Tasks 01-03 - Core Algebra & Exponents

Building Your Algebraic Foundation

A: Skills

Problem 1: Order of Operations (x)

Compute the following expressions without a calculator. Show each step.

  1. \(5 + 3^2 \times 4 - 18 \div 3\)

  2. \(2 \times [6 + 3 \times (4^2 - 14)]\)

  3. \(\frac{3^3 - 2 \times 4}{5 - 2}\)

  4. \((2 + 3)^3 - 5 \times (12 - 2 \times 4)\)

  5. \(\frac{4^2 + 2 \times 3}{2} - 3 \times (5 - 3)^2\)

  1. \(5 + 3^2 \times 4 - 18 \div 3\)
    • \(= 5 + 9 \times 4 - 18 \div 3\)
    • \(= 5 + 36 - 6\)
    • \(= 35\)
  2. \(2 \times [6 + 3 \times (4^2 - 14)]\)
    • \(= 2 \times [6 + 3 \times (16 - 14)]\)
    • \(= 2 \times [6 + 3 \times 2]\)
    • \(= 2 \times [6 + 6]\)
    • \(= 2 \times 12 = 24\)
  3. \(\frac{3^3 - 2 \times 4}{5 - 2}\)
    • \(= \frac{27 - 8}{3}\)
    • \(= \frac{19}{3}\)
  4. \((2 + 3)^3 - 5 \times (12 - 2 \times 4)\)
    • \(= 5^3 - 5 \times (12 - 8)\)
    • \(= 125 - 5 \times 4\)
    • \(= 125 - 20 = 105\)
  5. \(\frac{4^2 + 2 \times 3}{2} - 3 \times (5 - 3)^2\)
    • \(= \frac{16 + 6}{2} - 3 \times 2^2\)
    • \(= \frac{22}{2} - 3 \times 4\)
    • \(= 11 - 12 = -1\)

Problem 2: Exponent Laws (x)

Simplify the following expressions completely and write the results with positive exponents only.

  1. \((x^3)^2 \cdot x^{-4} \cdot x\)

  2. \(\frac{(2a^3b^2)^3}{8a^5b^4}\)

  3. \(\left(\frac{3x^2y}{z^2}\right)^2 \cdot \frac{z^3}{9xy^2}\)

  4. \((5^3)^2 \cdot 5^{-4} \div 5^2\)

  5. \(\frac{(m^2n^3)^2 \cdot m^{-3}n}{m^{-1}n^4}\)

  6. \(\left(\frac{2x^{-2}y^3}{z^{-1}}\right)^{-2}\)

  1. \((x^3)^2 \cdot x^{-4} \cdot x\)
    • \(= x^6 \cdot x^{-4} \cdot x^1\)
    • \(= x^{6-4+1} = x^3\)
  2. \(\frac{(2a^3b^2)^3}{8a^5b^4}\)
    • \(= \frac{8a^9b^6}{8a^5b^4}\)
    • \(= a^{9-5}b^{6-4}\)
    • \(= a^4b^2\)
  3. \(\left(\frac{3x^2y}{z^2}\right)^2 \cdot \frac{z^3}{9xy^2}\)
    • \(= \frac{9x^4y^2}{z^4} \cdot \frac{z^3}{9xy^2}\)
    • \(= \frac{9x^4y^2 \cdot z^3}{z^4 \cdot 9xy^2}\)
    • \(= \frac{x^3}{z}\)
  4. \((5^3)^2 \cdot 5^{-4} \div 5^2\)
    • \(= 5^6 \cdot 5^{-4} \cdot 5^{-2}\)
    • \(= 5^{6-4-2} = 5^0 = 1\)
  5. \(\frac{(m^2n^3)^2 \cdot m^{-3}n}{m^{-1}n^4}\)
    • \(= \frac{m^4n^6 \cdot m^{-3}n}{m^{-1}n^4}\)
    • \(= \frac{m^1n^7}{m^{-1}n^4}\)
    • \(= m^{1-(-1)}n^{7-4} = m^2n^3\)
  6. \(\left(\frac{2x^{-2}y^3}{z^{-1}}\right)^{-2}\)
    • \(= \left(\frac{2y^3z}{x^2}\right)^{-2}\)
    • \(= \frac{x^4}{4y^6z^2}\)

Problem 3: Scientific Notation (x)

Part A: Conversions

Give in scientific notation:

