IB DP Math MAA HL : IB Style Mock Exams – Set 10 Paper 2

Question

[Maximum mark: 6]
A police department has 4 male and 7 female officers. A special group of 5 officers is to be assembled for an undercover operation.

(a) Determine how many possible groups can be chosen.                      [2]

(b) Determine how many groups can be formed consisting of 2 males and 3 females.                      [2]

(c) Determine how many groups can be formed consisting of at least one male.                      [2]

Answer/Explanation

(a) We choose 5 people from $4+7=11$ males and females.
Hence the number of possible groups is
$
\begin{aligned}
\left(\begin{array}{c}
11 \\
5
\end{array}\right) & =\frac{11 !}{5 ! \cdot 6 !} \\
& =462
\end{aligned}
$

(b) We choose 2 officers from 4 males and then we choose 3 more officers from 7 females. Hence the number of groups consisting of 3 males and 2 females is
$
\begin{aligned}
\left(\begin{array}{l}
4 \\
2
\end{array}\right)\left(\begin{array}{l}
7 \\
3
\end{array}\right) & =\frac{4 !}{2 ! \cdot 2 !} \cdot \frac{7 !}{3 ! \cdot 4 !} \\
& =210
\end{aligned}
$

(c) The number of groups consisting of (0 males and 5 females is
$
\begin{aligned}
\left(\begin{array}{l}
4 \\
0
\end{array}\right)\left(\begin{array}{l}
7 \\
5
\end{array}\right) & =\frac{4 !}{0 ! \cdot 4 !} \cdot \frac{7 !}{5 ! \cdot 2 !} \\
& =21
\end{aligned}
$

Hence the number of groups consisting of at least 1 male is
$
462-21=441
$

Question

[Maximum mark: 7]
A triangle $\mathrm{ABC}$ has $a=10.2 \mathrm{~cm}, b=17.5 \mathrm{~cm}$ and area $32 \mathrm{~cm}^2$. Find the largest possible perimeter of triangle $\mathrm{ABC}$.

Answer/Explanation

Using the area of a triangle formula, we have
$
\begin{aligned}
A_{\triangle \mathrm{ABC}} & =\frac{1}{2}(a)(b) \sin \hat{\mathrm{C}} \\
32 & =\frac{1}{2}(10.2)(17.5) \sin \hat{\mathrm{C}} \\
\hat{\mathrm{C}} & \approx 21.0^{\circ}, 159^{\circ} \quad \quad \text { [by using G.D.C.] }
\end{aligned}
$

Hence, using the cosine rule, we get
$
\begin{aligned}
c^2 & =a^2+b^2-2(a)(b) \cos \hat{\mathrm{C}} \\
c & =\sqrt{(10.2)^2+(17.5)^2-2(10.2)(17.5) \cos \hat{\mathrm{C}}} \\
c & \approx 8.78 \mathrm{~cm}, 27.3 \mathrm{~cm} \quad\left[\text { for } \hat{\mathrm{C}} \approx 21.0^{\circ}, 159^{\circ}\right]
\end{aligned}
$

Hence the (largest possible) perimeter of triangle ABC is
$
\begin{aligned}
P_{\triangle \mathrm{ABC}} & =a+b+c_{\max } \\
& =10.2+17.5+27.3 \\
& =55 \mathrm{~cm}
\end{aligned}
$

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