WAEC 2016 · Paper 2 · Q11

  1. (a)

    If 3p+4q3p−4q=2\dfrac{3p + 4q}{3p - 4q} = 2, find p:qp : q.

    Show the answer

    4:14 : 1

  2. (b)

    The diagram shows the cross section PQRSTUPQRSTU of a bridge with a semicircular hollow TSTS in the middle. ∣PU∣=∣QR∣=4 m|PU| = |QR| = 4\text{ m} and ∣UT∣=∣SR∣=2 m|UT| = |SR| = 2\text{ m}. If the perimeter of the cross section is 34 m34\text{ m}, calculate the: (i) length PQPQ; (ii) area of the cross section. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

    4 m4 m2 m2 mPQRSTU

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

Worked solution (try it first)

(a)

  1. Cross-multiply: 3p+4q=2(3p−4q)=6p−8q3p + 4q = 2(3p - 4q) = 6p - 8q.
  2. So 12q=3p12q = 3p, p=4qp = 4q, and p:q=4:1p : q = 4 : 1.

(b)(i)

  1. Let the semicircle have radius rr.
  2. The width across the bottom is 2+2r+22 + 2r + 2, so ∣PQ∣=4+2r|PQ| = 4 + 2r.
  3. The perimeter goes round PQPQ, down QRQR (4), along RSRS (2), round the semicircle (227r\frac{22}{7}r), along TUTU (2) and up UPUP (4): (4+2r)+4+2+227r+2+4=34(4 + 2r) + 4 + 2 + \frac{22}{7}r + 2 + 4 = 34.
  4. So 16+367r=3416 + \frac{36}{7}r = 34, 367r=18\frac{36}{7}r = 18 and r=3.5r = 3.5 m.
  5. Then ∣PQ∣=4+7=11|PQ| = 4 + 7 = 11 m.

(ii)

  1. Area == rectangle −- semicircle =11×4−12×227×3.52= 11 \times 4 - \frac12 \times \frac{22}{7} \times 3.5^2
    =44−19.25= 44 - 19.25
    =24.75 m2= 24.75\text{ m}^2.

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