d[/tex] be the number of daytime calls Eshwa makes, and [tex]e[/tex] the number of evening calls. Then the cost [tex]C[/tex] (in pence) of making [tex]d+e[/tex] calls is
[tex]C = \boxed{50d + 40e}[/tex]
b. £1 = 100p, so the cost [tex]C'[/tex] (in £) is 1/100 of the cost found in part (a),
[tex]C' = \dfrac{50d + 40e}{100} = \boxed{\dfrac d2 + \dfrac{2e}5}[/tex]
c. If Eshwa makes 30 of each type of call in a month, then the total cost (in £) is
[tex]C' = \dfrac{30}2 + \dfrac{2\cdot30}5 = \boxed{27}[/tex]
c2. If Eshwa makes 20 daytime calls and 50 evening calls, then the total cost (in £) is
[tex]C' = \dfrac{20}2 + \dfrac{2\cdot50}5 = \boxed{30}[/tex]
d. Let [tex]e=40[/tex] and [tex]C'=42[/tex]. Solve for [tex]d[/tex].
[tex]42 = \dfrac d2 + \dfrac{2\cdot40}5[/tex]
[tex]42 = \dfrac d2 + 16[/tex]
[tex]\dfrac d2 = 26[/tex]
[tex]d = \boxed{52}[/tex]
