ESAT Mathematics 1 Mock 1

The total number of counters in the bag is: $\displaystyle \text{Total} = n + (n - 2) = 2n - 2$ The number of ways to choose 2 counters of different colours (one green and one red) is: $\displaystyle n \times (n - 2)$ The total number of ways to choose any 2 counters from the $\displaystyle 2n - 2$ counters is: $\displaystyle \binom{2n - 2}{2} = \dfrac{(2n - 2)(2n - 3)}{2} = (n - 1)(2n - 3)$ Setting the probability equal to $\displaystyle \dfrac{8}{15}$ gives: $\displaystyle \dfrac{n(n - 2)}{(n - 1)(2n - 3)} = \dfrac{8}{15}$ Cross-multiplying and expanding: $\displaystyle 15n(n - 2) = 8(n - 1)(2n - 3)$ $\displaystyle 15n^2 - 30n = 8(2n^2 - 5n + 3)$ $\displaystyle 15n^2 - 30n = 16n^2 - 40n + 24$ Rearranging into standard quadratic form: $\displaystyle n^2 - 10n + 24 = 0$ $\displaystyle (n - 4)(n - 6) = 0$ Thus, $\displaystyle n = 4$ or $\displaystyle n = 6$ . Both values satisfy the condition $\displaystyle n \ge 3$ .

The sum of all possible values of $\displaystyle n$ is: $\displaystyle 4 + 6 = 10$

Step 1:

First, determine the number of equations in each of the three categories.

Total number of equations is 200.
- Number of equations with two positive roots (PP): $\displaystyle 0.30 \times 200 = 60$ .
- Number of equations with two negative roots (NN): $\displaystyle 0.45 \times 200 = 90$ .
- The percentage of equations with one positive and one negative root (PN) is $\displaystyle 100\% - 30\% - 45\% = 25\%$ . So, the number of PN equations is $\displaystyle 0.25 \times 200 = 50$ .

Step 2:

Calculate the total number of positive roots. Each PP equation contributes 2 positive roots, and each PN equation contributes 1 positive root.

Total positive roots = $\displaystyle (60 \times 2) + (90 \times 0) + (50 \times 1) = 120 + 0 + 50 = 170$ .

Step 3:

Calculate the total number of negative roots. Each NN equation contributes 2 negative roots, and each PN equation contributes 1 negative root.

Total negative roots = $\displaystyle (60 \times 0) + (90 \times 2) + (50 \times 1) = 0 + 180 + 50 = 230$ .

Step 4:

The total number of positive roots is 170, and the total number of negative roots is 230.

The answer is $\displaystyle (170, 230)$ .

Since $\displaystyle DE$ is parallel to $\displaystyle BC$ , triangle $\displaystyle ADE$ is similar to triangle $\displaystyle ABC$ .

The total length of side $\displaystyle AB$ is: $\displaystyle AB = AD + DB = 4 + (x + 2) = x + 6$ Using the ratio of corresponding sides in similar triangles: $\displaystyle \dfrac{AD}{AB} = \dfrac{DE}{BC}$ $\displaystyle \dfrac{4}{x + 6} = \dfrac{x}{2x + 3}$ Cross-multiplying gives: $\displaystyle 4(2x + 3) = x(x + 6)$ $\displaystyle 8x + 12 = x^2 + 6x$ $\displaystyle x^2 - 2x - 12 = 0$ Solving for $\displaystyle x$ using the quadratic formula: $\displaystyle x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-12)}}{2(1)} = \dfrac{2 \pm \sqrt{4 + 48}}{2} = \dfrac{2 \pm \sqrt{52}}{2} = 1 \pm \sqrt{13}$ Since lengths must be positive, $\displaystyle x = 1 + \sqrt{13}$ .

