[NCS Numeracy Chapter 02] Practice Applied-Math Structures with Original Example Problems

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Applied-math questions can look very different on the surface, but many of them repeat the same small set of equation-building structures. For students, it is usually more effective to group those structures first and practice the solving order than to memorize many isolated questions.

Three Goals of This Post

  • Group the applied-math structures that appear repeatedly in NCS numeracy.
  • Explain what to identify first before you start calculating.
  • Let students read the explanation and immediately practice with an original problem set.

This page focuses on five recurring structures: distance-time relationships, mixtures with partial removal, work rates, base-fee equations, and weighted-response probability.

Suggested Study Order

Order Structure Difficulty Main Thing to Notice First
1 Base-fee equation Easy the fixed baseline value
2 Distance-time relationship Easy how distance is defined
3 Mixture with partial removal Medium what quantity stays conserved
4 Work rates Medium whether rates are added or subtracted
5 Weighted-response probability Hard how the rules split the cases

Structure Map for Applied Math

Recurring Structure What to Identify First Core Solving Logic
Distance-time relationship How distance is defined Fix the speed first by subtracting two equations
Mixture with partial removal Conserved solute mass Track the solute amount at each step
Working together Total work as 1 Add and subtract work rates
Base-fee equation The unknown fixed value Use the first case to lock the baseline
Weighted-response probability Rule-by-rule probability Separate weighted cases before solving the real rate

Four Lines to Write Before Solving

  1. What exactly am I solving for?
  2. What stays unchanged?
  3. Which quantity should I define with a variable?
  4. How many equations do I need?

Quick Recognition Hints

  • If you see seconds, kilometers, arrival time, or crossing time, check the distance-time structure first.
  • If you see percentages, mixing, removal, or evaporation, check the mixture structure.
  • If you see workers, pumps, together, or drains, check the work-rate structure.
  • If you see a base fee plus a per-item cost, check the base-fee structure.
  • If a survey forces certain answers based on a random rule, check the weighted-response probability structure.

1. Distance-Time: Fix the Speed First

Key formula: distance = speed x time. The main invariant here is that the speed stays the same across the two situations.

An inspection vehicle moves at a constant speed. It completely passes through a 320 m wind-shielded section in 11 seconds and a 560 m section in 17 seconds. What is the length of the vehicle in meters?

Let the vehicle length be LL and the speed be vv.

L+320v=11,L+560v=17\frac{L+320}{v} = 11, \qquad \frac{L+560}{v} = 17

Subtract the equations:

240v=6\frac{240}{v} = 6

So the speed is 40 m/s40\text{ m/s}. Plug that back in:

L+320=11×40=440L + 320 = 11 \times 40 = 440

Therefore,

L=120L = 120

The key move is always the same: use the difference in section length and time to fix the speed first.

2. Mixtures: Track the Solute, Not Just the Percentages

Key formula: solute mass = total mass x concentration. This is really a solute-conservation problem.

Mix 200 g of 15% concentrate with 200 g of 35% concentrate. Keep only 100 g of the mixture and discard the rest. Then add 200 g of 25% concentrate. What is the final concentration in percent?

First compute the solute mass.

  • 200×0.15=30200 \times 0.15 = 30
  • 200×0.35=70200 \times 0.35 = 70

So the first mixture has 100 g of solute in 400 g total. Keeping only 100 g means keeping one fourth, so the remaining solute is:

100×100400=25100 \times \frac{100}{400} = 25

Adding 200 g of 25% concentrate adds 50 g more solute, so the final mixture has 75 g of solute in 300 g total.

75300×100=25\frac{75}{300} \times 100 = 25

This is why mixture problems are easier when you follow the solute mass, not the percentages alone.

3. Work Rates: Separate Filling and Emptying

Key formula: work = rate x time. When possible, normalize the total work to 1.

Pump A fills a tank in 8 hours, Pump B fills it in 12 hours, and drain C empties it in 24 hours. First A and B run together for 2 hours. Then A and C run together. How many more hours are needed to fill the tank completely?

Treat the whole job as 1.

The combined rate of A and B is:

18+112=524\frac{1}{8} + \frac{1}{12} = \frac{5}{24}

So in 2 hours they complete:

2×524=5122 \times \frac{5}{24} = \frac{5}{12}

The remaining work is 712\frac{7}{12}. The net rate of A and C is:

18124=112\frac{1}{8} - \frac{1}{24} = \frac{1}{12}

So the remaining time is:

712÷112=7\frac{7}{12} \div \frac{1}{12} = 7

4. Base-Fee Equations: Lock the Unknown Baseline First

Key formula: total cost = fixed baseline + variable cost. The baseline does not change across cases.

Equipment rental is charged as a base fee FF won + 900 won for each standard item + 1,800 won for each special item. Team A rented 3 standard items and 4 special items for a total of 12,600 won. How much would Team B pay for 5 standard items and 2 special items?

From Team A:

F+3×900+4×1800=12600F + 3 \times 900 + 4 \times 1800 = 12600

So,

F=2700F = 2700

Now apply that to Team B:

2700+5×900+2×1800=108002700 + 5 \times 900 + 2 \times 1800 = 10800

This pattern is really about fixing the unknown baseline from the first case and transferring it to the second case.

5. Weighted-Response Probability: Split the Rules First

Key formula: overall probability = sum of (case probability x outcome probability inside that case).

In a sensitive survey, each participant rolls a die. If the result is 1 or 2, the participant answers truthfully. If the result is 3, 4, 5, or 6, the participant must answer “Yes.” If 70% of all responses were “Yes,” what is the actual experience rate in percent?

Let the real rate be p%p\%.

13p+23×100=70\frac{1}{3}p + \frac{2}{3} \times 100 = 70

So,

13p=10,p=10\frac{1}{3}p = 10, \qquad p = 10

The sentence can look long, but once you split the weighted rules, the equation becomes short.

Try the Original Practice Set

Original Applied-Math Practice

Five practice problems for distance-time, mixtures, work rates, equations, and probability

Original
문제 1 / 5
Original

Original Example 1. An inspection vehicle moves at a constant speed. It completely passes through a 320 m wind-shielded section in 11 seconds and a 560 m section in 17 seconds. What is the length of the vehicle in meters? Enter numbers only.

해설 보기

The section-length difference is 240 m and the time difference is 6 s, so the speed is 40 m/s. Then vehicle length + 320 = 11 x 40 = 440, so the vehicle length is 120 m.

How Students Should Study This Page

  1. Read the structure map first.
  2. Solve one example by hand without the component.
  3. Use the component to test whether you can repeat the same setup on a new surface story.
  4. If you get stuck, check whether you misread the variable, the conserved quantity, or the rate.

Check Yourself After Studying

  • Can you identify the right structure within 30 seconds?
  • Can you write what stays unchanged before you calculate?
  • Can you solve at least three of the five practice problems again without the solution?

Wrap-Up

The real goal in applied math is not seeing more numbers. It is learning to recognize what equation should be built first. This page is designed as a student practice page for exactly that habit. The next post separates data interpretation and trains index conversion, shares, weighted averages, and actual-value reconstruction in the same way.

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