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ESAT Mock Maths 1

27 questions27 marks40Updated July 2026

The ESAT Mock Maths 1 paper in full: all 27 questions, each with its answer and a worked solution that shows every step. ESAT is the Engineering and Science Admissions Test. Sit it cold under exam timing, mark it, then work back through anything you missed using the solutions below.

Question 1

1 mark
A rectangular water tank has a base area of 1.5 m21.5\,\text{m}^2. The tank is filled with water to a depth of 40 cm40\,\text{cm}. The density of water is 1.0 g cm−31.0\,\text{g cm}^{-3}.

What is the total mass of the water in the tank in kilograms?
  • A.6 kg6\,\text{kg}
  • B.60 kg60\,\text{kg}
  • C.600 kg600\,\text{kg}
  • D.6000 kg6000\,\text{kg}
  • E.60,000 kg60,000\,\text{kg}

Answer: C

Worked solution

To find the mass, we first calculate the volume of the water. We must ensure units are consistent.

The base area is A=1.5 m2A = 1.5\,\text{m}^2. Converting this to cm2\text{cm}^2:
1.5 m2=1.5×(100 cm)2=1.5×10,000 cm2=15,000 cm21.5\,\text{m}^2 = 1.5 \times (100\,\text{cm})^2 = 1.5 \times 10,000\,\text{cm}^2 = 15,000\,\text{cm}^2


The depth is d=40 cmd = 40\,\text{cm}. The volume VV is:
V=A×d=15,000 cm2×40 cm=600,000 cm3V = A \times d = 15,000\,\text{cm}^2 \times 40\,\text{cm} = 600,000\,\text{cm}^3


Using the density ρ=1.0 g cm−3\rho = 1.0\,\text{g cm}^{-3}, the mass mm in grams is:
m=V×ρ=600,000 cm3×1.0 g cm−3=600,000 gm = V \times \rho = 600,000\,\text{cm}^3 \times 1.0\,\text{g cm}^{-3} = 600,000\,\text{g}


Converting the mass to kilograms (1 kg=1000 g1\,\text{kg} = 1000\,\text{g}):
m=600,0001000 kg=600 kgm = \frac{600,000}{1000}\,\text{kg} = 600\,\text{kg}

Question 2

1 mark
A cylindrical water tank has a constant cross-sectional area of 0.5 m20.5\,\text{m}^2 and a total height of 2.0 m2.0\,\text{m}. The tank is initially empty and is being filled by a pipe at a constant rate of 2.5 litres per second2.5\,\text{litres per second}. The density of water is 1000 kg m−31000\,\text{kg m}^{-3}.

How many minutes will it take for the tank to reach 75%75\% of its maximum capacity?

(1 m3=1000 litres1\,\text{m}^3 = 1000\,\text{litres})
  • A.2.5
  • B.5.0
  • C.6.7
  • D.7.5
  • E.300

Answer: B

Worked solution

First, calculate the total volume of the tank: V=Area×height=0.5 m2×2.0 m=1.0 m3V = \text{Area} \times \text{height} = 0.5\,\text{m}^2 \times 2.0\,\text{m} = 1.0\,\text{m}^3.

The student wants to find the time to reach 75%75\% of this capacity: V75=0.75×1.0 m3=0.75 m3V_{75} = 0.75 \times 1.0\,\text{m}^3 = 0.75\,\text{m}^3.

Convert this volume from cubic metres to litres using the provided conversion (1 m3=1000 L1\,\text{m}^3 = 1000\,\text{L}):
V75=0.75×1000=750 litresV_{75} = 0.75 \times 1000 = 750\,\text{litres}.

Now, calculate the time in seconds using the flow rate R=2.5 L s−1R = 2.5\,\text{L s}^{-1}:
tseconds=V75R=7502.5=750025=300 secondst_{\text{seconds}} = \frac{V_{75}}{R} = \frac{750}{2.5} = \frac{7500}{25} = 300\,\text{seconds}.

Finally, convert the time from seconds to minutes:
tminutes=30060=5.0 minutest_{\text{minutes}} = \frac{300}{60} = 5.0\,\text{minutes}.

Question 3

1 mark
Let PP and QQ be defined by the following expressions:
P=(−23)2−56P = \left( -\frac{2}{3} \right)^2 - \frac{5}{6}

Q=12÷(−34)+13Q = \frac{1}{2} \div \left( -\frac{3}{4} \right) + \frac{1}{3}

Consider the following statements:
I. P<QP < Q
II. P+Q>−23P + Q > -\frac{2}{3}
III. PQ>1\frac{P}{Q} > 1
Which of these statements is/are true?
  • A.I only
  • B.II only
  • C.I and II only
  • D.I and III only
  • E.I, II and III

Answer: D

Worked solution

First, evaluate PP:
P=(−23)2−56=49−56P = (-\frac{2}{3})^2 - \frac{5}{6} = \frac{4}{9} - \frac{5}{6}
To subtract, find a common denominator (18):
P=818−1518=−718P = \frac{8}{18} - \frac{15}{18} = -\frac{7}{18}
Next, evaluate QQ:
Q=12×(−43)+13=−46+13=−23+13=−13Q = \frac{1}{2} \times (-\frac{4}{3}) + \frac{1}{3} = -\frac{4}{6} + \frac{1}{3} = -\frac{2}{3} + \frac{1}{3} = -\frac{1}{3}
Express QQ with denominator 18: Q=−618Q = -\frac{6}{18}
Now evaluate the statements:
I. P<QP < Q means −718<−618-\frac{7}{18} < -\frac{6}{18}. Since −7<−6-7 < -6, this is True.
II. P+Q=−718−618=−1318P + Q = -\frac{7}{18} - \frac{6}{18} = -\frac{13}{18}. We compare this to −23=−1218-\frac{2}{3} = -\frac{12}{18}. Since −13<−12-13 < -12, −1318<−1218-\frac{13}{18} < -\frac{12}{18}, so P+QP+Q is not greater than −23-\frac{2}{3}. This is False.
III. PQ=−7/18−1/3=718×31=2118=76\frac{P}{Q} = \frac{-7/18}{-1/3} = \frac{7}{18} \times \frac{3}{1} = \frac{21}{18} = \frac{7}{6}. Since 76>1\frac{7}{6} > 1, this is True.
Statements I and III are true.

