### Instructions

Pick the letter which represents the correct answer for each problem. Calculators are not allowed.

### Question 1

 2√12² - (√12)² √5
equals:

•  144 √5
•  12 5
•  12√5 5
•  0

#### Solution:

First, simplify the top square roots, (Review square rules)

Now √12² = 12 and (√12)² = 12

Substitute those answers back into the original equation

 2 × 12 - 12 = 12 √5 √5
but we are not done yet.

The last step is rationalising the denominator.

 12 √5
can be written as
 12 × √5 = 12√5 √5 × √5 5

### Question 2

 a − 1 b − c
can be simplified to

•  a + b − c b − c
•  a − 1 b − c
•  a − a −1 b c
•  a − b + c b − c

#### Solution:

 a − 1 b − c
is the original problem.

First you need to get the same denominator for both parts of the expression.

Write 1 as
 b − c b − c
and substitute back into the original equation for 1.

You now have
 a − b − c b − c b − c
with both parts having the same denominator.

We can now simply write the expression as
 a − (b − c) b − c

Note that the bracket is essential at this stage.

We now need to remove the bracket by distributing the – (minus) sign to the variables within the bracket (remember that the – outside the brackets changes the sings inside the bracket). (Review algebraic signs)

The simplified version of the equation becomes
 a − b + c b − c

### Question 3

 (42)−1 × 2−3 20 × 4−3
using the laws of indices equals:

•  4
•  ½
•  ¼
•  2

#### Solution:

 (42)−1 × 2−3 20 × 4−3
is the original problem.

We will use the laws of indices noting that (Review laws of indices):
 20 =1, 4 = 22, and 2−3 = 1 23

First we will simplify the bracket portion of the equation

(42)−1 = 4−2 = (22)−2 = 2−4

Now we will substitute this for the first numerator terms and then simplify

2−4 × 2−3 = 2−7

Now we will simplify the denominator

1 × (22)−3 = 2−6

Now we will substitute the simplified versions of the numerator and denominator into the equation

 2−7 2−6
We will now simplify so that the equation is stated with positive indices

 26 27
and then we will simplify to
 1 2

Note: Remember that is the same as writing 2 x 2 x 2 x 2 x 2 x 2. So when you simplify you cancel out all but one 2 on the bottom leaving
 1 2

### Question 4

Three members of a group are absent. If the absent members represent
 12 1 % 2
of the group, how many are in the group?

•  25
•  24
•  18
•  36

#### Solution:

Three members of a group are absent.

Let the number of members in the group be N, so
 12 1 % 2
of N = 3

Write
 12 1 % 2
as a fraction,
 12.5 × N = 3 100

(100 represents 100 % of the group members)

We now need to rearrange the equation so we are solving for N.
We multiply both sides by 100 to remove the denominator.

 100× 12.5 × N = 3×100 to get 12.5 × N = 3×100 100

(the 100s cancel each other out)

 1 × N = 3×100× 1 we get N = 3×100 = 24 12.5 12.5 12.5

(the 12.5s cancel each other out).

So there are 24 members in the group

### Question 5

 a4b-2 ÷ b-1 4a-2b 8a2
can be expressed with positive indices as:

•  4a6 b3
•  a4 2b2
•  2 a4 2 b
•  2a4 b3

#### Solution:

We do the division first by inverting the second fraction and changing the sign to multiplication, then multiple the terms (remember to add or subtract powers when you multiply terms. The power b is the same as b1) (Review exponential algebra rules).

 a4b-2 × 8a2 = 8a6b-2 4a-2b b-1 4a-2b0

Now we need to get rid of the negative indices by moving them to the top or bottom of the fraction and dividing the whole numbers if possible.
Then collect like terms.

 2a6×a2 = 2a8 b0×b2 b2

we can further simplify the answer to
 2 a4 2 b

### Question 6

If x = log3 and y = log5 then log45 is:

•  2x + y
•  x + 2y
•  2(2y + x)
•  3x + 2y

#### Solution:

Now log45 = log(5 × 9)

Which is log5 + log9 but 9 is the same as 32

So log9 = log32 we can rewrite this as 2 log3 (Review log exponential identities)

So log45 = log5 + 2 log3.

