(a) The first term of an Arithmetic Progression (AP) is 8, 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 the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.
(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 the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.
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Correct Answer: Option n
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.
(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.