ESAT Mathematics 1 Mock 3

The sum of the values in the first group is $\displaystyle n \times a = na$ , and the sum in the second group is $\displaystyle n \times b = nb$ .

Therefore, the total sum of all $\displaystyle 2n$ values is: $\displaystyle \text{Total sum} = na + nb = n(a + b)$ The sum of the $\displaystyle 2$ removed values is: $\displaystyle \text{Sum of removed values} = 2 \times c = 2c$ The sum of the remaining $\displaystyle (2n - 2)$ values can be expressed in two ways:
1. Total sum minus removed sum: $\displaystyle n(a + b) - 2c$ 2. Number of remaining values times their mean: $\displaystyle (2n - 2)m = 2nm - 2m$ Equating these two expressions: $\displaystyle n(a + b) - 2c = 2nm - 2m$ Rearranging to isolate terms with $\displaystyle n$ on the left-hand side: $\displaystyle n(a + b) - 2nm = 2c - 2m$ $\displaystyle n(a + b - 2m) = 2(c - m)$ $\displaystyle n = \dfrac{2(c - m)}{a + b - 2m}$ Hence, option A is correct.

Let $\displaystyle w_1$ and $\displaystyle w_2$ be the widths of the left and right sections respectively, and let $\displaystyle h_1$ and $\displaystyle h_2$ be the heights of the top and bottom sections respectively.

The area of the top-left rectangle is: $\displaystyle \text{Area}_{\text{TL}} = w_1 h_1 = 28\text{ cm}^2$ The area of the bottom-left rectangle is: $\displaystyle \text{Area}_{\text{BL}} = w_1 h_2 = 42\text{ cm}^2$ Taking the ratio of these areas gives the ratio of the heights: $\displaystyle \dfrac{h_1}{h_2} = \dfrac{w_1 h_1}{w_1 h_2} = \dfrac{28}{42} = \dfrac{2}{3}$ The combined area of the two rectangles on the right is: $\displaystyle \text{Area}_{\text{TR}} + \text{Area}_{\text{BR}} = w_2(h_1 + h_2) = 65\text{ cm}^2$ Because the two right-hand rectangles share the width $\displaystyle w_2$ , their areas are in the exact same ratio as their heights ( $\displaystyle h_1 : h_2 = 2 : 3$ ). Therefore, the area of the bottom-right rectangle is: $\displaystyle \text{Area}_{\text{BR}} = \dfrac{h_2}{h_1 + h_2} \times 65 = \dfrac{3}{2 + 3} \times 65 = \dfrac{3}{5} \times 65 = 39\text{ cm}^2$

Let the side length of the square be $\displaystyle x$ cm. Its area is $\displaystyle x^2$ .

The new rectangle has dimensions $\displaystyle 2x$ and $\displaystyle (x-4)$ , so its area is $\displaystyle 2x(x-4)$ . The problem states this area is 15 cm $\displaystyle ^2$ less than the area of the square, which gives the equation:Expanding and rearranging this leads to a quadratic equation:Factoring the quadratic gives $\displaystyle (x-3)(x-5) = 0$ , with solutions $\displaystyle x=3$ and $\displaystyle x=5$ .

We must check these solutions against the physical constraints of the problem. One of the rectangle's dimensions is $\displaystyle (x-4)$ , which must be positive. This requires $\displaystyle x > 4$ . Therefore, we must reject the solution $\displaystyle x=3$ .

The only valid side length for the square is $\displaystyle x=5$ cm.

The perimeter of the square is $\displaystyle 4x$ , so the perimeter is $\displaystyle 4 \times 5 = 20$ cm.

First, divide $\displaystyle 250$ by $\displaystyle 6$ to find its row and position within that row.This means that $\displaystyle 41$ full rows are completed, and $\displaystyle 250$ is the $\displaystyle 4$ th number in the $\displaystyle 42$ nd row.

Odd-numbered rows (like the 1st and 3rd) place integers from column $\displaystyle 1$ to $\displaystyle 6$ .
Even-numbered rows (like the 2nd and 42nd) place integers from column $\displaystyle 6$ down to $\displaystyle 1$ .

