[NCS Numeracy Chapter 01] Core Applied Math Archetypes

Applied math questions look numerous, but they collapse into a small set of repeating structures. Instead of memorizing every item from 1 through 68, you get more leverage by clustering similar questions, mastering one full walkthrough, and connecting the rest of the problem IDs to that pattern.

That is exactly how this chapter is organized:

• First, bundle problems into representative types.
• For each type, fix the solving order.
• Work through one anchor problem from start to finish.
• List the other problem numbers that share the same frame.

This chapter is written for learners who already know basic arithmetic and algebra but want a faster way to organize NCS applied-math problems.

What This Chapter Covers

Before you dive into the worked examples, keep these labels straight:

• Representative type: one recurring problem pattern that shows up in several question numbers
• Anchor problem: one fully worked example that teaches the pattern for that type
• Setup: the unknown, the conserved quantity, and the equation you need before calculating

Chapter 01 Map

Representative Type	Problem IDs	Core Judgment
Tunnel crossing, meeting, average speed	1–9	Can you align distance, speed, and time in the same unit set?
Mixing, evaporation, partial removal	10–18	Can you treat it as conservation of solute mass instead of chasing percentages?
Working together, filling and draining	19–21	Can you normalize total work to 1 and add/subtract rates?
Combinations, counting, conditional probability	22–32	Can you separate the full sample space from the conditional space?
Systems of equations, integer constraints, profit rate	33–57	Can you translate sentences into equations and keep integer conditions in view?
Max/min and ratio applications	58–68	Can you write the model and then handle vertices, ratio tables, or case splits?

What to Do First in Applied Math

Do not jump into arithmetic the moment you see a question. Build the habit of jotting down these three lines first:

1. What is the unknown?
2. What quantity is conserved?
3. What equations should you write?

Most mistakes in applied math come from setting up the wrong equation, not from calculation errors.

When a section says a quantity is conserved, it means that quantity stays unchanged while the problem unfolds. In this chapter, that is usually distance, total work, or solute mass.

1. Distance, Speed, and Time

This cluster covers problems 1–9. They may look different—trains, business trips, tracks, speed cameras—but the backbone is the same:

distance = speed × time

Tunnel problems include the train length in the distance term. Meeting problems add or subtract the speeds of the two moving objects.

1-1. Representative Type A: Fully Clearing a Tunnel

Problems 1, 2, 3 belong here. Let’s unpack problem 1.

A train moving at a constant speed completely passes through a 400 m tunnel in 13 s and a 800 m tunnel in 23 s. If it triples its speed, how long does it take to completely pass through a 960 m tunnel?

“Completely passing” means measuring from the moment the front enters to the moment the rear exits, so the distance traveled is train length + tunnel length.

Let the train length be $L$ and the original speed be $v$.

$$\frac{L+400}{v} = 13, \qquad \frac{L+800}{v} = 23$$

Subtract the equations:

$$\frac{400}{v} = 10$$

so

$$v = 40\text{ m/s}$$

Plug into the first equation:

$$L + 400 = 13 \times 40 = 520$$

Thus

$$L = 120\text{ m}$$

Tripling the speed gives $120\text{ m/s}$. Passing a 960 m tunnel covers

$$120 + 960 = 1080\text{ m}$$

and the time is

$$\frac{1080}{120} = 9\text{ s}$$

Answer: 9 s.

Frequent Pitfalls in This Type

• Forgetting to add train length to the tunnel length
• Solving both equations separately instead of subtracting them
• Thinking time also triples when speed triples

1-2. Representative Type B: Average Speed with Departure-Time Shifts

Problems 4, 5, 7, 8, 9 are linked. Consider problem 4.

Manager Kim left at 1

 p.m., arrived at 4

 p.m., and averaged 80 km/h. Manager Lee left 15 minutes later and arrived at the same time. The distance from the B branch to the C branch is 1.35× the distance from headquarters A to branch C. What is Lee’s average speed?

Kim traveled for 2 h 30 m, i.e., 2.5 h. Distance:

$$80 \times 2.5 = 200\text{ km}$$

So A→C is 200 km. B→C is 1.35 times that:

$$200 \times 1.35 = 270\text{ km}$$

Lee left 15 minutes later, so travel time was 2 h 15 m = 2.25 h.

