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Chapter 2 Revision [Exam style questions]
Question 1
Select the correct word to describe the following.
𝑃 = 2𝑙 + 2𝑤
Equation [ ]
Expression [ ]
Formula [ ]
Question 2
(b) Simplify
𝑦2 +𝑦2
…………………………
(1 mark)
Question 3
Simplify
7𝑧 + 8𝑧 + 9𝑧3 +19𝑞 + 8𝑧3
…………………………
Question 4
Amelie makes necklaces using red and green gems.
She uses this rule to work out how many red gems to use.
red = 3 × green + 10
Amelie uses 16 red gems to make a necklace.
Work out how many green gems she uses.
…………………………
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Question 5
Evaluate (4𝑎 − 𝑏)2 when 𝑎 = 1 and 𝑏 = −1
…………………………
Question 6
Work out the value of
2𝑦
𝑣
given that
1
2
𝑦 = 1 2 and 𝑣 = 3
Give your answer as a mixed fraction in its simplest form.
Question 7
Simplify
1
2
𝑦3 × 𝑦3
3
𝑦7
…………………………
Question 8
Simplify
2
(4𝑥5𝑝 )
…………………………
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Question 9
Simplify
12𝑚10𝑛2
2 2
(2𝑚 )
…………………………
Question 10
Write
14 +7𝑥
1
2𝑥3
in the form of 𝑎 + 𝑏𝑥𝑐, where 𝑎, 𝑏, and 𝑐 are constants to be found.
…………………………
Question 11
Given that
𝑞𝑎 × 𝑞10
= 𝑞8
𝑞9
Work out the value of 𝑎.
𝑎 = …………………………
Question 12
−2
Express ( 𝑡 )
4
in the form 𝑎𝑡𝑛 where 𝑎 and 𝑛 are integers to be found.
…………………………
Question 13
1
Given that 𝑝−2 = 25 find the possible values of 𝑝.
𝑝 = ...........
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or 𝑝 = . . . . . . . . . . .
Question 14
Given that 𝑝−3 = −0.008 find the value of 𝑝.
𝑝 = ...........
Question 15
Given that (53 ) 𝑦 = 0.008 find the value of 𝑦.
𝑦 = ...........
Question 16
Given that (𝑥2 ) −3 = 0.000001 find the possible values of 𝑥.
𝑥 = ...........
or 𝑥 = . . . . . . . . . . .
Question 17
Solve for 𝑥:
5𝑥+2 = 125𝑥+7 × 25𝑥+5
𝑥 = ...........
Question 18
Simplify
2
18
(27𝑚
−3
)
…………………………
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Question 19
Simplify fully
(
64
12 )
125𝑥
1
−3
…………………………
Question 20
Given that
3
√2
√2 ×
2
can be written in the form 2𝑛
Find the value of 𝑛.
𝑛 = …………………………
Question 21
Write
3
3
1
√𝑦 + 3 𝑦 + 𝑦4
√
as an expression where all terms are in the form 𝑎𝑥𝑏
…………………………
Question 22
Fully simplify
3
√27𝑝5𝑞
7𝑝4
,
writing your answer in the form 𝑎𝑝𝑏 𝑞𝑐 .
…………………………
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Question 23
Solve
3
𝑎2 = 27
…………………………
Question 24
Solve
2
𝑎3 = 12
giving your answer in the form of 𝑎 √3
…………………………
Question 25
Expand and simplify
5𝑎𝑏3 (3𝑐 − 2𝑐2 )
…………………………
Question 26
Expand and simplify
(𝑥 + 1)2 − 4𝑥(3𝑥 + 4)
…………………………
Question 27
Expand and simplify
(2𝑥 + 6)2 − (𝑥 − 1)2
…………………………
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Question 28
Expand and simplify
4 2
(3 + 𝑧 )
…………………………
Question 29
Expand and simplify
5𝑥(2𝑥 + 1) (3𝑥 + 5)
…………………………
Question 30
Expand and simplify:
(1 − 3𝑥)3 − (3 − 2𝑥)2
…………………………
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Mark scheme
Question 1
Formula
Question 2
2𝑦2
Question 3
17𝑧3 + 15𝑧 + 19𝑞
➀ Collect like terms.
7𝑧 + 8𝑧 + 9𝑧3 + 19𝑞 + 8𝑧3
= 7𝑧 + 8𝑧+9𝑧3 +19𝑞+8𝑧3
= 9𝑧3 + 8𝑧3 +7𝑧 + 8𝑧+19𝑞
= 17𝑧3 +15𝑧+19𝑞
Question 4
2
➀ Draw a function machine for the equation.