  1. 456,000,000
  2. 0.00000789
  3. \(23.45 \times 10^5\)

Give in standard form:

  1. \(3.67 \times 10^8\)
  2. \(9.2 \times 10^{-5}\)

Part B: Calculations

Compute and give the results in scientific notation:

  1. \((2.5 \times 10^6) \times (3.2 \times 10^{-3})\)

  2. \(\frac{9.6 \times 10^8}{3.2 \times 10^5}\)

  3. \((4.0 \times 10^4)^2\)

  4. \((3.6 \times 10^9) + (2.4 \times 10^8)\)

Hint for i): first write both numbers with the same power of ten.

Part A:

  1. \(4.56 \times 10^8\)
  2. \(7.89 \times 10^{-6}\)
  3. \(2.345 \times 10^6\)
  4. 367,000,000
  5. 0.000092

Part B:

  1. \((2.5 \times 10^6) \times (3.2 \times 10^{-3})\)
    • \(= 8.0 \times 10^3\)
  2. \(\frac{9.6 \times 10^8}{3.2 \times 10^5}\)
    • \(= 3.0 \times 10^3\)
  3. \((4.0 \times 10^4)^2\)
    • \(= 16 \times 10^8 = 1.6 \times 10^9\)
  4. \((3.6 \times 10^9) + (2.4 \times 10^8)\)
    • \(= (36 \times 10^8) + (2.4 \times 10^8)\)
    • \(= 38.4 \times 10^8 = 3.84 \times 10^9\)

Problem 4: Absolute Value (xx)

Solve the following equations and inequalities and give the solution sets in set or interval notation.

  1. \(|3x - 6| = 9\)

  2. \(|x - 4| < 3\)

  3. \(|2x + 8| \geq 6\)

  4. \(3|x - 2| - 4 = 5\)

  5. \(|4 - 2x| \leq 6\)

  1. \(|3x - 6| = 9\)
    • Case 1: \(3x - 6 = 9 \Rightarrow x = 5\)
    • Case 2: \(3x - 6 = -9 \Rightarrow x = -1\)
    • Solution: \(x \in \{-1, 5\}\)
  2. \(|x - 4| < 3\)
    • \(-3 < x - 4 < 3\)
    • \(1 < x < 7\)
    • Solution: \((1; 7)\)
  3. \(|2x + 8| \geq 6\)
    • \(2x + 8 \leq -6\) OR \(2x + 8 \geq 6\)
    • \(x \leq -7\) OR \(x \geq -1\)
    • Solution: \((-\infty; -7] \cup [-1; \infty)\)
  4. \(3|x - 2| - 4 = 5\)
    • \(3|x - 2| = 9\)
    • \(|x - 2| = 3\)
    • \(x - 2 = 3\) OR \(x - 2 = -3\)
    • Solution: \(x \in \{-1, 5\}\)
  5. \(|4 - 2x| \leq 6\)
    • \(-6 \leq 4 - 2x \leq 6\)
    • Subtract 4 from all parts: \(-10 \leq -2x \leq 2\)
    • Divide all parts by \(-2\), both signs flip: \(5 \geq x \geq -1\)
    • Solution: \([-1; 5]\)

Problem 5: Basic Factorization (xx)

Factor completely:

  1. \(15x^3 - 10x^2 + 5x\)

  2. \(x^2 - 144\)

  3. \(49x^2 - 64y^2\)

  4. \(x^2 + 16x + 64\)

  5. \(9x^2 - 12x + 4\)

  6. \(8x^3 - 32x\)

  7. \(3x^2 - 75\)

  8. \(x^2 - 18x + 81\)

  1. \(15x^3 - 10x^2 + 5x\)
    • \(= 5x(3x^2 - 2x + 1)\)
  2. \(x^2 - 144\)
    • \(= (x + 12)(x - 12)\)
  3. \(49x^2 - 64y^2\)
    • \(= (7x + 8y)(7x - 8y)\)
  4. \(x^2 + 16x + 64\)
    • \(= (x + 8)^2\)
  5. \(9x^2 - 12x + 4\)
    • \(= (3x - 2)^2\)
  6. \(8x^3 - 32x\)
    • \(= 8x(x^2 - 4)\)
    • \(= 8x(x + 2)(x - 2)\)
  7. \(3x^2 - 75\)
    • \(= 3(x^2 - 25)\)
    • \(= 3(x + 5)(x - 5)\)
  8. \(x^2 - 18x + 81\)
    • \(= (x - 9)^2\)

B: Set Up & Check

Problem 6: 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)

  1. “The filling weight of a 500 g bag of rice may deviate from 500 g by at most 1.5 g.”