Question 4

1 mark
Evaluate the following expression:

X=[(1−11−13)−1+(1+11+13)−1]−1X = \left[ \left( 1 - \frac{1}{1 - \frac{1}{3}} \right)^{-1} + \left( 1 + \frac{1}{1 + \frac{1}{3}} \right)^{-1} \right]^{-1}
  • A.-1.4
  • B.-0.7
  • C.0.7
  • D.1.4
  • E.7.0

Answer: B

Worked solution

To solve this, we evaluate the nested fractions from the inside out, following the order of operations.

1. First term inside the brackets:
Evaluate the denominator: 1−13=231 - \frac{1}{3} = \frac{2}{3}.
Taking the reciprocal: 12/3=32\frac{1}{2/3} = \frac{3}{2}.
Subtracting from 1: 1−32=−121 - \frac{3}{2} = -\frac{1}{2}.
Taking the outer reciprocal: (−12)−1=−2(-\frac{1}{2})^{-1} = -2.

2. Second term inside the brackets:
Evaluate the denominator: 1+13=431 + \frac{1}{3} = \frac{4}{3}.
Taking the reciprocal: 14/3=34\frac{1}{4/3} = \frac{3}{4}.
Adding to 1: 1+34=741 + \frac{3}{4} = \frac{7}{4}.
Taking the outer reciprocal: (74)−1=47(\frac{7}{4})^{-1} = \frac{4}{7}.

3. Combine the terms:
−2+47=−147+47=−107-2 + \frac{4}{7} = -\frac{14}{7} + \frac{4}{7} = -\frac{10}{7}.

4. Final reciprocal:
X=(−107)−1=−710=−0.7X = (-\frac{10}{7})^{-1} = -\frac{7}{10} = -0.7.

Thus, the correct answer is B.

Question 5

1 mark
Let f(n)=n3−1n3+1f(n) = \frac{n^3 - 1}{n^3 + 1} for integers n>1n > 1.

What is the value of the product P=f(2)×f(3)×f(4)×f(5)P = f(2) \times f(3) \times f(4) \times f(5)?
  • A.2845\frac{28}{45}
  • B.3145\frac{31}{45}
  • C.23\frac{2}{3}
  • D.1415\frac{14}{15}
  • E.124135\frac{124}{135}

Answer: B

Worked solution

We can factorise the numerator and denominator using the difference and sum of cubes:
n3−1=(n−1)(n2+n+1)n^3 - 1 = (n - 1)(n^2 + n + 1)
n3+1=(n+1)(n2−n+1)n^3 + 1 = (n + 1)(n^2 - n + 1)

Let g(n)=n2+n+1g(n) = n^2 + n + 1. Note that g(n−1)=(n−1)2+(n−1)+1=n2−2n+1+n−1+1=n2−n+1g(n-1) = (n-1)^2 + (n-1) + 1 = n^2 - 2n + 1 + n - 1 + 1 = n^2 - n + 1.
Thus, f(n)=n−1n+1⋅g(n)g(n−1)f(n) = \frac{n-1}{n+1} \cdot \frac{g(n)}{g(n-1)}.

The product PP can be written as:
P=(2−12+1⋅3−13+1⋅4−14+1⋅5−15+1)×(g(2)g(1)⋅g(3)g(2)⋅g(4)g(3)⋅g(5)g(4))P = \left( \frac{2-1}{2+1} \cdot \frac{3-1}{3+1} \cdot \frac{4-1}{4+1} \cdot \frac{5-1}{5+1} \right) \times \left( \frac{g(2)}{g(1)} \cdot \frac{g(3)}{g(2)} \cdot \frac{g(4)}{g(3)} \cdot \frac{g(5)}{g(4)} \right)


Notice the telescoping effect in both parts:
In the first bracket: 13⋅24⋅35⋅46=1⋅25⋅6=230=115\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6} = \frac{1 \cdot 2}{5 \cdot 6} = \frac{2}{30} = \frac{1}{15}.
In the second bracket: g(5)g(1)=52+5+112+1+1=313\frac{g(5)}{g(1)} = \frac{5^2 + 5 + 1}{1^2 + 1 + 1} = \frac{31}{3}.

Multiplying these together: P=115×313=3145P = \frac{1}{15} \times \frac{31}{3} = \frac{31}{45}.

Thus, the correct answer is B.

Question 6

1 mark
Given that xx is a positive real number satisfying the equation:

x+x+x+…=3x−x−x−…\sqrt{x + \sqrt{x + \sqrt{x + \dots}}} = 3\sqrt{x - \sqrt{x - \sqrt{x - \dots}}}


Where both sides represent infinite nested radicals, what is the value of xx?
  • A.0.25
  • B.0.50
  • C.0.75
  • D.1.50
  • E.2.00

Answer: C

Worked solution

Let A=x+x+…A = \sqrt{x + \sqrt{x + \dots}}. Squaring both sides gives A2=x+AA^2 = x + A, which can be rewritten as x=A2−Ax = A^2 - A.

Let B=x−x−…B = \sqrt{x - \sqrt{x - \dots}}. Squaring both sides gives B2=x−BB^2 = x - B, which can be rewritten as x=B2+Bx = B^2 + B.

The problem states A=3BA = 3B. Substituting this into our first expression for xx:
x=(3B)2−3B=9B2−3Bx = (3B)^2 - 3B = 9B^2 - 3B.

Since both expressions equal xx, we can set them equal to each other:
9B2−3B=B2+B9B^2 - 3B = B^2 + B
8B2−4B=08B^2 - 4B = 0
4B(2B−1)=04B(2B - 1) = 0

Since xx is positive, BB cannot be zero. Therefore, 2B−1=02B - 1 = 0, so B=0.5B = 0.5.

Now substitute B=0.5B = 0.5 back into the expression for xx:
x=B2+B=(0.5)2+0.5=0.25+0.5=0.75x = B^2 + B = (0.5)^2 + 0.5 = 0.25 + 0.5 = 0.75.

Thus, the correct answer is C.

Question 7

1 mark
A security code consists of 4 digits chosen from the set {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}. The digits in the code must be in non-decreasing order (from left to right).

