# CSEC Math Jan 2012 P2 Q6

CSEC Math Jan 2012 Paper 2 Question 6

Updated on April 16th, 2012: Added a video explanation

The table below shows corresponding values of $$x$$ and $$y$$ for the function $$y = x^2 – 2x – 3$$, for integer values of $$x$$ from -2 to 4.

A table showing the corresponding values of x and y for the given function

For part (a), we need to find the value of $$y$$ when $$x = -1$$ and when $$x = 2$$.

When $$x= -1$$,

$$\begin{eqnarray} y &=& (-1)^2 – 2(-1) – 3 \\ &=& 1 + 2 – 3 \\ &=& 0. \end{eqnarray}$$

When $$x= 2$$,

$$\begin{eqnarray} y &=& (2)^2 – 2(2) – 3 \\ &=& 4 + 4 – 3 \\ &=& -3. \end{eqnarray}$$

In part (b) we are asked plot the points whose $$x$$ and $$y$$ values are recorded in the table above, and to draw a smooth curve through the points. The graph below uses a scale of 2 cm to represent 1 unit on the $$x$$-axis, and 1 cm to represent 1 unit on the $$y$$-axis.

A graph showing a smooth curve through the points in the table

For part (c), using the graph we estimate that the value of $$y$$ when $$x = 3.5$$ is approximately $$2.2$$ (N.B. The exact answer is $$y = 2.25$$).

Using the graph to estimate the value of y at x = 3.5

Finally, in part (d) we make the following observations:

1. The equation of the axis of symmetry of the graph is $$x = 1$$.
2. $$-4$$ is the minimum value of the function $$y$$.
3. When $$x^2 – 2x – 3 = 0$$, i.e. $$y = 0$$ then from the table we see that $$x = -1$$ or $$x = 3$$.

# CSEC Math Jan 2012 P2 Q5

CSEC Math Jan 2012 Paper 2 Question 5

Updated on April 16th, 2012: Added some video explanations

In part (a), we given two triangles, $$JKL$$ and $$MLP$$, with $$JK$$ parallel to $$ML$$. $$LM = MP$$, $$KLP$$ is a straight line, angle $$JLM = 22^\circ$$ and angle $$LMP = 36^\circ$$.

Triangles JKL and MLP

Since $$LM$$ and $$MP$$ has the same length, it follows that $$\angle MLP = \angle MPL = \frac{180^\circ – 36^\circ}{2} = 72^\circ$$.

$$\angle LJK = \angle JLM = 22^\circ$$, since they are alternate angles.

$$\angle JKL = \angle MLP = 72^\circ$$, since they are corresponding angles.

$$\angle KLJ = 180^\circ – (72^\circ + 22^\circ) = 86^\circ$$.

In part (b), we have a triangle, $$PQR$$, and its image, $$P’Q'R’$$.

A transformation of triangle PQR

The coordinates of $$P$$ is $$(2, 1)$$ and the coordinates of $$Q$$ is $$(4, 3)$$.

The transformation that maps triangle $$PQR$$ onto triangle $$P’Q'R’$$ is a reflection in the x-axis.

Under the translation $$\begin{pmatrix}3 \\ -6\end{pmatrix}$$, P moves to the new point $$\begin{pmatrix}2 \\ 1\end{pmatrix} + \begin{pmatrix}3 \\ -6\end{pmatrix} = \begin{pmatrix}5 \\ -5\end{pmatrix}$$, i.e. $$(5, -5)$$ and Q moves to the new point $$\begin{pmatrix}4 \\ 3\end{pmatrix} + \begin{pmatrix}3 \\ -6\end{pmatrix} = \begin{pmatrix}7 \\ -3\end{pmatrix}$$, i.e. $$(7, -3)$$.

# CSEC Math Jan 2012 P2 Q4

CSEC Math Jan 2012 Paper 2 Question 4

Updated on April 13th, 2012: Added some video explanations

In part (a) we are given an extract from a bus schedule (for details, check the image above). The bus begins its journey at Belleview, travels to Chagville and ends its journey at St. Andrews.

A bus and its travel route

Since the bus arrived in Chagville at 7:35 a.m. and departed at 7:45 a.m., the bus spent 10 minutes at Chagville.