    Check: Is a bag with 498.6 g accepted?

  2. “A price is first raised by 8 % and then lowered by 8 %.”

    Check: Is the final price equal to the original price?

  3. “The profit is positive for every output above 3 QU, up to the capacity (the maximum possible output) of 10 QU.”

    Check: Does \(x = 3\) belong to this set?

  4. “Of 400 customers, 35 % shop online. 60 customers shop both online and in the store.”

    Check: What share of the online shoppers also shop in the store?

  5. “The number of users of a banking app grows by 3 % per month.”

    Check: Is the growth over two months exactly 6 %?

  1. With \(w\) the weight in g: \(|w - 500| \leq 1.5 \iff 498.5 \leq w \leq 501.5\).

    Check: \(|498.6 - 500| = 1.4 \leq 1.5\), so the bag is accepted.

    Typical mistake: \(|w - 500| < 1.5\). “At most” includes the boundary, so the sign is \(\leq\).

  2. With \(p\) the original price: \(p \cdot 1.08 \cdot 0.92 = 0.9936\,p\).

    Check: no, the final price is 0.64 % below the original price.

    Typical mistake: “+8 % − 8 % = 0 %”. Percentage changes multiply as growth factors (the factors by which a value is multiplied). They do not add.

  3. \((3; 10] = \{x \in \mathbb{R} \mid 3 < x \leq 10\}\)

    Check: no, “above 3” excludes 3 (round bracket), while “up to 10” includes 10 (square bracket).

  4. Online shoppers: \(0.35 \cdot 400 = 140\) customers. Online and in the store: \(60\) customers.

    Check: \(\frac{60}{140} \approx 42.9\,\%\).

    Typical mistake: \(\frac{60}{400} = 15\,\%\). “Of the online shoppers” makes the 140 online shoppers the base, not all 400 customers.

  5. With \(N_0\) users today, after \(t\) months: \(N_0 \cdot 1.03^t\).

    Check: no, \(1.03^2 = 1.0609\), so the growth over two months is 6.09 %.

    Typical mistake: \(2 \cdot 3\,\% = 6\,\%\). Each month’s growth also applies to the growth of the month before.

Problem 7: One Paragraph, All Conditions (xx)

A print shop has fixed costs of 450 EUR per day, which arise even if nothing is printed. Printing 100 flyers (1 QU) costs another 12 EUR. The shop sells flyers at 30 EUR per QU. Its machine prints at most 60 QU per day, and because of a standing order it always prints at least 5 QU.

List every condition in this paragraph as a mathematical statement, with \(x\) the daily output in QU. Do not solve anything.

Check: Which of your conditions rules out an output of 4 QU?

  1. Fixed costs: \(450\) EUR
  2. Variable costs (they grow with the output): \(12x\) EUR
  3. Total costs: \(12x + 450\) EUR
  4. Revenue: \(30x\) EUR
  5. Capacity: \(x \leq 60\)
  6. Standing order: \(x \geq 5\)

Together: \(x \in [5; 60]\).

Check: \(x = 4\) violates condition 6 (the standing order).

C: Applied

Problem 8: Quality Control at a Juice Bottler (xx)

A bottler fills 1-litre bottles. A bottle passes the check if its content deviates from 1000 ml by at most 4 ml.

  1. Formulate the condition as an absolute value inequality and give the accepted contents as an interval.

(for verification: \([996; 1004]\))

  1. A sample contains bottles with 995 ml, 1003 ml, 1004.5 ml and 998 ml. Decide for each bottle whether it passes.

  2. After a recalibration, the content deviates from 1000 ml by at most 0.25 %. Determine the new interval and explain whether the check has become stricter.

a) With \(c\) the content in ml:

\[|c - 1000| \leq 4 \iff -4 \leq c - 1000 \leq 4 \iff 996 \leq c \leq 1004\]

Accepted contents: \([996; 1004]\) ml.

b)

  • 995 ml: \(|995 - 1000| = 5 > 4\), fails
  • 1003 ml: \(|1003 - 1000| = 3 \leq 4\), passes
  • 1004.5 ml: \(|1004.5 - 1000| = 4.5 > 4\), fails
  • 998 ml: \(|998 - 1000| = 2 \leq 4\), passes

c) 0.25 % of 1000 ml: \(0.0025 \cdot 1000 = 2.5\) ml.

\[|c - 1000| \leq 2.5 \iff 997.5 \leq c \leq 1002.5\]

New interval: \([997.5; 1002.5]\) ml. The check has become stricter: the new interval lies inside the old one and is 5 ml wide instead of 8 ml.