How many such codes contain at least one repeated digit?
  • A.15
  • B.105
  • C.111
  • D.126
  • E.1281

Answer: C

Worked solution

First, we calculate the total number of codes with 4 digits from the set {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\} that are in non-decreasing order. This is equivalent to choosing 4 items from 6 with replacement, where the order is automatically determined by the non-decreasing constraint. Using the formula for combinations with repetition, (n+r−1r)\binom{n+r-1}{r}, with n=6n=6 and r=4r=4:
Total=(6+4−14)=(94)=9×8×7×64×3×2×1=126\text{Total} = \binom{6+4-1}{4} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126

Next, we identify the codes that have NO repeated digits. For the digits to be in non-decreasing order and not repeated, they must be in strictly increasing order. This is equivalent to choosing 4 distinct digits from 6:
Strictly increasing=(64)=(62)=6×52×1=15\text{Strictly increasing} = \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15

The number of codes with at least one repeated digit is the total number of non-decreasing codes minus the number of strictly increasing codes:
126−15=111126 - 15 = 111

Question 8

1 mark
How many 4-digit positive integers (from 1000 to 9999 inclusive) have the property that the product of their four digits is exactly 24?
  • A.44
  • B.52
  • C.60
  • D.64
  • E.72

Answer: D

Worked solution

Let the four digits be a,b,c,da, b, c, d. We require a×b×c×d=24a \times b \times c \times d = 24, where each digit is in the set {1,2,3,4,5,6,7,8,9}\{1, 2, 3, 4, 5, 6, 7, 8, 9\}. Note that no digit can be 0. We list the possible sets of four digits that multiply to 24 and calculate the permutations for each:
1. {8,3,1,1}\{8, 3, 1, 1\}: The number of permutations is 4!2!=12\frac{4!}{2!} = 12.
2. {6,4,1,1}\{6, 4, 1, 1\}: The number of permutations is 4!2!=12\frac{4!}{2!} = 12.
3. {6,2,2,1}\{6, 2, 2, 1\}: The number of permutations is 4!2!=12\frac{4!}{2!} = 12.
4. {4,3,2,1}\{4, 3, 2, 1\}: The number of permutations is 4!=244! = 24.
5. {3,2,2,2}\{3, 2, 2, 2\}: The number of permutations is 4!3!=4\frac{4!}{3!} = 4.

Summing these possibilities: 12+12+12+24+4=6412 + 12 + 12 + 24 + 4 = 64.
There are no other sets of single-digit integers whose product is 24 (for example, factors like 12 or 24 cannot be used as a single digit).

Question 9

1 mark
A square grid consists of 16 dots arranged in 4 rows and 4 columns. How many different squares can be formed such that all four vertices of the square are dots in the grid?
  • A.14
  • B.18
  • C.20
  • D.30
  • E.50

Answer: C

Worked solution

The dots form a grid with 3 units of distance between the outer boundaries. We can categorize squares by the size of the 'bounding box' they occupy.
1. 1×11 \times 1 bounding box: There are 3×3=93 \times 3 = 9 such boxes. Each box contains only 1 square (the axis-aligned one). Subtotal = 9.
2. 2×22 \times 2 bounding box: There are 2×2=42 \times 2 = 4 such boxes. Each box contains 2 squares: one axis-aligned (side length 2) and one tilted (vertices at the midpoints of the box sides, side length 2\sqrt{2}). Subtotal = 4×2=84 \times 2 = 8.
3. 3×33 \times 3 bounding box: There is 1×1=11 \times 1 = 1 such box. This box contains 3 squares: one axis-aligned (side length 3) and two tilted ones (one with vertices at (1,0),(3,1),(2,3),(0,2)(1,0), (3,1), (2,3), (0,2) and one at (2,0),(3,2),(1,3),(0,1)(2,0), (3,2), (1,3), (0,1) relative to the box corner). Subtotal = 1×3=31 \times 3 = 3.

Total number of squares = 9+8+3=209 + 8 + 3 = 20.

Question 10

1 mark
For a positive real number aa, the following relationship holds:
a23=1ak\sqrt[3]{a^2} = \sqrt{\frac{1}{a^k}}

What is the value of kk?
  • A.−43-\frac{4}{3}
  • B.−34-\frac{3}{4}
  • C.34\frac{3}{4}
  • D.43\frac{4}{3}
  • E.−13-\frac{1}{3}

Answer: A

Worked solution

To find kk, we first express both sides of the equation as powers of aa using the rules of indices:
1. The left-hand side is the cube root of a2a^2, which can be written as (a2)1/3=a2/3(a^2)^{1/3} = a^{2/3}.
2. The right-hand side is the square root of the reciprocal of aka^k. Recalling that 1ak=a−k\frac{1}{a^k} = a^{-k}, this becomes a−k=(a−k)1/2=a−k/2\sqrt{a^{-k}} = (a^{-k})^{1/2} = a^{-k/2}.

Setting the exponents equal to each other:
23=−k2\frac{2}{3} = -\frac{k}{2}


Multiply both sides by −2-2 to solve for kk:
k=−2×23=−43k = -2 \times \frac{2}{3} = -\frac{4}{3}


Therefore, the value of kk is −43-\frac{4}{3}, which corresponds to option A.

Question 11

1 mark
Which one of the following statements is NOT true for all real values of xx?
  • A.x2=∣x∣\sqrt{x^2} = |x|
  • B.(x3)3=x(\sqrt[3]{x})^3 = x
  • C.If xx is a square root of 4949, then x=49x = \sqrt{49}
  • D.−x3=−x3\sqrt[3]{-x} = -\sqrt[3]{x}
  • E.x4=x2\sqrt{x^4} = x^2

Answer: C

Worked solution

We evaluate each statement based on the definitions of square and cube roots:

A. This is true. The square root symbol \sqrt{} denotes the principal (non-negative) square root. For any real xx, x2x^2 is non-negative, and its positive root is the magnitude ∣x∣|x|.

B. This is true. The cube root of any real number is always defined, and cubing the cube root of a number returns the original number.

C. This is FALSE. A number yy has two square roots if y>0y > 0. The square roots of 4949 are 77 and −7-7. However, the notation 49\sqrt{49} refers specifically to the principal (positive) square root, which is 77. If x=−7x = -7, then xx is a square root of 4949, but x≠49x \neq \sqrt{49}.

