Class 11 Maths · Chapter 5
Short answer:
NCERT Class 11 Maths Chapter 5 Linear Inequalities sikhata hai ki jab do quantities barabar (=) nahi balki kam ya zyada (<, >, ≤, ≥) hoti hain to unhe kaise solve aur graph karte hain — pehle ek variable me number-line pe, phir do variables me coordinate plane pe. Ye is series ka wahi practical chapter hai jise students class 11 maths ncert solutions dhoondte waqt sabse pehle search karte hain kyunki iske rules (specially negative number se multiply/divide karne pe sign flip) exam me galti ka #1 source hain.
Chapter 5 Linear Inequalities us kaam ko aage badhata hai jo class 11 maths chapter 1 sets ncert solutions me shuru hua tha — interval notation (a,b), [a,b] jo Sets me define hui thi, yahan solution-sets likhne ka main tool ban jaati hai. Farak sirf itna hai: ab hum equation (=) nahi, inequality (<, >, ≤, ≥) solve kar rahe hain, aur inequality ke apne alag rules hain — jaise negative number se multiply/divide karte hi sign palat (flip ho) jaata hai, jo Class 10 tak equation-solving me kabhi zaroorat nahi padti thi.
Chapter teen hisso me banta hai: (1) ek variable wali inequality algebraically solve karna aur number-line pe dikhana, (2) do variable wali inequality ko coordinate plane pe graph karna, aur (3) do ya zyada inequalities ka system solve karke common (feasible) region nikalna — yahi tareeka Class 12 ke Linear Programming chapter ka seedha base banta hai. Latest class 11 maths syllabus 2026-27 pdf me ye chapter Algebra unit ke andar aata hai aur 14-chapter rationalised NCERT (Sets, Relations & Functions, Trigonometric Functions, Complex Numbers, Linear Inequalities, ...) ka Chapter 5 hai.
Chapter 5 Summary — 5 Minute Revision
Is chapter ka core idea simple hai: inequality solve karne ke rules almost equation jaise hi hain — dono taraf same number add/subtract karo, ya same positive number se multiply/divide karo, sign nahi badalta. Lekin jaise hi kisi negative number se multiply ya divide karte ho, inequality sign automatically reverse (flip) ho jaata hai — yahi is poore chapter ka sabse important aur sabse zyada bhoola jaane wala rule hai.
Ek-variable inequality ka solution number-line pe interval ki tarah dikhaya jaata hai (strict </> ke liye open circle, ≤/≥ ke liye closed/solid circle). Do-variable inequality (jaise 2x + y ≥ 6) ek poori half-plane represent karti hai — boundary line ke ek taraf ka poora region — jise seedhi line kheenchkar aur ek test point (aksar origin) daalkar shade kiya jaata hai. Jab do ya teen inequalities ek saath diye jaate hain (system), sab regions ka common (intersection) area hi final answer hota hai — kabhi-kabhi ye region khaali (empty) bhi ho sakta hai, matlab system ka koi solution nahi. Word problems (marks ka average, mixture/dilution, temperature conversion, IQ formula) is chapter ki real-life application dikhate hain aur board exam me high-scoring section hote hain.
In-Text Questions — Solutions
Exercise 5.1 – Q1. Solve 24x < 100, where x is a real number.
24x < 100
Dono taraf 24 se divide karo (24 positive hai, isliye sign flip nahi hoga):
x < 100/24 = 25/6
Answer: x ∈ (−∞, 25/6)
Exercise 5.1 – Q2. Solve −12x > 30, where x is a real number.
−12x > 30
Dono taraf −12 se divide karna hai — yahan number negative hai, isliye sign flip hoga (> ban jaayega <):
x < 30/(−12) = −5/2
Answer: x ∈ (−∞, −5/2)
Exercise 5.1 – Q3. Solve 5x − 3 < 7, where x is a real number.
5x − 3 < 7
Dono taraf 3 add karo:
5x < 10
Dono taraf 5 se divide karo (positive, sign same rahega):
x < 2
Answer: x ∈ (−∞, 2)
Exercise 5.1 – Q4. Solve 3x + 8 > 2, where x is a real number.
3x + 8 > 2
8 subtract karo:
3x > −6
3 se divide karo:
x > −2
Answer: x ∈ (−2, ∞)
Exercise 5.1 – Q5. Solve 3(x − 1) ≤ 2(x − 3).
