ESAT Mathematics 1 Mock 5

The point $\displaystyle P(a, b)$ lies on the curve $\displaystyle y^2 = kx$ . Substituting its coordinates into the equation gives:This is our first relationship.

The point $\displaystyle Q$ has coordinates $\displaystyle (9a, y_Q)$ , where $\displaystyle y_Q > 0$ . Since $\displaystyle Q$ also lies on the curve, its coordinates must also satisfy the equation:We can rearrange this equation as:Now, we can substitute the expression for $\displaystyle ka$ from our first relationship, $\displaystyle ka = b^2$ , into this equation for $\displaystyle Q$ :Taking the square root of both sides gives:The problem states that the $\displaystyle y$ -coordinate of $\displaystyle Q$ is positive, so we take the positive root.Therefore, the correct option is A.

First, simplify the second fraction by factoring the numerator and denominator: $\displaystyle \dfrac{x^2 + 3x}{9 - x^2} = \dfrac{x(x+3)}{(3-x)(3+x)} = \dfrac{x}{3-x} = -\dfrac{x}{x-3}$ Substitute this back into the original expression: $\displaystyle \dfrac{2x}{x-3} - \left(-\dfrac{x}{x-3}\right) - 1 = \dfrac{2x}{x-3} + \dfrac{x}{x-3} - 1 = \dfrac{3x}{x-3} - 1$ Combine the terms over a common denominator: $\displaystyle \dfrac{3x - (x - 3)}{x - 3} = \dfrac{2x + 3}{x - 3}$ Express this in mixed form: $\displaystyle \dfrac{2x + 3}{x - 3} = \dfrac{2(x - 3) + 9}{x - 3} = 2 + \dfrac{9}{x-3}$ Thus, the correct option is A.

First, write all bases as powers of $\displaystyle 2$ : $\displaystyle 4 = 2^2$ and $\displaystyle 16 = 2^4$ .

Substitute these into the equation:Simplify the exponents:Use the index law $\displaystyle a^m \times a^n = a^{m+n}$ on the left-hand side:Since the bases are equal, we can equate the exponents:Notice that the common denominator for the left-hand side is $\displaystyle (x-3)(x+3) = x^2-9$ . Combine the fractions:Multiply both sides by $\displaystyle x^2-9$ (assuming $\displaystyle x eq \pm 3$ ):Expand the brackets:Simplify the left-hand side:Rearrange to solve for $\displaystyle x$ :Therefore the correct answer is F.

Let $\displaystyle N_A$ and $\displaystyle N_B$ be the number of particles of type A and B, respectively.
Let $\displaystyle V_A$ and $\displaystyle V_B$ be the total volumes occupied by particles of type A and B.
Let $\displaystyle r_A$ and $\displaystyle r_B$ be the radii of single particles of type A and B.
Let $\displaystyle v_A$ and $\displaystyle v_B$ be the volumes of single particles of type A and B.

From the problem statement, we are given:
1. The ratio of the number of particles: $\displaystyle \dfrac{N_A}{N_B} = \dfrac{4}{1} = 4$ .
2. The ratio of the total volumes: $\displaystyle V_A = \dfrac{1}{2} V_B$ , which means $\displaystyle \dfrac{V_A}{V_B} = \dfrac{1}{2}$ .

The total volume for each type is the number of particles multiplied by the volume of a single particle: $\displaystyle V_A = N_A \times v_A$ $\displaystyle V_B = N_B \times v_B$ We can write the ratio of total volumes in terms of these:We want to find the ratio of the volumes of single particles, $\displaystyle \dfrac{v_A}{v_B}$ . We can rearrange the equation above:Substituting the given values:The volume of a sphere is given by the formula $\displaystyle v = \dfrac{4}{3}\pi r^3$ . Therefore, the volume of a particle is proportional to the cube of its radius ( $\displaystyle v \propto r^3$ ).

The ratio of the volumes of single particles is related to the ratio of their radii as follows:We have found that $\displaystyle \dfrac{v_A}{v_B} = \dfrac{1}{8}$ , so:To find the ratio of the radii, we take the cube root of both sides:Thus, the ratio of the radius of a particle of type A to that of type B is $\displaystyle \dfrac{1}{2}$ .
Therefore the correct answer is C.

