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Equation-based Word Problems (Advanced)

Introduction

Advanced word problems translate real-world situations into algebraic equations - often systems, higher-degree equations, or equations with parameters. Mastering this pattern helps you model relationships, set up correct equations, and solve efficiently.

These problems appear frequently in competitive exams and assessments where reasoning, formulation, and algebraic manipulation are tested together.

Pattern: Equation-based Word Problems (Advanced)

Pattern

Key concept: Convert the verbal statements into algebraic relations carefully (use variables, form equations, solve, and validate).

Typical steps:

  1. Assign variables clearly to unknown quantities.
  2. Translate phrases into algebraic expressions (rates, ratios, sums, differences, products, ages, work/time, mixtures, percentages).
  3. Form one or more equations as required (single equation, system of linear equations, or polynomial equation).
  4. Solve algebraically and check solutions in the original context (reject extraneous or infeasible results).

Step-by-Step Example

Question

Two trains start from stations A and B which are 540 km apart and move towards each other. The faster train covers 180 km more than the slower train in the same time. If their combined speed is 160 km/h, find the speeds of the two trains.

Solution

  1. Step 1: Assign variables

    Let the speed of the slower train be v km/h. Then the faster train's speed = v + 180/t is not ideal here because 180 is distance difference; instead use time variable.

    Use time: Let both trains travel for time t hours until they meet. Speeds: slower = s, faster = f.

  2. Step 2: Translate statements into equations

    They move towards each other and meet after time t, so distance covered together = 540 km:

    (s × t) + (f × t) = 540 → (s + f) t = 540.

    Given combined speed s + f = 160 km/h, substitute into above:

    160 × t = 540 ⇒ t = 540 / 160 = 27/8 hours = 3.375 hours.

    Also given: the faster train covers 180 km more than the slower in the same time:

    f t - s t = 180 ⇒ (f - s) t = 180.

  3. Step 3: Solve for (f - s)

    We have t = 27/8, so:

    f - s = 180 / t = 180 ÷ (27/8) = 180 × (8/27) = (180/27) × 8 = (20/3) × 8 = 160/3 ≈ 53.333... km/h.

  4. Step 4: Use system to find f and s

    We know:

    • f + s = 160
    • f - s = 160/3

    Add the two equations:

    2f = 160 + 160/3 = (480/3 + 160/3) = 640/3 ⇒ f = 320/3 ≈ 106.666... km/h.

    Then s = 160 - f = 160 - 320/3 = (480/3 - 320/3) = 160/3 ≈ 53.333... km/h.

  5. Step 5: Interpret and present final speeds

    Faster train speed = 320/3 km/h (≈ 106.67 km/h). Slower train speed = 160/3 km/h (≈ 53.33 km/h).

  6. Step 6: Quick Check

    1. Combined speed = 320/3 + 160/3 = 480/3 = 160 km/h ✓
    2. Time to meet = distance / combined speed = 540 / 160 = 27/8 hours ✓
    3. Distance faster covers = f × t = (320/3) × (27/8) = (320 × 27)/(24) = (320 × 9)/8 = 2880/8 = 360 km. Slower covers 540 - 360 = 180 km; difference = 180 ✓

Quick Variations

1. Work and time: use rates (work per hour) and add rates for combined work.

2. Mixture problems: use concentration × quantity balance to form linear equations.

3. Age problems: translate relative ages and form linear equations with time shifts.

4. Ratio and proportion: convert ratios into variable multiples and solve systems.

5. Parameterized problems: form equations containing parameter k and solve symbolically then apply constraints (e.g., integer roots, positivity).

Trick to Always Use

  • Step 1 → Define variables clearly (use separate variables for rates/time when distances differ).
  • Step 2 → Translate each sentence to an algebraic equation; keep units consistent (km, h, etc.).
  • Step 3 → If two quantities are compared over the same time, use time t as a linking variable to avoid misplacing constants.
  • Step 4 → Solve the resulting system (substitution or elimination). Always validate solutions in the problem context and discard infeasible ones.

Summary

Summary

Key takeaways for Equation-based Word Problems (Advanced):

  • Careful variable assignment is critical; choose variables that simplify translation.
  • Translate each phrase to an equation; check units and whether a parameter (time, rate) is shared.
  • Use systems of equations when multiple unknowns interact; prefer elimination/substitution based on convenience.
  • Always verify answers in the original context to catch extraneous or infeasible solutions.

Practice

(1/5)
1. The sum of two consecutive even numbers is 46. Find the numbers.
easy
A. 22 and 24
B. 20 and 26
C. 24 and 22
D. 18 and 28

Solution

  1. Step 1: Assign variables

    Let the first even number be x. The next consecutive even number is x + 2.