Dono taraf expand karo:
3x − 3 ≤ 2x − 6
Dono taraf se 2x subtract karo:
x − 3 ≤ −6
3 add karo:
x ≤ −3
Answer: x ∈ (−∞, −3]
Exercise 5.1 – Q6. Solve x/2 ≥ (5x − 2)/3 − (7x − 3)/5.
LCM(2,3,5) = 30 se poori inequality ko multiply karo:
15x ≥ 10(5x − 2) − 6(7x − 3)
15x ≥ 50x − 20 − 42x + 18 = 8x − 2
15x − 8x ≥ −2 ⟹ 7x ≥ −2
x ≥ −2/7
Answer: x ∈ [−2/7, ∞)
Exercise 5.1 – Q7. Solve (2x − 1)/3 ≥ (3x − 2)/4 − (2 − x)/5.
LCM(3,4,5) = 60 se multiply karo:
20(2x − 1) ≥ 15(3x − 2) − 12(2 − x)
40x − 20 ≥ 45x − 30 − 24 + 12x = 57x − 54
−20 + 54 ≥ 57x − 40x ⟹ 34 ≥ 17x
x ≤ 2
Answer: x ∈ (−∞, 2]
Exercise 5.1 – Q8. Solve 3x − 7 > 5x − 1 for real x.
3x − 7 > 5x − 1
−7 + 1 > 5x − 3x
−6 > 2x
2 se divide karo (positive):
x < −3
Answer: x ∈ (−∞, −3)
Exercise 5.1 – Q9. Solve 3(x − 2)/5 ≤ 5(2 − x)/3 for real x.
Dono taraf 15 se multiply:
9(x − 2) ≤ 25(2 − x)
9x − 18 ≤ 50 − 25x
9x + 25x ≤ 50 + 18 ⟹ 34x ≤ 68
x ≤ 2
Answer: x ∈ (−∞, 2]
Exercise 5.1 – Q10. Ravi ne pehle do unit tests me 70 aur 75 marks paaye. Third test me kam-se-kam kitne marks lene honge taaki teeno test ka average kam-se-kam 60 rahe (har test 100 marks ka hai)?
Maan lo third test ke marks = x
(70 + 75 + x)/3 ≥ 60
145 + x ≥ 180
x ≥ 35
Answer: Ravi ko kam-se-kam 35 marks chahiye, yaani x ∈ [35, 100]
Exercise 5.1 – Q11. Grade 'A' pane ke liye 5 papers (har ek 100 marks ka) me average kam-se-kam 90 chahiye. Sunita ke pehle 4 papers ke marks 87, 92, 94, 95 hain. 5ve paper me minimum kitne marks chahiye?
Pehle 4 papers ka sum = 87+92+94+95 = 368
Maan lo 5va paper = x
(368 + x)/5 ≥ 90
368 + x ≥ 450
x ≥ 82
Answer: Sunita ko kam-se-kam 82 marks chahiye, yaani x ∈ [82, 100]
Exercise 5.1 – Q12. Consecutive odd positive integers ki wo saari jodiyaan dhoondo jo dono 10 se chhoti hon aur unka sum 11 se zyada ho.
Maan lo chhota odd integer = x, agla = x + 2
Condition 1 (dono < 10): x + 2 < 10 ⟹ x < 8, aur x odd hai to x ∈ {1,3,5,7}
Condition 2 (sum > 11):
x + (x+2) > 11 ⟹ 2x + 2 > 11 ⟹ x > 4.5
Odd values me x > 4.5 satisfy karte hain x = 5, 7
Answer: Pairs hain (5, 7) aur (7, 9)
Exercise 5.2 – Q1. Solve graphically: x + y < 5.
Boundary line x + y = 5 kheencho — strict inequality (<) hai isliye line dashed hogi (line khud solution me include nahi).
Test point (0,0): 0 + 0 = 0 < 5 — True
Answer: Origin wali side shade karo — line ke neeche/left ka poora region (line exclude).
Exercise 5.2 – Q2. Solve graphically: 2x + y ≥ 6.
Boundary line 2x + y = 6 kheencho — ≥ hai isliye line solid hogi.
Test point (0,0): 0 + 0 = 0 ≥ 6 — False
Answer: Origin ki opposite side shade karo (line include, kyunki solid).
Exercise 5.2 – Q3. Solve graphically: 3x + 4y ≤ 12.
Boundary line 3x + 4y = 12 kheencho — ≤ hai isliye line solid hogi.
Test point (0,0): 0 ≤ 12 — True
Answer: Origin wali side shade karo, line included.
Exercise 5.2 – Q4. Solve graphically: y + 8 ≥ 2x.
