OP Malhotra Class 11 Maths Solutions Chapter 11 Inequalities Chapter Test

S Chand Class 11 ICSE Maths Solutions Chapter 11 Inequalities Chapter Test

Question 1. Solve : 24x < 100, when (i) x is a natural number (ii) x is an integer
Answer:
We are given the inequality \( 24x < 100 \).
First, we solve for x:
\( x < \frac { 100 }{ 24 } \)
\( x < \frac { 25 }{ 6 } \)
\( x < 4\frac { 1 }{6} \)

(i) When x is a natural number (N):
Natural numbers are \( \{1, 2, 3, \ldots\} \). Since x must be less than \( 4\frac{1}{6} \), the natural numbers that satisfy this are 1, 2, 3, and 4.
So, the solution set is \( \{1, 2, 3, 4\} \). This means x can be any of these whole positive numbers.

(ii) When x is an integer (Z):
Integers include all positive and negative whole numbers, including zero (\( \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\} \)). Since x must be less than \( 4\frac{1}{6} \), the integers that satisfy this condition are 4, 3, 2, 1, 0, -1, -2, -3, and so on, continuing infinitely in the negative direction.
So, the solution set is \( \{\ldots, -3, -2, -1, 0, 1, 2, 3, 4\} \).
In simple words: First, divide 100 by 24 to find the limit for x. Then, depending on whether x must be a natural number (counting numbers from 1 upwards) or an integer (all whole numbers including negative ones and zero), list all the numbers that are smaller than this limit.

🎯 Exam Tip: Always pay close attention to the specified domain for x (e.g., natural numbers, integers, real numbers) as it changes the possible values in the solution set. Write the solution set clearly using set notation.

Question 2. Solve the inequality \( \frac { 1 }{ 2 }[\frac { 3 }{ 5 }x + 4] \ge \frac { 1 }{ 3 }(x – 6) \)
Answer:
We are given the inequality: \( \frac { 1 }{ 2 }[\frac { 3 }{ 5 }x + 4] \ge \frac { 1 }{ 3 }(x - 6) \)
First, multiply both sides by the least common multiple of the denominators (2 and 3), which is 6, to remove fractions:
\( 6 \times \frac { 1 }{ 2 }[\frac { 3 }{ 5 }x + 4] \ge 6 \times \frac { 1 }{ 3 }(x - 6) \)
\( 3[\frac { 3 }{ 5 }x + 4] \ge 2(x - 6) \)
Next, distribute the numbers outside the brackets:
\( \frac { 9 }{ 5 }x + 12 \ge 2x - 12 \)
Now, group the x terms on one side and constant terms on the other. It's often easier to move x terms to the side where the coefficient will be positive, but here we'll follow the source steps:
\( 12 + 12 \ge 2x - \frac { 9 }{ 5 }x \)
\( 24 \ge (\frac { 10-9 }{ 5 })x \)
\( 24 \ge \frac { 1 }{ 5 }x \)
Multiply by 5:
\( 120 \ge x \)
This can also be written as:
\( x \le 120 \)
The solution set includes all real numbers less than or equal to 120.
So, the solution set in interval notation is \( (-\infty, 120] \).
In simple words: To solve this, first get rid of the fractions by multiplying everything by a common number (like 6). Then, spread out the numbers and move all the 'x' terms to one side and plain numbers to the other. Finally, divide to find what 'x' can be, remembering that the solution can include all numbers up to and including 120.

🎯 Exam Tip: When multiplying an inequality by a negative number, remember to reverse the inequality sign. Also, be careful with distributing numbers over terms in brackets.

Question 3. Solve : \( -12 \le 4 - \frac { 3x }{ -5 } < 2 \)
Answer:
We are given the compound inequality: \( -12 \le 4 - \frac { 3x }{ -5 } < 2 \)
First, simplify the term \( - \frac { 3x }{ -5 } \). Two negative signs make a positive, so it becomes \( + \frac { 3x }{ 5 } \):
\( -12 \le 4 + \frac { 3x }{ 5 } < 2 \)
Next, subtract 4 from all three parts of the inequality:
\( -12 - 4 \le \frac { 3x }{ 5 } < 2 - 4 \)
\( -16 \le \frac { 3x }{ 5 } < -2 \)
Now, multiply all three parts by 5 to remove the denominator:
\( -16 \times 5 \le 3x < -2 \times 5 \)
\( -80 \le 3x < -10 \)
Finally, divide all three parts by 3 to isolate x:
\( \frac { -80 }{ 3 } \le x < \frac { -10 }{ 3 } \)
This means x is greater than or equal to \( -\frac{80}{3} \) and strictly less than \( -\frac{10}{3} \).
So, the solution set in interval notation is \( [-\frac { 80 }{ 3 }, -\frac { 10 }{ 3 }) \). This shows the range of values x can take.
In simple words: To solve this, first clean up the signs (two negatives make a positive). Then, remove the plain number by subtracting it from all parts. Next, get rid of the fraction by multiplying all parts by the bottom number. Finally, divide by the number next to 'x' to find the range of x.

