ESAT Mathematics 1 Mock 4

Let the side length of the original square logo be $\displaystyle L$ , so its area is $\displaystyle A_0 = L^2$ .

1. Decreasing the side length by $\displaystyle 25\%$ gives a new side length of: $\displaystyle L_1 = L \left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4}L$ 2. Increasing this reduced side length by $\displaystyle 25\%$ gives: $\displaystyle L_2 = L_1 \left(1 + \dfrac{1}{4}\right) = \dfrac{3}{4}L \times \dfrac{5}{4} = \dfrac{15}{16}L$ 3. The area of the final square logo is: $\displaystyle A_2 = L_2^2 = \left(\dfrac{15}{16}L\right)^2 = \dfrac{225}{256}L^2 = \dfrac{225}{256}A_0$ 4. The difference in area is given as $\displaystyle 31\text{ cm}^2$ : $\displaystyle A_0 - A_2 = A_0 \left(1 - \dfrac{225}{256}\right) = \dfrac{31}{256}A_0 = 31\text{ cm}^2$ $\displaystyle \implies A_0 = 256\text{ cm}^2$ 5. The side length of the original square logo is: $\displaystyle L = \sqrt{256\text{ cm}^2} = 16\text{ cm}$ 6. The perimeter of the original square logo is: $\displaystyle \text{Perimeter} = 4L = 4 \times 16\text{ cm} = 64\text{ cm}$

Let $\displaystyle S = \{1, 2, 3, 4, 5\}$ . The set contains 5 numbers in total.

The odd numbers in $\displaystyle S$ are $\displaystyle \{1, 3, 5\}$ , so there are 3 odd numbers.
The even numbers in $\displaystyle S$ are $\displaystyle \{2, 4\}$ , so there are 2 even numbers.

When a number is chosen at random from $\displaystyle S$ :
The probability of choosing an odd number is $\displaystyle P(\text{odd}) = \dfrac{3}{5}$ .
The probability of choosing an even number is $\displaystyle P(\text{even}) = \dfrac{2}{5}$ .

Three numbers are chosen independently with replacement.
The product of these three numbers is even if at least one of the chosen numbers is even.
It is simpler to calculate the probability of the complement event: that the product is odd.
The product is odd if and only if all three chosen numbers are odd.

The probability that all three numbers are odd is:The probability that the product is even is 1 minus the probability that the product is odd:Therefore the correct answer is D.

Let the initial value of the asset be $\displaystyle V$ .

An increase of 60% corresponds to multiplying by $\displaystyle 1.6$ . A subsequent decrease of 25% corresponds to multiplying by $\displaystyle 1 - 0.25 = 0.75$ .

The combined multiplier for the two changes is the product of the individual multipliers. It is often easier to calculate this using fractions:So, the final value is $\displaystyle 1.2V$ . We are given that the final value is 1080, which gives the equation:Solving for the initial value $\displaystyle V$ :The initial value of the asset was 900.

Let the distance of the first stage be $\displaystyle d$ . The time taken is $\displaystyle t_1 = \dfrac{d}{30}$ .
The second stage has distance $\displaystyle 2d$ , so the time taken is $\displaystyle t_2 = \dfrac{2d}{90} = \dfrac{d}{45}$ .

The total distance is $\displaystyle d + 2d = 3d$ .
The total time is $\displaystyle t_1 + t_2 = \dfrac{d}{30} + \dfrac{d}{45}$ . Finding a common denominator of 90, this becomes:Average speed is total distance divided by total time:The variable $\displaystyle d$ cancels, as expected.

First, we express $\displaystyle y$ and $\displaystyle z$ in terms of the original quantity $\displaystyle x$ .
An increase of $\displaystyle 150\%$ corresponds to a multiplier of $\displaystyle 1 + 1.5 = 2.5$ , so $\displaystyle y = 2.5x$ .
A decrease of $\displaystyle 50\%$ corresponds to a multiplier of $\displaystyle 1 - 0.5 = 0.5$ , so $\displaystyle z = 0.5x$ .

We are told that $\displaystyle z$ is $\displaystyle q\%$ less than $\displaystyle y$ . This relationship can be written as:Substituting our expressions for $\displaystyle y$ and $\displaystyle z$ in terms of $\displaystyle x$ gives:Assuming $\displaystyle x eq 0$ , we can divide both sides by $\displaystyle 2.5x$ :This simplifies to $\displaystyle 0.2 = 1 - \dfrac{q}{100}$ .
Rearranging the equation to solve for $\displaystyle q$ gives $\displaystyle \dfrac{q}{100} = 1 - 0.2 = 0.8$ , which means $\displaystyle q = 80$ .