The bus departed Belleview at 6:40 a.m. and arrived in Chagville at 7:35 a.m. Hence, it took the bus 55 minutes to travel from Belleview to Chagville.

We are now told that the bus travelled at an average speed of 54 km/hour from Belleview to Chagville. In other words, in 1 hour (or 60 minutes) the bus travelled 54 km. But the bus didn’t travel for 60 minutes, it travelled for 55 minutes. So, if we can compute how far the bus can travel in 1 minute going at an average speed of 54 km/hour then we’d be able to determine how far it would have travelled in the 55 minutes going at the same average speed. Now, in 1 minute the bus would have travelled $$\frac{54}{60} \text{ km} = \frac{9}{10} \text{ km}$$. Therefore, in 55 minutes the bus would have travelled $$55 \times \frac{9}{10} \text{ km} = 49.5 \text{ km}$$. Hence, the distance from Belleview to Chagville is 49.5 km.

In part (b), 4.8 litres of water is poured into a cylindrical bucket with a base area of $$300 \text{ cm}^2$$. We have to find the height of the water in the bucket. Recall that the volume of a prism (a cylinder is a prism) is given by the product of its cross-sectional area times its perpendicular height.

A cylinder with water

In our case the cross-sectional area is given as the base area. Now, 4.8 litres is $$4800 \text{ cm}^3$$ since 1 litre is $$1000 \text{ cm}^3$$. Let $$h$$ represent the height of the water in the bucket. Then,

$$\begin{eqnarray} 300 \text{ cm}^2 \times h &=& 4800 \text{ cm}^3 \\ h &=& \frac{4800}{300} \text{ cm} \\ &=& 16 \text{ cm}. \end{eqnarray}$$

For part (c), the area of the shaded face, in terms of $$h$$, is $$4h \text{ cm}^2$$. The volume of the cuboid, in terms of $$h$$, is $$13 \text{ cm} \times 4h \text{ cm}^2 = 52h \text{ cm}^3$$. If the volume of the cuboid is $$286 \text{ cm}^3$$, then

$$\begin{eqnarray} 52h \text{ cm}^3 &=& 286 \text{ cm}^3 \\ h &=& \frac{286}{52} \text{ cm} \\ &=& 5.5 \text{ cm}. \end{eqnarray}$$

# CSEC Math Jan 2012 P2 Q3

CSEC Math Jan 2012 Paper 2 Question 3

Updated on April 12th, 2012: Added some video explanations

This question tests your knowledge of numbers, sets and geometric constructions.

In part (a) we are given a universal set $$U$$ defined as

$$U = \{ 51, 52, 53, 54, 55, 56, 57, 58, 59 \}.$$

Now, if $$A$$ and $$B$$ are subsets of $$U$$, such that

$$\begin{eqnarray} A &=& \{ \text{ odd numbers } \} \\ B &=& \{ \text{ prime numbers } \}. \end{eqnarray}$$

Then, the members of the sets $$A$$ and $$B$$ are

$$\begin{eqnarray} A &=& \{ 51, 53, 55, 57, 59 \} \\ B &=& \{ 53, 59 \}. \end{eqnarray}$$

Here is a Venn diagram that represents the sets $$A$$, $$B$$ and $$U$$.

A Venn diagram to represent the sets A, B and U

In part (b), we need to do some constructions using a pair of compasses, a ruler and a pencil.

Firstly, we are asked to construct a triangle $$CDE$$ in which $$DE = 10 \text{ cm}$$, $$DC = 8 \text{ cm}$$ and $$\angle CDE = 45^\circ$$.

Here is an image of the final construction along with the tools used:

A construction of triangle CDE

And here are the steps to reproduce it.