The question asks for the length of $\displaystyle BC$ : $\displaystyle BC = 2x + 3 = 2(1 + \sqrt{13}) + 3 = 2 + 2\sqrt{13} + 3 = 5 + 2\sqrt{13}\text{ cm}$

1. Calculate the sum of the original dataset: $\displaystyle \text{Sum}_{\text{original}} = 25 \times 40 = 1000$ 2. Calculate the sum of the removed subset: $\displaystyle \text{Sum}_{\text{removed}} = 5 \times 32 = 160$ 3. Calculate the sum and mean of the remaining $\displaystyle 20$ values: $\displaystyle \text{Sum}_{\text{remaining}} = 1000 - 160 = 840$ $\displaystyle \text{Mean}_{\text{remaining}} = \dfrac{840}{20} = 42$ 4. Deduce the bound on the range:
Let the maximum and minimum of the original dataset be $\displaystyle x_{\max}$ and $\displaystyle x_{\min}$ , so $\displaystyle x_{\max} - x_{\min} = 15$ .
For any subset of the dataset, the maximum cannot exceed $\displaystyle x_{\max}$ and the minimum cannot be less than $\displaystyle x_{\min}$ .
Therefore, the range of the remaining values satisfies: $\displaystyle \text{range}_{\text{remaining}} = x_{\text{rem, max}} - x_{\text{rem, min}} \le x_{\max} - x_{\min} = 15$ Thus, $\displaystyle \text{mean} = 42$ and $\displaystyle \text{range} \le 15$ .

Let the number of values in group P be $\displaystyle n$ . Since group Q has three times the number of values, it has $\displaystyle 3n$ values. The total number of values in the combined set is $\displaystyle n + 3n = 4n$ .

The combined mean is the weighted average of the means of the two groups. Let $\displaystyle \bar{x}_Q$ be the mean of group Q. We are given that the mean of group P is 20 and the combined mean is 32.The factor $\displaystyle n$ is common to all terms in the fraction and can be cancelled:Now we can solve for $\displaystyle \bar{x}_Q$ :The mean of group Q is 36.

Let a perpendicular be drawn from vertex $\displaystyle D$ to the side $\displaystyle BC$ , meeting $\displaystyle BC$ at point $\displaystyle E$ .

Since $\displaystyle \angle DAB = \angle ABC = 90^\circ$ and $\displaystyle DE \perp BC$ , quadrilateral $\displaystyle ABED$ is a rectangle. Therefore:
- $\displaystyle BE = AD = x + 1$ - $\displaystyle DE = AB = 3x - 1$ The segment $\displaystyle EC$ is: $\displaystyle EC = BC - BE = (2x + 4) - (x + 1) = x + 3$ Applying Pythagoras' theorem to the right-angled triangle $\displaystyle DEC$ : $\displaystyle DE^2 + EC^2 = CD^2$ $\displaystyle (3x - 1)^2 + (x + 3)^2 = (3x + 1)^2$ Expanding the brackets: $\displaystyle (9x^2 - 6x + 1) + (x^2 + 6x + 9) = 9x^2 + 6x + 1$ $\displaystyle 10x^2 + 10 = 9x^2 + 6x + 1$ $\displaystyle x^2 - 6x + 9 = 0$ $\displaystyle (x - 3)^2 = 0 \implies x = 3$ Using $\displaystyle x = 3$ , the lengths are:
- Parallel sides: $\displaystyle AD = 3 + 1 = 4\text{ cm}$ and $\displaystyle BC = 2(3) + 4 = 10\text{ cm}$ - Perpendicular height: $\displaystyle AB = 3(3) - 1 = 8\text{ cm}$ The area of the trapezium is: $\displaystyle \text{Area} = \dfrac{AD + BC}{2} \times AB = \dfrac{4 + 10}{2} \times 8 = 7 \times 8 = 56\text{ cm}^2$

First, we relate actual power to theoretical power using the efficiency at wind speed $\displaystyle v$ :This gives:Solving for the initial theoretical power:Since theoretical power is directly proportional to the cube of the wind speed ( $\displaystyle P_{\text{theo}} \propto v^3$ ), doubling the wind speed to $\displaystyle 2v$ increases the theoretical power by a factor of $\displaystyle 2^3 = 8$ :Finally, we calculate the new actual power using the new efficiency of $\displaystyle 15\%$ :Therefore the correct answer is D.