🎯 Exam Tip: When solving compound inequalities, perform the same operation on all three parts. Remember that dividing or multiplying by a negative number reverses the inequality signs, but here we only multiplied by a positive 5.

Question 4. Solve the following inequalities graphically
(i) 3x + 4y < 12
(ii) x > - 3
(iii) 5y – 3 ≤ 12
Answer:
(i) For the inequality \( 3x + 4y < 12 \):
First, we draw the graph of the line \( 3x + 4y = 12 \). This line passes through the points A(4, 0) (when \( y=0 \)) and B(0, 3) (when \( x=0 \)).
To determine the region for \( 3x + 4y < 12 \), we test a point, for example, the origin (0, 0).
\( 3(0) + 4(0) = 0 \). Since \( 0 < 12 \) is true, the region containing the origin is the solution set. We shade the area below the line \( 3x + 4y = 12 \). Since the inequality is strict (<), the line itself is not included in the solution and should be drawn as a dashed line.

(ii) For the inequality \( x > -3 \):
First, we draw the graph of the line \( x = -3 \). This is a vertical line passing through \( (-3, 0) \) and parallel to the y-axis.
To determine the region for \( x > -3 \), we test a point, such as the origin (0, 0).
\( 0 > -3 \) is true. Thus, the region containing the origin (to the right of the line \( x = -3 \)) is the solution set. We shade the area to the right of the line \( x = -3 \). Since the inequality is strict (>), the line itself is drawn as a dashed line.

(iii) For the inequality \( 5y - 3 \le 12 \):
First, simplify the inequality:
\( 5y \le 12 + 3 \)
\( 5y \le 15 \)
\( y \le 3 \)
Next, we draw the graph of the line \( y = 3 \). This is a horizontal line passing through \( (0, 3) \) and parallel to the x-axis.
To determine the region for \( y \le 3 \), we test a point, such as the origin (0, 0).
\( 0 \le 3 \) is true. Therefore, the region containing the origin (below the line \( y = 3 \)) is the solution set. We shade the area below the line \( y = 3 \). Since the inequality includes "equal to" (\( \le \)), the line itself is part of the solution and should be drawn as a solid line.

🎯 Exam Tip: Always remember to use a dashed line for strict inequalities (< or >) and a solid line for non-strict inequalities (≤ or ≥). A test point like the origin (0,0) is usually the easiest to check, unless the line passes through it.

Question 5. Solve graphically : \( 3x + 4y \le 60, x + 3y \le 30, x \ge 0, y \ge 0 \)
Answer:
We need to solve the following system of inequalities graphically:
1. \( 3x + 4y \le 60 \)
2. \( x + 3y \le 30 \)
3. \( x \ge 0, y \ge 0 \)

**Step 1: Draw the line for \( 3x + 4y = 60 \).**
When \( x=0 \), \( 4y = 60 \implies y = 15 \). So, point B(0, 15).
When \( y=0 \), \( 3x = 60 \implies x = 20 \). So, point A(20, 0).
Plot these points and draw a solid line through them. To check the inequality \( 3x + 4y \le 60 \), test (0,0): \( 3(0) + 4(0) = 0 \). Since \( 0 \le 60 \) is true, the region below the line towards the origin is the solution for this inequality.

**Step 2: Draw the line for \( x + 3y = 30 \).**
When \( x=0 \), \( 3y = 30 \implies y = 10 \). So, point D(0, 10).
When \( y=0 \), \( x = 30 \). So, point C(30, 0).
Plot these points and draw a solid line through them. To check the inequality \( x + 3y \le 30 \), test (0,0): \( 0 + 3(0) = 0 \). Since \( 0 \le 30 \) is true, the region below the line towards the origin is the solution for this inequality.

**Step 3: Consider \( x \ge 0, y \ge 0 \).**
These inequalities mean the solution must be in the first quadrant (where x and y values are both positive or zero).

**Step 4: Find the intersection point P.**
To find where \( 3x + 4y = 60 \) and \( x + 3y = 30 \) intersect, we can solve them as a system of equations.
From \( x + 3y = 30 \), we get \( x = 30 - 3y \).
Substitute this into the first equation:
\( 3(30 - 3y) + 4y = 60 \)
\( 90 - 9y + 4y = 60 \)
\( 90 - 5y = 60 \)
\( 30 = 5y \)
\( y = 6 \)
Now, find x: \( x = 30 - 3(6) = 30 - 18 = 12 \).
So, the intersection point is P(12, 6).

**Step 5: Identify the feasible region.**
The feasible region is the area that satisfies all inequalities simultaneously. This region is bounded by the x-axis, the y-axis, the line \( 3x + 4y = 60 \), and the line \( x + 3y = 30 \). The vertices of this polygon are O(0,0), A(20,0), P(12,6), and D(0,10). This shaded region represents all possible solutions.

🎯 Exam Tip: When graphing multiple inequalities, carefully determine the feasible region by considering where all shaded areas overlap. Clearly label all lines, axes, and intersection points to ensure full marks.