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Equation-based Word Problems (Advanced)

Introduction

Advanced word problems असली दुनिया की situations को algebraic equations में बदलते हैं - कई बार systems, higher-degree equations, या parameters वाली equations का उपयोग करना पड़ता है। इस pattern में mastery से आप relations को model कर सकते हैं, सही equations बना सकते हैं और efficiently solve कर सकते हैं।

ऐसे problems competitive exams और assessments में बहुत बार आते हैं, जहाँ reasoning, formulation और algebraic manipulation एक साथ test होते हैं।

Pattern: Equation-based Word Problems (Advanced)

Pattern

मुख्य concept: Verbal statements को ध्यान से algebraic relations में बदलें (variables लें, equations बनाएं, solve करें और validate करें)।

Typical steps:

  1. Unknown quantities के लिए variables को clearly assign करें।
  2. Phrases को algebraic expressions में बदलें (rates, ratios, sums, differences, products, ages, work/time, mixtures, percentages)।
  3. ज़रूरत के अनुसार equations बनाएं (single equation, system of linear equations, या polynomial equation)।
  4. Algebraically solve करें और original context में solutions check करें (extraneous या infeasible results को reject करें)।

Step-by-Step Example

Question

दो ट्रेनें स्टेशन A और B से एक-दूसरे की ओर चलती हैं। दोनों स्टेशनों के बीच 540 km की दूरी है। तेज ट्रेन, धीमी ट्रेन से उतने ही समय में 180 km ज़्यादा कवर करती है। अगर उनकी combined speed 160 km/h है, तो दोनों ट्रेनों की speed निकालें।

Solution

  1. Step 1: Variables assign करें

    मान लें धीमी ट्रेन की speed s km/h है और तेज ट्रेन की speed f km/h।

    दोनों एक ही समय t घंटों तक चलती हैं जब तक वे मिलती नहीं।

  2. Step 2: Statements को equations में बदलें

    वे एक-दूसरे की ओर चल रही हैं, इसलिए:

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

    दिया है कि combined speed = 160 km/h:

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

    दिया है कि तेज ट्रेन धीमी से 180 km ज़्यादा चलती है:

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

  3. Step 3: (f - s) निकालें

    t = 27/8 है, इसलिए:

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

  4. Step 4: System solve करें

    हमें पता है:

    • f + s = 160
    • f - s = 160/3

    दोनों को जोड़ें:

    2f = 160 + 160/3 = 640/3 ⇒ f = 320/3 ≈ 106.67 km/h.

    अब s = 160 - 320/3 = 160/3 ≈ 53.33 km/h.

  5. Step 5: Final speeds

    तेज ट्रेन की speed = 320/3 km/h (~106.67 km/h)। धीमी ट्रेन की speed = 160/3 km/h (~53.33 km/h)।

  6. Step 6: Quick Check

    1. Combined speed = 160 km/h ✓
    2. Time = 540 / 160 = 27/8 hours ✓
    3. Faster distance = 360 km, slower distance = 180 km; difference = 180 km ✓

Quick Variations

1. Work and time: rates (work per hour) का उपयोग करें और combined work के लिए rates जोड़ें।

2. Mixture problems: concentration × quantity balance से linear equations बनाएं।

3. Age problems: relative ages को equations में बदलें और time-shifts शामिल करें।

4. Ratio and proportion: ratios को variable multiples में बदलें और systems solve करें।

5. Parameterized problems: parameter k वाली equations बनाएं और बाद में constraints (जैसे integer roots, positivity) apply करें।

Trick to Always Use

  • Step 1 → Variables को clearly define करें (distances अलग हों तो rates/time अलग variables लें)।
  • Step 2 → हर sentence को algebraic equation में बदलें; units (km, h) consistent रखें।
  • Step 3 → अगर दो quantities एक ही time में compare हों, तो linking variable t का उपयोग करें ताकि constants गलत जगह न लगें।
  • Step 4 → System को solve करें (substitution या elimination)। हमेशा context में verify करें और infeasible solutions को reject करें।

Summary

Summary

Equation-based Word Problems (Advanced) के लिए मुख्य बातें:

  • Variables को ध्यान से assign करें; ऐसे variables चुनें जो translation आसान बनाएं।
  • हर phrase को equation में बदलें; units और shared parameters (time, rate) पर ध्यान दें।
  • Multiple unknowns हों तो systems of equations बनाएं; convenience के हिसाब से elimination/substitution चुनें।
  • Solutions को original context में verify करके extraneous या infeasible results हटाएं।

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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