FSP Training 01

Preparation for Mock Exam 02

Instructions

This training page covers all of Section 01 and prepares you for Mock Exam 02. Set A trains setting up: you translate a text into mathematics and check your translation. Set B is a practice mock in the format of Mock Exam 02: 90 minutes, 50 points, and all 3 problems must be completed. Write down every step as you would in the exam. The points show what each part is worth.

In the real FSP, decimals are written with a comma (1,25 means 1.25).

Set A: Set Up & Check (40 minutes)

From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works. (QU = quantity units, MU = monetary units)

1. “A car dealer sold 480 cars last year, 120 of them electric cars. Companies bought 200 of the 480 cars, among them 90 electric cars. We look at the share of electric cars among the cars bought by companies.”

   Check: Is this share the same as the share of company purchases among the electric cars?

2. “A coffee machine costs 214.20 EUR including 19 % VAT (value added tax), i.e. its gross price is the net price plus 19 % of the net price. In a sale, the shop takes 10 % off the net price and then adds the VAT.”

   Check: Does the customer pay the same as with 10 % off the gross price?

3. “The diameter \(d\) of a screw may deviate from 8 mm by at most 0.5 %.”

   Check: Is a screw with a diameter of 8.05 mm accepted?

4. “The cooling system of a server room raises an alarm as soon as the room temperature \(T\) deviates from 21 °C by more than 1.5 °C.”

   Check: Does the system raise an alarm at 19 °C?

5. “A square advertising banner has an area of 8 m². A smaller version of the banner, also square, has half of this area.”

   Check: Is the side length of the smaller banner half of the old side length?

6. “A company with the profit \((x - 2)(14 - x)\) (in 100 EUR) at an output of \(x\) QU and a capacity (the maximum possible output) of 12 QU makes a profit.”

   Check: Does \(x = 13\) belong to this set?

7. “The number of customers who still use an old version of an app falls by 20 % per month. The half-life \(T\) is the time after which only half of these customers are left.”

   Check: After 3 months, are slightly more than half of these customers left?

8. “A language-learning app has 1,800 users today, and the number of users grows by 6 % per month. The team wants to know after how many months \(t\) the app reaches 9,000 users.”

   Check: Does the app reach 9,000 users within two years?

9. “A garage mixes 25 litres of windscreen washer fluid from water and a concentrate that contains 60 % alcohol. The fluid must contain at least 36 % alcohol.”

   Check: Are 12 litres of concentrate enough?

10. “A print shop makes T-shirts. Its fixed costs are 3.6 MU per month. Producing 1 QU costs another 0.9 MU, and the shop sells 1 QU for 1.5 MU.” (1 QU = 100 T-shirts, 1 MU = 1,000 EUR)

    Check: What is the profit in EUR if the shop makes and sells 700 T-shirts?

CautionSolution

1. With \(E\) = electric cars and \(F\) = cars bought by companies: the share of electric cars among the company purchases has the base \(F\):

   \[\frac{\text{number of cars in } E \cap F}{\text{number of cars in } F} = \frac{90}{200} = 45\,\%\]

   Check: no. The share of company purchases among the electric cars has the base \(E\): \(\frac{90}{120} = 75\,\%\).

   Typical mistake: \(\frac{90}{120}\) or \(\frac{90}{480}\). The group named after “among” is the base, here the 200 cars bought by companies.

2. The gross price is 119 % of the net price: \(214.20 = 1.19 \cdot \text{net price}\), so the net price is \(\frac{214.20}{1.19} = 180\) EUR. Sale price: \(180 \cdot 0.90 \cdot 1.19 = 192.78\) EUR.

   Check: yes, \(214.20 \cdot 0.90 = 192.78\) EUR. Both ways multiply the net price by the factors \(0.90\) and \(1.19\), and the order of the factors does not matter.

   Typical mistake: net price \(= 214.20 - 0.19 \cdot 214.20 \approx 173.50\) EUR. The 19 % refer to the net price, not to the gross price.