Since the $\displaystyle 42$ nd row is even, the numbers are placed starting from column $\displaystyle 6$ and moving left.
The first number in this row ( $\displaystyle 247$ ) is in column $\displaystyle 6$ .
The second number ( $\displaystyle 248$ ) is in column $\displaystyle 5$ .
The third number ( $\displaystyle 249$ ) is in column $\displaystyle 4$ .
The fourth number ( $\displaystyle 250$ ) is in column $\displaystyle 3$ .

Alternatively, for an even row, the column can be found using $\displaystyle 7 - r$ , where $\displaystyle r$ is the remainder. Here, $\displaystyle 7 - 4 = 3$ .

Therefore the correct answer is C.

Let $\displaystyle F$ be the event that a fair die is chosen, and $\displaystyle B$ be the event that a biased die is chosen: $\displaystyle P(F) = \dfrac{5}{8}, \quad P(B) = \dfrac{3}{8}$ For each type of die, the probability of rolling a 6 on any given roll is: $\displaystyle P(6 \mid F) = \dfrac{1}{6}, \quad P(6 \mid B) = \dfrac{1}{2}$ Let $\displaystyle R_1 = 6$ and $\displaystyle R_2 = 6$ denote the events that the first and second rolls show a 6, respectively.

Using the law of total probability, the probability that the first roll is a 6 is: $\displaystyle P(R_1 = 6) = P(F)P(R_1 = 6 \mid F) + P(B)P(R_1 = 6 \mid B) = \dfrac{5}{8}\left(\dfrac{1}{6}\right) + \dfrac{3}{8}\left(\dfrac{1}{2}\right) = \dfrac{5}{48} + \dfrac{9}{48} = \dfrac{14}{48} = \dfrac{7}{24}$ Since the two rolls are conditionally independent given the chosen die, the probability that both rolls show a 6 is: $\displaystyle P(R_1 = 6 \cap R_2 = 6) = P(F)P(R_1 = 6 \mid F)^2 + P(B)P(R_1 = 6 \mid B)^2$ $\displaystyle = \dfrac{5}{8}\left(\dfrac{1}{6}\right)^2 + \dfrac{3}{8}\left(\dfrac{1}{2}\right)^2 = \dfrac{5}{288} + \dfrac{3}{32} = \dfrac{5}{288} + \dfrac{27}{288} = \dfrac{32}{288} = \dfrac{1}{9}$ By the definition of conditional probability: $\displaystyle P(R_2 = 6 \mid R_1 = 6) = \dfrac{P(R_1 = 6 \cap R_2 = 6)}{P(R_1 = 6)} = \dfrac{1/9}{7/24} = \dfrac{24}{63} = \dfrac{8}{21}$

The problem requires finding the total fuel consumed in litres. This involves converting units and applying the given rate of consumption.

Step 1: Convert the distance from miles to kilometres.
We are given that 5 miles = 8 km. Therefore, 1 mile = $\displaystyle \dfrac{8}{5}$ km.
The distance travelled is $\displaystyle D$ miles, so in kilometres this is:Step 2: Calculate the fuel consumption rate in litres per kilometre.
The consumption is given as $\displaystyle F$ litres per 100 km. To find the consumption per kilometre, we divide by 100:Step 3: Calculate the total fuel consumed.
Total fuel is the distance in kilometres multiplied by the rate in litres per kilometre.Step 4: Simplify the expression.
We can simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.Thus, the total volume of fuel consumed is $\displaystyle \dfrac{2DF}{125}$ litres.

First, find the gradient of the line $\displaystyle L$ passing through $\displaystyle (0,0)$ and $\displaystyle (4,12)$ :Second, find the $\displaystyle y$ -coordinate of the point $\displaystyle P$ . Since $\displaystyle P$ is in the first quadrant at a distance of $\displaystyle 18$ from the origin and at an angle of $\displaystyle 30^{\circ}$ to the positive $\displaystyle x$ -axis, its $\displaystyle y$ -coordinate is given by:Using the exact trigonometric value $\displaystyle \sin 30^{\circ} = \dfrac{1}{2}$ :Finally, calculate the sum of the gradient and the $\displaystyle y$ -coordinate:Therefore the correct answer is B.

