The first term of an Arithmetic Progression (AP) is 8, the ratio of the…

Explanation

(a) (T_{1} = a = -8)

(frac{T_{7}}{T_{9}} = frac{5}{8})

(frac{a + 6d}{a + 8d} = frac{5}{8} implies frac{-8 + 6d}{-8 + 8d} = frac{5}{8})

(-8 + 6d = frac{5}{8}(-8 + 8d) implies -8 + 6d = -5 + 5d)

(6d – 5d = -5 + 8)

(d = 3)

(b) Let :

the cost price (CP_{1}) of a basket of pawpaw be w;

the cost price (CP_{2}) of a basket of mangoes be m.

Then (30w + 100m = N2,450 ….. (1))

Also, let :

The selling price of the baskets of pawpaw be (SP_{1});

the selling price of the baskets of mangoes be (SP_{2}).

Then (SP_{1} + SP_{2} = N(2,450 + 855) = N3,305 …… (2))

%age profit on the sale of pawpaw = (frac{SP_{1} – CP_{1}}{CP_{1}} times 100%)

i.e. (40% = frac{SP_{1} – 30w}{30w} times 100%)

(implies 40 times 3w = (SP_{1} – 30w) times 10)

(12w = SP_{1} – 30w)

(SP_{1} = 12w + 30w = 42w)

%age profit on the sale of mangoes = (frac{SP_{2} – CP_{2}}{CP_{2}} times 100%)

i.e. (30% = frac{SP_{2} – 100m}{100m} times 100%)

(implies 30m = SP_{2} – 100m)

(SP_{2} = 30m + 100m = 130m)

But (SP_{1} + SP_{2} = N3,305)

(implies 42w + 130m = 3305 …. (2))

From (1), (m = frac{2450 – 30w}{100})

(m = frac{245 – 3w}{10})

Substitute (m = frac{245 – 3w}{10}) in (3)

(42w + 130(frac{245 – 3w}{10}) = 3305)

(42w + 3,185 – 39w = 3305)

(3w = 3305 – 3185 = 120)

(w = N40.00)

(i) Hence, the cost price of a basket of pawpaw = N40.00.

(ii) From (2),

(SP_{1} + SP_{2} = 3305)

(42w + 130m = 3305 implies 42 times 40 + 130m = 3305)

(1680 + 130m = 3305)

(130m = SP_{2} = 3305 – 1680)

= (N1,625.00) 

Hence, the selling point of 100 baskets of mangoes = N1,625.00.