Let the triangle be $\displaystyle ABC$ , with the right angle at $\displaystyle C$ . Let the required distance be $\displaystyle s$ . Let the feet of the perpendiculars from $\displaystyle P$ to $\displaystyle AC$ and $\displaystyle BC$ be $\displaystyle D$ and $\displaystyle E$ respectively. The quadrilateral $\displaystyle CDPE$ is a square of side length $\displaystyle s$ .

This construction forms two smaller right-angled triangles, $\displaystyle \triangle ADP$ and $\displaystyle \triangle PEB$ . Since $\displaystyle PD$ is parallel to $\displaystyle BC$ , these two triangles are similar.

We can set up a ratio of corresponding sides using the given lengths of the hypotenuse segments, $\displaystyle AP=3$ and $\displaystyle PB=4$ :Since $\displaystyle DP = PE = s$ , we can find expressions for the other segments of the legs: $\displaystyle AD = s \cdot \dfrac{AP}{PB} = s \cdot \dfrac{3}{4} = \dfrac{3s}{4}$ . $\displaystyle EB = s \cdot \dfrac{PB}{AP} = s \cdot \dfrac{4}{3} = \dfrac{4s}{3}$ .

The full lengths of the legs of $\displaystyle \triangle ABC$ are therefore: $\displaystyle AC = AD + DC = \dfrac{3s}{4} + s = \dfrac{7s}{4}$ $\displaystyle BC = EB + EC = \dfrac{4s}{3} + s = \dfrac{7s}{3}$ The hypotenuse $\displaystyle AB = AP + PB = 3+4=7$ . Applying Pythagoras' theorem to $\displaystyle \triangle ABC$ :Dividing the entire equation by $\displaystyle 7^2 = 49$ simplifies the calculation significantly:This gives $\displaystyle s^2 = \dfrac{144}{25}$ , and since $\displaystyle s$ must be a positive length, $\displaystyle s = \dfrac{12}{5}$ .

The process stops when the cuboid becomes a cube. Since the side lengths can only decrease, all three side lengths must become equal to the smallest initial side length, $\displaystyle 2^c$ . Therefore, the side length of the final cube is $\displaystyle S = 2^c$ .

To find the total number of cuts, $\displaystyle N$ , we can consider the number of halvings required for each dimension to reach the final length $\displaystyle 2^c$ .

1. The side with initial length $\displaystyle 2^a$ must be reduced to $\displaystyle 2^c$ . Each cut along this dimension reduces the exponent by 1. Thus, the number of cuts required for this dimension is $\displaystyle a-c$ .

2. The side with initial length $\displaystyle 2^b$ must be reduced to $\displaystyle 2^c$ . Similarly, the number of cuts required for this dimension is $\displaystyle b-c$ .

3. The side with initial length $\displaystyle 2^c$ is always the shortest (or equal shortest) and is therefore never cut. The number of cuts for this dimension is 0.

The total number of cuts $\displaystyle N$ is the sum of the cuts for each dimension. $\displaystyle N = (a-c) + (b-c) + 0 = a+b-2c$ Thus, the correct expressions are $\displaystyle N = a+b-2c$ and $\displaystyle S = 2^c$ .

To calculate the volume change, we must match the time units of the rates and durations.

For the inflow, the rate is given in litres per minute, so we convert the duration of $\displaystyle h$ hours into minutes:For the outflow, the rate is given in litres per hour, so we convert the duration of $\displaystyle m$ minutes into hours:The final volume is the initial volume plus the inflow minus the outflow:

Since $\displaystyle P$ lies on the line through $\displaystyle A$ and $\displaystyle B$ but not between them, it must be on the extension of the segment $\displaystyle AB$ . Given $\displaystyle AP = 3PB$ , the point $\displaystyle P$ is further from $\displaystyle A$ than from $\displaystyle B$ . This implies the order of points is $\displaystyle A$ , then $\displaystyle B$ , then $\displaystyle P$ .