We restrict our sample space to the outcomes where the sum of the dice is at least 9. Listing these systematically:

* Sum 9: $\displaystyle (3,6), (4,5), (5,4), (6,3)$ (4 outcomes)
* Sum 10: $\displaystyle (4,6), (5,5), (6,4)$ (3 outcomes)
* Sum 11: $\displaystyle (5,6), (6,5)$ (2 outcomes)
* Sum 12: $\displaystyle (6,6)$ (1 outcome)

The total number of valid outcomes is $\displaystyle 4 + 3 + 2 + 1 = 10$ .

We now count the favourable outcomes within this restricted set where the two numbers are equal. These are $\displaystyle (5,5)$ and $\displaystyle (6,6)$ , giving 2 favourable outcomes.

The conditional probability is therefore:

The core principle is that the volume of water flowing into the tank per unit time is constant. Let this rate be $\displaystyle R$ . We can express $\displaystyle R$ in two ways:

1. Based on the pipe: $\displaystyle R = (\text{cross-sectional area of pipe}) \times (\text{flow velocity})$ .
2. Based on the tank: $\displaystyle R = (\text{base area of tank}) \times (\text{rate of increase of depth})$ .

Let's calculate all quantities in a consistent system of units, for example, centimetres (cm) for length and seconds (s) for time.

First, convert the given dimensions:
- Pipe diameter = 200 mm = 20 cm. The radius is $\displaystyle r = \dfrac{20}{2} = 10$ cm.
- Tank base side length = 10 cm.
- Flow velocity = 1.2 m/s = $\displaystyle 1.2 \times 100$ cm/s = 120 cm/s.

Now, calculate the areas:
- Cross-sectional area of pipe: $\displaystyle A_{\text{pipe}} = \pi r^2 = \pi (10)^2 = 100\pi$ cm $\displaystyle ^2$ .
- Base area of tank: $\displaystyle A_{\text{tank}} = 10 \times 10 = 100$ cm $\displaystyle ^2$ .

Now calculate the volume flow rate, $\displaystyle R$ , in cm $\displaystyle ^3$ /s:Let the rate of increase of depth be $\displaystyle \dfrac{dh}{dt}$ . We have:Solving for $\displaystyle \dfrac{dh}{dt}$ in cm/s:The question asks for the rate in cm/min. We must convert from seconds to minutes. There are 60 seconds in a minute.

The maximum deviation from the central point is the amplitude, $\displaystyle A = 10 \text{ cm}$ .

In one complete oscillation, the particle travels from its central point to one extreme, back through the center to the other extreme, and finally returns to the center. The distance covered in one cycle is therefore $\displaystyle 4A$ .

Distance per cycle $\displaystyle = 4 \times 10 \text{ cm} = 40 \text{ cm}$ .

The time for one cycle is $\displaystyle 0.50 \text{ s}$ . In a total time of $\displaystyle 3.0 \text{ s}$ , the number of cycles is:The total distance is the distance per cycle multiplied by the number of cycles:

First, we find the radius for each shape by halving the given diameter.
- Sphere radius: $\displaystyle r_s = \dfrac{6}{2} = 3$ cm.
- Cylinder radius: $\displaystyle r_{cy} = \dfrac{8}{2} = 4$ cm.
- Cone radius: $\displaystyle r_{co} = \dfrac{2}{2} = 1$ cm.

Next, we calculate the volume of the sphere and the cylinder.

The volume of the sphere is given by the formula $\displaystyle V = \dfrac{4}{3}\pi r^3$ .The volume of the cylinder is given by the formula $\displaystyle V = \pi r^2 h$ .The total volume of metal is the sum of these two volumes.Now, we calculate the volume of a single cone using the formula $\displaystyle V = \dfrac{1}{3}\pi r^2 h$ .Finally, to find the number of cones, we divide the total volume of metal by the volume of one cone.

Let $\displaystyle E$ be the event that exactly two Heads are observed. The number of flips $\displaystyle N$ can be 2, 3, or 4, each with a probability of $\displaystyle \dfrac{1}{3}$ . We can find the total probability of $\displaystyle E$ by considering each case for $\displaystyle N$ .

The Law of Total Probability states:Since $\displaystyle P(N=2) = P(N=3) = P(N=4) = \dfrac{1}{3}$ , this simplifies to:Now we calculate the conditional probabilities using the binomial probability formula, $\displaystyle P(k \text{ successes in } n \text{ trials}) = \binom{n}{k}p^k(1-p)^{n-k}$ . For a fair coin, the probability of a Head is $\displaystyle p = \dfrac{1}{2}$ .