3. 0.5 % of 8 mm: \(0.005 \cdot 8 = 0.04\) mm. So \(|d - 8| \leq 0.04 \iff 7.96 \leq d \leq 8.04\) (in mm).

   Check: no, \(|8.05 - 8| = 0.05 > 0.04\).

   Typical mistake: \(|d - 8| \leq 0.5\). The tolerance is 0.5 % of 8 mm, not 0.5 mm. With this translation the screw with 8.05 mm would wrongly be accepted.

4. \(|T - 21| > 1.5 \iff T - 21 < -1.5\) or \(T - 21 > 1.5 \iff T < 19.5\) or \(T > 22.5\), i.e. \(T \in (-\infty; 19.5) \cup (22.5; \infty)\) (in °C).

   Check: yes, \(|19 - 21| = 2 > 1.5\).

   Typical mistake: \(19.5 < T < 22.5\). These are the temperatures without an alarm. “More than” gives two separate parts joined by “or”.

5. With \(s\) the old and \(t\) the new side length in m: \(s^2 = 8\), so \(s = \sqrt{8} = 2\sqrt{2} \approx 2.83\). \(t^2 = 4\), so \(t = 2\).

   Check: no. \(\frac{t}{s} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.707\): the new side is about 70.7 % of the old side.

   Typical mistake: halving the side, \(t = \sqrt{2} \approx 1.41\) m. Then the area is \((\sqrt{2})^2 = 2\) m², only a quarter of 8 m².

6. \((x - 2)(14 - x) > 0\) and \(0 \leq x \leq 12\). A product of two factors is positive if both factors are positive or both are negative. Both positive: \(x > 2\) and \(x < 14\). Both negative: \(x < 2\) and \(x > 14\) at the same time is impossible. With the capacity: \((2; 12] = \{x \in \mathbb{R} \mid 2 < x \leq 12\}\).

   Check: no. The term is positive at \(x = 13\), \((11)(1) = 11 > 0\), but 13 QU exceed the capacity of 12 QU.

   Typical mistake: \((2; 14)\). The profit term alone ignores the capacity. Outputs above 12 QU cannot be produced.

7. Each month, 80 % of the customers stay: after \(t\) months, the share left is \(0.8^t\). Half-life: \(0.8^T = 0.5 \iff T = \frac{\ln(0.5)}{\ln(0.8)} \approx 3.11\) months.

   Check: yes, \(0.8^3 = 0.512\), so 51.2 % are left after 3 months. This fits \(T \approx 3.11 > 3\).

   Typical mistake: \(T = \frac{\ln(0.5)}{\ln(0.2)} \approx 0.43\) months, using the rate 20 % instead of the factor \(1 - 0.20 = 0.8\). The check fails: \(0.8^{0.43} \approx 0.91\), so after 0.43 months about 91 % are still left.

8. \(1800 \cdot 1.06^t = 9000 \iff 1.06^t = 5 \iff t = \frac{\ln(5)}{\ln(1.06)} \approx 27.62\) months.

   Check: no, two years are 24 months, and \(1800 \cdot 1.06^{24} \approx 7{,}288\) users.

   Typical mistake: \((1800 \cdot 1.06)^t = 1908^t = 9000\), which gives \(t \approx 1.21\) months. Only the factor \(1.06\) has the exponent \(t\). Divide by 1,800 first.

9. With \(c\) the litres of concentrate: \(0 \leq c \leq 25\) and \(0.60c \geq 0.36 \cdot 25 = 9\), i.e. at least 9 litres of alcohol.

   Check: no, \(0.60 \cdot 12 = 7.2 < 9\) litres of alcohol (only 28.8 %). The condition needs \(c \geq 15\) litres.

   Typical mistake: \(c \geq 0.36 \cdot 25 = 9\). The concentrate is only 60 % alcohol, so 9 litres of concentrate contain only 5.4 litres of alcohol.

10. With \(x\) the output in QU: \(C = 0.9x + 3.6\), \(R = 1.5x\), \(P = R - C = 1.5x - (0.9x + 3.6) = 0.6x - 3.6\) (in MU).