From the graph, the point $\displaystyle (6, 12)$ lies on the curve $\displaystyle y = ax^2$ : $\displaystyle 12 = a(6^2) = 36a \implies a = \dfrac{12}{36} = \dfrac{1}{3}$ So $\displaystyle y = \dfrac{1}{3}x^2$ .

Since $\displaystyle x$ is inversely proportional to $\displaystyle t$ , we have: $\displaystyle x = \dfrac{k}{t}$ Using the given values $\displaystyle t = 4$ and $\displaystyle x = 3$ : $\displaystyle 3 = \dfrac{k}{4} \implies k = 12$ So $\displaystyle x = \dfrac{12}{t}$ .

Substituting $\displaystyle x = \dfrac{12}{t}$ into the expression for $\displaystyle y$ : $\displaystyle y = \dfrac{1}{3} \left(\dfrac{12}{t}\right)^2 = \dfrac{1}{3} \left(\dfrac{144}{t^2}\right) = \dfrac{48}{t^2}$

We treat the rod as two separate components: the inner core and the outer sleeve.

The inner core is a solid cylinder of radius $\displaystyle r$ and density $\displaystyle 3\rho$ . Its mass is:The outer sleeve is a hollow cylinder with inner radius $\displaystyle r$ and outer radius $\displaystyle 2r$ . Its volume is the difference between the total volume and the core volume:Given the sleeve density is $\displaystyle \rho$ , its mass is:The total mass is the sum of the components:

The problem asks for the total time to transmit a file, including overhead. This requires a series of unit conversions and a ratio calculation.

1. Convert file size to bits:
The file size is given in Megabytes (MB), but the speed is in Megabits per second (Mbps). We must convert the file size to bits.

File size in bytes = $\displaystyle 120 \text{ MB} \times 10^6 \text{ bytes/MB} = 120 \times 10^6$ bytes.

File size in bits = $\displaystyle (120 \times 10^6 \text{ bytes}) \times 8 \text{ bits/byte} = 960 \times 10^6$ bits.
This is the amount of 'data'.

2. Calculate total bits including overhead:
The overhead ratio is 1 part for every 15 parts of data. This means for every 15 bits of data, a total of $\displaystyle 15+1=16$ bits must be transmitted.
The scaling factor is therefore $\displaystyle \dfrac{16}{15}$ .

Total bits to transmit = (File size in bits) $\displaystyle \times \dfrac{16}{15}$We can simplify $\displaystyle 960/15$ : $\displaystyle 960/15 = (900+60)/15 = 60+4 = 64$ .3. Calculate transmission time:
The network speed is 32 Megabits per second, which is $\displaystyle 32 \times 10^6$ bits per second.

Time = $\displaystyle \dfrac{\text{Total bits}}{\text{Speed in bits per second}}$Recognising powers of 2: $\displaystyle 1024 = 2^{10}$ and $\displaystyle 32 = 2^5$ .Therefore the correct answer is 32 seconds.

Let the total capacity of the tank be 1 unit of work.

The rate of pipe P is the fraction of the tank it fills per hour, which is $\displaystyle R_P = \dfrac{1}{10}$ .
The rate of pipe Q is $\displaystyle R_Q = \dfrac{1}{T}$ .

Pipe P works alone for 2 hours, and then works with pipe Q for another 4 hours. So, the total time pipe P is active is $\displaystyle 2 + 4 = 6$ hours.
Pipe Q is active only during the second phase, which is 4 hours.

The total work done is the sum of the work done by each pipe. The work done by a pipe is its rate multiplied by the time it is active.

Total Work = (Work done by P) + (Work done by Q)

Since the tank is filled completely, the total work is 1.Substituting the values:Now, we solve for $\displaystyle T$ :Cross-multiplying gives:Thus, pipe Q can fill the tank alone in 10 hours.