Let the sum of values in set $\displaystyle S_1$ be $\displaystyle \Sigma_1 = 10n_1$ and in set $\displaystyle S_2$ be $\displaystyle \Sigma_2 = 30n_2$ .
The total number of values in the first combination is $\displaystyle n_1 + n_2$ .

The mean of the combined set is given as 14. We can set up an equation:Now, we solve for the ratio of $\displaystyle n_1$ to $\displaystyle n_2$ :This means the ratio $\displaystyle n_1:n_2$ is $\displaystyle 4:1$ .

For the second part of the problem, we form a new set by combining the numbers from $\displaystyle S_1$ with a set of $\displaystyle 2n_2$ values, each equal to 30.

The sum of values in this new combination is $\displaystyle 10n_1 + 30(2n_2) = 10n_1 + 60n_2$ .
The total number of values is $\displaystyle n_1 + 2n_2$ .

The new mean, $\displaystyle \bar{x}_{\text{new}}$ , is:We can now substitute the relationship $\displaystyle n_1 = 4n_2$ into this expression:Therefore the correct answer is B.

The curve has a maximum value of $\displaystyle 9$ at $\displaystyle x=2$ , so its vertex is at the point $\displaystyle (2, 9)$ .

This allows us to write the equation of the parabola in vertex form, $\displaystyle y = a(x-h)^2 + k$ , which becomes:We are also given that the curve passes through the point $\displaystyle (0, 1)$ . Substituting these coordinates into the equation allows us to solve for $\displaystyle a$ :The value of $\displaystyle a$ is $\displaystyle -2$ .

Step 1: Find the probabilities for the outcome of the die roll.
The possible outcomes are $\displaystyle \{1, 2, 3, 4, 5, 6\}$ .
The prime numbers are $\displaystyle \{2, 3, 5\}$ , so: $\displaystyle P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$ The non-prime numbers are $\displaystyle \{1, 4, 6\}$ (note that 1 is not prime), so: $\displaystyle P(\text{not prime}) = \dfrac{3}{6} = \dfrac{1}{2}$ Step 2: Calculate the conditional probabilities of obtaining exactly 2 heads.
- Case 1: Prime rolled (2 tosses of a biased coin with $\displaystyle P(H) = \dfrac{2}{3}$ ): $\displaystyle P(2\text{ heads} \mid \text{prime}) = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}$ - Case 2: Not prime rolled (3 tosses of a fair coin with $\displaystyle P(H) = \dfrac{1}{2}$ ): $\displaystyle P(2\text{ heads} \mid \text{not prime}) = \binom{3}{2} \left(\dfrac{1}{2}\right)^2 \left(\dfrac{1}{2}\right)^1 = 3 \times \dfrac{1}{8} = \dfrac{3}{8}$ Step 3: Apply the law of total probability: $\displaystyle P(\text{exactly 2 heads}) = P(\text{prime}) \times P(2\text{ heads} \mid \text{prime}) + P(\text{not prime}) \times P(2\text{ heads} \mid \text{not prime})$ $\displaystyle P(\text{exactly 2 heads}) = \dfrac{1}{2} \times \dfrac{4}{9} + \dfrac{1}{2} \times \dfrac{3}{8} = \dfrac{2}{9} + \dfrac{3}{16} = \dfrac{32 + 27}{144} = \dfrac{59}{144}$

The sum of the interior angles of an $\displaystyle n$ -sided polygon is given by $\displaystyle 180(n-2)^{\circ}$ .

The polygon has $\displaystyle n$ angles in total. The sum of the first $\displaystyle 5$ interior angles is:There are $\displaystyle n-5$ remaining interior angles, and their sum is:Equating the sum of these two groups to the total sum of interior angles gives:Expanding both sides:Simplifying the left hand side:Rearranging to solve for $\displaystyle n$ :(Alternatively, using exterior angles: the mean exterior angles are $\displaystyle 180 - 120 = 60^{\circ}$ and $\displaystyle 180 - 160 = 20^{\circ}$ . The equation $\displaystyle 5(60) + 20(n-5) = 360$ simplifies to $\displaystyle 300 + 20n - 100 = 360$ , yielding $\displaystyle 20n = 160$ and $\displaystyle n=8$ .)

