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WAEC 2014 · Paper 2 · Q6

  1. (a)

    If 32p−12=1314p+1\dfrac{3}{2p - \frac12} = \dfrac{\frac13}{\frac14p + 1}, find pp.

  2. (b)

    A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%20\%. The television set was actually sold at a discount of 5%5\%. Calculate, correct to 2 significant figures, the actual percentage profit.

Worked solution (try it first)

(a)

  1. One fraction equals another, so cross-multiply: 3(14p+1)=13(2p−12)3\left(\frac14p + 1\right) = \frac13\left(2p - \frac12\right).
  2. Expand both sides: 34p+3=23p−16\frac34p + 3 = \frac23p - \frac16.
  3. Multiply every term by 12 to clear the fractions: 9p+36=8p−29p + 36 = 8p - 2.
  4. So p=−38p = -38.

(b)

  1. The marked price, GH¢ 760.00, includes a 20%20\% profit, so it is 120%120\% of the cost price.
  2. Cost price =7601.2= \frac{760}{1.2}
    =19003= \frac{1900}{3}
    ≈\approx GH¢ 633.33.
  3. With a 5%5\% discount, the selling price is 95%95\% of the marked price: 0.95×760=0.95 \times 760 = GH¢ 722.00.
  4. Profit =722−19003= 722 - \frac{1900}{3}
    =2663= \frac{266}{3}
    ≈\approx GH¢ 88.67.
  5. Percentage profit =266/31900/3×100%= \frac{266/3}{1900/3} \times 100\%
    =2661900×100%= \frac{266}{1900} \times 100\%
    =14%= 14\%.

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