ESAT Mathematics 1 Mock 2

Let $\displaystyle Q$ represent the price of item Q in January.

In January, the price of item P was: $\displaystyle P_{\text{Jan}} = 1.25 Q$ In February: $\displaystyle P_{\text{Feb}} = 1.25 Q \times (1 - 0.20) = 1.25 Q \times 0.80 = 1.00 Q$ $\displaystyle Q_{\text{Feb}} = Q \times (1 + 0.10) = 1.10 Q$ In March: $\displaystyle P_{\text{Mar}} = 1.00 Q \times (1 + 0.15) = 1.15 Q$ $\displaystyle Q_{\text{Mar}} = 1.10 Q \times (1 - 0.05) = 1.10 Q \times 0.95 = 1.045 Q$ The difference in price in March is: $\displaystyle P_{\text{Mar}} - Q_{\text{Mar}} = 1.15 Q - 1.045 Q = 0.105 Q$ We are given that this difference is £6.30: $\displaystyle 0.105 Q = 6.30$ $\displaystyle Q = \dfrac{6.30}{0.105} = \dfrac{6300}{105} = 60$ Therefore, the price of item Q in January was £60.

To write the expression in the required form, we need to rationalise the denominator. We do this by multiplying the numerator and denominator by the conjugate of the denominator, which is $\displaystyle \sqrt{3} + 1$ .Expanding the brackets in the numerator gives $\displaystyle 3(3) + 3\sqrt{3} + 3\sqrt{3} + 3 = 9 + 6\sqrt{3} + 3 = 12 + 6\sqrt{3}$ .
The denominator becomes $\displaystyle (\sqrt{3})^2 - 1^2 = 3 - 1 = 2$ .

The expression simplifies to:Comparing this to the form $\displaystyle a + b\sqrt{3}$ , we can see that $\displaystyle a = 6$ and $\displaystyle b = 3$ .
Therefore, the value of $\displaystyle a + b$ is $\displaystyle 6 + 3 = 9$ .

For the quadratic equation $\displaystyle x^2 + ax + b = 0$ to have real roots, its discriminant must be non-negative: $\displaystyle \Delta = a^2 - 4(1)(b) \ge 0 \implies a^2 \ge 4b \implies b \le \dfrac{a^2}{4}$ The total number of equally likely outcomes is $\displaystyle 4 \times 6 = 24$ .

We systematically test each possible value of $\displaystyle a \in \{1, 2, 3, 4\}$ and count how many faces on the six-sided die ( $\displaystyle \{1, 1, 2, 3, 4, 6\}$ ) satisfy the condition:

- If $\displaystyle a = 1$ : $\displaystyle b \le \dfrac{1}{4} = 0.25$ . No faces satisfy this ( $\displaystyle 0$ outcomes).
- If $\displaystyle a = 2$ : $\displaystyle b \le \dfrac{4}{4} = 1$ . The two faces with ' $\displaystyle 1$ ' satisfy this ( $\displaystyle 2$ outcomes).
- If $\displaystyle a = 3$ : $\displaystyle b \le \dfrac{9}{4} = 2.25$ . The faces with ' $\displaystyle 1, 1, 2$ ' satisfy this ( $\displaystyle 3$ outcomes).
- If $\displaystyle a = 4$ : $\displaystyle b \le \dfrac{16}{4} = 4$ . The faces with ' $\displaystyle 1, 1, 2, 3, 4$ ' satisfy this ( $\displaystyle 5$ outcomes).

The total number of favourable outcomes is: $\displaystyle 0 + 2 + 3 + 5 = 10$ Therefore, the probability is: $\displaystyle P(\text{real roots}) = \dfrac{10}{24} = \dfrac{5}{12}$