Let $\displaystyle S$ be the event that a single quadratic equation of the given form has real roots.
The equation $\displaystyle x^2 + 2bx + c = 0$ has real roots if and only if its discriminant is non-negative.
The discriminant is $\displaystyle \Delta = (2b)^2 - 4(1)(c) = 4b^2 - 4c$ .
The condition for real roots is $\displaystyle 4b^2 - 4c \ge 0$ , which simplifies to $\displaystyle b^2 \ge c$ .

The coefficients $\displaystyle b$ and $\displaystyle c$ are chosen independently and uniformly at random from the set $\displaystyle \{1, 2, 3\}$ . The total number of possible pairs $\displaystyle (b, c)$ is $\displaystyle 3 \times 3 = 9$ .

We count the number of pairs $\displaystyle (b, c)$ that satisfy the condition $\displaystyle b^2 \ge c$ :
- If $\displaystyle b=1$ , $\displaystyle 1^2 \ge c \implies 1 \ge c$ . The only possible value for $\displaystyle c$ is $\displaystyle 1$ . This gives the pair $\displaystyle (1, 1)$ .
- If $\displaystyle b=2$ , $\displaystyle 2^2 \ge c \implies 4 \ge c$ . The possible values for $\displaystyle c$ are $\displaystyle 1, 2, 3$ . This gives the pairs $\displaystyle (2, 1), (2, 2), (2, 3)$ .
- If $\displaystyle b=3$ , $\displaystyle 3^2 \ge c \implies 9 \ge c$ . The possible values for $\displaystyle c$ are $\displaystyle 1, 2, 3$ . This gives the pairs $\displaystyle (3, 1), (3, 2), (3, 3)$ .

In total, there are $\displaystyle 1 + 3 + 3 = 7$ favourable pairs.
The probability that a single equation has real roots is $\displaystyle P(S) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}} = \dfrac{7}{9}$ .

The question asks for the probability that at least one of two such equations has real roots. Let the events for the two equations be $\displaystyle S_1$ and $\displaystyle S_2$ . The choices of coefficients are independent, so the events are independent. We want to find $\displaystyle P(S_1 \cup S_2)$ .

It is easiest to use the complement rule: $\displaystyle P(\text{at least one success}) = 1 - P(\text{no successes})$ . $\displaystyle P(\text{no successes}) = P(S_1' \cap S_2')$ .
Since the events are independent, $\displaystyle P(S_1' \cap S_2') = P(S_1') \times P(S_2')$ .

The probability of failure for a single equation is $\displaystyle P(S') = 1 - P(S) = 1 - \dfrac{7}{9} = \dfrac{2}{9}$ .
So, the probability that neither equation has real roots is: $\displaystyle P(S_1' \cap S_2') = \dfrac{2}{9} \times \dfrac{2}{9} = \dfrac{4}{81}$ The probability that at least one equation has real roots is: $\displaystyle P(S_1 \cup S_2) = 1 - P(S_1' \cap S_2') = 1 - \dfrac{4}{81} = \dfrac{77}{81}$

The region is bounded by two pairs of parallel lines. We find the vertices by intersecting non-parallel pairs. Adding and subtracting the equations simplifies the algebra:

- Intersection of $\displaystyle 3x+4y=22$ and $\displaystyle 3x-4y=14$ gives $\displaystyle 6x=36 \implies x=6$ and $\displaystyle 8y=8 \implies y=1$ . Vertex $\displaystyle (6,1)$ .
- Intersection of $\displaystyle 3x+4y=-2$ and $\displaystyle 3x-4y=-10$ gives $\displaystyle x=-2, y=1$ . Vertex $\displaystyle (-2,1)$ .
- Intersection of $\displaystyle 3x+4y=22$ and $\displaystyle 3x-4y=-10$ gives $\displaystyle x=2, y=4$ . Vertex $\displaystyle (2,4)$ .
- Intersection of $\displaystyle 3x+4y=-2$ and $\displaystyle 3x-4y=14$ gives $\displaystyle x=2, y=-2$ . Vertex $\displaystyle (2,-2)$ .

The vertices reveal that the diagonals are horizontal (from $\displaystyle x=-2$ to $\displaystyle x=6$ along $\displaystyle y=1$ ) and vertical (from $\displaystyle y=-2$ to $\displaystyle y=4$ along $\displaystyle x=2$ ).

