RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 11 questions · 26 marks
An equation pins a variable to a single value, but real life is usually about limits — a budget you must stay under, a mark you must exceed, an oven that can run only so long. Linear inequalities capture exactly that, and solving them is almost like solving equations, with one rule you must never forget about multiplying by a negative number. You will also learn to shade the half-plane described by an inequality in two variables, which is the first step towards linear programming.
The solution set of the inequality 2x + 5 ≤ 17, where x is a natural number, is:
Answer
The solution set is {1, 2, 3, 4, 5, 6}. Subtract 5 from both sides: 2x ≤ 12, then divide by the positive number 2 to get x ≤ 6. Since x must be a natural number, the admissible values are 1, 2, 3, 4, 5 and 6. The interval (-∞, 6] would be correct only if x were allowed to be any real number, and 0 is not a natural number so the third option is wrong.
If -4x ≥ 20, then the solution set for x is:
Answer
The correct statement is x ≤ -5. Divide both sides by -4. Because -4 is negative, the direction of the inequality must be reversed. So -4x ≥ 20 becomes x ≤ 20/(-4) = -5. Forgetting to flip the sign is what produces the incorrect option x ≥ -5.
For x ∈ R, the solution of 5x - 3 < 3x + 1 is the interval:
Answer
The solution is (-∞, 2). Bring the variable terms together: 5x - 3x < 1 + 3, so 2x < 4. Dividing by the positive number 2 gives x < 2, with the sign unchanged. The inequality is strict, so 2 itself is excluded and the interval is open at 2.
Which of the following points satisfies the inequality 3x + 4y ≥ 24?
Answer
The point (4, 3) satisfies it. Substituting, 3(4) + 4(3) = 12 + 12 = 24, and 24 ≥ 24 is true because the inequality is non-strict. Checking the others: (2, 3) gives 6 + 12 = 18, (1, 1) gives 3 + 4 = 7, and (0, 5) gives 0 + 20 = 20, all less than 24. So only (4, 3) lies in the required half-plane — in fact it lies exactly on the boundary line.
Assertion (A): The solution of the inequality -3x > 12 is x < -4. Reason (R): Multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states a correct rule of inequalities. Applying it, dividing -3x > 12 by -3 requires flipping the sign, giving x < 12/(-3) = -4, so Assertion A is true. A quick check confirms it: x = -5 gives -3(-5) = 15 > 12, while x = 0 gives 0, which is not greater than 12. Since the sign-reversal rule is exactly what produces the answer, R explains A.
Solve 7x + 3 < 5x + 9 for x ∈ R and describe how the solution is shown on a number line.
Answer
Collect like terms: 7x - 5x < 9 - 3, so 2x < 6. Dividing by the positive number 2 gives x < 3, i.e. the interval (-∞, 3). On the number line, draw a hollow (unfilled) circle at 3 to show that 3 is not included, and shade the whole line to the left of it, with the arrow continuing towards -∞.
Solve the system of inequalities 2x - 1 ≤ 9 and 3x + 4 > -5 for x ∈ R, and write the answer in interval form.
Answer
Solve the first inequality: 2x - 1 ≤ 9 gives 2x ≤ 10, so x ≤ 5, i.e. (-∞, 5]. Solve the second: 3x + 4 > -5 gives 3x > -9, so x > -3, i.e. (-3, ∞). The solution of the system is the intersection of the two sets, so x must satisfy both x > -3 and x ≤ 5. Hence the solution is -3 < x ≤ 5, written as the interval (-3, 5], open at -3 and closed at 5.
Solve (2x - 3)/4 ≥ (x + 1)/3 for x ∈ R.
Answer
Multiply both sides by 12, the LCM of 4 and 3. Since 12 is positive the inequality sign is unchanged. 3(2x - 3) ≥ 4(x + 1), so 6x - 9 ≥ 4x + 4. Bring like terms together: 6x - 4x ≥ 4 + 9, so 2x ≥ 13. Dividing by 2 gives x ≥ 13/2, i.e. the interval [13/2, ∞) or [6.5, ∞).
