WAEC 2015 · Paper 2 · Q9

  1. (a)

    The first term of an Arithmetic Progression (A.P.) is −8-8. If the ratio of the 7th term to the 9th term is 5:85 : 8, find the common difference of the A.P.

  2. (b)

    A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for ₦2,450.00. She sold the pawpaw at a profit of 40%40\% and the mangoes at a profit of 30%30\%. If her profit on the entire transaction was ₦855.00, find the: (i) cost price of a basket of pawpaw; (ii) selling price of the 100 baskets of mangoes.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. The nnth term of an A.P. is a+(n−1)da + (n - 1)d.
  2. With a=−8a = -8: the 7th term is −8+6d-8 + 6d and the 9th term is −8+8d-8 + 8d.
  3. Their ratio is 5:85 : 8: −8+6d−8+8d=58\frac{-8 + 6d}{-8 + 8d} = \frac58.
  4. Cross-multiply: 8(−8+6d)=5(−8+8d)8(-8 + 6d) = 5(-8 + 8d).
  5. So −64+48d=−40+40d-64 + 48d = -40 + 40d, 8d=248d = 24 and d=3d = 3.

(b)

  1. Let a basket of pawpaw cost ₦pp and a basket of mangoes cost ₦mm.
  2. Total cost: 30p+100m=245030p + 100m = 2450 (1).
  3. Profit on pawpaw is 40%40\% of 30p30p and on mangoes 30%30\% of 100m100m: 0.4×30p+0.3×100m=8550.4 \times 30p + 0.3 \times 100m = 855, which is 12p+30m=85512p + 30m = 855 (2).
  4. Multiply (1) by 0.3: 9p+30m=7359p + 30m = 735 (3).
  5. Take (3) from (2): 3p=1203p = 120, so p=40p = 40.

(i)

  1. A basket of pawpaw cost ₦40.00.

(ii)

  1. From (1), 100m=2450−1200=1250100m = 2450 - 1200 = 1250, so the mangoes cost ₦1,250.00.
  2. They sold at a 30%30\% profit: 1.3×1250=1.3 \times 1250 = ₦1,625.00.

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