We can now substitute in x and y to give log45 = y + 2x

### Question 7

If logaN2 = b then N is:

•  ab
•  a2 b
•  ba
• a

 b 2

#### Solution:

From the definition of a logarithm we can rewrite the equation as (Review logarithm relationship)

ab = N2. We now need to isolate N so we can solve for N.

We take the square root of both sides and get N = √ab.

Now we rewrite the square root as follows (Review square root rules):

(ab)
 1 2

this is simplified to

a

 b 2

### Question 8

Find the value of x in the following equation:
 2x − 1 + x = 1 3 6

•  8 5
•  5 8
•  3 2
•  2 3

#### Solution:

First we clear the fraction by multiplying all terms by the lowest common denominator, which is 6:

 6 2x − 1 + 6 x = 1 3 6
and this now becomes 2(2x−1)+x=6

We now multiply the bracketed term to get 4x−2+x=6

Now we collect like terms 5x−2=6, add 2 to each side 5x−2+2=6+2 and now we have 5x=8.

Now we divide both sides by 5 in order to solve for x,
 5x = 8 5 5
and we get
 x 8 5

### Question 9

A straight line passes through the points (0, 4) and (3, 0). The equation of the line is:

•  4y−3x+0
•  3y−4x−6=0
•  yy+4x−12=0
•  y−4x−12=0

#### Solution:

The equation of a straight line can be written as y = ax + b where x and y are the independent variable and dependent variable respectively, and a and b are numbers to be found. (Review linear equations)

We can substitute (0,4) into the equation and get 4 = 0a + b this means b = 4

The new equation now reads: y = ax + 4

We can substitute (3,0) into this equation and get 0 = 3a + 4 and solve for a

Subtract 4 from both sides and you get −4 = 3a. Now divide both sides by 3 and you get
 a = 4 3
now substitute the values of a and b back into the original equation.

 y = − 4 x + 4 3

We now simplify the equation by removing the fraction. Multiply both sides by 3.

3 × y = 3
 − 4 x 3
+ 4 × 3

and we get 3y = −4x + 12. We now need to set the whole equation equal to 0 for the final answer.

We move −4x + 12 to the other side and change their signs. This is 3y + 4x − 12 = 0.

### Question 10

If
 ƒ(x) = 4x2−9 2x−3
then ƒ(2a) is:

•  8a−9 2a−3
•  4a−3 2a−3
•  4a2−3 2
•  4a+3 1

#### Solution:

We substitute –a for x in ƒ(x) and get
 ƒ(−a)=(−a)2− 1 +2 −a

(Review algebraic functions).

We get rid of the signs because a square always give a positive number whether the number is negative or positive and the minus sign and a in the fraction cancel each other out using the rules of signs. (Review rules of signs)

Therefore the result is
 a2+ 1 +2 a

### Question 11

If sinZ = 4/5 and 0<Z<90 then tanZ is:

•  5/4
•  3/4
•  4/3
•  3/5

#### Solution:

We construct a right triangle with
 sin z = 4 5

(Review Trig. Ratios)

 sin z = a = opposite c hypotenuse

Pythagoras’s theorem states that a2 + b2 = c2 we can use this to solve for b or x in our problem (Review Pythagoras Theorem).

42 + x2 = 52 or 16 + x2 = 25 isolate b so that x2 = 25 ‐ 16 = 9

x2 = 9 so x = 3

We now need to solve for
 tanZ = Opposite = 4 = 4 Adjacent b 3

### Question 12

A ladder which is ȴ meters long leans against a vertical wall. The angle between the ladder and the horizontal ground is x degrees. The distance that the ladder reaches up the wall is given by?

•  ȴsec x
•  ȴcos x
•  ȴsin x
•  ȴtan x

#### Solution:

We construct a diagram which is a right triangle. Let the height the ladder reaches up the wall be h and the length of the ladder be ȴ.

Using sin x we can construct a trig ratio of the sides of the triangle as
 sin x = opposite = h hypotenuse ȴ

 sin x = h ȴ
we now multiply both sides by ȴ in order to isolate h so we get h = ȴ sin x

(Review Trig. Ratios)