A student scored 68 and 74 marks in the first two unit tests of a subject, each out of 100. (a) Find the minimum marks she must score in the third test so that her average across the three tests is at least 75. (b) Find the range of marks in the third test for which the average lies between 75 and 85, both inclusive, given that no test can be scored above 100.
Answer
(a) Let the third-test score be x marks. The average of the three tests is (68 + 74 + x)/3 = (142 + x)/3. We need (142 + x)/3 ≥ 75. Multiplying both sides by the positive number 3 gives 142 + x ≥ 225. So x ≥ 225 - 142 = 83. The minimum required score is 83 marks. (b) Now we need 75 ≤ (142 + x)/3 ≤ 85. Multiplying throughout by 3: 225 ≤ 142 + x ≤ 255. Subtracting 142 throughout: 83 ≤ x ≤ 113. But a test is out of 100, so x ≤ 100 is an additional constraint. Taking the intersection of 83 ≤ x ≤ 113 with x ≤ 100 gives 83 ≤ x ≤ 100. So she must score between 83 and 100 marks inclusive; note that the upper limit comes from the maximum mark, not from the average condition.
A tank holds 900 litres of a salt solution that is 12% salt by volume. A technician wants to add a 30% salt solution to it. How many litres of the 30% solution must be added so that the resulting mixture is more than 15% but less than 18% salt?
Answer
Step 1 — Let x litres of the 30% solution be added. Step 2 — Salt already present = 12% of 900 = 0.12 × 900 = 108 litres. Salt added = 0.30x litres, so total salt = 108 + 0.3x litres, and total volume = (900 + x) litres. Step 3 — The mixture must be more than 15% salt: 108 + 0.3x > 0.15(900 + x) = 135 + 0.15x. So 0.3x - 0.15x > 135 - 108, giving 0.15x > 27 and x > 180. Step 4 — The mixture must be less than 18% salt: 108 + 0.3x < 0.18(900 + x) = 162 + 0.18x. So 0.3x - 0.18x < 162 - 108, giving 0.12x < 54 and x < 450. Step 5 — Combining the two conditions, 180 < x < 450. So the technician must add more than 180 litres but less than 450 litres of the 30% solution. Check at x = 300: salt = 108 + 90 = 198 litres in 1200 litres, which is 16.5% — comfortably between 15% and 18%.
Read the following and answer the questions that follow: A small bakery bakes two kinds of cake in a single oven. Each chocolate cake needs 40 minutes of oven time and each vanilla cake needs 25 minutes. On a given day the oven is available for at most 600 minutes. Let x be the number of chocolate cakes and y the number of vanilla cakes baked that day. (a) Write the oven-time constraint as a linear inequality and simplify it to its smallest whole-number coefficients. (b) If the bakery bakes 6 chocolate cakes, what is the greatest number of vanilla cakes it can bake? (c) Is the plan (x, y) = (9, 10) possible? Justify with a calculation. (d) State the two additional constraints on x and y that the situation forces, and give the greatest number of vanilla cakes possible if no chocolate cake is baked.
Answer
(a) Total oven time is 40x + 25y minutes, which must not exceed 600, so 40x + 25y ≤ 600. Dividing throughout by 5 (a positive number, so the sign is unchanged) gives 8x + 5y ≤ 120. (b) Put x = 6 in 8x + 5y ≤ 120: 48 + 5y ≤ 120, so 5y ≤ 72 and y ≤ 14.4. Since y must be a whole number of cakes, the greatest value is y = 14. (c) Substitute x = 9, y = 10 into 8x + 5y: 72 + 50 = 122, and 122 > 120. So the plan needs more oven time than is available (610 minutes against 600) and is not possible. (d) The number of cakes cannot be negative, so x ≥ 0 and y ≥ 0; this restricts the feasible region to the first quadrant. With x = 0 the constraint becomes 5y ≤ 120, so y ≤ 24, and the greatest number of vanilla cakes is 24.
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