    Check: 700 T-shirts are 7 QU: \(P = 0.6 \cdot 7 - 3.6 = 0.6\) MU \(= 600\) EUR.

    Typical mistake: inserting \(x = 700\), which gives \(416.4\) MU. Here \(x\) counts QU, and the result is in MU, not in EUR.

One Paragraph, All Conditions (xx)

A kayak rental on a lake plans its July. Its fixed costs are 2,100 EUR for the month. They arise even if no kayak is rented. Each rental causes variable costs (they grow with the number of rentals) of 6 EUR for cleaning and wear. A customer pays 23.80 EUR per rental, including 19 % VAT. The rental keeps only the net price. With its boats, it can handle at most 600 rentals in July. The owner expects 450 rentals, but the actual number may deviate from this by at most 20 %.

List every condition in this paragraph and the terms for the total costs, the revenue and the profit, with \(x\) the number of rentals in July. Do not solve anything.

Check: Which of your conditions rules out 550 rentals, and which one does 550 rentals still meet?

CautionSolution

1. Fixed costs: \(2100\) EUR
2. Variable costs: \(6x\) EUR
3. Total costs: \(C = 6x + 2100\) EUR
4. Net price: \(\frac{23.80}{1.19} = 20\) EUR per rental
5. Revenue: \(R = 20x\) EUR
6. Profit: \(P = R - C = 20x - (6x + 2100) = 14x - 2100\) EUR
7. Capacity: \(x \leq 600\)
8. Expected number: \(|x - 450| \leq 0.20 \cdot 450 = 90\), i.e. \(360 \leq x \leq 540\)

Together: \(x \in [360; 540]\).

Check: 550 rentals violate condition 8, since \(|550 - 450| = 100 > 90\). They still meet the capacity (condition 7): \(550 \leq 600\).

Typical mistake: \(R = 23.80x\). The VAT goes to the tax office. The rental’s revenue is the net price times the number of rentals.

FSP Words So Far

The FSP problems use these words from Section 01. Learn what each one means and how to write it in mathematics.

Word	Meaning	Mathematics
quantity unit (QU)	a fixed number of items, e.g. 1 QU = 100 items	the output \(x\) is counted in QU
monetary unit (MU)	a fixed amount of money, e.g. 1 MU = 100 EUR	results in MU, convert to EUR at the end
fixed costs	costs that arise even if nothing is produced	the constant term of \(C(x)\)
variable costs	costs that grow with the output	output times costs per QU
total costs	fixed costs + variable costs	\(C(x)\)
revenue	quantity times price	\(R(x) = x \cdot \text{price}\)
profit	revenue − total costs	\(P(x) = R(x) - C(x)\)
break-even point	the output at which the profit changes from negative (a loss) to positive	\(P(x) = 0\), the smaller zero
profit limit (upper break-even point)	the larger output with zero profit: above it, the profit turns negative again	\(P(x) = 0\), the larger zero
profit zone	all outputs with a positive profit	\(P(x) > 0\): from the break-even point to the profit limit (or the capacity)
capacity	the maximum possible output	an upper bound: \(x \leq\) capacity
growth factor	the factor by which a quantity is multiplied each period	\(q\) in (starting value) \(\cdot\, q^t\)
growth rate	the change per period in percent	\(q = 1 + r\) with \(r\) as a decimal, a loss of 4 % gives \(q = 0.96\)
doubling time	the time after which a quantity has doubled	\(q^T = 2\)
half-life (halving time)	the time after which a quantity has halved	\(q^T = 0.5\)
“deviates from \(a\) by at most \(d\)” (tolerance)	the distance from \(a\) is at most \(d\)	\(\lvert x - a \rvert \leq d \iff a - d \leq x \leq a + d\)
“share of \(A\) among \(B\)”	the part of \(B\) that also belongs to \(A\): the base is \(B\)	\(\dfrac{\text{number in } A \cap B}{\text{number in } B}\)
“neither \(A\) nor \(B\)”	not in \(A\) and not in \(B\), i.e. outside both circles	\(\overline{A \cup B}\)
contingency table	a table of counts for two properties: the inner fields are the regions of the Venn diagram, the margins are the totals	inner fields \(A \cap B\), \(A \cap \overline{B}\), \(\overline{A} \cap B\), \(\overline{A} \cap \overline{B}\)
gross price, net price	gross price = net price + VAT	with 19 % VAT: gross \(= 1.19 \cdot\) net

In set-builder notation, \(\{x \in \mathbb{R} \mid \dots\}\) and \(\{x \in \mathbb{R} : \dots\}\) mean the same.

