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WAEC 2019 · Paper 2 · Q10

  1. (a)

    A twenty-kilogram bag of rice is consumed by mm boys in 10 days. When four more boys joined them, the same quantity of rice lasted only 8 days. If the rate of consumption is the same, find the value of mm.

  2. (b)

    If 56\frac56 of a number is 10 greater than 13\frac13 of it, find the number.

  3. (c)

    Find the equation of the line which passes through the points (2,12)\left(2, \frac12\right) and (−1,−12)\left(-1, -\frac12\right).

    Show the answer

    2x−6y−1=02x - 6y - 1 = 0

Worked solution (try it first)

(a)

  1. The same amount of rice is eaten either way, so the number of "boy-days" is the same: mm boys for 10 days equals (m+4)(m + 4) boys for 8 days.
  2. So 10m=8(m+4)10m = 8(m + 4).
  3. Then 10m=8m+3210m = 8m + 32, 2m=322m = 32 and m=16m = 16.

(b)

  1. Let the number be xx.
  2. 56\frac56 of it is 10 more than 13\frac13 of it: 56x=13x+10\frac56x = \frac13x + 10.
  3. Multiply by 6: 5x=2x+605x = 2x + 60.
  4. So 3x=603x = 60 and x=20x = 20.

(c)

  1. Gradient =−12−12−1−2= \frac{-\frac12 - \frac12}{-1 - 2}
    =−1−3= \frac{-1}{-3}
    =13= \frac13.
  2. Using the point (2,12)\left(2, \frac12\right): y−12=13(x−2)y - \frac12 = \frac13(x - 2).
  3. Multiply by 6: 6y−3=2x−46y - 3 = 2x - 4.
  4. So 2x−6y−1=02x - 6y - 1 = 0.

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