D. This is true. The cube root is an odd function, meaning −13=−1\sqrt[3]{-1} = -1, and more generally, −x3=−x3\sqrt[3]{-x} = -\sqrt[3]{x}.

E. This is true. Since x2x^2 is always non-negative for any real xx, x4=(x2)2=∣x2∣=x2\sqrt{x^4} = \sqrt{(x^2)^2} = |x^2| = x^2.

Since C is the only statement that is not true for all possible values of xx, the correct option is C.

Question 12

1 mark
For x>0x > 0, the expression (x34+x−14)2x12+x−12\frac{(x^{\frac{3}{4}} + x^{-\frac{1}{4}})^2}{x^{\frac{1}{2}} + x^{-\frac{1}{2}}} is equivalent to:
  • A.x+1x + 1
  • B.x−1x - 1
  • C.x12+1x^{\frac{1}{2}} + 1
  • D.x2+1x^2 + 1
  • E.x32+x12x^{\frac{3}{2}} + x^{\frac{1}{2}}

Answer: A

Worked solution

We first expand the numerator using the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. Let a=x34a = x^{\frac{3}{4}} and b=x−14b = x^{-\frac{1}{4}}.

a2=(x34)2=x32a^2 = (x^{\frac{3}{4}})^2 = x^{\frac{3}{2}}
b2=(x−14)2=x−12b^2 = (x^{-\frac{1}{4}})^2 = x^{-\frac{1}{2}}
2ab=2(x34⋅x−14)=2x34−14=2x122ab = 2(x^{\frac{3}{4}} \cdot x^{-\frac{1}{4}}) = 2x^{\frac{3}{4} - \frac{1}{4}} = 2x^{\frac{1}{2}}

So the numerator is x32+2x12+x−12x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + x^{-\frac{1}{2}}. We can factor out x12x^{\frac{1}{2}} or simply divide each term by the denominator. Alternatively, factor x−12x^{-\frac{1}{2}} out of the numerator:
x32+2x12+x−12=x−12(x2+2x+1)=x−12(x+1)2x^{\frac{3}{2}} + 2x^{\frac{1}{2}} + x^{-\frac{1}{2}} = x^{-\frac{1}{2}}(x^2 + 2x + 1) = x^{-\frac{1}{2}}(x+1)^2

Factor x−12x^{-\frac{1}{2}} out of the denominator:
x12+x−12=x−12(x+1)x^{\frac{1}{2}} + x^{-\frac{1}{2}} = x^{-\frac{1}{2}}(x + 1)

Dividing the numerator by the denominator gives:
x−12(x+1)2x−12(x+1)=x+1\frac{x^{-\frac{1}{2}}(x+1)^2}{x^{-\frac{1}{2}}(x+1)} = x + 1

Therefore, the correct option is A.

Question 13

1 mark
Two real numbers xx and yy are defined by x=423x = \sqrt[3]{4\sqrt{2}} and y=243y = \sqrt{2\sqrt[3]{4}}. The product xyxy can be expressed in the form 2k2^k for some constant kk. What is the value of kk?
  • A.56\frac{5}{6}
  • B.53\frac{5}{3}
  • C.73\frac{7}{3}
  • D.2536\frac{25}{36}
  • E.11

Answer: B

Worked solution

We convert the nested radicals into fractional powers using the rule an=a1/n\sqrt[n]{a} = a^{1/n}:

For xx:
x=(4⋅21/2)1/3=(22⋅21/2)1/3=(22+1/2)1/3=(25/2)1/3=25/6x = (4 \cdot 2^{1/2})^{1/3} = (2^2 \cdot 2^{1/2})^{1/3} = (2^{2 + 1/2})^{1/3} = (2^{5/2})^{1/3} = 2^{5/6}

For yy:
y=(2⋅41/3)1/2=(21⋅(22)1/3)1/2=(21⋅22/3)1/2=(21+2/3)1/2=(25/3)1/2=25/6y = (2 \cdot 4^{1/3})^{1/2} = (2^1 \cdot (2^2)^{1/3})^{1/2} = (2^1 \cdot 2^{2/3})^{1/2} = (2^{1 + 2/3})^{1/2} = (2^{5/3})^{1/2} = 2^{5/6}

Now find the product xyxy:
xy=25/6⋅25/6=25/6+5/6=210/6=25/3xy = 2^{5/6} \cdot 2^{5/6} = 2^{5/6 + 5/6} = 2^{10/6} = 2^{5/3}

Thus k=5/3k = 5/3, which is option B.

Question 14

1 mark
Two spherical planets, XX and YY, have surface areas SXS_X and SYS_Y respectively such that SX=1.6×107SYS_X = 1.6 \times 10^7 S_Y. The average density of planet XX is ρX=1.25×103 kg m−3\rho_X = 1.25 \times 10^3 \text{ kg m}^{-3} and the average density of planet YY is ρY=1.0×107 kg m−3\rho_Y = 1.0 \times 10^7 \text{ kg m}^{-3}. What is the ratio of the mass of planet XX to the mass of planet YY?

You may use the following formulae for a sphere of radius rr:
Surface Area=4πr2\text{Surface Area} = 4\pi r^2

Volume=43πr3\text{Volume} = \frac{4}{3}\pi r^3
  • A.8.0 ×\times 10^6
  • B.5.0 ×\times 10^{-1}
  • C.1.6 ×\times 10^7
  • D.8.0 ×\times 10^3
  • E.6.4 ×\times 10^{10}

Answer: A

Worked solution

First, find the ratio of the radii rXr_X and rYr_Y using the surface area ratio: SX/SY=(rX/rY)2=1.6×107=16×106S_X/S_Y = (r_X/r_Y)^2 = 1.6 \times 10^7 = 16 \times 10^6. Taking the square root gives rX/rY=16×106=4×103r_X/r_Y = \sqrt{16 \times 10^6} = 4 \times 10^3.
Next, calculate the volume ratio: VX/VY=(rX/rY)3=(4×103)3=64×109=6.4×1010V_X/V_Y = (r_X/r_Y)^3 = (4 \times 10^3)^3 = 64 \times 10^9 = 6.4 \times 10^{10}.
Now find the density ratio: ρX/ρY=(1.25×103)/(1.0×107)=1.25×10−4\rho_X/\rho_Y = (1.25 \times 10^3) / (1.0 \times 10^7) = 1.25 \times 10^{-4}.
Finally, the mass ratio MX/MYM_X/M_Y is the product of the density and volume ratios: MX/MY=(ρX/ρY)×(VX/VY)=(1.25×10−4)×(6.4×1010)M_X/M_Y = (\rho_X/\rho_Y) \times (V_X/V_Y) = (1.25 \times 10^{-4}) \times (6.4 \times 10^{10}).
Multiplying the coefficients: 1.25×6.4=1.25×(6+0.4)=7.5+0.5=8.01.25 \times 6.4 = 1.25 \times (6 + 0.4) = 7.5 + 0.5 = 8.0.
Combining with the powers of ten: 8.0×10−4+10=8.0×1068.0 \times 10^{-4 + 10} = 8.0 \times 10^6.