1. The area of the initial square is $\displaystyle A = x^2$ .

2. The dimensions of the bounding rectangle are: $\displaystyle b = x + 0.50x = \dfrac{3}{2}x$ $\displaystyle h = x - 0.20x = \dfrac{4}{5}x$ 3. The area of the bounding rectangle is: $\displaystyle A_{\text{rect}} = b \times h = \left(\dfrac{3}{2}x\right)\left(\dfrac{4}{5}x\right) = \dfrac{6}{5}x^2 = \dfrac{6}{5}A$ 4. The removed triangular corner has base $\displaystyle \dfrac{1}{3}b$ and height $\displaystyle \dfrac{1}{2}h$ , giving an area of: $\displaystyle A_{\text{tri}} = \dfrac{1}{2} \times \left(\dfrac{1}{3}b\right) \times \left(\dfrac{1}{2}h\right) = \dfrac{1}{12}bh = \dfrac{1}{12}\left(\dfrac{6}{5}A\right) = \dfrac{1}{10}A$ 5. The area of the resulting pentagon is: $\displaystyle A_{\text{pentagon}} = A_{\text{rect}} - A_{\text{tri}} = \dfrac{6}{5}A - \dfrac{1}{10}A = \dfrac{12}{10}A - \dfrac{1}{10}A = \dfrac{11}{10}A$

We must construct the expression for the total cost by converting units at each stage.

1. Convert volume to metric units:
The component has a volume of $\displaystyle V$ cubic inches. Since 1 inch = $\displaystyle L$ cm, it follows that 1 cubic inch = $\displaystyle (L \text{ cm})^3 = L^3 \text{ cm}^3$ . The volume of the component in cm³ is therefore $\displaystyle V \times L^3 = V L^3 \text{ cm}^3$ .

2. Calculate mass in grams:
The material density is $\displaystyle D \text{ g/cm}^3$ . The mass of the component is given by density multiplied by volume. Mass in grams = $\displaystyle (V L^3) \times D = V D L^3 \text{ g}$ .

3. Convert mass to kilograms:
There are 1000 grams in a kilogram. To convert the mass to kg, we divide by 1000. Mass in kg = $\displaystyle \dfrac{V D L^3}{1000} \text{ kg}$ .

4. Calculate the number of blocks required:
The material is supplied in blocks of mass $\displaystyle M$ kg. The number of blocks needed is the total mass divided by the mass per block: Number of blocks = $\displaystyle \dfrac{\text{Total mass (kg)}}{\text{Mass per block (kg)}} = \dfrac{V D L^3 / 1000}{M} = \dfrac{V D L^3}{1000 M}$ .

5. Calculate the total cost in pounds:
Each block costs $\displaystyle C$ pounds. The total cost in pounds is the number of blocks multiplied by the cost per block: Cost in £ = $\displaystyle \dfrac{V D L^3}{1000 M} \times C = \dfrac{V C D L^3}{1000 M}$ .

6. Convert the total cost to pence:
There are 100 pence in a pound. To convert the cost to pence, we multiply by 100: Cost in pence = $\displaystyle \dfrac{V C D L^3}{1000 M} \times 100 = \dfrac{100 V C D L^3}{1000 M} = \dfrac{V C D L^3}{10 M}$ .

Therefore, the correct expression is $\displaystyle \dfrac{V C D L^3}{10 M}$ .

The area $\displaystyle A$ of a trapezium with parallel sides $\displaystyle a$ and $\displaystyle b$ and perpendicular height $\displaystyle h$ is: $\displaystyle A = \dfrac{1}{2}(a + b)h$ Substitute the given side lengths and height: $\displaystyle a = 2^{x+1}$ $\displaystyle b = 2^{x+3}$ $\displaystyle h = 4^{2-x} = (2^2)^{2-x} = 2^{4-2x}$ Simplify the sum of the parallel sides by factoring out $\displaystyle 2^x$ : $\displaystyle a + b = 2^{x+1} + 2^{x+3} = 2^x \cdot 2^1 + 2^x \cdot 2^3 = 2^x(2 + 8) = 10 \cdot 2^x$ Compute the area expression: $\displaystyle A = \dfrac{1}{2} \cdot (10 \cdot 2^x) \cdot 2^{4-2x} = 5 \cdot 2^x \cdot 2^{4-2x} = 5 \cdot 2^{4-x}$ Express the given area $\displaystyle 40\sqrt{2}$ in terms of base 2: $\displaystyle 40\sqrt{2} = 5 \times 8 \times \sqrt{2} = 5 \times 2^3 \times 2^{1/2} = 5 \cdot 2^{3.5}$ Equating the two expressions: $\displaystyle 5 \cdot 2^{4-x} = 5 \cdot 2^{3.5} \implies 4 - x = 3.5 \implies x = 0.5$

For the graphs to intersect at exactly one point, the equation formed by setting them equal must have exactly one real solution for $\displaystyle x$ .Rearranging gives:There are two cases for this equation to have a unique solution.

