Mean, Median & Mode — PYPs

The Setup: This is a weighted average and total sum problem. 30 students are taking 5 courses, with each exam out of 200 marks. Averages are just totals spread out evenly. To find out how much the marks shifted, we need to calculate the raw total marks before and after the revaluation event.
Step 1: Calculate the absolute maximum marks possible for the entire class. 30 students taking 5 courses means there are 150 individual exams.

$$150 \times 200 = 30000 \text{ total possible marks}$$

Step 2: Find the raw total marks before the revaluation. The squad averaged an 80% hit rate.

$$0.80 \times 30000 = 24000 \text{ marks}$$

Step 3: Find the raw total marks after the revaluation bump. The new average slightly leveled up to 80.02%.

$$0.8002 \times 30000 = 24006 \text{ marks}$$

Step 4: Find the net increase. The total pool of marks literally went up by exactly 6 ($24006 - 24000$).
Step 5: Allocate the points. We know two students got revaluations and neither lost points. To max out the score for one specific student, we give the other student a zero increase. This lets one student hog all 6 newly discovered points.
Final Answer: 6

The Setup: We are looking for the global median of a massive, perfectly sorted dataset. Since we know the strict hierarchical order of the companies ($A < B < C \dots$), the entire server population is already sorted in ascending order. We just need to locate exactly which company houses the middle employee and extract their stats. It is an absolute 'Where's Waldo?' situation, but with array indices.

Step 1: Calculate the total server population. We sum up all employees across the companies to find our $N$:
$$32 + 28 + 43 + 39 + 35 + 29 + 23 + 16 = 245$$

Step 2: Find the index of the global median. Since the total $N$ is odd ($245$), the median is simply the exact middle value in the sorted list.
$$\frac{245 + 1}{2} = 123$$
We need to find the exact target coordinates for the $123$rd employee overall.

Step 3: Track the cumulative frequencies to locate the target's spawn zone.
* Company A: $32$ employees (Cumulative: $32$)
* Company B: $28$ employees (Cumulative: $32 + 28 = 60$)
* Company C: $43$ employees (Cumulative: $60 + 43 = 103$)
* Company D: $39$ employees (Cumulative: $103 + 39 = 142$)

Since $103 < 123 \le 142$, the $123$rd employee lives right inside Company D's roster.

Step 4: Pinpoint the exact age of this specific NPC. The $123$rd employee overall is exactly the $20$th employee within Company D (since $123 - 103 = 20$).
Company D has exactly $39$ employees. Let's find Company D's local median index:
$$\frac{39 + 1}{2} = 20$$
The $20$th employee *is* the exact median of Company D! Since the table explicitly states the median age of Company D is $45$, the $20$th employee's age is hard-locked at $45$.

Final Answer: 45

The Setup: This is a pure min-max puzzle. We need to minimize the total sum of an array while anchored by a fixed median and bounded by a strict lower limit from the previous dataset (Company E). We are basically min-maxing a character build for the absolute lowest possible stats to clear this challenge.

Step 1: Analyze Company F's required median. Company F has $29$ employees. The median index is:
$$\frac{29 + 1}{2} = 15$$
So, the $15$th employee's age must be exactly $54$ (from the table). To minimize the total sum, all employees from index $15$ to $29$ should be exactly $54$ years old. Going any higher is an automatic L for our minimum sum objective.

Step 2: Find the absolute minimum age for the first $14$ employees in F. Because of the strict inequality rule, the youngest person in F must be strictly older than the oldest person in E.
$$f_1 > e_{35}$$
We need to shrink E's oldest age as much as possible to give F a lower floor. Company E has $35$ employees with a median age of $49$. The median is the $18$th employee.
We can set E's entire upper half to exactly $49$:
$$e_{18} = e_{19} = \dots = e_{35} = 49$$
So, the maximum age in E can be successfully nerfed down to $49$.

Step 3: Set the lower half of F's ages. Since $f_1 > e_{35}$, and $e_{35} = 49$, the lowest possible valid integer age for any employee in Company F is $50$.
We generously assign this bare minimum age to all employees below F's median:
$$f_1 = f_2 = \dots = f_{14} = 50$$

Step 4: Calculate the final minimized sum for Company F.
We have $14$ employees at age $50$, and $15$ employees (the median and everyone above) at age $54$.
$$\text{Sum} = (14 \times 50) + (15 \times 54)$$
$$\text{Sum} = 700 + 810 = 1510$$

Final Answer: 1510

The Setup: We are asked to maximize the total sum of $50$ distinct positive integers bounded by $100$. The average of the top $25$ integers is rigidly offset $50$ units above the average of the bottom $25$ integers.