Set B: Practice Mock (90 minutes, 50 points)

Problem 1: Designer Chairs [20 pts. total]

A furniture workshop produces designer chairs. Its fixed costs amount to 75 monetary units (MU). If it produces \(x\) quantity units (QU), the variable costs amount to \(x^2 - 6x + 20\) MU per QU. To sell all \(x\) QU, the workshop can only charge a price of \(45 - 3x\) MU per QU.

(1 QU = 100 chairs, 1 MU = 1,000 EUR)

1. Give terms for the total costs \(C(x)\) and the revenue \(R(x)\).

(for verification: \(C(x) = x^3 - 6x^2 + 20x + 75\), \(\; R(x) = -3x^2 + 45x\))

[5 pts.]

2. Show that the profit function is given by \(P(x) = -x^3 + 3x^2 + 25x - 75\).

[3 pts.]

3. Factor \(P(x)\) completely by grouping. Determine all outputs at which the profit is exactly zero.

Note: a product is zero if one of its factors is zero. The smaller of these outputs is called the (lower) break-even point, the larger one the profit limit (upper break-even point).

(for verification: \(P(x) = (x - 3)(5 - x)(5 + x)\))

[5 pts.]

4. Calculate \(P(2)\), \(P(4)\) and \(P(6)\). Decide in which output range the workshop makes a profit and state it as an interval and in set-builder notation.

(for verification: the workshop makes a profit exactly for outputs between 3 QU and 5 QU)

[3 pts.]

5. Calculate the price per chair in EUR and the profit in EUR for an output of 450 chairs.

[2 pts.]

6. The workshop plans an output of 4.5 QU, but the actual output deviates by up to 0.6 QU. Formulate this as an absolute value inequality, solve it and decide whether a profit is guaranteed. Determine up to which deviation a profit is still guaranteed.

[2 pts.]

CautionSolutions

a) Total costs are variable costs plus fixed costs. The variable costs are the costs per QU times the number of QU:

\[C(x) = x \cdot (x^2 - 6x + 20) + 75 = x^3 - 6x^2 + 20x + 75 \quad \checkmark\]

Revenue is quantity times price:

\[R(x) = x \cdot (45 - 3x) = -3x^2 + 45x \quad \checkmark\]

Grading: 3 Pts. \(C(x)\) (1 for quantity \(\times\) cost per QU, 1 for the fixed costs, 1 for multiplying out), 2 Pts. \(R(x)\) (1 for quantity \(\times\) price, 1 for multiplying out).

b) Profit is revenue minus costs. The minus sign changes every sign in the bracket:

\[P(x) = (-3x^2 + 45x) - (x^3 - 6x^2 + 20x + 75) = -3x^2 + 45x - x^3 + 6x^2 - 20x - 75\]

Collect like terms:

\[P(x) = -x^3 + (-3 + 6)x^2 + (45 - 20)x - 75 = -x^3 + 3x^2 + 25x - 75 \quad \checkmark\]

Grading: 1 Pt. ansatz \(P = R - C\), 1 Pt. sign change in the bracket, 1 Pt. collecting like terms.

c) Group the first two and the last two terms:

\[P(x) = (-x^3 + 3x^2) + (25x - 75) = -x^2(x - 3) + 25(x - 3) = (x - 3)(25 - x^2)\]

\(25 - x^2\) is a difference of squares:

\[P(x) = (x - 3)(5 - x)(5 + x)\]

Zero-product rule: \(x = 3\), \(x = 5\) or \(x = -5\). The value \(x = -5\) is rejected, since the output cannot be negative.