Case 1: The equation is linear.
This occurs if the coefficient of $\displaystyle x^2$ is zero. $\displaystyle k-1=0 \implies k=1$ .
If $\displaystyle k=1$ , the equation becomes $\displaystyle 6x + (1+7) = 0$ , or $\displaystyle 6x+8=0$ , which has one solution. Thus, $\displaystyle k=1$ is a possible value.

Case 2: The equation is a quadratic with a single repeated root.
This requires $\displaystyle k-1 eq 0$ and the discriminant, $\displaystyle \Delta$ , to be zero.$\displaystyle 36 - 4(k^2 + 6k - 7) = 0$ Dividing by 4: $\displaystyle 9 - (k^2 + 6k - 7) = 0$ $\displaystyle 16 - k^2 - 6k = 0$ $\displaystyle k^2 + 6k - 16 = 0$ Factoring the quadratic in $\displaystyle k$ : $\displaystyle (k+8)(k-2) = 0$ This gives $\displaystyle k=-8$ and $\displaystyle k=2$ . Both are valid as neither is equal to 1.

The set of all possible values for $\displaystyle k$ is $\displaystyle \{-8, 1, 2\}$ .
The sum of these values is $\displaystyle -8 + 1 + 2 = -5$ .

We start by finding the ratio of the volumes of $\displaystyle Y$ and $\displaystyle X$ :For mathematically similar solids, the volume ratio is the cube of the linear scale factor, $\displaystyle k$ . Thus:The ratio of surface areas is the square of the linear scale factor ( $\displaystyle k^2$ ). We calculate:Since the surface area of $\displaystyle X$ is $\displaystyle S$ , the surface area of $\displaystyle Y$ is $\displaystyle \dfrac{9}{4}S$ .

Let $\displaystyle b$ be the number of black discs in the box. The number of white discs is $\displaystyle 15 - b$ .

Since there are more black discs than white discs, we have $\displaystyle b > 15 - b \implies 2b > 15 \implies b \ge 8$ .

Two discs are drawn without replacement. The total number of ways to draw two discs is: $\displaystyle \binom{15}{2} = \dfrac{15 \times 14}{2} = 105$ The probability that both discs drawn are of the same colour is the sum of the probabilities of drawing two black discs and drawing two white discs: $\displaystyle P(\text{same colour}) = \dfrac{b(b-1) + (15-b)(14-b)}{15 \times 14} = \dfrac{11}{21}$ Multiply both sides by $\displaystyle 210$ ( $\displaystyle 15 \times 14$ ): $\displaystyle b^2 - b + (210 - 29b + b^2) = \dfrac{11}{21} \times 210 = 110$ $\displaystyle 2b^2 - 30b + 210 = 110$ $\displaystyle 2b^2 - 30b + 100 = 0$ $\displaystyle b^2 - 15b + 50 = 0$ $\displaystyle (b - 10)(b - 5) = 0$ Since $\displaystyle b \ge 8$ , we have $\displaystyle b = 10$ (and the number of white discs is $\displaystyle 5$ ).

The probability that both discs drawn are black is: $\displaystyle P(\text{both black}) = \dfrac{b(b-1)}{15 \times 14} = \dfrac{10 \times 9}{210} = \dfrac{90}{210} = \dfrac{3}{7}$

Let's analyse the effect on the mean, range, and median in turn.

Mean:
The original mean of the 5 integers is 10. The integer being added, 8, is less than the mean. Adding a value below the current mean will always cause the mean to decrease. So, the mean decreases.

Range:
Let the original set in ascending order be $\displaystyle x_1, x_2, x_3, x_4, x_5$ . The median is $\displaystyle x_3=5$ , so the minimum value $\displaystyle x_1$ must be less than 5. The mean is 10. For the mean to be 10 with a median of 5, the maximum value $\displaystyle x_5$ must be greater than 10 (otherwise the sum would be too small). The value being added, 8, lies between the original minimum ( $\displaystyle x_1 < 5$ ) and maximum ( $\displaystyle x_5 > 10$ ). Therefore, the range, which is $\displaystyle x_5 - x_1$ , is unchanged.

