Unitary Method, Direct and Inverse Variation
Unitary Method, Direct and Inverse Variation
The unitary method finds the value of ONE unit and then scales up. Variation is the same idea written as an equation: in direct variation the ratio stays constant, in inverse variation the product stays constant. NTPC asks provisions-for-men questions, workers-and-days questions and combined variation such as "X ∝ A² and A ∝ 1/Y".
1. The Three Relationships
| Relationship | Meaning | Constant | Example |
|---|---|---|---|
| Direct (y ∝ x) | Both rise or fall together in the same ratio | y/x = k | Cost and number of items |
| Inverse (y ∝ 1/x) | One rises when the other falls | x × y = k | Workers and days; speed and time |
| Joint (y ∝ xz) | y varies directly as the product | y/(xz) = k | Work and (workers × days) |
| Square variation (y ∝ x²) | y changes by the square of the factor | y/x² = k | Area of a square and its side |
Direct: y₁/x₁ = y₂/x₂ Inverse: x₁y₁ = x₂y₂
2. Chain Rule (Men, Days, Hours, Work)
M₁ × D₁ × H₁ / W₁ = M₂ × D₂ × H₂ / W₂
Men, days and hours are inversely related to each other and directly to the work. Put the known values in and solve for the one unknown. Efficiency, if given, multiplies the men term.
3. Provisions Problems
Food is measured in man-days. Total stock = men × days. When men join or leave after some days, subtract the man-days already eaten, then divide the remainder by the new number of men.
Remaining days = (initial men × remaining days at that moment) / new number of men
4. Combined and Chain Variation
| Given | Result | Reason |
|---|---|---|
| X ∝ 1/Y and Y ∝ 1/Z | X ∝ Z | Two inversions cancel |
| X ∝ Y and Y ∝ Z | X ∝ Z | Direct chains stay direct |
| X ∝ A² and A ∝ 1/Y | X ∝ 1/Y² | Square the inverse relation |
| X ∝ Y and X ∝ 1/Z | X ∝ Y/Z | Joint variation |
✗ Speed up by 15 km/h saves time in the same proportion | ✓ Time is inversely proportional to speed: 75 → 90 km/h (20% faster) cuts time only by 1/6, not 20%
हिंदी नोट: प्रत्यक्ष समानुपात में दोनों राशियों का अनुपात स्थिर रहता है, जबकि व्युत्क्रम समानुपात में उनका गुणनफल स्थिर रहता है। भोजन वाले प्रश्नों में कुल भोजन = व्यक्ति × दिन मानकर चलिए।
Exam Pointer: Verified NTPC patterns: provisions for men with people leaving (Graduate CBT-1, March 2026), workers and days in inverse proportion (UG CBT-1, May 2026), x inversely proportional to y (Graduate CBT-2, October 2025) and combined or chain variation (Graduate CBT-1, June 2025, two shifts). The plain unitary method and the four-variable chain rule had no verified NTPC shift in our check, so they are tagged syllabus-based; the chain rule returns in full in the Time and Work chapter.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Unitary method and its two-item version
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. 12 pens cost ₹222. What will 20 pens cost?
One pen costs 222 ÷ 12 = ₹18.50, so 20 pens cost 18.50 × 20 = ₹370. Faster: 222 × 20/12 = 222 × 5/3 = ₹370.
Answer: ₹370
EXAMATLAS LEVEL
Q. 4 chairs and 3 tables cost ₹13,500, while 2 chairs and 5 tables cost ₹15,500. Find the cost of one chair and one table together.
Double the second statement: 4 chairs and 10 tables cost ₹31,000. Subtract the first: 7 tables cost 31,000 − 13,500 = ₹17,500, so one table = ₹2,500. Then 4 chairs = 13,500 − 7,500 = ₹6,000, one chair = ₹1,500. Chair + table = ₹4,000.
Answer: ₹4,000
Pattern 2: Provisions for a group when people join or leave
[PYQ: NTPC Graduate CBT-1 16-Mar-2026 Shift-2]
EXAM LEVEL
Q. A camp has food for 400 soldiers for 36 days. How long will the food last if there are 480 soldiers?
Food = 400 × 36 = 14,400 soldier-days. For 480 soldiers: 14,400 ÷ 480 = 30 days.