Problem 9: Cloud Storage Costs (xxx)

A cloud provider stores \(3.2 \cdot 10^{15}\) bytes of customer data. An average user stores \(4 \cdot 10^{9}\) bytes. Storage costs \(2.5 \cdot 10^{-11}\) EUR per byte and month.

  1. Compute the number of users.

(for verification: \(8 \cdot 10^5\))

  1. Compute the monthly storage costs in total and per user.

  2. The stored data grows by 5 % per month. Determine the stored volume after one year and give it in scientific notation with three significant digits (i.e. round the leading number to two decimal places, e.g. \(4.26 \cdot 10^{12}\)).

Hint: Without a calculator, use \(1.05^{12} \approx 1.7959\).

a)

\[\frac{3.2 \cdot 10^{15}}{4 \cdot 10^{9}} = 0.8 \cdot 10^{6} = 8 \cdot 10^{5} \text{ users}\]

b) Total:

\[3.2 \cdot 10^{15} \cdot 2.5 \cdot 10^{-11} = 8.0 \cdot 10^{4} \text{ EUR} = 80{,}000 \text{ EUR per month}\]

Per user:

\[\frac{8 \cdot 10^{4}}{8 \cdot 10^{5}} = 0.1 \text{ EUR} = 10 \text{ cents per month}\]

c) Growth factor 1.05 per month, 12 months:

\[3.2 \cdot 10^{15} \cdot 1.05^{12} \approx 3.2 \cdot 10^{15} \cdot 1.7959 \approx 5.75 \cdot 10^{15} \text{ bytes}\]

Problem 10: Profit of a Car Wash (xxx)

A car wash models its daily profit (in 100 EUR) at an output of \(x\) QU (1 QU = 10 washes), with \(0 \leq x \leq 10\), by the expression \(9 - (x - 5)^2\).

  1. Compute the profit for \(x = 2\), \(x = 5\) and \(x = 9\). Interpret the result for \(x = 9\).

  2. Show with a difference of squares that the profit can be written as \((8 - x)(x - 2)\). Hint: \(9 = 3^2\). Treat \((x - 5)\) as one term.

  3. Explain why the profit can never exceed 900 EUR, and give the output with the highest profit.

  4. Determine all outputs with a positive profit.

Note: a product of two factors is positive if both factors are positive or both are negative.

a)

  • \(x = 2\): \(9 - (2 - 5)^2 = 9 - 9 = 0\)
  • \(x = 5\): \(9 - 0 = 9\), i.e. 900 EUR
  • \(x = 9\): \(9 - (9 - 5)^2 = 9 - 16 = -7\)

At 9 QU the car wash makes a loss of 700 EUR per day.

b) \(9 = 3^2\), so the difference of squares \(a^2 - b^2 = (a - b)(a + b)\) with \(a = 3\) and \(b = x - 5\) gives

\[9 - (x - 5)^2 = \big(3 - (x - 5)\big)\big(3 + (x - 5)\big) = (8 - x)(x - 2) \quad \checkmark\]

c) A square is never negative: \((x - 5)^2 \geq 0\). Hence \(9 - (x - 5)^2 \leq 9\), so the profit is at most 900 EUR. It equals 9 only if \((x - 5)^2 = 0\), i.e. at \(x = 5\) QU (50 washes).

d) Both factors positive: \(8 - x > 0\) and \(x - 2 > 0\), so \(2 < x < 8\). Both negative: \(x > 8\) and \(x < 2\) at the same time is impossible.

Positive profit for \(x \in (2; 8)\).

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Sourdough Bakery [17 pts. total]

A bakery bakes sourdough loaves. Its fixed costs, which arise even when no bread is baked, amount to 600 EUR per day. Each loaf causes variable costs of 1.10 EUR for flour, energy and packaging. The bakery sells each loaf for 3.50 EUR. The oven can bake at most 800 loaves per day.