We can express the distance $\displaystyle AP$ as $\displaystyle AB + BP$ . Substituting the given ratio:We calculate the vector $\displaystyle \vec{AB}$ :Therefore, $\displaystyle \vec{BP} = \dfrac{1}{2}\begin{pmatrix} 4 \\ -8 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix}$ . Adding this to the position vector of $\displaystyle B$ gives the coordinates of $\displaystyle P$ :Thus, $\displaystyle P$ is at $\displaystyle (8, -6)$ .

The digit 6 is required but restricted to the internal positions (2nd, 3rd, or 4th). This gives 3 possible positions for the digit 6.

Once 6 is placed, there are 4 remaining positions to fill. We must choose and arrange 4 digits from the remaining 5 available digits ( $\displaystyle \ \{1, 2, 3, 4, 5\} \$ ).

The number of ways to arrange these remaining digits is:To find the total number of valid passcodes, we multiply the number of valid positions for 6 by the arrangements of the other digits:

For the final position to be due north, the net horizontal displacement must be zero. We resolve each displacement horizontally.

The first displacement has magnitude $\displaystyle d$ on a bearing of $\displaystyle 030^\circ$ . The horizontal (East) component is:The second displacement has magnitude $\displaystyle 6\sqrt{3}$ on a bearing of $\displaystyle 300^\circ$ , which is $\displaystyle 60^\circ$ West of North. The horizontal (West) component is:Equating the East and West components:

We first observe that the densities are given in different units. Since $\displaystyle 1 \, \text{g cm}^{-3} = 1000 \, \text{kg m}^{-3}$ , the density of material Y is:Let the volume of material Y be $\displaystyle V$ . The volume of material X is $\displaystyle 2V$ , giving a total volume of $\displaystyle 3V$ . We calculate the total mass by summing the mass of each component:The density of the composite object is the total mass divided by the total volume:

The total surface area of the cylinder is the sum of the areas of its two circular bases and its curved surface: $\displaystyle A_{\text{cylinder}} = 2\pi r^2 + 2\pi r h$ .

The cube has six faces, each with an area of $\displaystyle (2r)^2 = 4r^2$ . Its total surface area is $\displaystyle A_{\text{cube}} = 6 \times 4r^2 = 24r^2$ .

We are given that these areas are equal, so we can set up an equation:To solve for $\displaystyle h$ , we can first divide the entire equation by $\displaystyle 2\pi r$ (since $\displaystyle r eq 0$ ):Now, we rearrange to make $\displaystyle h$ the subject and factor out $\displaystyle r$ :

The total number of possible outcomes for three dice is $\displaystyle 6^3 = 216$ .

First, we find the number of outcomes where the sum of the scores is 5. The possible sets of numbers are \{1, 1, 3\} and \{1, 2, 2\}.
- The set \{1, 1, 3\} can be rolled in $\displaystyle \dfrac{3!}{2!} = 3$ distinct ways.
- The set \{1, 2, 2\} can also be rolled in $\displaystyle \dfrac{3!}{2!} = 3$ distinct ways.
This gives a total of $\displaystyle 3+3=6$ successful outcomes, so $\displaystyle p_S = \dfrac{6}{216}$ .

Next, we find the number of outcomes where the product of the scores is 6. The possible sets of numbers are \{1, 1, 6\} and \{1, 2, 3\}.
- The set \{1, 1, 6\} can be rolled in $\displaystyle \dfrac{3!}{2!} = 3$ distinct ways.
- The set \{1, 2, 3\} can be rolled in $\displaystyle 3! = 6$ distinct ways.
This gives a total of $\displaystyle 3+6=9$ successful outcomes, so $\displaystyle p_P = \dfrac{9}{216}$ .

The required difference is:

Let the exterior angles of polygons P and Q be $\displaystyle e_P$ and $\displaystyle e_Q$ respectively. Let their interior angles be $\displaystyle i_P$ and $\displaystyle i_Q$ .

From the problem statement, we have two conditions:
1) The exterior angle of P is three times the exterior angle of Q: $\displaystyle e_P = 3e_Q$ .
2) The sum of their interior angles is $\displaystyle 300^{\circ}$ : $\displaystyle i_P + i_Q = 300$ .

For any regular polygon, the interior and exterior angles sum to $\displaystyle 180^{\circ}$ . So, $\displaystyle i_P = 180 - e_P$ and $\displaystyle i_Q = 180 - e_Q$ .