Median:
The original set has 5 elements, so the median is the 3rd value, which is 5. After adding 8, the new set has 6 elements. The new median is the average of the 3rd and 4th values in the new sorted list. Since 8 is greater than 5, the 3rd value in the new sorted list is still 5. The 4th value is either the original $\displaystyle x_4$ or 8. As the original integers are distinct, $\displaystyle x_4 > 5$ , so the 4th value is definitely greater than 5. The new median is $\displaystyle \dfrac{5 + k}{2}$ for some integer $\displaystyle k>5$ . This value must be greater than 5. So, the median increases.

Combining these results: the mean decreases, the median increases, and the range is unchanged.

Let G be a gold coin and S be a silver coin. We are given $\displaystyle P(G) = \dfrac{1}{3}$ and so $\displaystyle P(S) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ .

When creating a box by selecting two coins, the four possible outcomes for its contents have the following probabilities:

- $\displaystyle P(GG) = P(G) \times P(G) = \dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}$ - $\displaystyle P(GS) = P(G) \times P(S) = \dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{2}{9}$ - $\displaystyle P(SG) = P(S) \times P(G) = \dfrac{2}{3} \times \dfrac{1}{3} = \dfrac{2}{9}$ - $\displaystyle P(SS) = P(S) \times P(S) = \dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{4}{9}$ The event that a box contains one of each coin is GS or SG, so $\displaystyle P(\text{one of each}) = P(GS) + P(SG) = \dfrac{2}{9} + \dfrac{2}{9} = \dfrac{4}{9}$ .

The condition is that each box contains 'at least one gold coin'. This event, let's call it $\displaystyle A$ , eliminates the SS outcome. The probability of this condition being met for a single box is:We now find the conditional probabilities for the composition of a box, given event $\displaystyle A$ .Let $\displaystyle D_G$ be the event of drawing a gold coin from a conditioned box. We use the law of total probability:If the box is GG, the probability of drawing a gold coin is 1.
If the box is GS or SG, the probability of drawing a gold coin is $\displaystyle \dfrac{1}{2}$ .

So, the probability of drawing a gold coin from one box is:This is the probability for Box 1. By symmetry, it is also the probability for Box 2.
Since the two draws are independent, the probability that both coins drawn are gold is:

The solution involves tracking the quantities produced through each stage.

1. Calculate the number of batches:
The total mass of raw material is 1200 kg, and each batch is 20 kg.
Number of batches = $\displaystyle \dfrac{1200}{20} = 60$ batches.

2. Calculate substances produced in Stage 1:
Each of the 60 batches produces 10 units of P and 5 units of G.
Total substance P produced = $\displaystyle 60 \times 10 = 600$ units.
Total substance G from Stage 1 = $\displaystyle 60 \times 5 = 300$ units.

3. Calculate substance G produced in Stage 2:
All 600 units of substance P are processed. Each unit of P produces 0.5 units of G.
Total substance G from Stage 2 = $\displaystyle 600 \times 0.5 = 300$ units.

4. Calculate the total amount of substance G:
The total amount of G is the sum from both stages.
Total G = (G from Stage 1) + (G from Stage 2) = $\displaystyle 300 + 300 = 600$ units.

5. Calculate the total amount of final product F:
3 units of G are required to produce 2 units of F. This is a ratio of $\displaystyle \dfrac{2 \text{ units F}}{3 \text{ units G}}$ .
Total F produced = $\displaystyle 600 \text{ units of G} \times \dfrac{2}{3} = \dfrac{1200}{3} = 400$ units.

6. Calculate the total time:
The product F is produced at a rate of 5 units per minute.
Total time = $\displaystyle \dfrac{\text{Total units of F}}{\text{Rate}} = \dfrac{400}{5} = 80$ minutes.

Let the volume of the inner core be $\displaystyle V_c = \pi R^2 L$ . Its mass is given as $\displaystyle M$ , which establishes the relationship $\displaystyle M = \rho V_c$ .