Inequality ko rewrite karo: 2x − y ≤ 8. Boundary line 2x − y = 8 kheencho — solid line (≤).
Test point (0,0): 2(0) − 0 = 0 ≤ 8 — True
Answer: Origin wali side shade karo, line included.
Exercise 5.3 – Q1. Solve graphically: x + y ≤ 4, x ≥ 0, y ≥ 0.
x ≥ 0, y ≥ 0 se region first quadrant tak seemit ho jaata hai. Line x + y = 4 kheencho (solid), origin test point (0,0): 0 ≤ 4 True — origin wali side shade.
Answer: Feasible region ek triangle hai jiske vertices (0,0), (4,0), (0,4) hain, boundaries solid (included).
Exercise 5.3 – Q2. Solve graphically: 2x + y ≥ 8, x + 2y ≥ 10.
Dono lines ka intersection nikalo: 2x + y = 8 se y = 8 − 2x. Isse x + 2y = 10 me daalo:
x + 2(8 − 2x) = 10 ⟹ x + 16 − 4x = 10 ⟹ −3x = −6 ⟹ x = 2, y = 4
Test point (0,0) dono me False hai (0 ≥ 8 galat, 0 ≥ 10 galat), isliye dono taraf origin se door wali side shade hogi.
Answer: Common region unbounded hai, dono solid lines origin se door ki side pe milte hue, intersection point (2, 4) se guzarti hui.
Exercise 5.3 – Q3. Solve graphically: 2x − y > 1, x − 2y < −1.
Dono boundary lines 2x − y = 1 aur x − 2y = −1 (dono dashed, strict inequality) ka intersection:
y = 2x − 1 ko x − 2y = −1 me daalo: x − 2(2x−1) = −1 ⟹ −3x + 2 = −1 ⟹ x = 1, y = 1
Test (0,0): 2(0)−0=0 > 1 False; 0−0=0 < −1 False — dono taraf origin wali side EXCLUDE hogi.
Answer: Common region dono lines ki origin-wali opposite side ka intersection hai, dashed boundaries, meeting point (1,1) se.
Exercise 5.3 – Q4. Solve graphically: x + y ≤ 6, x + y ≥ 4.
Ye do parallel lines hain: x + y = 6 aur x + y = 4, dono solid (≤ aur ≥).
Answer: Common region in dono parallel lines ke beech ki ek 'strip' (band) hai, dono boundaries included.
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q10)
Misc. Exercise – Q1. Solve 2 ≤ 3x − 4 ≤ 5, x real.
Poori inequality me 4 add karo:
6 ≤ 3x ≤ 9
3 se divide karo:
2 ≤ x ≤ 3
Answer: x ∈ [2, 3]
Misc. Exercise – Q2. Solve 6 ≤ −3(2x − 4) < 12.
Pehle expand karo: −3(2x−4) = −6x + 12
6 ≤ −6x + 12 < 12
Left part: 6 ≤ −6x + 12 ⟹ −6 ≤ −6x ⟹ (−6 se divide, sign flip) x ≤ 1
Right part: −6x + 12 < 12 ⟹ −6x < 0 ⟹ (flip) x > 0
Answer: 0 < x ≤ 1, yaani x ∈ (0, 1]
Misc. Exercise – Q3. Solve −3 ≤ (4 − 7x)/2 ≤ 18.
Poori inequality ko 2 se multiply karo:
−6 ≤ 4 − 7x ≤ 36
4 subtract karo:
−10 ≤ −7x ≤ 32
−7 se divide karo (sign flip + order reverse):
−32/7 ≤ x ≤ 10/7
Answer: x ∈ [−32/7, 10/7]
Misc. Exercise – Q4. Solve −12 < 4 − 3x/(−5) ≤ 2.
4 − 3x/(−5) ko simplify karo = 4 + 3x/5
−12 < 4 + 3x/5 ≤ 2
4 subtract karo:
−16 < 3x/5 ≤ −2
5/3 se multiply karo:
−80/3 < x ≤ −10/3
Answer: x ∈ (−80/3, −10/3]
Misc. Exercise – Q5. Solve 7 ≤ (3x + 11)/2 ≤ 11.
2 se multiply karo:
14 ≤ 3x + 11 ≤ 22
11 subtract karo:
3 ≤ 3x ≤ 11
3 se divide karo:
1 ≤ x ≤ 11/3
Answer: x ∈ [1, 11/3]
Misc. Exercise – Q6. Ek solution ko 68°F aur 77°F ke beech rakhna hai. Formula C = (5/9)(F − 32) use karke Celsius me range batao.