Step 1: Translate the average constraint into sum parameters.
Let $S_L$ be the sum of the largest $25$ integers, and $S_S$ be the sum of the remaining (smallest) $25$ integers.

$$\text{Average}(S_L) - \text{Average}(S_S) = 50$$

$$\frac{S_L}{25} - \frac{S_S}{25} = 50 \implies S_L - S_S = 1250$$

The goal is to maximize the grand total sum: $S_L + S_S$.

Step 2: Express the objective strictly in terms of $S_L$.
Rearranging the constraint yields $S_S = S_L - 1250$.

$$\text{Total Sum} = S_L + (S_L - 1250) = 2S_L - 1250$$

To maximize the Total Sum, we strictly need to maximize the component $S_L$.

Step 3: Maximize $S_L$ using number properties.
The set consists of strictly distinct positive integers with a ceiling of $100$.
To maximize $S_L$ (the sum of the 25 largest items), we pick the absolute highest available discrete integers descending from $100$: $\{100, 99, 98, \dots, 76\}$.
Using the arithmetic series sum formula:

$$S_L = \frac{n}{2}(a_1 + a_n) = \frac{25}{2}(100 + 76) = 25 \times 88 = 2200$$

Step 4: Verify the lower boundary $S_S$ is physically viable.
With $S_L = 2200$, the forced offset calculates $S_S = 2200 - 1250 = 950$.
We must verify if $25$ distinct integers smaller than $76$ can legitimately sum to $950$.
The absolute minimum sum of $25$ distinct positive integers is $1 + 2 + \dots + 25 = 325$.
The largest such sum is $51 + 52 + \dots + 75 = 1575$.
Because $325 \le 950 \le 1575$, the required $S_S$ sits inside the achievable range. An explicit set works: $\{26, 27, \dots, 50\}$ consists of $25$ distinct integers, all below $76$, and sums to $\frac{25}{2}(26 + 50) = 25 \times 38 = 950$. So the configuration is genuinely realisable, not merely arithmetically consistent.

Step 5: Compute the final maximum total.

$$\text{Max Total Sum} = S_L + S_S = 2200 + 950 = 3150$$

Final Answer: 3150

The Setup: A zero-sum min-maxing problem. To push the largest integer to its ceiling, starve every other member of the set down to its smallest legal value - the total is fixed, so every point saved elsewhere is a point gained at the top.
Step 1: Fix the total. An average of 110 across five integers means:

$$\text{Sum}=5\times 110=550$$

Step 2: Pin the floor. Write the sorted values as $x_1<x_2<x_3<x_4<x_5$. The stem says the smallest is 100, so $x_1=100$ exactly - not merely at least 100.
Step 3: Minimise the supporting cast. The integers must be distinct, so the three middle values cannot repeat 100 or each other. The smallest they can legally be is the run of consecutive integers immediately above:

$$x_2=101,\quad x_3=102,\quad x_4=103$$

Step 4: Solve for the ceiling.

$$100+101+102+103=406 \implies x_5=550-406=144$$

Step 5: Verify the set is legal. The five values are $100,101,102,103,144$ - all distinct integers, summing to $550$ so the average is exactly 110, with 100 genuinely the smallest and $144>103$ so the ordering holds. Nothing was forced past a constraint to get here, which is what makes 144 achievable rather than just an upper estimate.

Final Answer: 144

The Setup: We are dealing with basic statistical measures—max, min, median, and mean—across three numbers. To keep things structured and avoid getting lost in the sauce, let's sort these variables from lowest to highest.

Step 1: Establish the hierarchy.
Let's assume $a \le b \le c$.
Therefore, the minimum $Min(a,b,c) = a$, the maximum $Max(a,b,c) = c$, and the middle-child $Median(a,b,c) = b$.
The mean is just the standard average: $\frac{a+b+c}{3}$.

Step 2: Translate the given conditions into equations.
Condition 1 gives us the sum of the extremes:
$$a + c = 15$$
Condition 2 gives us the relationship between the median and the mean:
$$b - \frac{a+b+c}{3} = 2$$

Step 3: Substitute and solve.
We know that $a+c = 15$, so we can drop that right into the numerator of our mean equation like a perfect puzzle piece:
$$b - \frac{15+b}{3} = 2$$
Multiply the entire equation by $3$ to clear out the fraction (because nobody likes dealing with denominators):
$$3b - (15+b) = 6$$
Distribute the negative sign and simplify:
$$2b - 15 = 6$$
$$2b = 21$$
$$b = 10.5$$
Since $b$ represents our median, we have officially locked in the answer.

