a) A twenty - kilogram bag of rice is consumed by m number of boys in 10 days. When four more boys joined them, the same quantity of rice lasted only 8 days...
Question 1 Report
a) A twenty - kilogram bag of rice is consumed by m number of 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 m.
(b) If \(\frac{5}{6}\) of a number is 10 greater than \(\frac{1}{3}\) of it. find the number
(c) Find the equation of the line which passes through the points (2, \(\frac{1}{2}\)) and (-1, -\(\frac{1}{2}\)
Answer Details
(a) Value of m. The same quantity of rice is fixed, so boy-days are equal:
\(m\times 10 = (m+4)\times 8 \Rightarrow 10m = 8m+32 \Rightarrow 2m=32 \Rightarrow m=\mathbf{16}\).
(b) The number. Let the number be \(n\):
\(\dfrac{5}{6}n = \dfrac{1}{3}n + 10 \Rightarrow \left(\dfrac{5}{6}-\dfrac{2}{6}\right)n = 10 \Rightarrow \dfrac{1}{2}n = 10 \Rightarrow n=\mathbf{20}\).
(c) Equation of the line through \(\left(2,\tfrac{1}{2}\right)\) and \(\left(-1,-\tfrac{1}{2}\right)\).
Gradient \(m=\dfrac{\frac{1}{2}-(-\frac{1}{2})}{2-(-1)}=\dfrac{1}{3}\).
\(y-\dfrac{1}{2}=\dfrac{1}{3}(x-2)\Rightarrow y=\dfrac{1}{3}x-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{1}{3}x-\dfrac{1}{6}\).
\(y=\dfrac{1}{3}x-\dfrac{1}{6}\), i.e. \(2x-6y-1=0\).
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