The volume of the outer shell, $\displaystyle V_s$ , is the difference between the volume of the entire cylinder (radius $\displaystyle 2R$ ) and the volume of the core.The shell has density $\displaystyle 2\rho$ , so its mass $\displaystyle M_s$ is:Since we know $\displaystyle M = \rho V_c$ , we can express the shell's mass in terms of $\displaystyle M$ : $\displaystyle M_s = 6(\rho V_c) = 6M$ .

The total mass of the rod is the sum of the core mass and the shell mass: $\displaystyle M_{\text{total}} = M + M_s = M + 6M = 7M$ .

We consider the symmetry of the situation. In any random arrangement of the $\displaystyle n$ distinct items, there are only two possibilities for the relative ordering of $\displaystyle X$ and $\displaystyle Y$ :

1. $\displaystyle X$ appears earlier than $\displaystyle Y$ .
2. $\displaystyle Y$ appears earlier than $\displaystyle X$ .

For every arrangement where $\displaystyle X$ precedes $\displaystyle Y$ , there is a corresponding arrangement formed by swapping the positions of $\displaystyle X$ and $\displaystyle Y$ where $\displaystyle Y$ precedes $\displaystyle X$ . Since the ordering is random, these two cases are equally likely and cover all possibilities.

Therefore, the probability that $\displaystyle X$ appears earlier than $\displaystyle Y$ is independent of $\displaystyle n$ :This result holds for any $\displaystyle n \geq 2$ .

The given quadratic is $\displaystyle x^2 + bx + c = 0$ , where $\displaystyle a = 1$ , $\displaystyle b = 4^{2-2k} \times 8^k$ , and $\displaystyle c = 16^{2-k}$ .

First, express $\displaystyle b$ and $\displaystyle c$ as powers of $\displaystyle 2$ :For exactly one real root, the discriminant must be zero, so $\displaystyle b^2 = 4ac$ :Equating the two expressions gives:Since the bases are equal, we can equate the exponents:Rearranging this linear equation yields $\displaystyle 2k = 2$ , so $\displaystyle k = 1$ .

Therefore the correct answer is D.

The age difference between P and Q is constant. We are told that in 12 years, P will be 21 years older than Q. This means P is always 21 years older than Q.

Five years ago, the ratio of their ages was $\displaystyle 5:2$ . Let their ages at that time be $\displaystyle 5x$ and $\displaystyle 2x$ . The difference in their ages was $\displaystyle 5x - 2x = 3x$ .

We can equate this difference to the known age gap:So, five years ago, P's age was $\displaystyle 5x = 5 \times 7 = 35$ .

P's current age is therefore $\displaystyle 35 + 5 = 40$ .

The calculation proceeds in three main stages: finding the length of the filament, finding its volume, and then finding its mass.

1. Calculate the total length of the filament.
The printer runs for 10 minutes. We need to convert this to seconds to match the feed rate unit.
Time in seconds = $\displaystyle 10 \text{ minutes} \times 60 \dfrac{\text{seconds}}{\text{minute}} = 600$ seconds.
The total length $\displaystyle L$ is the feed rate multiplied by the time: $\displaystyle L = 5 \dfrac{\text{mm}}{\text{s}} \times 600 \text{ s} = 3000$ mm.

2. Calculate the volume of the filament.
The density is given in g/cm $\displaystyle ^3$ , so it is best to convert all dimensions to cm.
Length $\displaystyle L = 3000$ mm $\displaystyle = 300$ cm.
The diameter is 4 mm, so the radius is $\displaystyle r = 2$ mm $\displaystyle = 0.2$ cm.
The filament is a cylinder, so its volume $\displaystyle V$ is given by the formula $\displaystyle V = \pi r^2 L$ .3. Calculate the mass of the plastic.
The relationship between mass, density ( $\displaystyle \rho$ ), and volume is Mass $\displaystyle = \rho \times V$ .To calculate $\displaystyle 2.5 \times 12$ without a calculator, we can write $\displaystyle 2.5$ as $\displaystyle \dfrac{5}{2}$ :Thus, the total mass of plastic used is 30 g.

