Solve 24x < 100, when (i) x is a natural number, (ii) x is an integer.
Hint. Divide both sides by 24 (positive, no sign flip) to isolate x, then list the values allowed by each number system.
24x<100 gives x<100/24=25/6, approximately 4.17. (i) Natural numbers less than 25/6: 1,2,3,4. (ii) Integers less than 25/6: ...,-2,-1,0,1,2,3,4.
✦ Working through each part gives: (i) {1,2,3,4}. (ii) {...,-2,-1,0,1,2,3,4}.
Solve -12x > 30, when (i) x is a natural number, (ii) x is an integer.
Hint. Dividing by -12 flips the inequality sign.
-12x>30, dividing by -12 (flip): x<-30/12=-5/2. (i) No natural number is less than -5/2, since all natural numbers are positive -- no solution. (ii) Integers less than -5/2: ...,-5,-4,-3.
✦ Working through each part gives: (i) No solution (empty set). (ii) {...,-5,-4,-3}.
Solve 5x - 3 < 7, when (i) x is an integer, (ii) x is a real number.
Hint. Isolate x using addition and division by a positive number -- no sign flip needed.
5x<10, so x<2. (i) Integers less than 2: ...,-1,0,1. (ii) Real numbers less than 2: the interval (-infinity,2).
✦ Working through each part gives: (i) {...,-1,0,1}. (ii) (-infinity, 2).
Solve 3x + 8 > 2, when (i) x is an integer, (ii) x is a real number.
Hint. Isolate x using subtraction and division by a positive number.
3x>-6, so x>-2. (i) Integers greater than -2: -1,0,1,2,.... (ii) Real numbers greater than -2: the interval (-2,infinity).
✦ Working through each part gives: (i) {-1,0,1,2,...}. (ii) (-2, infinity).
Solve 4x + 3 < 5x + 7 for real x.
Hint. Collect x terms on one side and constants on the other.
4x-5x<7-3, so -x<4, i.e. x>-4 (dividing by -1 flips the sign).
✦ Working through each part gives: x > -4, i.e. the interval (-4, infinity).
Solve 3x - 7 > 5x - 1 for real x.
Hint. Collect x terms on one side; dividing by a negative coefficient will flip the sign.
3x-5x>-1+7, so -2x>6, i.e. x<-3 (dividing by -2 flips the sign).
✦ Working through each part gives: x < -3, i.e. the interval (-infinity, -3).
Solve 3(x-1) <= 2(x-3) for real x.
Hint. Expand both sides first, then collect x terms on one side.
3x-3<=2x-6, so 3x-2x<=-6+3, giving x<=-3.
✦ Working through each part gives: x <= -3, i.e. the interval (-infinity, -3].
Solve 3(2-x) >= 2(1-x) for real x.
Hint. Expand both sides, then collect x terms carefully since both sides have a negative x term.
6-3x>=2-2x, so 6-2>=-2x+3x, giving 4>=x, i.e. x<=4.
✦ Working through each part gives: x <= 4, i.e. the interval (-infinity, 4].
Solve x + x/2 + x/3 < 11 for real x.
Hint. Multiply through by the LCM of the denominators (6) to clear fractions in one step.
Multiplying by 6: 6x+3x+2x<66, so 11x<66, giving x<6.
✦ Working through each part gives: x < 6, i.e. the interval (-infinity, 6).
Solve x/3 > x/2 + 1 for real x.
Hint. Multiply through by 6 (LCM of 3 and 2) to clear both fractions at once.
Multiplying by 6: 2x>3x+6, so 2x-3x>6, giving -x>6, i.e. x<-6.
✦ Working through each part gives: x < -6, i.e. the interval (-infinity, -6).
Solve 3(x-2)/5 <= 5(2-x)/3 for real x.
Hint. Multiply by 15 (LCM of 5 and 3) to clear both denominators simultaneously.
Multiplying by 15: 9(x-2)<=25(2-x), so 9x-18<=50-25x, giving 34x<=68, i.e. x<=2.
✦ Working through each part gives: x <= 2, i.e. the interval (-infinity, 2].
Solve (1/2)(3x/5 + 4) >= (1/3)(x-6) for real x.
Hint. Multiply through by 30 (LCM of 2, 5, and 3) to clear every fraction at once.
Multiplying by 30: 15(3x/5+4) x 2... more directly, multiplying both sides by 30 gives 9x+60 >= 10(x-6), i.e. 9x+60>=10x-60, so 120>=x, i.e. x<=120.
✦ Working through each part gives: x <= 120, i.e. the interval (-infinity, 120].
Solve 2(2x+3) - 10 < 6(x-2) for real x.
Hint. Expand and simplify both sides fully before collecting x terms.
4x+6-10<6x-12, so 4x-4<6x-12, giving -4+12<6x-4x, i.e. 8<2x, so x>4.
✦ Working through each part gives: x > 4, i.e. the interval (4, infinity).
Solve 37 - (3x+5) > 9x - 8(x-3) for real x.
Hint. Expand both sides fully, being careful with the distributed negative signs.
37-3x-5>9x-8x+24, so 32-3x>x+24, giving 32-24>x+3x, i.e. 8>4x, so x<2.
✦ Working through each part gives: x < 2, i.e. the interval (-infinity, 2).
Solve x/4 < (5x-2)/3 - (7x-3)/5 for real x.
Hint. Multiply through by 60 (LCM of 4, 3, and 5) to clear every denominator in a single step.