Check: \(P(3) = -27 + 27 + 75 - 75 = 0 \; \checkmark\)

Break-even point at \(x = 3\) QU (300 chairs), profit limit at \(x = 5\) QU (500 chairs).

Grading: 2 Pts. grouping with the common factor \((x - 3)\), 1 Pt. difference of squares, 1 Pt. zero-product rule, 1 Pt. rejecting \(x = -5\) and naming both points.

d) One output from each range between the zeros:

\[P(2) = (2-3)(5-2)(5+2) = -21 < 0, \quad P(4) = (1)(1)(9) = 9 > 0, \quad P(6) = (3)(-1)(11) = -33 < 0\]

The profit is zero only at \(x = 3\) and \(x = 5\), so its sign cannot change in between.

Profit zone: \((3; 5) = \{x \in \mathbb{R} \mid 3 < x < 5\}\). The interval is open, because the profit is exactly zero at \(x = 3\) and \(x = 5\).

Grading: 1 Pt. three values, 1 Pt. decision with reference to the zeros from c), 1 Pt. open interval in both notations.

e) 450 chairs \(= 4.5\) QU.

Price: \(45 - 3 \cdot 4.5 = 45 - 13.5 = 31.5\) MU per QU \(= 31{,}500\) EUR per 100 chairs, i.e. 315 EUR per chair.

Profit: \(P(4.5) = (1.5)(0.5)(9.5) = 7.125\) MU \(=\) 7,125 EUR.

Grading: 1 Pt. price per chair with unit conversion, 1 Pt. profit in EUR.

f) “Deviates by up to 0.6” means the distance from \(4.5\) is at most \(0.6\):

\[|x - 4.5| \leq 0.6 \iff -0.6 \leq x - 4.5 \leq 0.6 \iff 3.9 \leq x \leq 5.1\]

Since \(5.1 > 5\), the interval \([3.9; 5.1]\) is not a subset of the profit zone \((3; 5)\). A profit is not guaranteed. For example, \(P(5.1) \approx -2.12 < 0\).

The planned output \(4.5\) is \(4.5 - 3 = 1.5\) away from the break-even point and \(5 - 4.5 = 0.5\) away from the profit limit. The smaller distance limits the deviation: a profit is guaranteed for every deviation smaller than 0.5 QU (50 chairs).

Grading: 1 Pt. inequality and interval, 1 Pt. decision with reason and the deviation limit of 0.5 QU.

Problem 2: Company Vans [16 pts. total]

A delivery company buys a van for 48,000 EUR (\(t = 0\), \(t\) in years). Two years later, the van is worth 34,680 EUR. The van loses the same percentage of its value every year.

1. Determine the yearly growth factor \(q\) and the percentage by which the value decreases per year. Give the function \(V(t)\) for the value of the van.

(for verification: \(V(t) = 48000 \cdot 0.85^t\))

[3 pts.]

2. Compute the value after ten years. Give the result in scientific notation with three significant digits (i.e. round the leading number to two decimal places, e.g. \(1.23 \cdot 10^{5}\)).

[2 pts.]

3. Determine the time after which the van has lost half of its value, exactly as a logarithm and rounded to two decimal places.

[3 pts.]

4. A second van costs 60,000 EUR at \(t = 0\) and loses 20 % of its value every year. Give the function \(W(t)\) for its value. Show that both vans have the same value at

\[t = \frac{\ln 5 - \ln 4}{\ln 17 - \ln 16}\]

and determine in which year this happens.

[5 pts.]

5. The accountant prefers the model \(V(t) = 48000 \cdot e^{-kt}\). Determine \(k\) exactly so that both models describe the same loss of value. Explain why \(k\) is not equal to \(0.15\).

[3 pts.]

