Question 1
Solve 24x<100 when (i) x is a natural number, (ii) x is an integer.
Steps only
Divide by 24: x<25/6≈4.17. Restrict this interval to the requested number system.
Answer
- (i) x∈{1,2,3,4};
- (ii) x∈Z and x≤4.
CLASS 11 · MATHEMATICS · CHAPTER 5
Linear Inequalities
Solve linear inequalities carefully, interpret solution intervals and apply them to marks, lengths and number patterns.
40 worked answersAll 1 exercise setsAll 14 end-of-chapter questions
What you will learn
- Solve linear inequalities in one variable.
- Respect sign reversal when dividing by a negative number.
- Represent solutions as intervals or on number lines.
- Translate practical constraints into inequalities.
40 worked answers · short steps first, explanations when you need them
Exercise 5.1
Textbook pages 95–96
Question 2
Solve −12x>30 when (i) x is a natural number, (ii) x is an integer.
Steps only
Divide by −12 and reverse the sign: x<−30/12=−5/2. No positive natural number is below −2.5; integers below it end at −3.
Answer
- (i) No natural-number solutions;
- (ii) x∈Z and x≤−3.
Question 3
Solve 5x−3<7 when (i) x is an integer, (ii) x is real.
Steps only
Add 3: 5x<10. Divide by 5: x<2. Then apply the requested domain.
Answer
- (i) x∈Z and x≤1;
- (ii) x<2.
Question 4
Solve 3x+8>2 when (i) x is an integer, (ii) x is real.
Steps only
Subtract 8: 3x>−6. Divide by 3: x>−2.
Answer
- (i) x∈Z and x≥−1;
- (ii) x>−2.
Question 5
Solve 4x+3<5x+7 for real x.
Steps only
Subtract 4x: 3<x+7. Subtract 7: −4<x.
Answer
x>−4.
Question 6
Solve 3x−7>5x−1 for real x.
Steps only
Subtract 3x: −7>2x−1. Add 1: −6>2x. Divide by 2: x<−3.
Answer
x<−3.
Question 7
Solve 3(x−1)≤2(x−3).
Steps only
Expand: 3x−3≤2x−6. Subtract 2x, then add 3: x≤−3.
Answer
x≤−3.
Question 8
Solve 3(2−x)≥2(1−x).
Steps only
Expand: 6−3x≥2−2x. Subtract 2: 4−3x≥−2x. Add 3x: 4≥x.
Answer
x≤4.
Question 9
Solve x + x/2 + x/3 < 11.
Steps only
Multiply every term by 6: 6x+3x+2x<66.
Combine: 11x<66.
Divide by 11: x<6.
Answer
x < 6.
Question 10
Solve x/3 > x/2 + 1.
Steps only
Multiply by 6: 2x>3x+6.
Subtract 3x: −x>6.
Divide by −1 and reverse the sign: x<−6.
Answer
x < −6.
Question 11
Solve 3(x−2)/5 ≤ 5(2−x)/3.
Steps only
Multiply by 15: 9(x−2)≤25(2−x).
Expand and collect: 34x≤68.
Divide by 34: x≤2.
Answer
x ≤ 2.
Question 12
Solve ½(3x/5+4) ≥ ⅓(x−6).
Steps only
Multiply by 30: 15(3x/5+4)≥10(x−6).
Simplify: 9x+60≥10x−60.
Rearrange: 120≥x.
Answer
x ≤ 120.
Question 13
Solve 2(2x+3)−10 < 6(x−2).
Steps only
Expand: 4x−4<6x−12.
Add 12 and subtract 4x: 8<2x.
Divide by 2: x>4.
Answer
x > 4.
Question 14
Solve 37−(3x+5) ≥ 9x−8(x−3).
Steps only
Simplify: 32−3x≥x+24.
Subtract 24 and add 3x: 8≥4x.
Divide by 4: x≤2.
Answer
x ≤ 2.
Question 15
Solve x/4 < (5x−2)/3 − (7x−3)/5.
Steps only
Multiply by 60: 15x<20(5x−2)−12(7x−3).
Simplify: 15x<16x−4.
Subtract 16x: −x<−4; reverse when dividing by −1, so x>4.
Answer
x > 4.
Question 16
Solve (2x−1)/3 ≥ (3x−2)/4 − (2−x)/5.
Steps only
Multiply by 60: 20(2x−1)≥15(3x−2)−12(2−x).
Simplify: 40x−20≥57x−54.
Rearrange: 34≥17x, so x≤2.
Answer
x ≤ 2.
Question 17
Solve 3x−2<x+1 and show its number-line solution.
Steps only
Subtract x and add 2: 2x<3. Divide by 2. The endpoint is not included because the inequality is strict.
Answer
x<3/2; open circle at 1.5 and shade left.
Question 18
Solve 5x−3>3x−5 and show its number-line solution.
Steps only
Subtract 3x: 2x−3>−5. Add 3 and divide by 2: x>−1.
Answer
x>−1; open circle at −1 and shade right.
Question 19
Solve 3(1−x)<2(x+4) and show its number-line solution.
Steps only
Expand: 3−3x<2x+8. Subtract 3 and add 3x: −5<5x, so −1<x.
Answer
x>−1; open circle at −1 and shade right.
Question 20
Solve x/2 ≥ (5x−2)/3 − (7x−3)/5 and show the solution on a number line.
Steps only
Multiply by 30: 15x≥10(5x−2)−6(7x−3).
Simplify: 15x≥8x−2.
Thus 7x≥−2 and x≥−2/7.
Answer
x ≥ −2/7; closed endpoint at −2/7, shade to the right.
Question 21
Ravi scored 70 and 75 in two tests. What minimum score in test 3 gives an average of at least 60?