Therefore the correct answer is B.

Simplify each term using standard form and index rules:

1. First term: $\displaystyle \dfrac{x \times 10^3 + 4x \times 10^2}{7 \times 10^{-1}} = \dfrac{1400x}{0.7} = 2000x = 2 \times 10^3 x$ Squaring this gives: $\displaystyle (2 \times 10^3 x)^2 = 4 \times 10^6 x^2$ 2. Second term: $\displaystyle \dfrac{x \times 10^4 + 2x \times 10^3}{3 \times 10^{-2}} = \dfrac{12000x}{0.03} = 400000x = 4 \times 10^5 x$ 3. Substitute back into the equation: $\displaystyle 4 \times 10^6 x^2 - 4 \times 10^5 x = 2.4 \times 10^7$ Divide the entire equation by $\displaystyle 10^5$ : $\displaystyle 40x^2 - 4x = 240$ Divide by $\displaystyle 4$ : $\displaystyle 10x^2 - x - 60 = 0$ 4. Solve the quadratic equation using the quadratic formula: $\displaystyle x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(10)(-60)}}{2(10)} = \dfrac{1 \pm \sqrt{1 + 2400}}{20} = \dfrac{1 \pm \sqrt{2401}}{20}$ Since $\displaystyle 49^2 = 2401$ , $\displaystyle \sqrt{2401} = 49$ : $\displaystyle x_1 = \dfrac{1 + 49}{20} = \dfrac{50}{20} = 2.5$ $\displaystyle x_2 = \dfrac{1 - 49}{20} = \dfrac{-48}{20} = -2.4$ 5. The positive difference between the two solutions is: $\displaystyle x_1 - x_2 = 2.5 - (-2.4) = 4.9$

The density of the composite material is calculated by summing the individual densities weighted by their respective volume fractions.

We calculate:Simplifying the terms gives:

Let $\displaystyle d_p$ and $\displaystyle d_t$ be the internal diameters of the pipe and the tank, and let $\displaystyle v_p$ and $\displaystyle v_t$ be the speed of the water in the pipe and the rate of depth increase in the tank, respectively. By equating the volume flow rates, we have:where $\displaystyle A_p$ and $\displaystyle A_t$ are the cross-sectional areas.

First, convert all measurements to metres and seconds to match the requested final units. The radius of the tank is $\displaystyle r_t = 1 \text{ m}$ , so its cross-sectional area is $\displaystyle A_t = \pi (1)^2 = \pi \text{ m}^2$ .

The radius of the pipe is $\displaystyle r_p = 2.5 \text{ cm} = 0.025 \text{ m} = \dfrac{1}{40} \text{ m}$ , so its cross-sectional area is $\displaystyle A_p = \pi \left(\dfrac{1}{40}\right)^2 = \dfrac{\pi}{1600} \text{ m}^2$ .

The rate of depth increase in the tank is:Substitute these values into the flow rate equation:which simplifies to:Therefore the correct answer is C.

Since $\displaystyle \angle SPT = \angle QPR$ (common angle) and $\displaystyle \angle PST = \angle PRQ$ (given), triangle $\displaystyle PST$ is similar to triangle $\displaystyle PRQ$ by AA similarity ( $\displaystyle \triangle PST \sim \triangle PRQ$ ).

The ratio of corresponding sides is: $\displaystyle \dfrac{PS}{PR} = \dfrac{PT}{PQ}$ From the given segment lengths:
- $\displaystyle PQ = PS + SQ = x + 1$ - $\displaystyle PR = PT + TR = 6 + (x - 2) = x + 4$ Substituting into the ratio: $\displaystyle \dfrac{x}{x + 4} = \dfrac{6}{x + 1}$ Cross-multiplying gives: $\displaystyle x(x + 1) = 6(x + 4)$ $\displaystyle x^2 + x = 6x + 24$ $\displaystyle x^2 - 5x - 24 = 0$ $\displaystyle (x - 8)(x + 3) = 0$ Since lengths must be positive, $\displaystyle x = 8$ .