Diya hai: 68 < F < 77, aur F = (9/5)C + 32
Substitute karo:
68 < (9/5)C + 32 < 77
32 subtract karo:
36 < (9/5)C < 45
5/9 se multiply karo:
20 < C < 25
Answer: Temperature 20°C se 25°C ke beech honi chahiye.
Misc. Exercise – Q7. 8% boric acid ke 640 litres solution me 2% boric acid solution mix karna hai taaki final mixture 4% se zyada par 6% se kam ho. 2% solution kitna add karna hoga?
Maan lo x litres 2% solution add kiya. Total volume = (640 + x)
Total acid = 0.08(640) + 0.02x = 51.2 + 0.02x
Concentration condition:
0.04 < (51.2 + 0.02x)/(640 + x) < 0.06
Left part: 51.2 + 0.02x > 0.04(640+x) = 25.6 + 0.04x ⟹ 25.6 > 0.02x ⟹ x < 1280
Right part: 51.2 + 0.02x < 0.06(640+x) = 38.4 + 0.06x ⟹ 12.8 < 0.04x ⟹ x > 320
Answer: 320 litres se zyada aur 1280 litres se kam 2% solution add karna hoga, yaani x ∈ (320, 1280)
Misc. Exercise – Q8. IQ ka formula hai IQ = (MA/CA) × 100, jahan MA = mental age, CA = chronological age. Agar 12-saal ke bachcho ke group ke liye 80 ≤ IQ ≤ 140 ho, to unki mental age ki range batao.
CA = 12, isliye IQ = (100 × MA)/12
80 ≤ (100 MA)/12 ≤ 140
12/100 se multiply karo:
9.6 ≤ MA ≤ 16.8
Answer: Mental age 9.6 se 16.8 saal ke beech hogi.
Misc. Exercise – Q9. Solve graphically: x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.
y ≥ 1 aur x + 2y ≤ 3 se: x ≤ 3 − 2y ≤ 3 − 2(1) = 1, yaani x ≤ 1 hamesha rahega jab y ≥ 1.
Doosri taraf 3x + 4y ≥ 12 se: x ≥ (12 − 4y)/3. y = 1 pe x ≥ 8/3 ≈ 2.67 chahiye.
Lekin pehli condition se x ≤ 1 hi allowed hai — 1 aur 2.67 ek saath satisfy nahi ho sakte, aur y badhne se ye gap aur badhta hai.
Answer: In char conditions ko satisfy karne wala koi common point exist nahi karta — feasible region empty (khaali) hai, yaani system ka koi solution nahi.
Misc. Exercise – Q10. Solve graphically: 3x + 2y ≤ 12, x ≥ 1, y ≥ 2.
Teeno boundary lines ke intersection points nikalo:
x = 1 aur y = 2 ka intersection: (1, 2) — check: 3(1)+2(2) = 7 ≤ 12 ✓
Line 3x+2y=12 aur x=1 ka intersection: 3+2y=12 ⟹ y = 4.5 → (1, 4.5)
Line 3x+2y=12 aur y=2 ka intersection: 3x+4=12 ⟹ x = 8/3 → (8/3, 2)
Answer: Feasible region ek bounded triangle hai jiske vertices (1, 2), (1, 4.5), aur (8/3, 2) hain.
Important Equations — Ek Nazar Me
| Rule / Concept | Statement |
|---|---|
| Addition/Subtraction Rule | Agar a > b, to a + c > b + c aur a − c > b − c (dono taraf same number add/subtract karne se sign nahi badalta) |
| Multiplication/Division by Positive | Agar a > b aur c > 0, to ac > bc aur a/c > b/c (sign same rehta hai) |
| Multiplication/Division by Negative | Agar a > b aur c < 0, to ac < bc aur a/c < b/c (sign reverse (flip) ho jaata hai) |
| Interval Notation | (a, b) = strict {x : a < x < b}; [a, b] = inclusive {x : a ≤ x ≤ b}; (a, b] aur [a, b) mixed |
| Number Line Representation | Strict inequality (<, >) → open (khula) circle; Non-strict (≤, ≥) → closed/solid circle |
| Graphing Two-Variable Inequality | Boundary line kheencho — strict (<,>) ho to dashed, non-strict (≤,≥) ho to solid. Test point (aksar origin) se decide karo kaunsi side shade karni hai |
| Test Point Method | Koi point boundary line par na ho (jaise (0,0)) — use inequality me daalo. True aayega to us point wali side shade karo, False aayega to opposite side |
| System of Inequalities | Do ya zyada inequalities ka solution = sab individual regions ka intersection (common area) — ye area empty bhi ho sakta hai |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Negative number se multiply/divide karte waqt inequality sign flip karna bhool jaana — ye is chapter ki sabse common exam galti hai.