Average speed:

$$\frac{270}{2.25} = 120\text{ km/h}$$

Answer: 120 km/h.

Always convert minutes to hours first; failing to turn 15 minutes into 0.25 hours derails the math.

1-3. Representative Type C: Meeting on a Circular Track

Problem 6 stands alone but is a staple.

On a 400 m circular track, two people meet for the first time after 100 s when running in the same direction, and after 50 s when running in opposite directions. If person A is faster, what is person B’s speed?

Meeting in the same direction means the faster runner gains exactly one lap, so

$$\text{speed difference} = \frac{400}{100} = 4\text{ m/s}$$

Meeting in opposite directions means their combined distance equals one lap:

$$\text{speed sum} = \frac{400}{50} = 8\text{ m/s}$$

Let $a$ be the faster speed and $b$ the slower speed.

$$a-b = 4, \qquad a+b = 8$$

Add to obtain $2a = 12 \Rightarrow a = 6$, so $b = 2$.

Answer: 2 m/s.

2. Saline-Solution Concentration

Problems 10–18 live here. Percentages invite mistakes; look at mass of salt and mass of water instead.

2-1. Representative Type A: Using Fixed Amounts and Finding the Remaining Concentration

Problems 10, 15 fit this type. Take problem 10.

There are 300 g of water and 75 g of bamboo salt. You mix 122 g of water with 23 g of bamboo salt to gargle once, and repeat the same once more. What is the concentration of the remaining solution?

Initial amounts:

• Water: 300 g
• Salt: 75 g

Each gargle consumes 122 g of water and 23 g of salt. Twice means

• Water used: $122 \times 2 = 244$ g
• Salt used: $23 \times 2 = 46$ g

Leftover:

• Water: $300 - 244 = 56$ g
• Salt: $75 - 46 = 29$ g

Concentration:

$$\frac{29}{56+29} \times 100 = \frac{29}{85} \times 100 \approx 34.1\%$$

Answer: 34.1 %.

This is not really a mixing problem—it is a remaining-amount problem. Track the leftover water and salt separately first.

2-2. Representative Type B: Mix, Keep Part, Then Mix Again

Problem 13 is prototypical.

Mix 300 g of an 8 % solution with 200 g of a 13 % solution, then discard everything except 300 g. Add 700 g of a 20 % solution, then again discard everything except 300 g. How much salt remains in the final 300 g?

First mix:

• Salt from 8 % × 300 g: $24$ g
• Salt from 13 % × 200 g: $26$ g

Total salt = $24 + 26 = 50$ g in 500 g of solution (10 %). Keeping 300 g means keeping $\tfrac{3}{5}$ of everything, so salt becomes

$$50 \times \frac{3}{5} = 30\text{ g}$$

Adding 700 g of 20 % solution introduces

$$700 \times 0.2 = 140\text{ g}$$

Total salt now: $30 + 140 = 170$ g in 1,000 g. Keeping 300 g again keeps $\tfrac{3}{10}$, so final salt mass is

$$170 \times \frac{3}{10} = 51\text{ g}$$

Answer: 51.0 g.

Key idea: when you keep only part of a well-mixed solution, the concentration stays the same, so the solute amount shrinks in the same ratio.

2-3. Representative Type C: Raising Concentration via Evaporation

Problems 11, 12, 16, 17, 18 belong together.

Use the template: if the initial solute mass is $S$ and the target concentration is $p$ written as a decimal, the final total mass is $\tfrac{S}{p}$. This works because evaporation removes only the solvent, so the solute mass stays unchanged.

Stick to this order every time:

1. Compute the initial solute mass.
2. Compute the total mass at the target concentration.
3. Subtract to find how much liquid must evaporate.

3. Work-Rate Problems

Problems 19–21 are all work-rate questions. Once you set the total work to 1, the arithmetic is straightforward.

3-1. Representative Type: Combined Rate with Multiple Workloads per Day

Consider problem 19.