Question 15

1 mark
A laser pulse has a total energy of E=9.9×10−4 JE = 9.9 \times 10^{-4} \text{ J}. Each photon in the pulse has an energy EphE_{ph} given by the formula Eph=hcλE_{ph} = \frac{hc}{\lambda}.

Take the following values:
- Planck's constant h=6.6×10−34 J sh = 6.6 \times 10^{-34} \text{ J s}
- Speed of light c=3.0×108 m s−1c = 3.0 \times 10^8 \text{ m s}^{-1}
- Wavelength λ=6.0×10−7 m\lambda = 6.0 \times 10^{-7} \text{ m}

How many photons are contained in this laser pulse?
  • A.3.0 ×\times 10^{15}
  • B.3.0 ×\times 10^{16}
  • C.5.0 ×\times 10^{14}
  • D.3.3 ×\times 10^{15}
  • E.3.0 ×\times 10^{14}

Answer: A

Worked solution

First, calculate the energy of a single photon EphE_{ph}: Eph=(6.6×10−34)×(3.0×108)6.0×10−7E_{ph} = \frac{(6.6 \times 10^{-34}) \times (3.0 \times 10^8)}{6.0 \times 10^{-7}}.
Eph=19.8×10−266.0×10−7=3.3×10−19 JE_{ph} = \frac{19.8 \times 10^{-26}}{6.0 \times 10^{-7}} = 3.3 \times 10^{-19} \text{ J}.
Next, calculate the total number of photons NN by dividing the total pulse energy by the energy of one photon: N=EEph=9.9×10−43.3×10−19N = \frac{E}{E_{ph}} = \frac{9.9 \times 10^{-4}}{3.3 \times 10^{-19}}.
Dividing the coefficients: 9.9/3.3=3.09.9 / 3.3 = 3.0.
Subtracting the exponents: −4−(−19)=15-4 - (-19) = 15.
Thus, N=3.0×1015N = 3.0 \times 10^{15}.

Question 16

1 mark
A rectangular plot of land has an area of 1.0×1013 m21.0 \times 10^{13} \text{ m}^2 and a perimeter of 2.2×107 m2.2 \times 10^7 \text{ m}. What is the positive difference between the length and the width of the rectangle, expressed in standard form?
  • A.9.0 ×\times 10^6  m\text{ m}
  • B.1.0 ×\times 10^6  m\text{ m}
  • C.1.1 ×\times 10^7  m\text{ m}
  • D.1.0 ×\times 10^7  m\text{ m}
  • E.8.1 ×\times 10^6  m\text{ m}

Answer: A

Worked solution

Let the length and width be LL and WW. We are given:
1) LW=1.0×1013LW = 1.0 \times 10^{13}
2) 2(L+W)=2.2×107  ⟹  L+W=1.1×1072(L+W) = 2.2 \times 10^7 \implies L+W = 1.1 \times 10^7.
We need to find L−WL-W. Recall the identity (L−W)2=(L+W)2−4LW(L-W)^2 = (L+W)^2 - 4LW.
Substituting the given values: (L−W)2=(1.1×107)2−4(1.0×1013)(L-W)^2 = (1.1 \times 10^7)^2 - 4(1.0 \times 10^{13}).
(L−W)2=1.21×1014−4.0×1013(L-W)^2 = 1.21 \times 10^{14} - 4.0 \times 10^{13}.
Convert 4.0×10134.0 \times 10^{13} to the same power as 101410^{14} to allow subtraction: 4.0×1013=0.4×10144.0 \times 10^{13} = 0.4 \times 10^{14}.
(L−W)2=1.21×1014−0.4×1014=0.81×1014(L-W)^2 = 1.21 \times 10^{14} - 0.4 \times 10^{14} = 0.81 \times 10^{14}.
Taking the square root: L−W=0.81×1014=0.81×1014=0.9×107L-W = \sqrt{0.81 \times 10^{14}} = \sqrt{0.81} \times \sqrt{10^{14}} = 0.9 \times 10^7.
In standard form, 0.9×107=9.0×106 m0.9 \times 10^7 = 9.0 \times 10^6 \text{ m}.

Question 17

1 mark
Let x=0.25˙x = 0.2\dot{5} and y=0.2˙5˙y = 0.\dot{2}\dot{5}. What is the value of x−yx - y expressed as a fraction in its simplest form?
  • A.1330\frac{1}{330}
  • B.133\frac{1}{33}
  • C.1110\frac{1}{110}
  • D.5198\frac{5}{198}
  • E.190\frac{1}{90}

Answer: A

Worked solution

First, we convert both recurring decimals to fractions.

For x=0.25˙=0.2555…x = 0.2\dot{5} = 0.2555\dots:
10x=2.555…10x = 2.555\dots
100x=25.555…100x = 25.555\dots
100x−10x=25−2=23100x - 10x = 25 - 2 = 23
90x=23  ⟹  x=239090x = 23 \implies x = \frac{23}{90}

For y=0.2˙5˙=0.2525…y = 0.\dot{2}\dot{5} = 0.2525\dots:
100y=25.2525…100y = 25.2525\dots
100y−y=25100y - y = 25
99y=25  ⟹  y=259999y = 25 \implies y = \frac{25}{99}

Now find the difference x−yx - y:
x−y=2390−2599x - y = \frac{23}{90} - \frac{25}{99}

The lowest common multiple of 9090 and 9999 is 990990.
x−y=23×11990−25×10990x - y = \frac{23 \times 11}{990} - \frac{25 \times 10}{990}

x−y=253−250990=3990x - y = \frac{253 - 250}{990} = \frac{3}{990}

Simplifying the fraction:
3990=1330\frac{3}{990} = \frac{1}{330}

Question 18

1 mark
A chemical process involves mixing three solutions of acid: XX, YY, and ZZ.

