Solve 2 <= 3x - 4 <= 5.
Hint. Add 4 across all three parts of the chain, then divide by 3.
Adding 4: 6<=3x<=9. Dividing by 3: 2<=x<=3.
✦ Working through each part gives: 2 <= x <= 3, i.e. the interval [2, 3].
Solve 6 <= -3(2x-4) < 12.
Hint. Expand the bracket first, then isolate x across the whole chain -- dividing by a negative will flip both inequality directions.
-3(2x-4) = -6x+12, so the chain is 6<=-6x+12<12. Subtracting 12: -6<=-6x<0. Dividing by -6 (flips both signs and reverses the chain order): 1>=x>0, i.e. 0<x<=1.
✦ Working through each part gives: 0 < x <= 1, i.e. the interval (0, 1].
Solve -3 <= 4 - 7x/2 <= 18.
Hint. Subtract 4 from all three parts first, then multiply by a negative number, remembering to flip both inequality signs.
Subtracting 4: -7<=-7x/2<=14. Multiplying by -2/7 (flips both signs and reverses order): 2>=x>=-4, i.e. -4<=x<=2.
✦ Working through each part gives: -4 <= x <= 2, i.e. the interval [-4, 2].
Solve -15 < 3(x-2)/5 <= 0.
Hint. Multiply through by 5 first to clear the denominator, then isolate x.
Multiplying by 5: -75<3(x-2)<=0. Dividing by 3: -25<x-2<=0. Adding 2: -23<x<=2.
✦ Working through each part gives: -23 < x <= 2, i.e. the interval (-23, 2].
Solve -12 < 4 - 3x/(-5) <= 2.
Hint. Simplify 3x/(-5) to -3x/5 first, so the middle expression becomes 4 + 3x/5, then isolate x across the chain.
4 - 3x/(-5) = 4 + 3x/5. Chain: -12<4+3x/5<=2. Subtracting 4: -16<3x/5<=-2. Multiplying by 5/3 (positive, no flip): -80/3<x<=-10/3.
✦ Working through each part gives: -80/3 < x <= -10/3.
Solve 7 <= (3x+11)/2 <= 11.
Hint. Multiply through by 2 first to clear the denominator.
Multiplying by 2: 14<=3x+11<=22. Subtracting 11: 3<=3x<=11. Dividing by 3: 1<=x<=11/3.
✦ Working through each part gives: 1 <= x <= 11/3.
Solve the system: 5x + 1 > -24, 5x - 1 < 24. Show the solution on a number line.
Hint. Solve each inequality separately, then intersect the two solution sets.
First: 5x>-25, so x>-5. Second: 5x<25, so x<5. Intersection: -5<x<5. On a number line: open circles at -5 and 5, shading between them.
✦ Working through each part gives: -5 < x < 5, shown with open circles at both -5 and 5, shaded between.
Solve the system: 2(x-1) < x+5, 3(x+2) > 2-x. Show the solution on a number line.
Hint. Solve each inequality separately, then intersect.
First: 2x-2<x+5, so x<7. Second: 3x+6>2-x, so 4x>-4, i.e. x>-1. Intersection: -1<x<7. On a number line: open circles at -1 and 7, shading between.
✦ Working through each part gives: -1 < x < 7, shown with open circles at both -1 and 7, shaded between.
Solve the system: 3x - 7 > 2(x-6), 6 - x > 11 - 2x. Show the solution on a number line.
Hint. Solve each inequality separately; notice one solution set turns out to be strictly stronger than the other.
First: 3x-7>2x-12, so x>-5. Second: 6-x>11-2x, so x>5. Since x>5 already forces x>-5, the intersection is simply x>5. On a number line: open circle at 5, shaded to the right.
✦ Working through each part gives: x > 5, shown with an open circle at 5 and shading to the right.
Solve the system: 5(2x-7) - 3(2x+3) <= 0, 2x + 19 <= 6x + 47. Show the solution on a number line.
Hint. Expand and simplify each inequality separately before intersecting.
First: 10x-35-6x-9<=0, so 4x<=44, i.e. x<=11. Second: 2x+19<=6x+47, so -28<=4x, i.e. x>=-7. Intersection: -7<=x<=11. On a number line: filled circles at -7 and 11, shading between.
✦ Working through each part gives: -7 <= x <= 11, shown with filled circles at both -7 and 11, shaded between.
A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (C) if F = (9/5)C + 32?
Hint. Substitute the formula into the compound inequality 68<F<77 and solve for C across the whole chain.
68<(9/5)C+32<77. Subtracting 32: 36<(9/5)C<45. Multiplying by 5/9: 20<C<25.
✦ Working through each part gives: the temperature must be kept between 20°C and 25°C.
A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
Hint. Let x be the litres of 2% solution added; write the acid content as more than 4% and less than 6% of the total mixture, as two separate inequalities.
Acid content: 8%(640)+2%(x) = 51.2+0.02x. Total mixture: 640+x. First: 51.2+0.02x>0.04(640+x), giving 25.6>0.02x, so x<1280. Second: 51.2+0.02x<0.06(640+x), giving 12.8<0.04x, so x>320.
✦ Working through each part gives: between 320 litres and 1280 litres of the 2% solution must be added.
How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
Hint. Let x be the litres of water added; the acid amount itself stays fixed at 45% of 1125 litres, since water contributes no acid, while the total volume grows.
Fixed acid amount: 45%(1125)=506.25 litres. Total after adding water: 1125+x. Requiring this between 25% and 30% of the total: 0.25(1125+x)<506.25 gives x<900, and 506.25<0.30(1125+x) gives x>562.5.
✦ Working through each part gives: between 562.5 litres and 900 litres of water must be added.
IQ of a person is given by IQ = (MA/CA) x 100, where MA is mental age and CA is chronological age. If 80 <= IQ <= 140 for a group of 12-year-old children, find the range of their mental age.
Hint. Substitute CA=12 into the IQ formula and solve the resulting compound inequality for MA.
80<=(MA/12)x100<=140. Dividing by 100: 0.8<=MA/12<=1.4. Multiplying by 12: 9.6<=MA<=16.8.
✦ Working through each part gives: the mental age ranges from 9.6 to 16.8 years.
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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