Using carriers A and B together takes 7 days; B and C together take 8 days; C and A together take 4 days. If A, B, and C work together and complete two transport cycles per day, what is the minimum number of days required?

Let total work be 1 per transport cycle. Daily rates per single run are

$$A+B = \frac{1}{7}, \quad B+C = \frac{1}{8}, \quad C+A = \frac{1}{4}$$

Add the equations:

$$2(A+B+C) = \frac{1}{7} + \frac{1}{8} + \frac{1}{4}$$

Common denominator 56 yields

$$\frac{8+7+14}{56} = \frac{29}{56}$$

so

$$A+B+C = \frac{29}{112}$$

With two runs per day, double the rate:

$$2 \times \frac{29}{112} = \frac{29}{56}$$

Time to finish work 1:

$$\frac{1}{29/56} = \frac{56}{29} \approx 1.93$$

Therefore, minimum days: 2.

Problem 20 follows the same template by adding fill rates and subtracting drain rates; problem 21 reconstructs individual rates through simultaneous equations.

4. Probability and Counting

Problems 22–32 benefit from splitting into “plain combinations” and “conditional probability”.

4-1. Representative Type A: Counting Without Conditions

Problems 22, 25, 26 fit here. Look at problem 22.

Among nine points on a semicircle, how many triangles can you form by choosing any three points?

Every choice of three points yields a triangle, so

$$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$

Answer: 84 combinations.

Even with geometric wording, do not hunt for exceptions unless the prompt states one. If no special rule exists, start with combinations.

4-2. Representative Type B: Conditional Probability and Bayes Thinking

Problems 28, 29, 30, 31, 32 share this flow. Take problem 30.

Cheolsu correctly identifies a marble’s color 80 % of the time. There are 120 red and 80 yellow marbles. If he says a marble is red, what is the probability that it is actually yellow?

True proportions:

• Red: $120/200 = 0.6$
• Yellow: $80/200 = 0.4$

Accuracy:

• If red, he says red with probability 0.8.
• If yellow, he still says red with probability 0.2.

Probability that he says “red”:

$$0.6 \times 0.8 + 0.4 \times 0.2 = 0.48 + 0.08 = 0.56$$

Case where it is actually yellow but he says red:

$$0.4 \times 0.2 = 0.08$$

Conditional probability:

$$\frac{0.08}{0.56} = \frac{1}{7}$$

Answer: $1/7$.

Before writing fractions, check whether the sample space changed. Here the new sample space is “cases where he said red.”

5. Equation Applications

Problems 33–57 hinge on translating prose into equations.

5-1. Representative Type A: Price × Quantity = Total

Problems 33, 34, 39, 40, 41, 43 match this template. Consider problem 33.

Pencils cost 300 won and pens cost 400 won. If you buy 27 items total for 9,000 won, how many pens are there?

Let $x$ be the pen count; pencils are $27 - x$.

$$400x + 300(27-x) = 9000$$

Expand:

$$400x + 8100 - 300x = 9000 \Rightarrow 100x = 900$$

Thus $x = 9$.

Answer: 9 pens.

This structure repeats: pick a variable, then write total quantity and total price.

5-2. Representative Type B: Integer Constraints in Purchase Problems

Problems 35, 36, 47, 48, 49 repeatedly enforce integer or range constraints. After solving the equations, double-check “at least one each”, “at least 20 in total”, “45 or more”, etc. The goal is not just solving but filtering valid integer solutions.

5-3. Representative Type C: Translating Averages and Scores

Problems 42, 43, 46 belong here. Look at problem 43.

Across nine tests, the only scores were 79 or 85, and the average was 81. How many times did the student score 85?

Let $x$ be the number of 85s. Then 79 occurred $9 - x$ times. Total score:

$$81 \times 9 = 729$$

Actual total:

$$85x + 79(9-x)$$

Set equal:

$$85x + 79(9-x) = 729$$

Expand:

$$85x + 711 - 79x = 729 \Rightarrow 6x = 18$$

Therefore $x = 3$.

Answer: scored 85 three times.

When you see an average, immediately convert it to “average × count = total”.