CautionSolutions

a) Ansatz \(V(t) = V_0 \cdot q^t\) with \(V_0 = 48000\):

\[48000 \cdot q^2 = 34680 \implies q^2 = \frac{34680}{48000} = 0.7225 \implies q = \sqrt{0.7225} = 0.85\]

Only the positive root makes sense for a growth factor. A factor of \(0.85 = 1 - 0.15\) means that the value decreases by \(15\,\%\) per year:

\[V(t) = 48000 \cdot 0.85^t\]

Grading: 1 Pt. ansatz \(48000 \cdot q^2 = 34680\), 1 Pt. square root, 1 Pt. rate and function.

b) \[V(10) = 48000 \cdot 0.85^{10} \approx 48000 \cdot 0.19687 \approx 9449.97\]

\[V(10) \approx 9.45 \cdot 10^3 \text{ EUR}\]

Grading: 1 Pt. calculation, 1 Pt. scientific notation.

c) Half of the value means \(V(t) = 24000\):

\[48000 \cdot 0.85^t = 24000 \implies 0.85^t = 0.5\]

Take the logarithm of both sides and use the power law \(\ln(a^t) = t \cdot \ln(a)\):

\[t \cdot \ln(0.85) = \ln(0.5) \implies t = \frac{\ln(0.5)}{\ln(0.85)} = \log_{0.85}(0.5) \approx 4.27 \text{ years}\]

Both logarithms are negative, so \(t\) is positive \(\checkmark\)

Grading: 1 Pt. equation \(0.85^t = 0.5\), 1 Pt. logarithm law, 1 Pt. exact and rounded result.

d) A loss of 20 % means a factor of \(1 - 0.20 = 0.8\):

\[W(t) = 60000 \cdot 0.8^t\]

Set both functions equal:

\[60000 \cdot 0.8^t = 48000 \cdot 0.85^t\]

Divide by \(48000\) and by \(0.8^t\) (always positive):

\[\frac{60000}{48000} = \frac{0.85^t}{0.8^t} \implies \frac{5}{4} = \left(\frac{0.85}{0.8}\right)^t = \left(\frac{17}{16}\right)^t\]

Take the logarithm and use the quotient law \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\):

\[\ln 5 - \ln 4 = t \cdot (\ln 17 - \ln 16) \implies t = \frac{\ln 5 - \ln 4}{\ln 17 - \ln 16} \quad \checkmark\]

\(t \approx 3.68\), so both vans have the same value (about 26,391 EUR) during the fourth year. After that, the second van is worth less.

Grading: 1 Pt. \(W(t)\), 1 Pt. equation, 1 Pt. power-quotient rule \(\frac{a^t}{b^t} = \left(\frac{a}{b}\right)^t\), 1 Pt. logarithm and quotient law, 1 Pt. year.

e) Both models agree if \(e^{-kt} = 0.85^t\) for all \(t\), i.e. \(e^{-k} = 0.85\):

\[-k = \ln(0.85) \implies k = -\ln(0.85) = \ln\left(\frac{20}{17}\right) \approx 0.1625\]

Explanation: \(15\,\%\) is the loss per year with yearly compounding. With base \(e\), the loss of value acts continuously, so the loss compounds within the year. With \(k = 0.15\) we would get \(e^{-0.15} \approx 0.8607\), i.e. a loss of only about \(13.9\,\%\) per year. To reach the factor \(0.85\), a larger continuous rate is necessary.

Grading: 1 Pt. \(e^{-k} = 0.85\), 1 Pt. \(k = -\ln(0.85)\), 1 Pt. explanation.

Problem 3: Commuter Survey and Cleaning Agent [14 pts. total]

A logistics company with 1,200 employees asks how they get to work. 45 % of all employees use a company bike (\(B\)), 660 employees use a public transport ticket (\(T\)), and 300 employees use neither.

1. Draw a Venn diagram and complete the following table.

	ticket (\(T\))	no ticket (\(\bar{T}\))	sum
bike (\(B\))
no bike (\(\bar{B}\))
sum			1200

(for verification: 300 employees use both offers)

[5 pts.]

2. Argue whether the following statements are true or false.

1. “Every employee with a ticket also uses a company bike.” (\(x \in T \Rightarrow x \in B\))
2. \(\bar{B} \cap \bar{T} = \overline{B \cup T}\)
3. More than half of all employees use exactly one of the two offers.

[3 pts.]

3. Compute the percentage of ticket holders who also use a company bike and the percentage of company bike users who do not have a ticket.