The lengths of these diagonals are $\displaystyle 8$ and $\displaystyle 6$ . Since they are perpendicular, the area is:

We are given the ratio:This implies that for some constant of proportionality $\displaystyle c$ , we have $\displaystyle 3x=2c$ , $\displaystyle 4y=3c$ , and $\displaystyle 5z=4c$ . We can express $\displaystyle x, y, z$ in terms of $\displaystyle c$ :The ratio $\displaystyle x:y:z$ is therefore:To find the simplest integer ratio, we multiply by the lowest common multiple of the denominators (3, 4, 5). The LCM is $\displaystyle 3 \times 4 \times 5 = 60$ .The integers 40, 45, and 48 have no common factor strictly greater than 1 (prime factors of $\displaystyle 40$ are $\displaystyle 2, 5$ ; of $\displaystyle 45$ are $\displaystyle 3, 5$ ; of $\displaystyle 48$ are $\displaystyle 2, 3$ ). So we can set $\displaystyle x=40$ , $\displaystyle y=45$ , and $\displaystyle z=48$ .

Now, we substitute these values into the given linear equation:

We need to identify quadrilaterals P and Q that satisfy the given symmetry conditions.

Analysis of Quadrilateral P:
P must have rotational symmetry of order greater than 1, but no line symmetry.
Let's consider the options for P:
- A Square has rotational symmetry of order 4 and 4 lines of symmetry. It fails the 'no line symmetry' condition.
- A Rectangle (non-square) has rotational symmetry of order 2 and 2 lines of symmetry. It fails the 'no line symmetry' condition.
- A Rhombus (non-square) has rotational symmetry of order 2 and 2 lines of symmetry. It fails the 'no line symmetry' condition.
- A Parallelogram (non-rhombus, non-rectangle) has rotational symmetry of order 2 about the intersection of its diagonals. It has no lines of symmetry. This is a valid choice for P.
- A Kite (non-rhombus) has rotational symmetry of order 1. It fails the 'order greater than 1' condition.
- An Isosceles Trapezium has rotational symmetry of order 1. It fails the 'order greater than 1' condition.

So, P must be a parallelogram.

Analysis of Quadrilateral Q:
Q must have at least one line of symmetry, but rotational symmetry of order 1 (i.e., no rotational symmetry other than the identity).
Let's consider the options for Q:
- A Square, Rectangle, or Rhombus all have rotational symmetry of order greater than 1. They fail the 'order 1' condition.
- A Parallelogram has no lines of symmetry. It fails the 'one or more lines of symmetry' condition.
- A Kite (non-rhombus) has one line of symmetry (along its main diagonal) and has rotational symmetry of order 1. This is a valid choice for Q.
- An Isosceles Trapezium has one line of symmetry and has rotational symmetry of order 1. This is also a valid choice for Q.

Conclusion:
We need to find a pair (P, Q) from the options where P is a parallelogram and Q is either a kite or an isosceles trapezium.

The option $\displaystyle (\text{Parallelogram, Kite})$ matches our findings.

Therefore, the correct option is A.

The equation of the curve is $\displaystyle y = (x-k)(x^2 - 4)$ .

First, we find the intercepts.

The $\displaystyle x$ -intercepts occur when $\displaystyle y=0$ :This gives $\displaystyle x-k=0$ or $\displaystyle x^2-4=0$ . The solutions are $\displaystyle x=k$ , $\displaystyle x=2$ , and $\displaystyle x=-2$ .

The $\displaystyle y$ -intercept occurs when $\displaystyle x=0$ :The dataset is therefore $\displaystyle \{ -2, 2, k, 4k \}$ .

Next, we find the mean and median of this dataset. We are given the condition $\displaystyle k > 1$ .

Since $\displaystyle k > 1$ , we know that $\displaystyle 4k > 4$ . This means the smallest value in the dataset is $\displaystyle -2$ and the largest is $\displaystyle 4k$ .
The two middle values are $\displaystyle 2$ and $\displaystyle k$ . The order of these two does not matter for the calculation of the median of four numbers, which is the average of the two middle terms.

Median = $\displaystyle \dfrac{2+k}{2}$ .

The mean of the dataset is the sum of the values divided by the number of values (which is 4).

Sum = $\displaystyle -2 + 2 + k + 4k = 5k$ .

Mean = $\displaystyle \dfrac{5k}{4}$ .

The problem states that the mean is equal to the median:To solve for $\displaystyle k$ , we can multiply both sides by 4:This value satisfies the condition $\displaystyle k > 1$ , since $\displaystyle \dfrac{4}{3} \approx 1.33$ . Therefore, the value of $\displaystyle k$ is $\displaystyle \dfrac{4}{3}$ .