Substitute these into the second condition:Now we have a simple system of two linear equations:Substitute the first equation into the second:The number of sides of a regular polygon, $\displaystyle n$ , is given by the formula $\displaystyle n = \dfrac{360}{e}$ , where $\displaystyle e$ is the exterior angle in degrees.

For polygon Q, the number of sides $\displaystyle n_Q$ is:Thus, polygon Q has 24 sides.

The total area of the four shaded rectangles is equal to the area of the outer square minus the area of the central unshaded square: $\displaystyle \text{Shaded Area} = \left(\sqrt{21} + \sqrt{3}\right)^2 - \left(\sqrt{21} - \sqrt{3}\right)^2$ Using the identity $\displaystyle (u+v)^2 - (u-v)^2 = 4uv$ : $\displaystyle \text{Shaded Area} = 4\left(\sqrt{21}\right)\left(\sqrt{3}\right) = 4\sqrt{63} = 4\sqrt{9 \times 7} = 12\sqrt{7}$ Alternatively, if each rectangle has length $\displaystyle x$ and width $\displaystyle y$ , then $\displaystyle x + y = \sqrt{21} + \sqrt{3}$ and $\displaystyle x - y = \sqrt{21} - \sqrt{3}$ . Adding and subtracting gives $\displaystyle x = \sqrt{21}$ and $\displaystyle y = \sqrt{3}$ . The total shaded area is $\displaystyle 4xy = 4(\sqrt{21})(\sqrt{3}) = 12\sqrt{7}$ .

Let $\displaystyle O$ be the vertex of the sector and $\displaystyle C$ be the centre of the inscribed circle.

By symmetry, the line $\displaystyle OC$ bisects the central angle of the sector, so the angle between $\displaystyle OC$ and each straight boundary is $\displaystyle \theta$ .

The inscribed circle is tangent to the circular arc of radius $\displaystyle R$ , so the distance from $\displaystyle O$ to the contact point on the arc is $\displaystyle R$ . Therefore, the distance from $\displaystyle O$ to the centre of the inscribed circle is: $\displaystyle OC = R - r$ Let $\displaystyle T$ be the point where the inscribed circle touches one of the straight edges of the sector. The line segment $\displaystyle CT$ is perpendicular to the straight edge and has length $\displaystyle CT = r$ .

In the right-angled triangle $\displaystyle \triangle OTC$ : $\displaystyle \sin \theta = \dfrac{CT}{OC} = \dfrac{r}{R - r}$ Rearranging to solve for $\displaystyle r$ : $\displaystyle r = (R - r)\sin \theta$ $\displaystyle r + r\sin \theta = R\sin \theta$ $\displaystyle r(1 + \sin \theta) = R\sin \theta$ $\displaystyle r = \dfrac{R\sin \theta}{1 + \sin \theta}$ Hence, the correct option is A.

Let $\displaystyle P$ be the initial pressure. A $\displaystyle 25\%$ increase corresponds to a multiplier of $\displaystyle 1.25$ or $\displaystyle \dfrac{5}{4}$ . We have:Isolating $\displaystyle P$ yields:We could also observe that an increase of $\displaystyle \dfrac{1}{4}$ is reversed by a decrease of $\displaystyle \dfrac{1}{5}$ from the final value ( $\displaystyle 120 - 24 = 96$ ).

Let the original list of 6 numbers be $\displaystyle x_1, x_2, \dots, x_6$ . Let their sum be $\displaystyle S = \sum_{i=1}^6 x_i$ . The mean of the original list is $\displaystyle \mu = \dfrac{S}{6}$ .

Let the new number added to the list be $\displaystyle y$ . The new list has 7 numbers, and their sum is $\displaystyle S+y$ . The mean of the new list is $\displaystyle \dfrac{S+y}{7}$ .

We are given that the means are identical:This shows that the number added, $\displaystyle y$ , must be equal to the mean of the original list, $\displaystyle \mu$ .

Now, let's consider the range. The range of a list is the difference between its maximum and minimum values. Let the minimum and maximum values of the original list be $\displaystyle x_{\min}$ and $\displaystyle x_{\max}$ , respectively. The original range is $\displaystyle R_1 = x_{\max} - x_{\min}$ .