Solution XX contains 40%40\% acid by volume.

Solution YY contains 14\frac{1}{4} acid by volume.

Solutions XX and YY are mixed in the ratio 2:32 : 3 by volume to create a base mixture MM.

A final mixture is then formed by combining mixture MM with solution ZZ in the ratio 5:25 : 2 by volume.

If the final mixture is 25%25\% acid by volume, what is the acid concentration of solution ZZ as a percentage?
  • A.10%10\%
  • B.12.5%12.5\%
  • C.15%15\%
  • D.17.5%17.5\%
  • E.20%20\%

Answer: A

Worked solution

Let us consider specific volumes to simplify the arithmetic. First, we mix XX and YY in the ratio 2:32 : 3. Let the volume of XX be 22 litres and the volume of YY be 33 litres, giving 55 litres of mixture MM.

Since XX is 40%40\% acid, it contributes 2×0.40=0.802 \times 0.40 = 0.80 litres of acid. Since YY is 14=25%\frac{1}{4} = 25\% acid, it contributes 3×0.25=0.753 \times 0.25 = 0.75 litres of acid. Thus, 55 litres of MM contains 0.80+0.75=1.550.80 + 0.75 = 1.55 litres of acid.

Next, we mix mixture MM with solution ZZ in the ratio 5:25 : 2. Since we have 55 litres of MM, we mix it with 22 litres of ZZ. Let the acid concentration of ZZ be zz.

The total volume of the final mixture is 5+2=75 + 2 = 7 litres. We are told the final concentration is 25%25\%, so the total acid in the final mixture is 7×0.25=1.757 \times 0.25 = 1.75 litres.

The acid from MM plus the acid from ZZ must equal the total acid: 1.55+2z=1.751.55 + 2z = 1.75. This gives 2z=0.202z = 0.20, so z=0.10z = 0.10. In percentage terms, this is 10%10\%.

Question 19

1 mark
The wholesale cost of a machine is first increased by 25%25\% to determine its retail price.

During a promotion, the retail price is discounted to a 'special price' which is 10%10\% less than the original wholesale cost.

The special price is subsequently increased by 3313%33\frac{1}{3}\% to give a final price.

What is the ratio of the final price to the retail price?
  • A.24:2524 : 25
  • B.25:2425 : 24
  • C.4:54 : 5
  • D.1:11 : 1
  • E.5:65 : 6

Answer: A

Worked solution

Let the original wholesale cost be WW.

The retail price is R=W×(1+0.25)=1.25W=54WR = W \times (1 + 0.25) = 1.25W = \frac{5}{4}W.

The special price SS is 10%10\% less than the original wholesale cost, so S=W×(1−0.10)=0.90W=910WS = W \times (1 - 0.10) = 0.90W = \frac{9}{10}W.

The special price is increased by 3313%33\frac{1}{3}\%. Note that 3313%=1333\frac{1}{3}\% = \frac{1}{3}, so an increase by this amount is equivalent to multiplying by (1+13)=43(1 + \frac{1}{3}) = \frac{4}{3}.

The final price FF is S×43=910W×43=3630W=65W=1.2WS \times \frac{4}{3} = \frac{9}{10}W \times \frac{4}{3} = \frac{36}{30}W = \frac{6}{5}W = 1.2W.

The ratio of the final price to the retail price is FR=1.2W1.25W=1.21.25\frac{F}{R} = \frac{1.2W}{1.25W} = \frac{1.2}{1.25}.

To simplify the fraction 1.21.25\frac{1.2}{1.25}, we multiply the numerator and denominator by 100100: 120125\frac{120}{125}. Dividing both by 55 gives 2425\frac{24}{25}.

Thus, the ratio is 24:2524 : 25.

Question 20

1 mark
Simplify the following expression as far as possible:

105+15−2\frac{10}{\sqrt{5}} + \frac{1}{\sqrt{5} - 2}


  • A.25+22\sqrt{5} + 2
  • B.35−23\sqrt{5} - 2
  • C.35+23\sqrt{5} + 2
  • D.115+211\sqrt{5} + 2
  • E.5+2\sqrt{5} + 2

Answer: C

Worked solution

To simplify the expression, we address each term individually.

For the first term, 105\frac{10}{\sqrt{5}}, we rationalise the denominator by multiplying the numerator and denominator by 5\sqrt{5}:
105=1055=25\frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}


For the second term, 15−2\frac{1}{\sqrt{5} - 2}, we rationalise the denominator by multiplying by the conjugate (5+2)(\sqrt{5} + 2):
15−2=5+2(5−2)(5+2)\frac{1}{\sqrt{5} - 2} = \frac{\sqrt{5} + 2}{(\sqrt{5} - 2)(\sqrt{5} + 2)}

Using the difference of squares in the denominator (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2, we get:
5+25−4=5+21=5+2\frac{\sqrt{5} + 2}{5 - 4} = \frac{\sqrt{5} + 2}{1} = \sqrt{5} + 2


Adding the two simplified terms together gives:
25+(5+2)=35+22\sqrt{5} + (\sqrt{5} + 2) = 3\sqrt{5} + 2


Therefore, the correct answer is C.

Question 21

1 mark
A car travels a distance of 120 km120\text{ km}, measured correct to the nearest 10 km10\text{ km}. The time taken for the journey is 2.0 hours2.0\text{ hours}, measured correct to the nearest 0.1 hours0.1\text{ hours}. Which one of the following expressions gives the difference between the maximum possible average speed and the minimum possible average speed for this journey?
  • A.128001599 km/h\frac{12800}{1599}\text{ km/h}
  • B.122001599 km/h\frac{12200}{1599}\text{ km/h}
  • C.128001600 km/h\frac{12800}{1600}\text{ km/h}
  • D.5 km/h5\text{ km/h}
  • E.120001599 km/h\frac{12000}{1599}\text{ km/h}

Answer: A

Worked solution

The average speed vv is given by v=dtv = \frac{d}{t}. To find the maximum average speed vmaxv_{\text{max}}, we use the maximum distance dmaxd_{\text{max}} and the minimum time tmint_{\text{min}}. To find the minimum average speed vminv_{\text{min}}, we use the minimum distance dmind_{\text{min}} and the maximum time tmaxt_{\text{max}}.