  2. Step 2: Form the equation

    x + (x + 2) = 46 ⇒ 2x + 2 = 46.

  3. Step 3: Solve for x

    2x = 44 ⇒ x = 22. So the numbers are 22 and 24.

  4. Final Answer:

    22 and 24 → Option A.

  5. Quick Check:

    22 + 24 = 46 ✅

Hint: For consecutive even numbers use x and x + 2.
Common Mistakes: Using x and x + 1 (which are consecutive integers, not even numbers).
2. A number is 6 less than twice another number. If their sum is 24, find the numbers.
easy
A. 12 and 6
B. 10 and 14
C. 8 and 16
D. 9 and 15

Solution

  1. Step 1: Assign variables

    Let the smaller number be x. Then the other number = 2x - 6 (since it is 6 less than twice the smaller).

  2. Step 2: Form the equation

    x + (2x - 6) = 24 ⇒ 3x - 6 = 24.

  3. Step 3: Solve

    3x = 30 ⇒ x = 10. The other number = 2×10 - 6 = 14.

  4. Final Answer:

    10 and 14 → Option B.

  5. Quick Check:

    10 + 14 = 24 and 14 = 2×10 - 6 ✅

Hint: Translate 'a number is 6 less than twice another' as y = 2x - 6.
Common Mistakes: Reversing the relation and writing 2x = y - 6 instead of y = 2x - 6.
3. A pipe can fill a tank in 12 hours, while another can empty it in 18 hours. If both are opened together, how long will it take to fill the tank?
easy
A. 24 hours
B. 10.8 hours
C. 14.4 hours
D. 36 hours

Solution

  1. Step 1: Define rates

    Filling pipe rate = 1/12 tank per hour. Emptying pipe rate = 1/18 tank per hour.

  2. Step 2: Net rate when both are open

    Net fill rate = 1/12 - 1/18 = (3 - 2)/36 = 1/36 tank per hour.

    3. Step 3: Time to fill

    Time = 1 / (1/36) = 36 hours.

  3. Final Answer:

    36 hours → Option D.

  4. Quick Check:

    In 36 hours, filler supplies 36/12 = 3 tanks, emptier removes 36/18 = 2 tanks → net = 1 tank filled ✅

Hint: Net rate = (1/T_fill - 1/T_empty); total time = 1 / net rate.
Common Mistakes: Adding rates when one pipe empties instead of subtracting.
4. The sum of a number and its reciprocal is 10/3. Find the number.
medium
A. 3 or 1/3
B. 2 or 1/2
C. 5 or 1/5
D. 4 or 1/4

Solution

  1. Step 1: Assign variable

    Let the number be x. Given x + 1/x = 10/3.

  2. Step 2: Clear denominator

    Multiply both sides by x: x² + 1 = (10/3)x ⇒ 3x² - 10x + 3 = 0.

  3. Step 3: Factor or use quadratic formula

    Factor: (3x - 1)(x - 3) = 0 ⇒ x = 1/3 or x = 3.

  4. Final Answer:

    3 or 1/3 → Option A.

  5. Quick Check:

    3 + 1/3 = 10/3 ✅

Hint: Multiply x + 1/x = k by x to form a quadratic: x² - kx + 1 = 0 (after clearing denominators appropriately).
Common Mistakes: Forgetting to multiply the whole equation by x and losing the reciprocal term.
5. A train travels 240 km at a certain speed. If its speed were 20 km/h more, it would take 1 hour less. Find the speed.
medium
A. 40 km/h
B. 50 km/h
C. 60 km/h
D. 80 km/h

Solution

  1. Step 1: Let speed be x km/h

    Then time = 240/x hours. If speed is x + 20, new time = 240/(x + 20).

  2. Step 2: Form the time-difference equation

    240/x - 240/(x + 20) = 1.

  3. Step 3: Simplify

    240[(x + 20 - x)/(x(x + 20))] = 1 ⇒ 240×20 = x(x + 20) ⇒ x² + 20x - 4800 = 0.

  4. Step 4: Solve the quadratic

    Discriminant = 400 + 19200 = 19600, √D = 140. x = (-20 ± 140)/2 ⇒ x = 60 or x = -80 (reject negative).

  5. Final Answer:

    60 km/h → Option C.

  6. Quick Check:

    240/60 = 4 hr; 240/80 = 3 hr; difference = 1 hr ✅

Hint: Use distance = speed × time and set up time difference equation for 'takes 1 hour less'.
Common Mistakes: Algebraic sign errors when simplifying the rational equation.

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