Steps only
Let the third score be x. (70+75+x)/3≥60, so 145+x≥180 and x≥35.
Answer
At least 35 marks.
Question 22
Sunita needs an average of at least 90 over five tests. Her first four marks are 87,92,94,95. Find the minimum fifth score.
Steps only
The first four total 368. Require (368+x)/5≥90, so 368+x≥450 and x≥82.
Answer
82 marks.
Question 23
Find consecutive positive odd pairs, both below 10, whose sum exceeds 11.
Steps only
Possible pairs are (1,3),(3,5),(5,7),(7,9). Their sums are 4,8,12,16; only the last two exceed 11.
Answer
(5,7) and (7,9).
Question 24
Find consecutive positive even pairs, both above 5, whose sum is below 23.
Steps only
List pairs in order and test sums: 14,18,22,26. The first three are below 23.
Answer
(6,8),(8,10),(10,12).
Question 25
A triangle’s longest side is three times its shortest side x; the third side is 2 cm shorter than the longest. Its perimeter is at least 61 cm. Find the minimum x.
Steps only
The sides are x,3x,3x−2. Thus 7x−2≥61, so x≥9. At x=9, the sides 9,27,25 satisfy the triangle inequalities, so the minimum is attainable.
Answer
9 cm.
Question 26
A 91 cm board is cut into x, x+3 and 2x cm. The third piece is at least 5 cm longer than the second. Find possible x.
Steps only
Length constraint: x+(x+3)+2x≤91 gives x≤22. Length comparison: 2x≥(x+3)+5 gives x≥8. Combine them.
Answer
8≤x≤22 cm.
Selected Miscellaneous Exercise
Textbook pages 96–98
Question 1
Solve 2≤3x−4≤5.
Steps only
Add 4 throughout: 6≤3x≤9. Divide all parts by positive 3: 2≤x≤3.
Answer
2≤x≤3.
Question 2
Solve 6≤−3(2x−4)<12.
Steps only
Expand: 6≤−6x+12<12. The left part gives x≤1 after division by −6 and reversing sign. The right part gives x>0. Intersect: 0<x≤1.
Answer
0<x≤1.
Question 3
Solve −3 ≤ 4−7x/2 ≤ 18.
Steps only
Subtract 4: −7≤−7x/2≤14.
Multiply by −2/7 and reverse both inequality signs: 2≥x≥−4.
Write in increasing order: −4≤x≤2.
Answer
−4 ≤ x ≤ 2.
Question 4
Solve −15 < 3(x−2)/5 ≤ 0.
Steps only
Multiply all parts by 5/3: −25<x−2≤0.
Add 2 throughout: −23<x≤2.
Answer
−23 < x ≤ 2.
Question 5
Solve −12 < 4−3x/(−5) ≤ 2.
Steps only
Simplify: −12<4+3x/5≤2.
Subtract 4: −16<3x/5≤−2.
Multiply by 5/3: −80/3<x≤−10/3.
Answer
−80/3 < x ≤ −10/3.
Question 6
Solve 7 ≤ (3x+11)/2 ≤ 11.
Steps only
Multiply by 2: 14≤3x+11≤22.
Subtract 11: 3≤3x≤11.
Divide by 3: 1≤x≤11/3.
Answer
1 ≤ x ≤ 11/3.
Question 7
Solve the simultaneous inequalities 5x+1>−24 and 5x−1<24.
Steps only
The first gives 5x>−25, hence x>−5. The second gives 5x<25, hence x<5. Both must hold.
Answer
−5<x<5.
Question 8
Solve 2(x−1)<x+5 and 3(x+2)>2−x simultaneously.
Steps only
First inequality gives x<7. Second gives 3x+6>2−x, so 4x>−4 and x>−1. Take the overlap.
Answer
−1<x<7.
Question 9
Solve 3x−7>2(x−6) and 6−x>11−2x simultaneously.
Steps only
The first gives x>−5. The second gives x>5. Their common solution is x>5.
Answer
x>5.
Question 10
Solve 5(2x−7)−3(2x+3)≤0 and 2x+19≤6x+47 together; show the solution on a number line.
Steps only
First inequality simplifies to 4x−44≤0, so x≤11.
Second gives 2x+19≤6x+47, hence −28≤4x and x≥−7.
Their overlap is [−7,11], including both endpoints.
Answer
−7 ≤ x ≤ 11.
Question 11
A solution must stay between 68°F and 77°F. Find its Celsius range if F=9C/5+32.
Steps only
Translate “between” as 68< F <77.
Substitute: 68<9C/5+32<77.
Subtract 32, then multiply by 5/9: 20<C<25.
Answer
20°C < C < 25°C.
Question 12
How many litres of 2% boric acid solution should be added to 640 L of 8% solution so the mixture is more than 4% but less than 6% acid?
Steps only
Let x litres be added. Acid amount is 51.2+0.02x litres; total volume is 640+x.
Require 0.04(640+x)<51.2+0.02x<0.06(640+x).
Solving the left side gives x<1280; the right side gives x>320.
Answer
More than 320 L and less than 1280 L.
Question 13
How much water must be added to 1125 L of 45% acid so the mixture has more than 25% but less than 30% acid?
Steps only
The amount of acid stays 0.45×1125=506.25 L. Let x be the water added.
Require 0.25(1125+x)<506.25<0.30(1125+x).
The left inequality gives x<900; the right gives x>562.5.
Answer
More than 562.5 L and less than 900 L.
Question 14
For a 12-year-old group, IQ=MA/CA×100 and 80≤IQ≤140. Find the mental-age range.
Steps only
Here CA=12, so 80≤100MA/12≤140.
Multiply by 12/100, a positive number, so the signs stay the same.
This gives 9.6≤MA≤16.8.
Answer
9.6 years ≤ MA ≤ 16.8 years.