1. Draw line $$DE$$ of length 10 cm.
2. Construct angle $$CDE$$ of $$45^\circ$$, such that $$CD$$ has length 8 cm.
1. Using $$D$$ as center and any suitable radius, draw an arc to cut at $$a$$.
2. Then with $$a$$ as center and radius $$Da$$, draw an arc to cut at $$b$$.
3. Next using $$Da$$ again, with $$b$$ as center, draw another arc to cut at $$c$$.
4. Now with $$b$$ and $$c$$ as centers and any convenient radius, draw arcs to intersect at $$d$$.
5. Draw a line from $$D$$ through $$d$$ intersecting the arc $$a$$ at $$e$$.
6. Using $$a$$ and $$e$$ as centers, draw equal arcs to intersect at $$f$$.
7. Draw a line from $$D$$ through $$f$$.
8. Draw line $$DC$$ of length 8 cm on line $$Df$$. Angle $$CDE$$ is equal to $$45^\circ$$.
3. Draw line $$CE$$ to complete the construction of triangle $$CDE$$.

Secondly, we are asked to construct a line, $$CF$$, perpendicular to $$DE$$ such that $$F$$ lies on $$DE$$.

Here is an image of the final construction:

A construction of the line CF perpendicular to DE

And here are the steps to reproduce it.

1. Using $$C$$ as center, draw an arc to cut $$DE$$ at $$g$$ and $$h$$.
2. Now with $$g$$ and $$h$$ as centers, draw arcs of equal radii to intersect at $$i$$.
3. Draw a line from $$C$$ through $$i$$ cutting $$DE$$ at $$F$$. $$CF$$ is perpendicular to $$DE$$ as required.

Finally, using a protractor I have measured that $$\angle DCE$$ has a size of approximately $$83^\circ$$.

A protractor being used to measure the size of angle DCE

# CSEC Math Jan 2012 P2 Q2

CSEC Math Jan 2012 Paper 2 Question 2

Updated on April 11th, 2012: Added some video explanations

In part (a) we want to solve the pair of simultaneous equations

$$\begin{eqnarray} 3x + 2y &=& 13 \\ x – 2y &=& -1. \end{eqnarray}$$

You may notice that in the second equation we can solve for $$x$$ in terms of $$y$$ to get $$x = 2y – 1$$. Substituting for $$x$$ in the first equation gives us

$$\begin{eqnarray} 3(2y – 1) + 2y &=& 13 \\ 6y – 3 + 2y &=& 13 \\ 8y – 3 &=& 13 \\ 8y &=& 16 \\ y &=& 2. \end{eqnarray}$$

If $$y = 2$$ then $$x = 2(2) – 1 = 4 – 1 = 3$$. Therefore, $$x = 3$$ and $$y = 2$$ is the solution for the pair of simultaneous equations.

An alternative approach is to start by adding both equations since this would eliminate the term in $$y$$. Doing so, we get

$$\begin{eqnarray} (3x + 2y) + (x – 2y) &=& 13 + (-1) \\ 3x + x + 2y – 2y &=& 13 – 1 \\ 4x &=& 12 \\ x &=& 3. \end{eqnarray}$$

And now, if $$x = 3$$, then using the first or second equation we can solve for $$y$$ to get $$y = 2$$. Hence, $$x = 3$$ and $$y = 2$$ is the solution just as before.

In part (b) we have some factorisation to do.

$$\begin{eqnarray} x^2 – 16 &=& x^2 – 4^2 \\ &=& (x + 4)(x – 4). \end{eqnarray}$$

And,

$$\begin{eqnarray} 2x^2 – 3x + 8x – 12 &=& x(2x – 3) + 4(2x – 3) \\ &=& (2x – 3)(x + 4). \end{eqnarray}$$

Finally, in part (c) we are told that tickets for a football match are sold at $30 per adult and$15 per child and that a company bought 28 tickets.

If $$x$$ of the tickets were for adults, then $$28 – x$$ tickets were for children. It means then that $$\30x$$ was spent on tickets for adults and $$\15(28 – x)$$ was spent on tickets for children.

Therefore, the total amount spent on the 28 tickets is

$$\begin{eqnarray} & & \30x + \15(28 – x) \\ &=& \(30x + 420 – 15x) \\ &=& \(15x + 420). \end{eqnarray}$$

Given that the cost of the 28 tickets was \$660, it follows that

$$\begin{eqnarray} 15x + 420 &=& 660 \\ 15x &=& 240 \\ x &=& 16. \end{eqnarray}$$

Since $$x$$ represents the number of tickets bought by adults, it means then that 16 tickets were bought by adults.