Since $\displaystyle AB$ is a diameter of the circle, the angle subtended in the semicircle is a right angle: $\displaystyle \angle ACB = 90^\circ$ Because $\displaystyle ACD$ is a straight line, $\displaystyle \angle BCD = 180^\circ - 90^\circ = 90^\circ$ .

In right-angled triangle $\displaystyle ABC$ : $\displaystyle \angle ABC = 180^\circ - 90^\circ - 30^\circ = 60^\circ$ Since $\displaystyle BD$ is tangent to the circle at $\displaystyle B$ , the diameter $\displaystyle AB$ is perpendicular to $\displaystyle BD$ , so $\displaystyle \angle ABD = 90^\circ$ . Therefore: $\displaystyle \angle CBD = \angle ABD - \angle ABC = 90^\circ - 60^\circ = 30^\circ$ In right-angled triangle $\displaystyle BCD$ : $\displaystyle \tan(\angle CBD) = \dfrac{CD}{BC}$ $\displaystyle \tan 30^\circ = \dfrac{CD}{x}$ $\displaystyle CD = x \tan 30^\circ = x \left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\sqrt{3}}{3}x$

Let the first cone have base radius $\displaystyle r$ and height $\displaystyle h_1$ . Its volume is:Let the second cone have base radius $\displaystyle r_2$ and height $\displaystyle h_2$ . We are given that $\displaystyle h_2 = 2h_1$ and $\displaystyle V_1 = V_2$ . Therefore:Dividing both sides by $\displaystyle \dfrac{1}{3}\pi h_1$ gives:Rearranging to solve for $\displaystyle r_2^2$ :Taking the positive square root, we find the base radius of the second cone:Therefore the correct answer is A.

First, find the maximum point $\displaystyle M$ of the curve $\displaystyle C$ .
The equation can be written as $\displaystyle y = 6x - x^2 = 9 - (x - 3)^2$ .
The maximum occurs at $\displaystyle x = 3$ , giving $\displaystyle y = 9$ . So, $\displaystyle M$ has coordinates $\displaystyle (3, 9)$ , and $\displaystyle \vec{OM} = \begin{pmatrix} 3 \\ 9 \end{pmatrix}$ .

Next, use the vector equation to find $\displaystyle \vec{OQ}$ :Substitute the known vectors:So, $\displaystyle Q$ is the point $\displaystyle (1, 12)$ .

Now, find where $\displaystyle C$ intersects the positive $\displaystyle x$ -axis by setting $\displaystyle y = 0$ :The roots are $\displaystyle x = 0$ and $\displaystyle x = 6$ . The intersection on the positive $\displaystyle x$ -axis is $\displaystyle R(6, 0)$ .

Finally, calculate the distance from $\displaystyle Q(1, 12)$ to $\displaystyle R(6, 0)$ using the distance formula:Therefore the correct answer is A.

Let $\displaystyle \rho$ , $\displaystyle V$ , and $\displaystyle M$ represent density, volume, and mass respectively. We know that $\displaystyle M = \rho V$ .

We are given the properties of the blend of liquid X and mixture Y: $\displaystyle \rho_X = 4$ $\displaystyle \dfrac{V_X}{V_Y} = \dfrac{1}{3}$ $\displaystyle \dfrac{M_X}{M_Y} = \dfrac{2}{5}$ We can find the density of mixture Y, $\displaystyle \rho_Y$ , using the relationship between these ratios:Substituting the given values:Now, consider mixture Y, which is made of liquids A and B. $\displaystyle \rho_A = 5$ $\displaystyle \rho_B = 2$ Let the volumes of A and B in mixture Y be $\displaystyle V_A$ and $\displaystyle V_B$ . The total volume of Y is $\displaystyle V_Y = V_A + V_B$ . The total mass of Y is $\displaystyle M_Y = M_A + M_B = \rho_A V_A + \rho_B V_B$ .

The density of mixture Y is its total mass divided by its total volume:Substitute the known densities:To find the ratio $\displaystyle \dfrac{V_A}{V_B}$ , let's cross-multiply:Therefore, the ratio of the volume of liquid A to the volume of liquid B in mixture Y is $\displaystyle 4:5$ .