Case 1: $\displaystyle N=2$ The probability of exactly 2 Heads in 2 flips is:Case 2: $\displaystyle N=3$ The probability of exactly 2 Heads in 3 flips is:Case 3: $\displaystyle N=4$ The probability of exactly 2 Heads in 4 flips is:Now, substitute these back into the total probability formula:To sum the fractions, we find a common denominator, which is 8:Thus, the overall probability is $\displaystyle \dfrac{1}{3}$ .

The relationship between volume flow rate ( $\displaystyle R$ ), cross-sectional area ( $\displaystyle A$ ), and linear speed ( $\displaystyle v$ ) is $\displaystyle R = A \times v$ , so $\displaystyle v = \dfrac{R}{A}$ .

First, convert the volume flow rate, $\displaystyle R$ , into units of cm $\displaystyle ^3$ /s.We know that $\displaystyle 1\text{ m} = 100\text{ cm}$ , so $\displaystyle 1\text{ m}^3 = (100\text{ cm})^3 = 1,000,000\text{ cm}^3 = 10^6\text{ cm}^3$ .
We also know that $\displaystyle 1\text{ hour} = 60 \times 60 = 3600\text{ s}$ .

Substituting these conversions:Next, calculate the cross-sectional area, $\displaystyle A$ , in cm $\displaystyle ^2$ .
The diameter is given as $\displaystyle d = 40\text{ mm}$ .
Converting to cm: $\displaystyle d = 4\text{ cm}$ .
The radius is half the diameter: $\displaystyle r = \dfrac{d}{2} = 2\text{ cm}$ .
The area of the circular cross-section is $\displaystyle A = \pi r^2$ .Finally, calculate the linear speed $\displaystyle v = \dfrac{R}{A}$ .

The correct expression for the average price is $\displaystyle \dfrac{3pqr}{pq+qr+rp}$ .

Let $\displaystyle t$ be the time elapsed from the start until the specified conditions are met.

First, consider the motion of point P. It starts at $\displaystyle (R, 0)$ and moves anticlockwise to the positive $\displaystyle y$ -axis. The point on the positive $\displaystyle y$ -axis on the circle $\displaystyle x^2 + y^2 = R^2$ is $\displaystyle (0, R)$ .
This movement corresponds to a rotation of $\displaystyle \dfrac{\pi}{2}$ radians, which is one quarter of a full circle ( $\displaystyle 2\pi$ radians).
The period for a full revolution is $\displaystyle T$ . Therefore, the time taken for P to complete a quarter of a revolution is:Next, consider the motion of point Q. It starts at $\displaystyle (R, H)$ and moves vertically downwards to the $\displaystyle x$ -axis (where $\displaystyle y=0$ ). The distance travelled by Q is $\displaystyle H$ .
The speed of Q is $\displaystyle v$ . The time taken for this motion is given by the formula time = distance / speed:The problem states that these two events happen at the same instant, so the times are equal: $\displaystyle t_P = t_Q$ .We need to find $\displaystyle v$ . Rearranging the equation:Therefore, the correct answer is A.

1. Since $\displaystyle y$ is inversely proportional to $\displaystyle x^2$ , we have: $\displaystyle y = \dfrac{k}{x^2}$ From the graph, the curve passes through the point $\displaystyle (2, 3)$ : $\displaystyle 3 = \dfrac{k}{2^2} = \dfrac{k}{4} \implies k = 12$ So, $\displaystyle y = \dfrac{12}{x^2}$ .

2. Since $\displaystyle x$ is directly proportional to $\displaystyle z^3$ , we have: $\displaystyle x = c z^3$ When $\displaystyle z = 2$ , $\displaystyle x = 4$ : $\displaystyle 4 = c \times 2^3 = 8c \implies c = \dfrac{1}{2}$ So, $\displaystyle x = \dfrac{1}{2} z^3$ .