(Could not load Desmos image)
➁ Substitute the value red = 16 into the function machine.
(Could not load Desmos image)
➂ Work backwards along the function machine, performing the inverse of each operation.
(Could not load Desmos image)
➃ Conclude that green = 2.
Question 5
25
➀ Substitute 𝑎 = 1 and 𝑏 = −1
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(4𝑎 − 𝑏)2 = (4 × 1 −× −1)2
= (4 + 1)2
= (5)2
= 25
Question 6
1
42
➀ Write the fractions in the expression as divisions.
(2𝑦) ÷ 𝑣
➁ Substitute the value of 𝑦 and 𝑣 into the expression.
1
2
= ( 2 × 1 2) ÷ 3
➂ Write the mixed numbers as improper fractions.
3
2
= ( 2 × 2) ÷ 3
➃ Write the division by a fraction as a multiplication by its reciprocal.
3
3
= ( 2 × 2) × 2
➄ Work out the value of the expression.
3
=3 × 2
9
= 2
1
= 42
Question 7
4
𝑦7
➀ Use laws of indices to simplify.
1
2
𝑦3 × 𝑦3
3
1
=
𝑦7
2
+
𝑦3 3
3
𝑦7
=
𝑦1
3
𝑦7
3
= 𝑦1−7
4
= 𝑦7
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Question 8
16𝑥10 𝑝2
➀ Raise each factor in the bracket to the power of 2
2
2
(4𝑥5𝑝 ) = 42 (𝑥5 ) (𝑝)2
= 42 𝑥 (5×2) 𝑝 (1×2)
= 16𝑥10 𝑝2
Question 9
3𝑚6 𝑛2
➀ Use laws of indices to simplify the denominator.
12𝑚10𝑛2
2
(2𝑚2 )
=
12𝑚10 𝑛2
2
22 (𝑚2 )
12𝑚10 𝑛2
= 4𝑚(2×2)
=
12𝑚10 𝑛2
4𝑚4
➁ Use the laws of indices for division to simplify the fraction.
12𝑚10𝑛2
4𝑚4
123
= 4 𝑚(10−4) 𝑛2
= 3𝑚6 𝑛2
Question 10
7
7 +2𝑥
➀ Split the fraction.
14 +7𝑥
1
2𝑥3
=
14
1
2𝑥3
+
7𝑥
1
2𝑥3
➁ Simplify using the laws of indices for division.
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1
5
14𝑥3
7𝑥3
+
1
1
2𝑥3
2𝑥3
147 (1−1)
7 5 1
= 2 𝑥 3 3 + 2𝑥(3−3)
7
= 7 +2𝑥
Question 11
7
➀ Use laws of indices to simplify the left hand side.
𝑞𝑎 × 𝑞10
𝑞𝑎+10
= 𝑞9
𝑞9
= 𝑞𝑎+10 − 9
= 𝑞𝑎+1
➁ Equate the powers to find 𝑎.
𝑞𝑎+1 = 𝑞8
⇒ 𝑎+1 = 8
−1 ↓
↓ −1
𝑎=7
Question 12
16𝑡−2 or 16𝑡−2
➀ Take the reciprocal of
𝑡
−2
(4)
𝑡
due to the negative power.
4
4 2
= (𝑡)
➁ Raise each part of the term to the power of 2 .
4 2
42
𝑡2
16
= 𝑡2
(𝑡) =
= 16𝑡−2
Question 13
𝑝 = 5 or 𝑝 = −5
➀ Use index laws to rewrite 𝑝−2
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1
𝑝−2 = 25
1
1
⇒ 𝑝2 = 25
⇒ 𝑝2 = 25
➁ Solve to find 𝑝.
⇒ 𝑝 = ±√25
⇒ 𝑝 = ±5
Question 14
𝑝 = −5
➀ Write −0.008 as a fraction and use index laws to rewrite 𝑝−3
𝑝−3 = −0.008
1
1
⇒ 𝑝3 = − 125
⇒ 𝑝3 = −125
➁ Solve to find 𝑝.
3
⇒ 𝑝 = √−125
⇒ 𝑝 = −5
Question 15
𝑦 = −1
Convert into a fraction.