Answer: 30 days
EXAMATLAS LEVEL
Q. A hostel has food for 250 students for 40 days. After 10 days, 50 more students join. Five days later 100 students leave. For how many days in total did the food last from the start?
Stock = 250 × 40 = 10,000 student-days. First 10 days use 2,500, leaving 7,500. Next 5 days with 300 students use 1,500, leaving 6,000. Now 200 students remain, so the food lasts 6,000 ÷ 200 = 30 more days. Total = 10 + 5 + 30 = 45 days.
Answer: 45 days
Pattern 3: Inverse proportion: workers and days, speed and time
[PYQ: NTPC UG CBT-1 8-May-2026 Shift-2 | NTPC Graduate CBT-2 13-Oct-2025 Shift-2]
EXAM LEVEL
Q. 12 workers can build a wall in 15 days. In how many days will 9 workers build it?
Workers × days is constant: 12 × 15 = 180 worker-days. 180 ÷ 9 = 20 days. Fewer workers, more days, so the answer must exceed 15.
Answer: 20 days
EXAMATLAS LEVEL
Q. A car covers a distance in 6 hours at 75 km/h. If its speed is increased by 15 km/h, how much time is saved? By what percentage must the original speed rise to save 2 hours instead?
Distance = 450 km. At 90 km/h the time is 5 hours, so 1 hour is saved. To save 2 hours the time must be 4 hours, needing 450 ÷ 4 = 112.5 km/h, which is 37.5 km/h more, a 50% rise. Inverse proportion: cutting time by one-third needs speed × 3/2.
Answer: 1 hour; 50%
Pattern 4: Direct, combined and chain variation
[PYQ: NTPC Graduate CBT-1 10-Jun-2025 Shift-1 | NTPC Graduate CBT-1 23-Jun-2025 Shift-2]
EXAM LEVEL
Q. X varies inversely as Y, and Y varies inversely as Z. If X = 12 when Z = 4, find X when Z = 10.
Two inverse relations combine into a direct one: X ∝ Z, so X/Z is constant = 12/4 = 3. When Z = 10, X = 30.
Answer: 30
EXAMATLAS LEVEL
Q. X varies directly as the square of A, and A varies inversely as Y. If X = 50 when Y = 6, find X when Y = 3.
A ∝ 1/Y gives A² ∝ 1/Y², so X ∝ 1/Y² and X × Y² is constant = 50 × 36 = 1,800. When Y = 3, X = 1,800 ÷ 9 = 200. Halving Y multiplies X by 4, a quick check.
Answer: 200
Pattern 5: Chain rule with men, days, hours and work
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. 12 workers build a wall in 15 days working 8 hours a day. In how many days will 16 workers build the same wall working 6 hours a day?
12 × 15 × 8 = 16 × D × 6, so D = 1,440 ÷ 96 = 15 days.
Answer: 15 days
EXAMATLAS LEVEL
Q. 24 men working 8 hours a day dig a 60 m long trench in 15 days. How many men are needed to dig a 90 m trench in 12 days working 10 hours a day?
M₁D₁H₁/W₁ = M₂D₂H₂/W₂: (24 × 15 × 8)/60 = (M × 12 × 10)/90. Left side = 48, so M × 120 = 48 × 90 = 4,320 and M = 36.
Answer: 36 men
60-Second Revision
- Direct: ratio constant (y₁/x₁ = y₂/x₂); inverse: product constant (x₁y₁ = x₂y₂).
- Provisions: total = men × days; subtract what is eaten before men join or leave.
- Speed and time are inverse: 1/3 less time needs 50% more speed.
- Two inverse variations give a direct one; X ∝ A² with A ∝ 1/Y gives X ∝ 1/Y².
- Chain rule: M₁D₁H₁/W₁ = M₂D₂H₂/W₂.
Next Step: Ratio and Proportion done. Attempt the ratio, proportion and variation questions in the ExamAtlas RRB NTPC 2026 mock tests next, and map every mistake back to its pattern on these four pages.