(1 QU = 100 loaves, total costs = fixed costs + variable costs, profit = revenue − total costs)

Part a)

Give terms for the total costs \(C\) and the revenue \(R\) per day if the bakery bakes and sells \(x\) QU. [3 pts.]

(for verification: \(C = 110x + 600\), \(\; R = 350x\))

Convert to QU. Variable costs per QU: \(100 \cdot 1.10 = 110\) EUR. Price per QU: \(100 \cdot 3.50 = 350\) EUR.

Total costs are variable costs plus fixed costs:

\[C = 110x + 600 \quad \checkmark\]

Revenue is quantity times price:

\[R = 350x \quad \checkmark\]

Grading: 1 Pt. conversion to EUR per QU, 1 Pt. \(C\), 1 Pt. \(R\).

Part b)

Show that the daily profit is given by \(P = 240x - 600\). [2 pts.]

\[P = R - C = 350x - (110x + 600)\]

Remove the bracket. The minus sign changes every sign inside it:

\[P = 350x - 110x - 600 = 240x - 600 \quad \checkmark\]

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

Part c)

Determine all outputs for which the bakery makes a profit. State this profit zone as an interval and in set-builder notation. [3 pts.]

Capacity. 800 loaves = 8 QU, so \(0 \leq x \leq 8\).

Profit means \(P > 0\):

\[240x - 600 > 0 \iff 240x > 600 \iff x > 2.5\]

Profit zone:

\[(2.5; 8] = \{x \in \mathbb{R} \mid 2.5 < x \leq 8\}\]

At \(x = 2.5\) the profit is exactly zero, so 2.5 is excluded. 8 QU is the capacity limit and is included.

Grading: 1 Pt. inequality \(P > 0\), 1 Pt. solving it, 1 Pt. zone including the capacity, in both notations.

Part d)

The daily demand deviates from the planned 5 QU by at most 2.8 QU. Formulate this condition as an absolute value inequality and solve it. Decide whether the bakery is guaranteed to make a profit every day. [3 pts.]

\[|x - 5| \leq 2.8 \iff -2.8 \leq x - 5 \leq 2.8 \iff 2.2 \leq x \leq 7.8\]

So \(x \in [2.2; 7.8]\).

Compare with c): \([2.2; 7.8]\) is not inside \((2.5; 8]\). For outputs from 2.2 to 2.5 QU there is no profit, e.g. at \(x = 2.2\): \(240 \cdot 2.2 - 600 = -72\) EUR.

Decision: No. On days with weak demand the bakery can make a loss.

Grading: 1 Pt. inequality and interval, 1 Pt. comparison with the profit zone, 1 Pt. decision with reason.

Part e)

At the planned output of 5 QU, the bakery expects its sales to rise by 12 %. Compute the percentage increase of the profit and explain why it is larger than 12 %. [3 pts.]

Old profit at \(x = 5\): \(240 \cdot 5 - 600 = 600\) EUR

New output: \(5 \cdot 1.12 = 5.6\) QU, new profit \(240 \cdot 5.6 - 600 = 1344 - 600 = 744\) EUR

Change:

\[\frac{744 - 600}{600} = 0.24 = 24\,\%\]

Explanation: the fixed costs of 600 EUR stay the same. The additional 0.6 QU add \(0.6 \cdot 240 = 144\) EUR, all of which goes to profit, and that is 24 % of the old profit.

Grading: 1 Pt. both profits, 1 Pt. percentage change, 1 Pt. explanation referring to the fixed costs.

Part f)

The owner wants a daily profit of at least 1,000 EUR. Determine the minimum number of loaves the bakery has to sell and decide whether this is possible with the oven. [3 pts.]

\[240x - 600 \geq 1000 \iff 240x \geq 1600 \iff x \geq \frac{20}{3} \approx 6.667 \text{ QU}\]

In loaves: \(6.667 \cdot 100 \approx 666.7\). Only whole loaves are sold, so round up: at least 667 loaves (with 666 loaves the profit is only \(240 \cdot 6.66 - 600 = 998.40\) EUR).

Decision: 667 loaves \(\leq\) 800 loaves, so this is possible with the oven.

Grading: 1 Pt. inequality and solution, 1 Pt. rounding up to whole loaves with reason, 1 Pt. decision with capacity.