3
(5 ) 𝑦 = 0.008
1
⇒ (53 ) 𝑦 = 125
➀ Write the right hand side as a power of 5.
1
⇒ (53 ) 𝑦 = 3
5
➁ Use index laws and compare powers.
⇒ 53𝑦 = 5−3
⇒ 3𝑦 = −3
⇒ 𝑦 = −1
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Question 16
𝑥 = 10 or 𝑥 = −10
➀ Convert 0.000001 into a fraction.
(𝑥2 ) −3 = 0.000001
1
⇒ (𝑥2 ) −3 = 1000000
➁ Use index laws to rewrite (𝑥2 ) −3
1
⇒ 𝑥−6 = 1000000
1
1
⇒ 𝑥6 = 1000000
⇒ 𝑥6 = 1000000
➂ Solve to find 𝑥.
6
⇒ 𝑥 = ± √1000000
⇒ 𝑥 = ±10
Question 17
29
𝑥 = −4
➀ Start by writing the equation in the form 5𝑎 = 5𝑏 and then equate the powers.
5𝑥+2 = 125𝑥+7 × 25𝑥+5
5𝑥+2 = (53 )𝑥+7 × (52 )𝑥+5
5𝑥+2 = 53𝑥+21 × 52𝑥+10
𝑥+2
5𝑥+31
5
= 5
𝑥 + 2 = 5𝑥 + 31
29
∴ 𝑥 = −4
Question 18
1
9𝑚12
1
or 9 𝑚−12
➀ As the power is negative, take the reciprocal of 27𝑚18
2
(27𝑚18 )
−3
=
1
(27𝑚
2
18 3
)
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2
➁ Raise each term inside the bracket to the power of 3
1
(
2
27𝑚18 3
=
)
=
=
=
1
2
2
273 (𝑚18 )3
1
2
3
( √27) 𝑚12
1
2
3 𝑚12
1
9𝑚12
Question 19
5𝑥4
4
➀ As the power is negative, take the reciprocal of
(
64
12 )
125𝑥
1
−3
= (
12
125𝑥
64
64
125𝑥12
1
) 3
1
➁ Raise each term inside the bracket to the power of 3
1
1
=
=
=
1253 (𝑥12 )3
1
643
3
√125𝑥4
3
√64
5𝑥4
4
Question 20
1
−6
➀ Express all roots in index form.
3
√2 ×
√2
2
1
= 22 ×
1
23
21
➁ Multiply the lone term with the fraction.
1
=
1
22 × 23
21
➂ Simplify the numerator.
5
=
26
21
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➃ Subtract the powers.
1
= 2−6
Question 21
1
1
𝑦3 +3𝑦−3 +𝑦−4
➀ Simplify each term using index laws.
3
3
1
√𝑦 + 3 𝑦 + 𝑦4
√
1
3
3
1
= 𝑦 + 1 + 𝑦4
𝑦3
1
1
3
= 𝑦 +3𝑦−3 +𝑦−4
Question 22
3 −7 1
𝑝 3 𝑞3
7
5
1/3
&#9312Write the cube root as a power: (27𝑝 𝑞4) &#9313Simplify the numerator
7𝑝
using index laws:Using
𝑝𝑎
3𝑝
(𝑝𝑎 )𝑏 = 𝑝𝑎𝑏 ,
5/3 1/3
𝑞
7𝑝4
&#9314Simplify the fraction and write in
3
the correct form:Using 𝑝𝑏 = 𝑝𝑎−𝑏 , 7 𝑝−7/3 𝑞1/3
Question 23
9
➀ Raise both sides to the reciprocal of the power.
3
𝑎2 = 27
2
𝑎 = 273
3
𝑎 = ( √27 ) 2
𝑎 = 32
𝑎 = 9
Question 24
24√3
➀ Raise both sides to the reciprocal of the power.
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2
𝑎3 = 12
3
𝑎 = 122
𝑎 = (√12 ) 3
𝑎 = (2 √3 ) 3
➁ Raise each term in the bracket to the power of 3.
𝑎 = 2 3 ( √3 ) 3
𝑎 = 8 ( √3 ) 3
𝑎 = 8 ( √ 3 √3 × √3 )
𝑎 = 8 ( 3 × √3 )
𝑎 = 8 × 3 √3
𝑎 = 24√3
Question 25
15𝑎𝑏3𝑐 − 10𝑎𝑏3𝑐2
➀ Multiply all terms in the bracket by 5𝑎 𝑏3
5𝑎 𝑏3 (3𝑐 − 2𝑐2 )
= 5𝑎 𝑏3 × 3𝑐 + 5𝑎 𝑏3 × −2𝑐2
➁ Simplify each term by multiplying the numbers and adding the powers for each letter.
