We continue now with the solutions to the CXC general proficiency math (sample) test. We left off at No. 6.

6)(a) Given that: U = {a, b, c, d, e, f, g} where U is the universal set, and

L = {a, b, c, d, e}

M= {a, c, e, g}

N = {b, e, f, g}

(i)Draw a Venn diagram showing the sets U, L, M, N and their elements

The solution to this is shown in the accompanying image, for Problem 6.

(ii)List the members of the set represented by the union of N with the intersection of L and M

From the Venn diagram solution in (a) we obtain: {a, b, c, e, f, g}

b) If:

(a)

(b) =

(2...3)(-3)

(-1..2)(1)

determine the values of a and b

This is done by simple matrix multiplication, i.e. multiplying the 2 x 2 matrix on the left side by the column matrix on the right. We get:

(2 x (-3) + 3 x 1)

(-3 x -1 + 2 x 1) =

(-6 + 3)

(3 + 2) =

(-3)

(5)

so that a = -3 and b = 5

c) Given the vector:

A =

(4)

(7)

Calculate:

i) ‖A‖, the length of vector A

This is simply: A = [4^2 + 7^2]^½ = [65]^½ = 8.06

ii) the size of the angle made by the vector A and the x-axis.

We need to recognize here that the tangent of the angle is 7/4 since tan Θ = y/x and for A, x = 4 and y = 7. Then: Θ = arctan (y/x) = arctan (7/4) = arctan(1.75) and thence, Θ = 60. 2 degrees.

7)The diagram represents the Earth and shows the equator and the Greenwich meridian. Town I is located at (16 N, 30 W), and Town J is at (16 N, 45 W).

(i) Copy the diagram (or print it out) and show the positions of Towns I, J.

The solution for this part is provided in the graphic referenced to Problem (7). The key point is to note that town I must be closer to the Greenwich meridian (at 0 degrees longitude) and town J 15 degrees further to the west. Both are on the same latitude circle 16 degrees north of the equator (0 degrees latitude).

(ii) Calculate the radius of the circle of latitude on which Towns I and J are situated.

From simple solid geometry one can show that the radius (r') of 16N latitude circle is reduced by a factor equal to the cosine of the latitude, or 16 degrees. This should be evident from inspection of the same solution diagram, from which we can ascertain: cos (16 deg) = r'/r where r is given as 6400 km. Therefore:

r' = r (cos 16) = 6400 km (0.96) = 6150 km.

(iii) Calculate the shortest distance, measured along the Earth's surface, between the two towns.

This would be along the latitude circle of 16 degrees N, which separated the towns by 45 - 30 = 15 degrees. So the distance will be the fraction of this separation over 360 degrees, multiplied by the circumference for the 16 N latitude circle. Call the latter C(16N) and C(16N) = 2π (r') = 2π (6150 km) = 2(3.14)(6150km) = 38620 km

Then: d (I-J) = 15/ 360 (38620 km) = 1, 600km

8)The matrix R =

(cos(Θ)......-sin(Θ))

(sin (Θ)......cos(Θ))

a) Determine the coordinates of the image (1, 2) under the transformation R when Θ = 90 degrees.

This is done via simple matrix multiplication and when substituting for Θ in the R-transformation matrix (e.g. Θ = 90) we find, R(1 2) =

(0...-1)(1)

(1....0)(2) = (-2 ...1)

b) If the point (p, 3) is on the line (L)given by: x + 2y = 5, calculate the value of p.

This is straightforward. We have x = p and y = 3, so p + 2(3) = 5, and p = 5 -6 = -1.

c) Given the point (1,2) is on L, determine the image of L (L') of the line L under the transformation R.

This is more involved but we already know from part (a) how the image of the point is produced via the R-transformation matrix. Since: R(1 2) = (-2 1) and y = -x/2 + 5/2 then we can use the point slope formula: y - y1 = m(x - x1) to obtain the new line which is really L' the image of L:

Then: y - 1 = -½ (x - (-2)) = -x/2 + 2 or y = -x/2 + 3 or (re-arranging):

x = 2y -6 or x - 2y = -6

d) Write the matrix equation to represent the pair of simultaneous equations given by L and L'.

This is straightforward since we know the two equations are: x + 2y = 5 and x - 2y = -6. Then:

(1...2)(x)

(1..-2)(y) =

(5)

(-6)

9) This refers to Fig. 3 (of the original problem set blog). The diagram shows a rectangular sheet of metal ABCD supported by a vertical wall (shaded) at right angles to the level ground OX. AB measures 3 meters and AD measures 10 meters.

a) Calculate the size of the angle ODA.

From the geometry we recognize that sin (ODA) = opp./hyp. = OA/ AD = 4/10. Then:

angle ODA = arc sin (0.4) = 23.5 degrees.

b) Hence, calculate the size of the angle CDX.

From inspection and basic knowledge of geometry (i.e. the contributions to a straight angle must total 180 degrees) we see:

angle CDX = 180 deg - (90 deg + 23.5 deg) = 180 deg - 113.5 deg = 66.5 deg.

c) If CX represents the length in meters of C above the ground, calculate CX.

We use, after noting the geometrical relations:

sin (66.5) = CX/ CD, but CD = AB = 3m, (by def.) so sin(66.5) = CX/ 3 m

CX = 3 m (sin 66.5) = 2.75m

10)This refers to the Venn diagram in Fig. 4 (of original problem blog) and the information therein. Assume the same number, x, play football only and tennis only.

(a) Calculate the number who play football.

From the Venn diagram and relations shown therein, we have:

U = 40 = F + C + T + 8

amd F + T = 2x (x = those who only play football, and tennis). From the diagram we see those who play cricket C = 15, so:

F + T = 32 - C = 32 - 15 = 17. (This number refers to those who play both football and tennis). Since the

*intersection*of F and T shows '7' then it follows that:

2x + 7 = 17, and 2x = 10, with x = 5 is the number who play football only and tennis only. But the number that

*must logically include not only those who play football only but those who play football PLUS tennis, as well as those who play cricket plus football, hence the answer is: 7 + 15 + 5 = 27. To check this, add to this those who play only tennis (5) plus those who play neither football, tennis or cricket (8) and you obtain: 27 + 8 + 5 = 40, the total of members.*

**play football**(b) State the information represented by the shaded portion of the Venn diagram.

This refers to the numbers that play both football and tennis.

(c)State the relationship between the members of the sets C and F, and between the sets C and T.

In the first case, since C is

*a subset of F*, then all the cricketers also play football, however in the second case, the result is the empty set since there is a disjoint relation between C and T.

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