The distance d=120 kmd = 120\text{ km} to the nearest 10 km10\text{ km} gives the bounds 115≤d<125115 \le d < 125.
The time t=2.0 ht = 2.0\text{ h} to the nearest 0.1 h0.1\text{ h} gives the bounds 1.95≤t<2.051.95 \le t < 2.05.

vmax=1251.95=12500195=250039 km/hv_{\text{max}} = \frac{125}{1.95} = \frac{12500}{195} = \frac{2500}{39}\text{ km/h}
vmin=1152.05=11500205=230041 km/hv_{\text{min}} = \frac{115}{2.05} = \frac{11500}{205} = \frac{2300}{41}\text{ km/h}

The difference is:
vmax−vmin=250039−230041=2500×41−2300×3939×41v_{\text{max}} - v_{\text{min}} = \frac{2500}{39} - \frac{2300}{41} = \frac{2500 \times 41 - 2300 \times 39}{39 \times 41}

Calculating the numerator:
2500×41=1025002500 \times 41 = 102500
2300×39=2300×(40−1)=92000−2300=897002300 \times 39 = 2300 \times (40 - 1) = 92000 - 2300 = 89700
102500−89700=12800102500 - 89700 = 12800

Calculating the denominator:
39×41=(40−1)(40+1)=402−12=1600−1=159939 \times 41 = (40 - 1)(40 + 1) = 40^2 - 1^2 = 1600 - 1 = 1599

Thus, the difference is 128001599 km/h\frac{12800}{1599}\text{ km/h}, which is option A.

Question 22

1 mark
A rectangular block has dimensions 20 cm20\text{ cm} by 30 cm30\text{ cm} by 40 cm40\text{ cm}, each measured correct to the nearest 2 cm2\text{ cm}. What is the difference between the maximum possible volume and the minimum possible volume of the block?
  • A.5202 cm35202\text{ cm}^3
  • B.5200 cm35200\text{ cm}^3
  • C.10416 cm310416\text{ cm}^3
  • D.2601 cm32601\text{ cm}^3
  • E.4800 cm34800\text{ cm}^3

Answer: A

Worked solution

The volume VV of a rectangular block is V=a×b×cV = a \times b \times c.
Measurements correct to the nearest 2 cm2\text{ cm} imply an uncertainty of ±1 cm\pm 1\text{ cm}.
a∈[19,21]a \in [19, 21], b∈[29,31]b \in [29, 31], c∈[39,41]c \in [39, 41].

Vmax=21×31×41V_{\text{max}} = 21 \times 31 \times 41
Vmin=19×29×39V_{\text{min}} = 19 \times 29 \times 39

Using the expansion (x+δ)(y+δ)(z+δ)−(x−δ)(y−δ)(z−δ)=2δ(xy+yz+zx)+2δ3(x+\delta)(y+\delta)(z+\delta) - (x-\delta)(y-\delta)(z-\delta) = 2\delta(xy + yz + zx) + 2\delta^3:
Here x=20,y=30,z=40x=20, y=30, z=40 and δ=1\delta=1.
xy=600xy = 600
yz=1200yz = 1200
zx=800zx = 800
Sum =600+1200+800=2600= 600 + 1200 + 800 = 2600.

Difference =2(1)(2600)+2(13)=5200+2=5202 cm3= 2(1)(2600) + 2(1^3) = 5200 + 2 = 5202\text{ cm}^3.

Alternatively, by direct multiplication:
Vmax=21×31×41=651×41=26691V_{\text{max}} = 21 \times 31 \times 41 = 651 \times 41 = 26691
Vmin=19×29×39=551×39=21489V_{\text{min}} = 19 \times 29 \times 39 = 551 \times 39 = 21489
26691−21489=520226691 - 21489 = 5202.

Question 23

1 mark
A cylindrical copper pipe has an outer diameter of 100 mm100\,\text{mm} and an inner diameter of 90 mm90\,\text{mm}. The pipe is 2 m2\,\text{m} long. Given that the density of copper is 8960 kg m−38960\,\text{kg}\,\text{m}^{-3}, which of the following is the best estimate for the mass of the pipe?
  • A.3 kg3\,\text{kg}
  • B.9 kg9\,\text{kg}
  • C.27 kg27\,\text{kg}
  • D.54 kg54\,\text{kg}
  • E.110 kg110\,\text{kg}

Answer: C

Worked solution

First, we calculate the volume of the copper used in the pipe. The pipe is a cylindrical shell.
Outer radius R=50 mm=0.05 mR = 50\,\text{mm} = 0.05\,\text{m}.
Inner radius r=45 mm=0.045 mr = 45\,\text{mm} = 0.045\,\text{m}.
Length L=2 mL = 2\,\text{m}.
Cross-sectional area A=π(R2−r2)=π(0.052−0.0452)A = \pi(R^2 - r^2) = \pi(0.05^2 - 0.045^2).
Using the difference of two squares: A=π(0.05−0.045)(0.05+0.045)=π(0.005)(0.095)=0.000475π m2A = \pi(0.05 - 0.045)(0.05 + 0.045) = \pi(0.005)(0.095) = 0.000475\pi\,\text{m}^2.
Total volume V=A×L=0.000475π×2=0.00095π m3V = A \times L = 0.000475\pi \times 2 = 0.00095\pi\,\text{m}^3.
Mass M=density×V=8960×0.00095πM = \text{density} \times V = 8960 \times 0.00095\pi.
To estimate, we use π≈3.14\pi \approx 3.14 and density≈9000 kg m−3\text{density} \approx 9000\,\text{kg}\,\text{m}^{-3}.
M≈9000×0.00095×3.14=8.55×3.14M \approx 9000 \times 0.00095 \times 3.14 = 8.55 \times 3.14.
8×3=248 \times 3 = 24 and 9×3=279 \times 3 = 27.
Specifically, 8.55×3.14≈26.85 kg8.55 \times 3.14 \approx 26.85\,\text{kg}.
The closest estimate among the options is 27 kg27\,\text{kg}.

