Worksheet – Algebra
1. Simplify by expressing as a single power:
a. 83 ÷ 25
b. 167 × 25
c. 165 ÷ 84
2. [IMC 2000] The ratio a:b = 2:3 and the ratio a:c = 3:4. What is the ratio b:c?
A) 1:8
B) 1:2
C) 8:9
D) 9:8
E) 2:1
1
1
1
1
3. [IMC 1999] When exactly is the value of the product (1 + 2) (1 + 3) (1 + 4) … (1 + 𝑛)
equal to an integer? (Hint: put the result of each of the brackets as a improper fraction,
allowing you to apply Tip #6)
A)
when n is odd
B)
when n is even
C)
when n is a multiple of 3
D)
always
E)
never
4. [IMC 2004] In a maths exam with N questions, you score m marks for a correct answer to
each of the first q questions and m+2 marks for a correct answer to each of the remaining
questions. What is the maximum possible score?
A) (m+2)N – 2q
B) Nm
C) mq + (m+2)q
D) N(m+1)
E) Nm + q(m+2)
5. [Hamilton 2008] Find all real solutions for x and y to the equations:
x2 – y2 = -5
2x2 + xy – y2 = 5
(You can use a method similar to the slides)
1
1
1
1
6. Simplify (1 − 4) (1 − 9) (1 − 16) … (1 − 𝑛2 ) as fully as possible. (Hint: Each bracket has a
particular form that you’ve seen many times before!)
For more resources see www.drfrostmaths.com/rzc
Algebra Tips & Tricks
1. Be adept with the laws of indices.
a. Remember for example that 3 = 31.
b. 𝑎𝑏 × 𝑎𝑐 = 𝑎𝑏+𝑐
c. 𝑎𝑏 ÷ 𝑎𝑐 = 𝑎𝑏−𝑐
d. (𝑎𝑏 )𝑐
e. If the bases are different, make them consistent. e.g. To change 43 to make it in
terms of base 2, then 43 = (22 )3 = 26
Square rooting halves the power. √96 = √(93 )2 = 93 . Cube rooting similarly
thirds the power, etc.
2. Always look out for the difference of two squares! It’ll often make your life much easier.
3. To deal with more difficult simultaneous equations involving two variables (e.g. one
quadratic equation and one linear equation), and adding/subtracting them won’t help:
for the simpler equation, make one of the variables the subject of the formula, then
substitute this expression into the more complex one so that it’s now in terms of one
variable, and solve.
4. Introduce variables to represent unknown quantities.
f.
𝑦
𝑏
5. If x:y is the same ratio as a:b, then 𝑥 = 𝑎. And thus 𝑏𝑥 = 𝑎𝑦.
This allows you to form equations for ratios, e.g. when dealing with lengths and areas. If
the ratio of lengths x and y is a:b, then form the equation bx = ay.
6. When you multiply lots of terms of add lots of terms, see if you can simplify or cancel
terms, e.g. by looking at pairs.
2
3
2
4
2
5
2
𝑛
1
3
2
4
3
5
4
6
e.g. (1 − ) (1 − ) (1 − ) … (1 − ) = × × × × … ×
𝑛−3
𝑛−1
×
𝑛−2
𝑛
We have the numbers 3 to n-2 both in the numerators and denominators, so we’re left
with
2
.
𝑛(𝑛−1)
For more resources see www.drfrostmaths.com/rzc
Algebra Worksheet Answers
1. Simplify by expressing as a single power:
𝟑
a. 83 ÷ 25 = (𝟐𝟑 ) ÷ 𝟐𝟓 = 𝟐𝟗 ÷ 𝟐𝟓 = 𝟐𝟒
𝟕
b. 167 × 25 = (𝟐𝟒 ) × 𝟐𝟓 = 𝟐𝟐𝟖 × 𝟐𝟓 = 𝟐𝟑𝟑
𝟓
𝟓
c. 165 ÷ 84 = (𝟐𝟒 ) ÷ (𝟐𝟑 ) = 𝟐𝟐𝟎 ÷ 𝟐𝟏𝟓 = 𝟐𝟓
2. [IMC 2000] The ratio a:b = 2:3 and the ratio a:c = 3:4. What is the ratio b:c?