- Strict inequality (<, >) ke solution ko closed circle/solid line se dikhana, jabki open circle/dashed line sahi hai.
- Do-variable inequality graph karte waqt test point check na karna aur galat side shade kar dena.
- Compound inequality (jaise 2 ≤ 3x−4 ≤ 5) solve karte waqt beech ke sabhi teeno parts pe same operation apply na karna.
- System of inequalities me sirf ek inequality ka region final answer maan lena, jabki intersection (common region) hi correct solution hai.
- Word problems me 'kam-se-kam' (≥) aur 'zyada se zyada' (≤) ko ulta likh dena — jaise average ≥60 ki jagah ≤60 likh dena.
Board-Style Important Questions
- Solve 3x + 8 > 2, jahan x ek real number hai, aur solution ko interval notation me likho.
- Jab inequality ke dono taraf ek negative number se multiply karte hain, to sign ka kya hota hai? Ek example ke saath samjhao.
- Solve 24x < 100, agar x ke liye solution set (i) natural numbers (N) aur (ii) real numbers (R) me alag-alag batao.
- Ek student ke pehle do unit tests me marks 62 aur 48 hain (har test 100 marks ka). Third test me kam-se-kam kitne marks chahiye taaki average kam-se-kam 60 rahe?
- Solve graphically: 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0. Feasible region describe karo aur boundary lines ka intersection point nikalo.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 5 Linear Inequalities ke solutions kahan milte hain aur exercises kitne hain?
Is chapter me teen exercises hain — 5.1 (algebraic solution, one variable), 5.2 (graphical solution, two variables) aur 5.3 (system of inequalities), plus ek Miscellaneous Exercise. Step-by-step CBSE marking-scheme style class 11 maths ncert solutions yahan is page pe hi cover ki gayi hain — 5.1 ka word-problem section (marks average) exam me sabse zyada pucha jaata hai.
Linear Inequalities aur Chapter 1 Sets ka kya connection hai?
Interval notation — (a,b), [a,b] — jo aap class 11 maths chapter 1 sets ncert solutions me set-builder form ke saath seekhte ho, wahi yahan inequality ka solution likhne ke liye direct use hoti hai. Isliye Sets chapter ka interval-notation part revise kar lena is chapter ko easy bana deta hai.
Chapter ke saare formulas aur rules ka ek short revision chart kahan milega?
Is page ke formulas table me saare core rules (addition/subtraction, positive/negative multiplication, interval notation, test-point method) ek jagah diye hain. Agar poori kitaab ke liye chahiye to class 11 maths formulas pdf download karke sab chapters ka ek saath revision kar sakte ho.
NCERT Exemplar ke Linear Inequalities questions NCERT textbook se kaise alag hote hain?
class 11 maths exemplar solutions me is chapter ke questions thode zyada tricky hote hain — jaise multiple compound inequalities ek saath, ya system of inequalities jisme feasible region empty nikalta hai (jaisa Miscellaneous Exercise Q9 me dikhaya gaya). Textbook se practice ke baad hi Exemplar attempt karna better rehta hai.
Linear Inequalities chapter ka syllabus 2026-27 me exact weightage kya hai?
Exact number-of-marks weightage har saal CBSE ki official marking scheme se change ho sakti hai, isliye yahan koi specific figure claim nahi kar rahe — latest ke liye class 11 maths chapter wise weightage 2026-27 CBSE ke official curriculum document se hi verify karo. Itna zaroor pakka hai ki ye Algebra unit ka core, high-frequency chapter hai.
Chapter 5 ke baad revision ke liye kaunse aur chapters/resources cover karne chahiye?
Sequence me aage class 11 maths chapter 3 trigonometric functions ncert solutions, class 11 maths straight lines ncert solutions, class 11 maths chapter 8 sequences and series important questions, class 11 maths conic sections notes, class 11 maths limits and derivatives ncert solutions aur class 11 maths probability important questions revise karna helpful rahega — inequality ka logic (interval, sign flip) inme baar-baar wapas aata hai. Poori kitaab ke liye class 11 maths ncert book pdf download aur last-minute practice ke liye class 11 maths important questions with solutions bhi use kar sakte ho.
Class 11 Maths — Saare Chapters
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Class 11 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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