QMI

QMI 1500
PRACTICE QUESTIONS &
SOLUTIONS

,Question 1
'
If x = 4, find the value of the sum &()(5 + 𝑥&)

Solution

• Substitute x = 4 into this equation: '&()(5 + 4& )
• When t = 0, you get (5 + 4) ) = 5 + 1) = 5 + 1 = 5
• When t = 1, you get (5 + 4- ) = 5 + 4 = 5 + 2 = 7
• When t= 2, you get (5 + 4. ) = 5 + 16 = 5 + 4 = 9
• When t= 3, you get (5 + 4' ) = 5 + 64 = 5 + 8 = 13
• The sum of all these gives 5 + 7 + 9 + 13 = 34


Question 2

Solve for x in the following equation
'01. - -130
- =
2 2 .

Solution
'01.1- '01'
• Left Hand Side gives: =
2 2
-130
• Right Hand Side gives:
.
.∗(-130)
o Multiplying top and bottom by 2 gives Right Hand Side gives: =
.∗.
.1-)0
2
'01' .1-)0
Now equate the left to the right and you get: =
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
3
And therefore x =
-'

Question 3

Find r if 4𝑟 + 3 − 2 = 3

Solution

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

4𝑟 + 3 = 3 + 2 = 5
.
Square both sides to give: 4𝑟 + 3 = 5. = 25

,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 = =
2 .

Question 4

Find the value of the following expression:
3 - -
𝟒 - 3 + 1
; 2 3

Solution

• Convert the mixed fractions into improper fractions (i.e. where the numerator is
bigger than the denominator)
3 '<
o 𝟒 becomes
; ;
- -'
o 3 becomes
2 2
- =
o 1 becomes
3 3
• 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:

80 ∗ 8'0
2'0 ∗ 40?.
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
Numerator

• same base, therefore can add exponents (i.e. add x and 3x to give 4x)
• this gives 820
• we know that 8 = 2*2*2 = 23
• therefore 84x = (23)4x = 212x


So you now have the following

2-.0
230?2
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

2<012
This can be re-written as: 27x * 2-4

Remember that a negative exponent means the inverse function
-
• Therefore 2-4 =
.@

.AB
You now have:
.@

.AB
Since 24 = 2*2*2*2 = 16, this gives a final answer of
-=

Question 6

Simplify the following expression:
. .
𝑥3 ∗ 𝑥'
.
𝑥 12
Solution

Denominator
. -
• =
2 .