= 15𝑎 𝑏3 𝑐 − 10𝑎 𝑏3 𝑐2
Question 26
−11𝑥2 − 14𝑥 + 1
➀ Expand (𝑥 + 1)2
(𝑥 + 1)(𝑥 + 1)
= 𝑥×𝑥+𝑥×1+1×𝑥+1×1
= 𝑥2 + 2𝑥 + 1
➁ Expand −4𝑥(3𝑥 + 4)
4𝑥(3𝑥 + 4)
= 4𝑥 × 3𝑥 + 4 × 4𝑥
= −12𝑥2 − 16𝑥
➂ Simplify.
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(𝑥 + 1)2 − 4𝑥(3𝑥 + 4)
= ( 𝑥2 + 2𝑥 + 1 ) − ( −12𝑥2 − 16𝑥 )
= −11𝑥2 − 14𝑥 + 1
Question 27
3𝑥2 + 26𝑥 + 35
➀ Expand (2𝑥 + 6)2
(2𝑥 + 6)(2𝑥 + 6)
= 2𝑥 × 2𝑥 + 2𝑥 × 6 + 6 × 2𝑥 + 6 × 6
= 4𝑥2 + 24𝑥 + 36
➁ Expand (𝑥 − 1)2
(𝑥 − 1)(𝑥 − 1)
= 𝑥×𝑥−𝑥×1−1×𝑥+1×1
= 𝑥2 − 2𝑥 + 1
➂ Simplify.
(2𝑥 + 6)2 − (𝑥 − 1)2
= ( 4𝑥2 + 24𝑥 + 36 ) − ( 𝑥2 − 2𝑥 + 1 )
= 3𝑥2 + 26𝑥 + 35
Alternatively, use the difference of two squares.
(Could not display math)
Question 28
24
16
9 + 𝑧 + 𝑧2
➀ Expand the brackets and simplify.
4
𝑧
4
𝑧
(3 + ) (3 + )
4
4
4
4
= 3×3+3×𝑧+𝑧×3+𝑧×𝑧
12
12
24
16
16
= 9 + 𝑧 + 𝑧 + 𝑧2
= 9 + 𝑧 + 𝑧2
Question 29
30𝑥3 + 65𝑥2 + 25𝑥
➀ Expand 5𝑥(2𝑥 + 1)
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5𝑥 ( 2𝑥 + 1 ) ( 3𝑥 + 5)
= ( 10𝑥2 + 5𝑥 ) (3𝑥 + 5)
➁ Expand and simplify the remaining brackets.
= 10𝑥2 × 3𝑥 + 10𝑥2 × 5 + 5𝑥 × 3𝑥 + 5𝑥 × 5
= 30𝑥3 + 50𝑥2 + 15𝑥2 + 25𝑥
= 30𝑥3 + 65𝑥2 + 25𝑥
Question 30
−27𝑥3 + 23𝑥2 + 3𝑥 − 8
➀ Write the cubed term as the product of 3 brackets and expand the first two.
(1 − 3𝑥) 3
= (1 − 3𝑥) (1 − 3𝑥) (1 − 3𝑥)
= ( 1 − 3𝑥 − 3𝑥 + 9𝑥2 )(1 − 3𝑥)
= ( 1 − 6𝑥 + 9𝑥2 )(1 − 3𝑥)
➁ Expand the remaining two brackets and simplify this expression.
= 1 − 6𝑥 + 9𝑥2 − 3𝑥 + 18𝑥2 − 27𝑥3
= 1 − 9𝑥 + 27𝑥2 − 27𝑥3
➂ Write the squared term as the product of 2 brackets and expand them.
(3 − 2𝑥) 2
= (3 − 2𝑥) (3 − 2𝑥)
= 9 − 6𝑥 − 6𝑥 + 4𝑥2
= 9 − 12𝑥 + 4𝑥2
➃ Subtract the second expanded expression from the first.
(1 − 3𝑥)3 − (3 − 2𝑥)2
= (1 − 9𝑥 + 27𝑥2 − 27𝑥3) − (9 − 12𝑥 + 4𝑥2 )
= −27𝑥3 + 23𝑥2 + 3𝑥 − 8
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