Let the given expression be $\displaystyle N$ . We can rewrite it using algebraic manipulation, which is equivalent to polynomial long division:For $\displaystyle N$ to be an integer, the term $\displaystyle \dfrac{30}{3x + 1}$ must be an integer. This requires that $\displaystyle 3x + 1$ is an integer factor of 30.

Furthermore, since $\displaystyle x$ must be an integer, $\displaystyle 3x$ is a multiple of 3. This means that $\displaystyle 3x + 1$ must be a number that is 1 more than a multiple of 3. In modular arithmetic terms, we require $\displaystyle 3x + 1 \equiv 1 \pmod 3$ .

The integer factors of 30 are $\displaystyle \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30$ .

We now check which of these factors satisfy the condition $\displaystyle k \equiv 1 \pmod 3$ :

- Positive factors: $\displaystyle 1$ and $\displaystyle 10$ satisfy the condition.
- Negative factors: $\displaystyle -2$ and $\displaystyle -5$ satisfy the condition (since $\displaystyle -2 = 3(-1) + 1$ and $\displaystyle -5 = 3(-2) + 1$ ).

The possible values for $\displaystyle 3x + 1$ are therefore $\displaystyle 1, 10, -2, -5$ .

Each of these leads to a valid integer value for $\displaystyle x$ :
- $\displaystyle 3x + 1 = 1 \implies 3x = 0 \implies x=0$ - $\displaystyle 3x + 1 = 10 \implies 3x = 9 \implies x=3$ - $\displaystyle 3x + 1 = -2 \implies 3x = -3 \implies x=-1$ - $\displaystyle 3x + 1 = -5 \implies 3x = -6 \implies x=-2$ There are 4 such integer values of $\displaystyle x$ .

We first determine the equations of the lines $\displaystyle L_1$ and $\displaystyle L_2$ .

The gradient of the segment $\displaystyle AB$ is $\displaystyle \dfrac{4-2}{5-1} = \dfrac{1}{2}$ . Since $\displaystyle L_1$ is perpendicular to $\displaystyle AB$ and passes through $\displaystyle A(1, 2)$ , its gradient is $\displaystyle -2$ . Its equation is:The line $\displaystyle x + 3y = 0$ has gradient $\displaystyle -\dfrac{1}{3}$ . Since $\displaystyle L_2$ is perpendicular to this and passes through $\displaystyle B(5, 4)$ , its gradient is $\displaystyle 3$ . Its equation is:We find the intersection point $\displaystyle C$ by equating the expressions for $\displaystyle y$ :Substituting $\displaystyle x=3$ into either equation gives $\displaystyle y = -2$ . Thus, $\displaystyle C$ is at $\displaystyle (3, -2)$ .

Since $\displaystyle L_1$ is perpendicular to $\displaystyle AB$ , the triangle $\displaystyle ABC$ is right-angled at $\displaystyle A$ . We calculate the side lengths:The area of triangle $\displaystyle ABC$ is:

Let $\displaystyle R$ be the radius of the circle in which both polygons are inscribed.

The area of a regular $\displaystyle n$ -sided polygon inscribed in a circle of radius $\displaystyle R$ can be found by dividing it into $\displaystyle n$ congruent isosceles triangles. Each triangle has two sides of length $\displaystyle R$ and the angle between them is $\displaystyle \dfrac{360^\circ}{n}$ . The area of one such triangle is $\displaystyle \dfrac{1}{2}R^2 \sin(\dfrac{360^\circ}{n})$ .

For the equilateral triangle ( $\displaystyle n=3$ ), the central angle is $\displaystyle \dfrac{360^\circ}{3} = 120^\circ$ . The total area is:We know that $\displaystyle \sin(120^\circ) = \sin(60^\circ) = \dfrac{\sqrt{3}}{2}$ .We are given that the area of the triangle is $\displaystyle 27\sqrt{3}$ .Dividing both sides by $\displaystyle 3\sqrt{3}$ gives:Now, we find the area of the regular 12-sided polygon (dodecagon, $\displaystyle n=12$ ). The central angle is $\displaystyle \dfrac{360^\circ}{12} = 30^\circ$ . The total area is:We know that $\displaystyle \sin(30^\circ) = \dfrac{1}{2}$ .Substituting the value of $\displaystyle R^2 = 36$ :