QMI1500 EXAM PACK 2025 [DETAILED QUESTIONS AND ANSWERS]

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QMI1500
Question 1

’
If x = 4, find the value of the sum &()!" #$ %&*

Solution

¥ Substitute x = 4 into this equation: ’&()!" #$ +&*
¥ When t = 0, you get (5 + +)) = 5 + ,) = 5 + 1 = 5 ¥ When t = 1, you
get (5 + +-) = 5 + + = 5 + 2 = 7
¥ When t= 2, you get (5 + +.) = 5 + ,/ = 5 + 4 = 9
¥ When t= 3, you get (5 + +’) = 5 + /+ = 5 + 8 = 13 ¥ The sum of all
these gives 5 + 7 + 9 + 13 = 34


Question 2

Solve for x in the following equation

- $$$- =$$
2


Solution


¥ Left Hand Side gives: ’01.1-2 4$ ’01’2 ¥ Right Hand Side gives:

o Multiplying top and bottom by 2 gives Right Hand Side gives: 4$

Now equate the left to the right and you get: ’01’ 4$ .1-)0
2 2


Since the denominators are equal, it means that we can also equate the numerators. This gives 3x -3 = 2 -
10x

Manipulate the equation to get 13x = 5 And therefore x

3
=

-’ Question 3

Find r if +6 # 7$ 8 9 4 7

Solution

Manipulate the equation to isolate the variable that you are trying to solve for

+6 # 7$ 4 7 # 9 4 "

.
. 4 9"$
Square both sides to give: +6 # 7$$ 4"


2
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QMI1500
On the left: the square root and the power of 2 cancel each other out. We are left with just 4r + 3

Equating both sides gives 4r + 3 = 25

More manipulation gives: 4r = 25 — 3 = 22 Divide both sides

by 4 and get r = .. 4$

2

Question 4

Find the value of the following expression:

: 3 -$$$7 - + $$, -
; 2 3


Solution

¥ Convert the mixed fractions into improper fractions (i.e. where the numerator is bigger than

,-
the denominator) o : 3; becomes ’< ; o 7 -2 becomes -’ o 3$becomes

¥ Find common denominator o 8 and 4 already have a common denominator between them
which is 8.
Multiply this by 5 to get a common denominator amongst all 3 — you get 40
¥ Adjust numerators accordingly before adding them o Multiply
37 by 5, to get 185 o Multiply 13 by 10, to get 130 o
Multiply 6 by 6, to get 48
o Adding these up gives a numerator of 103 (185 — 130 + 48)
¥ The new expression of -)2)’

¥ Simplifying this gives us 2 .’2)

Question 5

Simplify the following:

>0 5 >’0 ’0 5 +0?.

9
Solution

Denominator

¥ need the same base before you can add exponents
¥ tip: 4 = 22
¥ therefore: 4x+2 = (22)x+2 = 22x+4

¥ now that you have the same base, you can add the exponents (i.e. 3x + 2x + 4 = 5x +4)
¥ the denominator is therefore equals 25x+4
3
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QMI1500
Numerator

¥ same base, therefore can add exponents (i.e. add x and 3x to give 4x)
¥ this gives >20
¥ we know that 8 = 2*2*2 = 23
¥ therefore 84x = (23)4x = 212x


So you now have the following

9-.0 30?2

9
This can be further simplified since you have the same base

¥ Tip: division means you have to subtract the exponents
¥ 12x — (5x + 4) = 12x -5x — 4 = 7x — 4


So you now have the following

9<012

This can be re-written as: 27x * 2-4

Remember that a negative exponent means the inverse function

¥ Therefore 2-4 = . -
@




You now have: . .AB@ Since 24 = 2*2*2*2 = 16, this gives a final


answer of . AB




-=

Question 6

Simplify the following expression:
. .
%3 5 %’ .


%12

Solution Denominator

¥ .2 = -.

¥ Therefore x-2/4 = x-1/2

4
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