Ans: (a) $8$ (b) (i) $45^\circ$ (ii) No
-
$\begin{array}{rcl}
\dfrac{6}{6+11+5+k+10} & = & \dfrac{3}{20} \\
120 & = & 96 + 3k \\
k & = & 8
\end{array}$ -
- The required angle
$\begin{array}{cl}
= & \dfrac{5}{6+11+5+8+10}\times 360^\circ \\
= & 45^\circ
\end{array}$ - Let $n$ be the number of students newly join the group. To double the angle of the sector,
$\begin{array}{rcl}
\dfrac{5+n}{40+n} & = & \dfrac{1}{4} \\
20+4n & = & 40+n \\
3n & = & 20 \\
n & = & \dfrac{20}{3}
\end{array}$Since $n$ must be a positive integer, then it is impossible to find an $n$ such that the angle of the sector representing the most favourite fruit is orange be double.
- The required angle
