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WAEC 2008 · Paper 2 · Q5

In a hotel the breakfast menu is a choice between yam (YY) or plantain (PP) or both. The Venn diagram shows the choices made by 25 guests of the hotel.

PY(2x + 1)x(x − 2)2
  1. (a)

    Find the value of xx.

  2. (b)

    What is the probability that a guest chosen at random chose only one of the two?

Worked solution (try it first)

(a)

  1. Every guest chose yam, plantain or both, so the three regions add up to 25: (2x+1)+x+(x−2)2=25(2x + 1) + x + (x - 2)^2 = 25.
  2. Expand (x−2)2=x2−4x+4(x - 2)^2 = x^2 - 4x + 4: x2−x+5=25x^2 - x + 5 = 25.
  3. Rearrange: x2−x−20=0x^2 - x - 20 = 0, which factorises as (x−5)(x+4)=0(x - 5)(x + 4) = 0.
  4. A number of guests cannot be negative, so x=5x = 5.

(b)

  1. Only plantain: 2(5)+1=112(5) + 1 = 11 guests.
  2. Only yam: (5−2)2=9(5 - 2)^2 = 9 guests.
  3. So 11+9=2011 + 9 = 20 guests chose only one.
  4. Probability =2025=45= \frac{20}{25} = \frac45.

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