The problem consists of two parts. (a) Given an arithmetic progression (AP) with the first term $a_1 = -8$, and the ratio of the 7th term to the 9th term is $5:8$, find the common difference of the AP. (b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find (i) the cost price of a basket of pawpaw; (ii) the selling price of the 100 baskets of mangoes.

1. Problem Description

The problem consists of two parts.

(a) Given an arithmetic progression (AP) with the first term $a_1 = -8$, and the ratio of the 7th term to the 9th term is $5:8$, find the common difference of the AP.

(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.

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0. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find

(i) the cost price of a basket of pawpaw;

(ii) the selling price of the 100 baskets of mangoes.

2. Solution Steps

(a)

The general formula for the nth term of an AP is:

$a_n = a_1 + (n-1)d$

where $a_1$ is the first term and $d$ is the common difference.

We are given that $a_1 = -8$.

The 7th term is $a_7 = a_1 + (7-1)d = -8 + 6d$

The 9th term is $a_9 = a_1 + (9-1)d = -8 + 8d$

We are given that $\frac{a_7}{a_9} = \frac{5}{8}$, so:

$\frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}$

Cross-multiplying gives:

$8(-8 + 6d) = 5(-8 + 8d)$

$-64 + 48d = -40 + 40d$

$48d - 40d = -40 + 64$

$8d = 24$

$d = \frac{24}{8} = 3$

(b)

Let $x$ be the cost price of a basket of pawpaw and $y$ be the cost price of a basket of mangoes.

We have the equation:

$30x + 100y = 2450$ (1)

The profit on pawpaw is 40%, so the profit is $0.40 \times 30x = 12x$.

The profit on mangoes is 30%, so the profit is $0.30 \times 100y = 30y$.

The total profit is N855.00, so:

$12x + 30y = 855$ (2)

We can solve this system of equations.

Multiply equation (2) by $\frac{10}{3}$ to get:

$40x + 100y = 2850$ (3)

Subtract equation (1) from equation (3):

$(40x + 100y) - (30x + 100y) = 2850 - 2450$

$10x = 400$

$x = \frac{400}{10} = 40$

Substitute $x = 40$ into equation (1):

$30(40) + 100y = 2450$

$1200 + 100y = 2450$

$100y = 2450 - 1200$

$100y = 1250$

$y = \frac{1250}{100} = 12.50$

(i) The cost price of a basket of pawpaw is $x = N40.00$.

(ii) The selling price of 100 baskets of mangoes is:

Cost price + profit = $100y + 30y = 130y = 130 \times 12.50 = 1625$

The selling price of 100 baskets of mangoes is N1625.

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3. Final Answer

(a) The common difference of the AP is

3. (b)

(i) The cost price of a basket of pawpaw is N40.

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0. (ii) The selling price of 100 baskets of mangoes is N1625.00.