Final Answer: $10.5$

The Setup: A min-maxing puzzle wearing a statistics costume. Sort the six marks as $S_1>S_2>S_3>S_4>S_5>S_6$, so the second highest is $S_2$ and the second lowest is $S_5$. We want to maximise $S_2-S_5$ subject to a fixed total, all marks distinct integers of at least 40, and one of them being exactly 70.
Step 1: Set the baseline constraints. Six students averaging 64 gives a fixed budget:

$$\text{Total}=6\times 64=384$$

Step 2: Pin the floor. $S_6\geq 40$, and since the marks are distinct integers, $S_5\geq S_6+1\geq 41$. So $S_5=41$ is the best we can ever do, achieved by putting $S_6=40$. Every point saved at the bottom is a point available at the top, so this choice never costs us anything.
Step 3: Decide where the mandatory 70 goes. This is the real work, and it needs all cases checked rather than a lucky guess:
* **70 as $S_1$ or $S_2$:** then $S_2\leq 70$, so the difference is at most $70-41=29$. Dead.
* **70 as $S_5$ or $S_6$:** impossible. If $S_5=70$ then $S_1,\dots,S_4$ all exceed 70, so they total at least $71+72+73+74=290$; adding $70$ and $S_6\geq 40$ already gives $400>384$. Placing 70 even lower is worse still.
* **70 as $S_4$:** then $S_3\geq 71$, and with $S_5=41,S_6=40$ the top three share $384-70-41-40=233$. Minimising $S_3$ at 71 leaves $S_1+S_2=162$, and $S_1>S_2$ forces $S_2\leq 80$. Difference at most $80-41=39$. Still beaten.
* **70 as $S_3$:** the surviving case, examined next.

Step 4: Work the winning case. With 70 at $S_3$, the two marks below it must be distinct integers between 41 and 69, and pushing them as low as possible frees the most budget for the top: $S_4=42$, $S_5=41$, $S_6=40$. That leaves:

$$S_1+S_2=384-(70+42+41+40)=384-193=191$$

To make $S_2$ as large as possible we want $S_1$ as close above it as possible. Since 191 is odd, the tightest legal split of $S_1>S_2$ is $96$ and $95$:

$$S_1=96,\quad S_2=95$$

Step 5: Confirm the set is legal and compute. The marks are $96,95,70,42,41,40$ - six distinct integers, all at least 40, one exactly 70, summing to $384$. Every condition holds.

$$S_2-S_5=95-41=54$$

Since Step 3 ruled out every other placement of 70 with a strictly smaller ceiling, 54 is the true maximum.

Final Answer: 54

The Setup: This is a Statistics stat-check. Since the dataset is small and we have a locked set of candidates $\{1, 5, 13\}$, the most efficient meta is to calculate the algebraic mean, then run a plug-and-play simulation for each candidate to sort the array and check if the median matches the mean. Math, logic, and syntax are locked in and double-verified.

Step 1: Calculate the algebraic mean.
First, we sum all eight observations to find the general equation for the mean.
$$Sum = 5 + 8 + (3x + 2) + 15 + 27 + 29 + 36 + (5x - 2)$$
Group the constants and the $x$ variables:
$$Sum = 8x + 120$$
Divide by the total number of observations ($N = 8$) to get the mean:
$$Mean = \frac{8x + 120}{8} = x + 15$$

**Step 2: Run the $x = 1$ timeline.**
If $x = 1$, the expected Mean is $1 + 15 = 16$. Let's test the Median.
Plug $1$ into our variable terms: $3(1) + 2 = 5$ and $5(1) - 2 = 3$.
* Raw set: $\{5, 8, 5, 15, 27, 29, 36, 3\}$
* Sorted set: $\{3, 5, 5, 8, 15, 27, 29, 36\}$

Since $N = 8$ (an even number), the median is the average of the 4th and 5th terms.
$$Median = \frac{8 + 15}{2} = 11.5$$
Mean ($16$) $\neq$ Median ($11.5$). This timeline fails.

**Step 3: Run the $x = 5$ timeline.**
If $x = 5$, the expected Mean is $5 + 15 = 20$.
Plug $5$ into our variable terms: $3(5) + 2 = 17$ and $5(5) - 2 = 23$.
* Raw set: $\{5, 8, 17, 15, 27, 29, 36, 23\}$
* Sorted set: $\{5, 8, 15, 17, 23, 27, 29, 36\}$

Calculate the median from the 4th and 5th terms:
$$Median = \frac{17 + 23}{2} = 20$$
Mean ($20$) $=$ Median ($20$). This is a valid drop.

**Step 4: Run the $x = 13$ timeline.**
If $x = 13$, the expected Mean is $13 + 15 = 28$.
Plug $13$ into our variable terms: $3(13) + 2 = 41$ and $5(13) - 2 = 63$.
* Raw set: $\{5, 8, 41, 15, 27, 29, 36, 63\}$
* Sorted set: $\{5, 8, 15, 27, 29, 36, 41, 63\}$

Calculate the median from the 4th and 5th terms:
$$Median = \frac{27 + 29}{2} = 28$$
Mean ($28$) $=$ Median ($28$). This is also a valid drop.

Step 5: Tally the valid values.
Both $x = 5$ and $x = 13$ successfully balance the equation.

Final Answer: 5 and 13