Multiplying by 60: 15x < 20(5x-2) - 12(7x-3), so 15x < 100x-40-84x+36 = 16x-4, giving 15x-16x<-4, i.e. -x<-4, so x>4.
✦ Working through each part gives: x > 4, i.e. the interval (4, infinity).
Solve (2x-1)/3 >= (3x-2)/4 - (2-x)/5 for real x.
Hint. Multiply through by 60 (LCM of 3, 4, and 5) to clear every denominator at once.
Multiplying by 60: 20(2x-1) >= 15(3x-2) - 12(2-x), so 40x-20 >= 45x-30-24+12x = 57x-54, giving 40x-57x>=-54+20, i.e. -17x>=-34, so x<=2 (dividing by -17 flips the sign).
✦ Working through each part gives: x <= 2, i.e. the interval (-infinity, 2].
Solve 3x - 2 < 2x + 1 for real x and show the solution on a number line.
Hint. Collect x terms on one side.
3x-2x<1+2, so x<3. On a number line: an open circle at 3, shaded to the left.
✦ Working through each part gives: x < 3, shown with an open circle at 3 and shading to the left.
Solve 5x - 3 >= 3x - 5 for real x and show the solution on a number line.
Hint. Collect x terms on one side; the result is a non-strict inequality.
5x-3x>=-5+3, so 2x>=-2, giving x>=-1. On a number line: a filled circle at -1, shaded to the right.
✦ Working through each part gives: x >= -1, shown with a filled circle at -1 and shading to the right.
Solve 3(1-x) < 2(x+4) for real x and show the solution on a number line.
Hint. Expand both sides first, then collect x terms.
3-3x<2x+8, so 3-8<2x+3x, giving -5<5x, i.e. x>-1. On a number line: an open circle at -1, shaded to the right.
✦ Working through each part gives: x > -1, shown with an open circle at -1 and shading to the right.
Solve x/2 >= (5x-2)/3 - (7x-3)/5 for real x and show the solution on a number line.
Hint. Multiply through by 30 (LCM of 2, 3, and 5) to clear every denominator at once.
Multiplying by 30: 15x >= 10(5x-2) - 6(7x-3), so 15x >= 50x-20-42x+18 = 8x-2, giving 7x>=-2, so x>=-2/7. On a number line: a filled circle at -2/7, shaded to the right.
✦ Working through each part gives: x >= -2/7, shown with a filled circle at -2/7 and shading to the right.
Ravi obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Hint. Write the average of all three scores as an inequality with the unknown third score, requiring it to be at least 60.
Let x be the third score. (70+75+x)/3>=60, so 145+x>=180, giving x>=35.
✦ Working through each part gives: a minimum of 35 marks in the third test.
To receive Grade A, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in the first four examinations are 87, 92, 94, and 95, find the minimum marks she must obtain in the fifth examination to get grade A.
Hint. Set up the average of all five scores as an inequality requiring it to be at least 90.
Let x be the fifth score. (87+92+94+95+x)/5>=90, so 368+x>=450, giving x>=82.
✦ Working through each part gives: a minimum of 82 marks in the fifth examination.
Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.
Hint. Let x be the smaller odd integer; write one inequality for 'both smaller than 10' and one for 'sum more than 11', then intersect.
Let x be the smaller odd integer, so the other is x+2. Conditions: x+2<10 (so x<8) and x+(x+2)>11 (so 2x>9, x>4.5). Combined with x odd and positive: x can be 5 or 7.
✦ Working through each part gives: the pairs are (5,7) and (7,9).
Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Hint. Let x be the smaller even integer; write one inequality for 'both larger than 5' and one for the sum condition, then intersect.
Let x be the smaller even integer, so the other is x+2. Conditions: x>5 (so x>=6, being even) and x+(x+2)<23 (so 2x<21, x<10.5, i.e. x<=10). Even values from 6 to 10: 6, 8, 10.
✦ Working through each part gives: the pairs are (6,8), (8,10), and (10,12).
The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter is at least 61 cm, find the minimum length of the shortest side.
Hint. Express all three sides in terms of the shortest side, then write the perimeter as an inequality.
Let the shortest side be x. Longest = 3x, third = 3x-2. Perimeter = x+3x+(3x-2) = 7x-2 >= 61, so 7x>=63, giving x>=9.
✦ Working through each part gives: the shortest side must be at least 9 cm.
A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest, and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?
Hint. Let x be the shortest length; the second is x+3 and the third is 2x. Write one inequality for the total board length and one for the 'third at least 5cm longer than second' condition.
Total length: x+(x+3)+2x<=91, so 4x+3<=91, giving x<=22. Third at least 5cm longer than second: 2x>=(x+3)+5, so x>=8. Combined: 8<=x<=22.
✦ Working through each part gives: the shortest board can be between 8 cm and 22 cm.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh105.pdf) — one numbered exercise (5.1, 26 questions) plus the chapter's Miscellaneous Exercise (14 questions); the chapter has no Exercise 5.2/5.3 and no two-variable graphical-inequality content, matching CBSE's own 'formative-only' classification for that topic. Several nested-fraction questions in Exercise 5.1 (Q9-16, Q20) and the Miscellaneous Exercise (Q2-6) were re-rendered directly from the PDF at 300dpi after raw text extraction garbled the stacked fractions. Questions are referenced from the NCERT textbook for identification.
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