The mean of a set of numbers must lie between (or be equal to) the minimum and maximum values of that set. Since the numbers are distinct, the mean must be strictly between the minimum and maximum. So, $\displaystyle x_{\min} < \mu < x_{\max}$ .

The new number added is $\displaystyle y = \mu$ . So, we know that $\displaystyle x_{\min} < y < x_{\max}$ .

When this new number $\displaystyle y$ is added to the list, the original minimum value $\displaystyle x_{\min}$ is still the minimum value of the new list. Similarly, the original maximum value $\displaystyle x_{\max}$ is still the maximum value of the new list.

Therefore, the new range is $\displaystyle R_2 = x_{\max} - x_{\min} = R_1$ . The range must be unchanged.

Let's check the other options with a counterexample. Consider the original list of 6 distinct numbers: $\displaystyle \{0, 1, 8, 9, 10, 16\}$ .

- The sum is $\displaystyle 0+1+8+9+10+16 = 44$ . This is not great for no-calc. Let's pick better numbers.
Consider the list $\displaystyle \{0, 1, 2, 9, 10, 14\}$ .
- The sum is $\displaystyle S = 0+1+2+9+10+14 = 36$ .
- The mean is $\displaystyle \mu = 36/6 = 6$ . So the number added is $\displaystyle y=6$ .
- The original sorted list is $\displaystyle \{0, 1, 2, 9, 10, 14\}$ . The original median is the average of the two middle terms: $\displaystyle M_1 = \dfrac{2+9}{2} = 5.5$ .
- The new sorted list is $\displaystyle \{0, 1, 2, 6, 9, 10, 14\}$ . The new median is the 4th term: $\displaystyle M_2 = 6$ .
- The original range is $\displaystyle 14-0=14$ . The new range is $\displaystyle 14-0=14$ . The range is unchanged.

- Comparing medians: $\displaystyle M_2 = 6$ and $\displaystyle M_1 = 5.5$ . The median has increased. This single example shows that 'The median is unchanged' (B) and 'The median decreases' (F) are not `must be true` statements.
- Is the new median equal to the mean? Yes, in this case $\displaystyle M_2=6$ and $\displaystyle \mu=6$ . But is this always true?

Let's try another list: $\displaystyle \{0, 1, 8, 9, 10, 14\}$ .
- The sum is $\displaystyle S = 0+1+8+9+10+14 = 42$ .
- The mean is $\displaystyle \mu = 42/6 = 7$ . So the number added is $\displaystyle y=7$ .
- The original median is $\displaystyle M_1 = \dfrac{8+9}{2} = 8.5$ .
- The new sorted list is $\displaystyle \{0, 1, 7, 8, 9, 10, 14\}$ . The new median is the 4th term: $\displaystyle M_2 = 8$ .
- In this case, the median has decreased ( $\displaystyle 8 < 8.5$ ). This shows 'The median increases' (E) is not a `must be true` statement.
- Also, the new median ( $\displaystyle M_2=8$ ) is not equal to the mean ( $\displaystyle \mu=7$ ). This shows 'The median is now equal to the mean' (D) is not a `must be true` statement.

Only statement A, 'The range is unchanged', must be true.

We are given two equations:From the first equation, we can express $\displaystyle q$ in terms of $\displaystyle p$ : $\displaystyle q = 1 - p$ .

Substitute this into the second equation:To solve this, we should express the numbers 8, 27, and 12 as powers of their prime factors, which are 2 and 3. $\displaystyle 8 = 2^3$ $\displaystyle 27 = 3^3$ $\displaystyle 12 = 4 \times 3 = 2^2 \times 3^1$ Substituting these into the equation gives:Using the index law $\displaystyle (a^m)^n = a^{mn}$ , we get:For this equality to hold, the powers of each unique prime base on the left-hand side must equal the corresponding powers on the right-hand side.

Equating the powers of base 2:As a check, we can also equate the powers of base 3:Both methods give the same result. Therefore, the value of $\displaystyle p$ is $\displaystyle \dfrac{2}{3}$ .