5-4. Representative Type D: Discounts and Profit Rates

Problems 50–57 test the “cost → list price → sale price → profit” chain. Problem 56 illustrates it.

Both products A and B set list prices by adding 40 % to cost. During a sale, A is discounted 10 % and B is discounted 40 %, resulting in the same sale price. If B’s cost is 60,000 won, what is A’s profit per unit?

For B:

$$60,000 \times 1.4 = 84,000 \quad \text{(list price)}$$

Discount 40 %:

$$84,000 \times 0.6 = 50,400$$

Let $x$ be A’s cost. Its list price is $1.4x$, and its 10 % discount price is $1.4x \times 0.9 = 1.26x$. Since sale prices match:

$$1.26x = 50,400 \Rightarrow x = 40,000$$

Profit:

$$50,400 - 40,000 = 10,400$$

Answer: 10,400 won.

Do not skip stages. Write the sequence cost → list → sale → profit every time.

6. Max/Min and Ratio Applications

Problems 58–68 resemble earlier sections until the finale, where you must interpret maximum/minimum conditions or ratio constraints.

6-1. Representative Type A: Maximum Area with Fixed Perimeter

Problems 58, 59 fall here. Problem 58 is classic.

Using ropes of 48 m and 36 m to form separate rectangles, what is the maximum sum of their areas?

For a fixed perimeter, the rectangle with the largest area is a square.

• Perimeter 48 m → side 12 m → area $12^2 = 144$
• Perimeter 36 m → side 9 m → area $9^2 = 81$

Sum:

$$144 + 81 = 225$$

Answer: 225.

6-2. Representative Type B: Maximizing Revenue = Price × Quantity

Problems 60, 61 share the same quadratic revenue curve. Consider problem 60.

At a price of 3,000 won, 6,000 units sell. Lowering the price by 100 won increases sales by 300 units. At what sales quantity is revenue maximized?

If the price is lowered by 100 won $x$ times:

• Price: $3000 - 100x$
• Quantity: $6000 + 300x$

Revenue:

$$R(x) = (3000 - 100x)(6000 + 300x)$$

Expand:

$$R(x) = -30,000x^2 + 300,000x + 18,000,000$$

The quadratic opens downward, so the vertex yields the maximum. Vertex $x$:

$$x = -\frac{b}{2a} = -\frac{300,000}{2(-30,000)} = 5$$

Sales quantity at $x = 5$:

$$6000 + 300 \times 5 = 7,500$$

Answer: 7,500 units.

6-3. Representative Type C: Gender Ratios and Acceptance Rates

Problems 65–68 are best solved with tables. Draw rows for categories (e.g., male/female or preference/non-preference), fill in actual counts using a common multiple, and plug the ratio constraints. Instead of rushing to equations, follow this order:

1. Choose a convenient total (a common multiple).
2. Split into accepted vs. rejected or preferred vs. non-preferred.
3. Apply each ratio to concrete numbers in the table.

Handling Table-Based Items

Problems 37, 38, 42, 44, 45, 46, 50–55 require tables or supplementary data to provide complete walkthroughs. Here we only mapped where they belong. Once the original tables are on hand, you can extend this framework with the same representative-problem approach.

Recommended Order for Studying This Chapter

Rather than tackling all 68 problems sequentially, study them in small groups that reinforce one idea at a time:

1. Problems 1, 4, 6 to lock in the three main distance-speed-time patterns
2. Problems 10, 13 to practice tracking solute mass in concentration problems
3. Problem 19 to review how combined work rates are built from unit work
4. Problems 22, 30 to separate plain counting from conditional probability
5. Problems 33, 43, 56 to connect equations, averages, and profit-rate setups
6. Problems 58, 60 to learn the standard max/min models

If you are studying over multiple sessions, treat steps 1-2 as the first block, steps 3-4 as the second block, and steps 5-6 as the third block. That way, each session mixes one calculation-heavy topic with one interpretation-heavy topic.

Once this order feels natural, the 68 problems stop looking isolated—you start to see just a few archetypes repeating.

The next milestone is to expand each cluster from one anchor problem to full 3–4 problem solution sets so the pattern survives any wording change.