A) 1:8
B) 1:2
C) 8:9
D) 9:8
E) 2:1
𝒃
𝒂
𝒃
𝟑
𝟑
𝟗
Answer: D. 𝒄 = 𝒄 × 𝒂 = 𝟒 × 𝟐 = 𝟖 .
Alternatively, we could scale the ratios so the value of a is the same. a:b = 6:9 and a:c =
6:8. At that point, we can easily see that b:c = 9:8.
1
1
1
1
3. [IMC 1999] When exactly is the value of the product (1 + 2) (1 + 3) (1 + 4) … (1 + 𝑛)
equal to an integer? (Hint: put the result of each of the brackets as a improper fraction,
allowing you to apply Tip #6)
A)
when n is odd
B)
when n is even
C)
when n is a multiple of 3
D)
always
E)
never
𝟏
𝟐
𝟏
𝟑
𝟏
𝟒
𝟏
𝒏
𝟑
𝟒
𝟓
𝟐
𝟑
𝟒
𝒏+𝟏
. This is
𝟐
Answer: A. (𝟏 + ) (𝟏 + ) (𝟏 + ) … (𝟏 + ) = × × × … ×
common terms in the diagonals, we’re left with
multiples of 3 aren’t necessarily odd).
𝒏
𝒏+𝟏
×
.
𝒏−𝟏
𝒏
By cancelling
an integer when n is odd (and
4. [IMC 2004] In a maths exam with N questions, you score m marks for a correct answer to
each of the first q questions and m+2 marks for a correct answer to each of the remaining
questions. What is the maximum possible score?
A) (m+2)N – 2q
B) Nm
C) mq + (m+2)q
D) N(m+1)
E) Nm + q(m+2)
Answer: A. This is just a case of being careful about how you form equations! For the first
q questions there’s mq marks on offer. And for the remaining N-q questions, there’s
(N-q)(m+2) = Nm + 2N – mq – 2q marks on offer. In total, that’s mq + Nm + 2N – mq – 2q =
Nm + 2m – 2q = (m+2)N – 2q.
For more resources see www.drfrostmaths.com/rzc
5. [Hamilton 2008] Find all real solutions for x and y to the equations:
x2 – y2 = -5
2x2 + xy – y2 = 5
(You can use a method similar to the slides)
Factorising each:
(x+y)(x-y) = -5
2x2 + xy – y2 = x2 + xy + x2 – y2 = x(x+y) + (x-y)(x+y) = (2x-y)(x+y) = 5
Dividing one by the other, we get
𝟐𝒙−𝒚
𝒙−𝒚
= −𝟏. Rearranging this gives 2x-y = y-x. So 3x = 2y,
𝟑
and thus 𝒚 = 𝒙.
𝟐
But this just tells us how x and y are related – it doesn’t tell us how each is restricted.
Substituting it back into the first equation, we get:
𝟗
𝟓
𝒙𝟐 − 𝒙𝟐 = − 𝒙𝟐 = −𝟓
𝟒
𝟒
𝟑
Therefore x2 = 4, and thus x = -2 or +2. Since 𝒚 = 𝟐 𝒙, we get the pairs of solutions x=2, y =
3, and x=-2, y = -3.
1
1
1
1
6. Simplify (1 − 4) (1 − 9) (1 − 16) … (1 − 𝑛2 ) as fully as possible.
𝟏
𝟏
𝟏
𝟏
(𝟏 − ) (𝟏 − ) (𝟏 − ) … (𝟏 − 𝟐 )
𝟒
𝟗
𝟏𝟔
𝒏
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
= (𝟏 − ) (𝟏 + ) (𝟏 − ) (𝟏 + ) … (𝟏 − ) (𝟏 + )
𝟐
𝟐
𝟑
𝟑
𝒏
𝒏
𝟏 𝟑 𝟐 𝟒 𝟑 𝟓
𝒏−𝟏 𝒏+𝟏
= × × × × × × …×
×
𝟐 𝟐 𝟑 𝟑 𝟒 𝟒
𝒏
𝒏
Notice that all the pairs after the first fraction cancel, just leaving the first and the last
𝒏+𝟏
fraction. Thus the whole expression simplifies to
.
𝟐𝒏
For more resources see www.drfrostmaths.com/rzc