Question 24

1 mark
A rectangular nature reserve has an area of 8 cm28\,\text{cm}^2 on a map with a scale of 1:25,0001 : 25,000. What is the actual area of the reserve in square kilometres (km2)(\text{km}^2)?
  • A.0.2
  • B.0.5
  • C.2.0
  • D.5.0
  • E.20.0

Answer: B

Worked solution

To find the actual area, we must use the square of the linear scale factor. The scale is 1:25,0001 : 25,000, which means 1 cm1\,\text{cm} on the map represents 25,000 cm25,000\,\text{cm} in reality.

1. Convert the linear scale to km: 25,000 cm=250 m=0.25 km25,000\,\text{cm} = 250\,\text{m} = 0.25\,\text{km}.
2. Determine the area scale factor: (0.25 km/1 cm)2=0.0625 km2/cm2(0.25\,\text{km}/1\,\text{cm})^2 = 0.0625\,\text{km}^2/\text{cm}^2.
3. Multiply the map area by the area scale factor:
Actual Area=8 cm2×0.0625 km2/cm2\text{Actual Area} = 8\,\text{cm}^2 \times 0.0625\,\text{km}^2/\text{cm}^2

Actual Area=8×116=0.5 km2\text{Actual Area} = 8 \times \frac{1}{16} = 0.5\,\text{km}^2

Alternatively, using powers of 10:
8×(2.5×104)2=8×6.25×108=50×108=5×109 cm28 \times (2.5 \times 10^4)^2 = 8 \times 6.25 \times 10^8 = 50 \times 10^8 = 5 \times 10^9\,\text{cm}^2.
Since 1 km2=(105 cm)2=1010 cm21\,\text{km}^2 = (10^5\,\text{cm})^2 = 10^{10}\,\text{cm}^2, we have 5×1091010=0.5 km2\frac{5 \times 10^9}{10^{10}} = 0.5\,\text{km}^2.

Question 25

1 mark
The distance between two cities XX and YY is measured on two different maps. On Map A, with a scale of 1:n1 : n, the distance is 12 cm12\,\text{cm}. On Map B, with a scale of 1:(n+5000)1 : (n + 5000), the distance is 10 cm10\,\text{cm}. What is the actual distance between the two cities in kilometres?
  • A.0.3
  • B.2.5
  • C.3.0
  • D.6.0
  • E.30.0

Answer: C

Worked solution

Let DD be the actual distance in cm\text{cm}.
From Map A: D=12×nD = 12 \times n
From Map B: D=10×(n+5000)D = 10 \times (n + 5000)

Equating the two expressions for DD:
12n=10(n+5000)12n = 10(n + 5000)

12n=10n+50,00012n = 10n + 50,000

2n=50,0002n = 50,000

n=25,000n = 25,000


Now, substitute nn back into either equation to find DD:
D=12×25,000=300,000 cmD = 12 \times 25,000 = 300,000\,\text{cm}


Convert the result to kilometres:
300,000 cm=3000 m=3 km300,000\,\text{cm} = 3000\,\text{m} = 3\,\text{km}.

Question 26

1 mark
A construction company has two teams, Alpha and Beta. Team Alpha can complete a specific project in 1010 days. When Team Alpha and Team Beta work together, they can complete the same project in 44 days. Assuming both teams work at constant daily rates, express the daily work rate of Team Beta as a fraction of the daily work rate of Team Alpha.
  • A.25\frac{2}{5}
  • B.23\frac{2}{3}
  • C.32\frac{3}{2}
  • D.35\frac{3}{5}
  • E.52\frac{5}{2}

Answer: C

Worked solution

Let the total work required for the project be 11 unit. We define the rates as project units completed per day.

Team Alpha's rate, RA=110R_A = \frac{1}{10} project/day.

The combined rate of Team Alpha and Team Beta, RA+B=14R_{A+B} = \frac{1}{4} project/day.

To find Team Beta's rate (RBR_B), we subtract Alpha's rate from the combined rate:
RB=RA+B−RAR_B = R_{A+B} - R_A
RB=14−110R_B = \frac{1}{4} - \frac{1}{10}
Finding a common denominator of 2020:
RB=520−220=320R_B = \frac{5}{20} - \frac{2}{20} = \frac{3}{20} project/day.

The question asks for the rate of Team Beta as a fraction of the rate of Team Alpha:
Fraction =RBRA=3/201/10= \frac{R_B}{R_A} = \frac{3/20}{1/10}
Fraction =320×101=3020=32= \frac{3}{20} \times \frac{10}{1} = \frac{30}{20} = \frac{3}{2}

Question 27

1 mark
An alloy consists of three metals: copper, zinc, and nickel. The ratio of the mass of copper to the mass of zinc is 3:23:2. The ratio of the mass of zinc to the mass of nickel is 4:34:3. What fraction of the total mass of the alloy is copper?
  • A.613\frac{6}{13}
  • B.35\frac{3}{5}
  • C.313\frac{3}{13}
  • D.47\frac{4}{7}
  • E.14\frac{1}{4}

Answer: A

Worked solution

To find the fraction of the total mass that is copper, we first need to express the ratios of all three metals relative to each other.

Let the mass of copper be mCum_{Cu}, the mass of zinc be mZnm_{Zn}, and the mass of nickel be mNim_{Ni}. We are given:
mCu:mZn=3:2m_{Cu} : m_{Zn} = 3 : 2
mZn:mNi=4:3m_{Zn} : m_{Ni} = 4 : 3

To combine these ratios into a single continuous ratio mCu:mZn:mNim_{Cu} : m_{Zn} : m_{Ni}, we must find a common value for the zinc component. The least common multiple (LCM) of 2 and 4 is 4.

Scaling the first ratio by a factor of 2, we get:
mCu:mZn=(3×2):(2×2)=6:4m_{Cu} : m_{Zn} = (3 \times 2) : (2 \times 2) = 6 : 4

Now that the zinc component is consistent across both ratios, we can combine them:
mCu:mZn:mNi=6:4:3m_{Cu} : m_{Zn} : m_{Ni} = 6 : 4 : 3

The total mass corresponds to the sum of these parts:
Total parts =6+4+3=13= 6 + 4 + 3 = 13

The mass of copper represents 6 out of these 13 total parts. Therefore, the fraction of the total mass that is copper is:
Fraction=613\text{Fraction} = \frac{6}{13}

This matches option A.