FIX KELUAR! BAHAS TKA 2025 FULL PEMBAHASAN | Mantappu Academy

Main ideas / concepts covered

1) Mean, Median, Mode (basic statistics)

Mean / Average

• Defined as: ((\text{sum of all values}) \div (\text{number of values}))
• Often symbolized as (\bar{x}) (x-bar)
• Example (values: 7, 8, 5, 9, 6):
  • Mean = ((7+8+5+9+6) \div 5 = 35 \div 5 = 7)

Median

• The middle value after sorting the data from smallest to largest.

Mode

• The value that appears most frequently.
• If all values appear equally often, then there is no mode.

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2) Percentages and “how many do not…?” problems

Scenario: A survey of 200 young people; 35% choose reading.

Method 1: Count those who like, then subtract

• (35\%) of 200 = (0.35 \times 200 = 70)
• Those who do not like reading = (200 - 70 = 130)

Method 2: Use the complement

• Not reading = (100\% - 35\% = 65\%)
• (65\%) of 200 = (0.65 \times 200 = 130)

Either approach gives the same result.

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3) Finding a missing value using the mean (algebra)

Scenario: An average is given and one value (x) is missing. There are 5 numbers total.

• Mean: (\dfrac{4 + 5 + x + 7 + 8}{5})
• Given mean = 6

Algebra shown:

• (6 = \dfrac{24 + x}{5})
• Multiply by 5: (30 = 24 + x)
• Solve: (x = 6)

Lesson: turn the mean into an equation, then solve for the missing term.

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4) Combined average from two groups

Scenario: “Bread score” for boys and girls.

• 12 male students average = 70
• 18 female students average = 80

Steps:

• Total male score = (70 \times 12 = 840)
• Total female score = (80 \times 18 = 1440)
• Combined mean:
  • Total students = (12 + 18 = 30)
  • Combined mean = ((840 + 1440) \div 30 = 2280 \div 30 = 76)

Combined mean = (sum of all group totals) ÷ (total number of people).

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5) Pie chart / sector angle from counts (percent ↔ degrees)

Scenario: Parents’ occupations in a circle.

Given counts:

• Civil servants = 15
• Private sector workers = 20
• Self-employed = 10
• Others = 5
• Total = (15 + 20 + 10 + 5 = 50)

Core rule:

• Full circle = 360° = 100% of people

Find angle for private sector (20 people):

• (50 \text{ people} \leftrightarrow 360^\circ)
• (20 \text{ people} \leftrightarrow x)
• (x = (360 \times 20) \div 50 = 144^\circ)

Extra idea:

• You can use “per one person” reasoning:
  • (360^\circ \div 50) per person, then multiply by 20

Lesson: use proportional comparisons to convert counts/percentages into angles.

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Probability & combinatorics lessons (levels)

6) Level 1: Sample space in probability (two coins)

Scenario: Toss two coins simultaneously.

• Sample space (S): all possible cases/outcomes of the experiment.
• Each coin has 2 possibilities (e.g., number/picture).
• Total outcomes:
  • (2 \times 2 = 4)

If you understand it, sample space size comes from multiplying independent possibilities.

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7) Level 2: Permutations vs combinations (management positions)

Scenario: 6 candidates: A, B, C, D, E, F Need:

• 1 chairman
• 1 secretary
• 1 treasurer

Since the roles are ordered (different positions), use permutations.

Permutation formula used:

• (P(6,3) = \dfrac{6!}{(6-3)!} = \dfrac{6!}{3!})
• Simplifies to: (6 \times 5 \times 4 = 120)

“Slot filling” counting trick:

• Chairman: 6 ways
• Secretary: 5 ways
• Treasurer: 4 ways
• Total = (6 \times 5 \times 4 = 120)

Lesson: when roles are distinct, use permutations/ordered counting.

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8) Level 3 (higher thinking): Probability of drawing two red balls (without replacement)

Scenario:

• Box contains:
  • 4 red bicycles (interpreted as 4 red items)
  • 6 white balls
  • Total = 10
• Draw 2 balls one by one without replacement
• Find probability both are red

Method / steps

• For both draws to be red:
  • First draw must be red
  • Second draw must also be red

First draw:

• (P(\text{red first}) = \dfrac{4}{10} = \dfrac{2}{5})

Second draw (given the first was red):

• Remaining: 3 red and 6 white → total 9
• (P(\text{red second} \mid \text{red first}) = \dfrac{3}{9} = \dfrac{1}{3})

Multiply dependent events:

• (P(\text{both red}) = \dfrac{2}{5} \times \dfrac{1}{3} = \dfrac{2}{15})

Lesson: without replacement, the second probability changes (use conditional probability).

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Instructional / methodology bullet list (all explicit methods)

Mean / average

• Add all values.
• Divide by the number of values:
  • [ \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} ]

Median

• Sort values from smallest to largest.
• Select the middle value.

Mode

• Find the value that appears most often.
• If no value repeats more than others → no mode.

Percent “do not choose…”

• Convert percent to number:

  • [ \% \text{ of total} = (\text{percent} \div 100) \times \text{total} ]

• Use either:

  • Like-reading count then subtract, or
  • Complement: ((100\% - \text{given \%}))

Missing value in mean (algebra)

• Write mean equation:
  • given mean = (sum including (x)) ÷ (number of terms)
• Multiply both sides by the number of terms.
• Solve the linear equation for (x).

Combined average of two groups

• Compute each group’s total:
  • ( \text{Total}_1 = \text{mean}_1 \times \text{count}_1 )
  • ( \text{Total}_2 = \text{mean}_2 \times \text{count}_2 )
• Combined mean:
  • [ \frac{\text{Total}_1 + \text{Total}_2}{\text{count}_1 + \text{count}_2} ]

Pie chart angle (count ↔ degrees)

• Use:
  • 50 people = 360°

• Proportion:

  • [ \text{angle} = \frac{360 \times \text{category_count}}{\text{total_count}} ]

• Equivalent comparative method:

  • [ \left(\frac{360}{\text{total_count}}\right) \times \text{category_count} ]

Probability using sample space

• Identify sample space (S) = all possible outcomes.

• Use:

  • [ P = \frac{NA}{NS} ] where:

  • (NA) = outcomes meeting the condition

  • (NS) = total outcomes

Permutations for ordered roles

• If roles are distinct (chairman/secretary/treasurer):

  • [ P(n,r) = \frac{n!}{(n-r)!} ]

• Or “slot filling”:

  • First role: (n) choices
  • Next role: (n-1) choices
  • Continue until (r) roles: multiply choices

Dependent probability (without replacement)

• Compute:
  • (P(\text{first condition})) using current totals
• Update counts after the first draw.
• Compute:
  • (P(\text{second condition} \mid \text{first condition})) with new totals
• Multiply:
  • [ P(\text{both}) = P(\text{first}) \times P(\text{second given first}) ]

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Speakers / sources featured

• Billy (presenter/host)
• Eva (co-presenter)
• “Mantap u Academy” / “Mantapu Akademi Tutor Team” (institution/channel mentioned)
• Evan (tutor mentioned; “taught by me, Evan…”)