[2 pts.]

4. For 60 litres of cleaning agent, the company mixes a 15 % concentrate (2 EUR per litre) with a 35 % concentrate (5 EUR per litre). The mixture must contain at least 20 % active ingredient, and the budget is at most 210 EUR. Determine all possible amounts of the 15 % concentrate.

[4 pts.]

CautionSolutions

a) Given: \(B\) contains \(0.45 \cdot 1200 = 540\) employees, \(T\) contains 660, and \(\overline{B \cup T}\) (neither offer) contains 300.

At least one offer: \(B \cup T\) contains \(1200 - 300 = 900\) employees. The sum \(540 + 660\) counts the employees in \(B \cap T\) twice, so

\[\text{number in } B \cap T = 540 + 660 - 900 = 300\]

Venn regions: only \(B\): \(540 - 300 = 240\), both: \(300\), only \(T\): \(660 - 300 = 360\), outside: \(300\).

	\(T\)	\(\bar{T}\)	sum
\(B\)	300	240	540
\(\bar{B}\)	360	300	660
sum	660	540	1200

Check: \(300 + 240 + 360 + 300 = 1200 \; \checkmark\)

Grading: 1 Pt. \(B\) with 540 employees, 1 Pt. \(B \cup T\) with 900 employees, 1 Pt. \(B \cap T\) with 300 employees, 1 Pt. Venn diagram, 1 Pt. complete table.

b)

1. False. The statement would mean \(T \subseteq B\), but 360 employees have a ticket and no bike.
2. True. “Not in \(B\) and not in \(T\)” is the same as “not in \(B \cup T\)”: both sets are the region outside both circles and contain the 300 employees who use neither.
3. False. Exactly one offer: \(240 + 360 = 600\), which is exactly half of 1200, not more than half.

Grading: 1 Pt. per statement (decision and reason).

c) The base is the group named after “of”:

\[\frac{300}{660} \approx 45.5\,\% \;\; (\text{base } T) \qquad \frac{240}{540} \approx 44.4\,\% \;\; (\text{base } B)\]

Grading: 1 Pt. each (correct base).

d) Variables: \(x\) = litres of the 15 % concentrate, \(60 - x\) = litres of the 35 % concentrate, \(0 \leq x \leq 60\).

Concentration: active ingredient \(\geq 20\,\%\) of 60 litres:

\[0.15x + 0.35(60 - x) \geq 12 \implies -0.2x + 21 \geq 12 \implies -0.2x \geq -9 \implies x \leq 45\]

Budget:

\[2x + 5(60 - x) \leq 210 \implies -3x + 300 \leq 210 \implies -3x \leq -90 \implies x \geq 30\]

Both times we divide by a negative number, so the inequality sign reverses. Both conditions must hold, so we intersect the solution sets:

\[x \in [30; 45]\]

The company must use between 30 and 45 litres of the 15 % concentrate (and between 15 and 30 litres of the 35 % concentrate).

Check: \(x = 30\): cost \(60 + 150 = 210\) EUR \(\checkmark\). \(x = 45\): \(\frac{6.75 + 5.25}{60} = 20\,\%\) \(\checkmark\)

Grading: 1 Pt. variables, 1 Pt. concentration inequality, 1 Pt. budget inequality with sign reversal, 1 Pt. intersection \([30; 45]\).

Exam Preparation Checklist

Before the exam, ensure you can:

• Work with intervals, set operations and Venn diagrams
• Decide whether statements and implications are true
• Compute percentages with the correct base and chain growth factors
• Apply exponent laws and scientific notation
• Factor with common factors, grouping, the AC method, substitution and special products
• Simplify radicals and rationalize denominators
• Apply logarithm laws, the change of base formula and \(\ln\)
• Solve exponential equations with logarithms and convert to continuous growth \(e^{kt}\)
• Solve absolute value equations and inequalities
• Translate word problems into linear equations and inequalities (IDEA)
• Set up cost, revenue and profit functions and find break-even points

Remember: Show all work clearly, label units, and always interpret your results in the business context!