Therefore, the length of $\displaystyle PR$ is: $\displaystyle PR = x + 4 = 8 + 4 = 12\text{ cm}$

Let $\displaystyle k = x - y$ , which means $\displaystyle x = y + k$ .
Substitute this into the given equation:Expand the terms:Simplify and collect like terms with respect to $\displaystyle y$ :For $\displaystyle y$ to be a real number, this quadratic equation must have real roots, so its discriminant must be non-negative ( $\displaystyle \Delta \ge 0$ ):Taking the square root gives $\displaystyle -2 \le k \le 2$ .
Thus, the maximum possible value of $\displaystyle k = x - y$ is $\displaystyle 2$ .
Therefore the correct answer is B.

The relationship between volume rate, cross-sectional area, and speed is:

Volume Rate = Area × Speed

First, calculate the cross-sectional area of the cylinder in cm². The diameter is given as 4 cm, so the radius $\displaystyle r$ is 2 cm.Next, use the given volume rate to find the speed in cm per minute.Finally, convert the units of the speed from cm/min to the required mm/s.

To convert cm to mm, multiply by 10:To convert minutes to seconds, divide by 60:Thus, the speed is 10 mm/s.

We analyse each statement in turn:

1. The median of the 7 ordered numbers is the middle (4th) value, $\displaystyle x_4 = m$ . After removing the smallest and largest, five ordered numbers remain: $\displaystyle x_2, x_3, x_4, x_5, x_6$ . Their median is the middle (3rd) value, which is also $\displaystyle x_4 = m$ . Therefore, statement 1 must be true.

2. The original range is $\displaystyle R = x_7 - x_1$ . The range of the remaining five numbers is $\displaystyle x_6 - x_2$ . Since the numbers are strictly increasing, $\displaystyle x_6 < x_7$ and $\displaystyle x_2 > x_1$ , so $\displaystyle x_6 - x_2 < x_7 - x_1$ . Therefore, statement 2 must be true.

3. The sum of the seven numbers is $\displaystyle 7M$ . The sum of the remaining five is $\displaystyle 7M - (x_1 + x_7)$ , so their mean is $\displaystyle \dfrac{7M - (x_1 + x_7)}{5}$ . For this to equal $\displaystyle M$ , we need $\displaystyle x_1 + x_7 = 2M$ . Knowing only that $\displaystyle M = m$ does not guarantee this.

For example, take the numbers $\displaystyle 1, 2, 3, 10, 11, 12, 31$ .
- The median is $\displaystyle m = 10$ .
- The mean is $\displaystyle M = \dfrac{70}{7} = 10$ , so $\displaystyle M = m$ .
- After removing $\displaystyle 1$ and $\displaystyle 31$ , the remaining numbers are $\displaystyle 2, 3, 10, 11, 12$ , with mean $\displaystyle \dfrac{38}{5} = 7.6 eq 10$ .

Thus, statement 3 is not necessarily true.

Hence, only statements 1 and 2 must be true.

1. Find the initial area $\displaystyle A_1$ : $\displaystyle A_1 = \dfrac{2x + 3x}{2} \times h = \dfrac{5}{2}xh = 2.5xh$ 2. Determine the modified dimensions:
- New longer parallel side: $\displaystyle 3x \times \left(1 + \dfrac{40}{100}\right) = 3x \times 1.4 = 4.2x$ - New shorter parallel side: $\displaystyle 2x \times \left(1 - \dfrac{35}{100}\right) = 2x \times 0.65 = 1.3x$ - Sum of new parallel sides: $\displaystyle 4.2x + 1.3x = 5.5x$ - New vertical height: $\displaystyle h \times \left(1 + \dfrac{20}{100}\right) = 1.2h$ 3. Calculate the new area $\displaystyle A_2$ : $\displaystyle A_2 = \dfrac{5.5x}{2} \times 1.2h = 2.75x \times 1.2h = 3.3xh$ 4. Find the ratio $\displaystyle k = \dfrac{A_2}{A_1}$ : $\displaystyle k = \dfrac{3.3xh}{2.5xh} = \dfrac{3.3}{2.5} = \dfrac{33}{25}$ Hence, the correct option is D.