3. Substituting $\displaystyle x$ into the equation for $\displaystyle y$ : $\displaystyle y = \dfrac{12}{\left(\dfrac{1}{2} z^3\right)^2} = \dfrac{12}{\dfrac{1}{4} z^6} = \dfrac{48}{z^6}$

First, find the distance $\displaystyle L$ between the two given points $\displaystyle (-1, 2)$ and $\displaystyle (3, 6)$ : $\displaystyle L^2 = (3 - (-1))^2 + (6 - 2)^2 = 4^2 + 4^2 = 32$ For a regular hexagon of side length $\displaystyle s$ , the area is: $\displaystyle \text{Area} = 6 \times \left(\dfrac{\sqrt{3}}{4}s^2\right) = \dfrac{3\sqrt{3}}{2}s^2$ There are three possible geometric relationships between any two vertices of a regular hexagon:
1. The two points are adjacent vertices (sharing an edge), so $\displaystyle s = L$ . In this case, $\displaystyle s^2 = L^2 = 32$ , giving the maximum possible area: $\displaystyle \text{Area}_{\max} = \dfrac{3\sqrt{3}}{2}(32) = 48\sqrt{3}$ 2. The two points are separated by one vertex (short diagonal), so $\displaystyle L = s\sqrt{3}$ , giving $\displaystyle s^2 = \dfrac{L^2}{3} = \dfrac{32}{3}$ . $\displaystyle \text{Area} = \dfrac{3\sqrt{3}}{2}\left(\dfrac{32}{3}\right) = 16\sqrt{3}$ 3. The two points are opposite vertices (main diagonal), so $\displaystyle L = 2s$ , giving $\displaystyle s^2 = \dfrac{L^2}{4} = \dfrac{32}{4} = 8$ . This yields the minimum possible area: $\displaystyle \text{Area}_{\min} = \dfrac{3\sqrt{3}}{2}(8) = 12\sqrt{3}$ The difference between the largest and smallest possible areas is: $\displaystyle \text{Area}_{\max} - \text{Area}_{\min} = 48\sqrt{3} - 12\sqrt{3} = 36\sqrt{3}$

Let the set be $\displaystyle \{1, 2, \dots, n\}$ . The sum of two chosen integers is odd if and only if one is even and one is odd.

Case 1: $\displaystyle n$ is even. Let $\displaystyle n = 2k$ .
There are $\displaystyle k$ even integers and $\displaystyle k$ odd integers.
The number of ways to choose one even and one odd integer is $\displaystyle k \times k = k^2$ .
The total number of ways to choose two distinct integers is $\displaystyle \dfrac{2k(2k-1)}{2} = k(2k-1)$ .
The probability of an odd sum is:Setting this equal to $\displaystyle \dfrac{12}{23}$ :So one possible value is $\displaystyle n = 2(12) = 24$ .

Case 2: $\displaystyle n$ is odd. Let $\displaystyle n = 2k + 1$ .
There are $\displaystyle k$ even integers and $\displaystyle k+1$ odd integers.
The number of ways to choose one even and one odd integer is $\displaystyle k(k+1)$ .
The total number of ways to choose two distinct integers is $\displaystyle \dfrac{(2k+1)(2k)}{2} = k(2k+1)$ .
The probability of an odd sum is:Setting this equal to $\displaystyle \dfrac{12}{23}$ :So another possible value is $\displaystyle n = 2(11) + 1 = 23$ .

The possible values for $\displaystyle n$ are $\displaystyle 23$ and $\displaystyle 24$ .
The sum of all possible values of $\displaystyle n$ is $\displaystyle 23 + 24 = 47$ .
Therefore the correct answer is E.

At the point of intersection $\displaystyle P(x, k)$ , the coordinates satisfy the equation of the curve with $\displaystyle y = k$ : $\displaystyle k = \dfrac{3x^2 - 4}{x^2 + 2}$ Multiply both sides by $\displaystyle (x^2 + 2)$ : $\displaystyle k(x^2 + 2) = 3x^2 - 4$ $\displaystyle kx^2 + 2k = 3x^2 - 4$ Rearrange to group all terms containing $\displaystyle x^2$ on one side and constant terms on the other: $\displaystyle 2k + 4 = 3x^2 - kx^2$ $\displaystyle 2k + 4 = x^2(3 - k)$ Divide both sides by $\displaystyle (3 - k)$ : $\displaystyle x^2 = \dfrac{2k + 4}{3 - k}$ Since $\displaystyle x \ge 0$ as shown on the graph, taking the positive square root gives: $\displaystyle x = \sqrt{\dfrac{2k + 4}{3 - k}}$