Anand Classes provides comprehensive and well-explained Linear Inequalities NCERT Solutions Miscellaneous Exercise Class 11 Maths Chapter 5 (Set-2) to help students gain a clear understanding of linear inequalities and their applications. These solutions are carefully prepared as per the latest NCERT and CBSE syllabus, featuring detailed, step-by-step explanations for every question. Perfect for Class 11 students preparing for exams, these solutions simplify complex concepts and enhance problem-solving confidence. Click the print button to download study material and notes.
NCERT Question 8 : Solve the inequalities and represent the solution on the number line:
$$2(x – 1) < x + 5 \quad \text{and} \quad 3(x + 2) > 2 – x$$
Solution :
Given:
$$2(x – 1) < x + 5 \quad \text{and} \quad 3(x + 2) > 2 – x$$
Solving First Inequality:
$$2(x – 1) < x + 5$$
$$2x – 2 < x + 5$$
$$2x – x < 5 + 2$$
$$x < 7 \quad \dots (i)$$
Solving Second Inequality:
$$3(x + 2) > 2 – x$$
$$3x + 6 > 2 – x$$
$$3x + x > 2 – 6$$
$$4x > -4$$
$$x > -1 \quad \dots (ii)$$
Combined Solution:
From (i) and (ii):
$$-1 < x < 7$$
So,
$$x \in (-1, 7)$$
Number Line Representation:
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NCERT Question 9 : Solve the inequalities and represent the solution on the number line:
$$3x – 7 > 2(x – 6) \quad \text{and} \quad 6 – x > 11 – 2x$$
Solution :
Given:
$$3x – 7 > 2(x – 6) \quad \text{and} \quad 6 – x > 11 – 2x$$
Solving First Inequality:
$$3x – 7 > 2(x – 6)$$
$$3x – 7 > 2x – 12$$
$$3x – 2x > -12 + 7$$
$$x > -5 \quad \dots (i)$$
Solving Second Inequality:
$$6 – x > 11 – 2x$$
$$-x + 2x > 11 – 6$$
$$x > 5 \quad \dots (ii)$$
Combined Solution:
From (i) and (ii):
$$x > 5$$
So,
$$x \in (5, \infty)$$
Number Line Representation:
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NCERT Question 10 : Solve the inequalities and represent the solution on the number line:
$$5(2x-7)-3(2x+3)\le 0,\qquad 2x+19\le 6x+47$$
Solution:
Solve the first inequality:
$$5(2x-7)-3(2x+3)\le 0$$
$$10x-35-6x-9\le 0$$
$$4x-44\le 0$$
$$4x\le 44$$
$$x\le 11$$
Solve the second inequality:
$$2x+19\le 6x+47$$
$$2x-6x\le 47-19$$
$$-4x\le 28$$
$$x\ge -7$$
Intersection of the two solutions:
$$-7\le x\le 11$$
So,
$$x\in[-7, 11]$$
Number line representation :
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NCERT Question 11 : A solution is to be kept between $68^\circ F$ and $77^\circ F$. What is the range in temperature in degree Celsius (°C) if
$$F = \frac{9}{5}C + 32$$
Solution:
Given:
$$68^\circ < F < 77^\circ$$
Substitute $F = \frac{9}{5}C + 32$ :
$$68 < \frac{9}{5}C + 32 < 77$$
Subtract $32$ from each side:
$$68 – 32 < \frac{9}{5}C < 77 – 32$$
$$36 < \frac{9}{5}C < 45$$
Multiply by $\frac{5}{9}$:
$$36 \cdot \frac{5}{9} < C < 45 \cdot \frac{5}{9}$$
$$20 < C < 25$$
Thus, the temperature in degree Celsius must lie in:
$$C \in (20^\circ, 25^\circ)$$
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Question 12. 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?
Solution:
It is given that we have $640$ litres of $8\%$ boric acid solution.
Let $x$ be the number of litres of $2\%$ boric acid solution to be added.
Total mixture $=640+x$ litres.
The resulting concentration must satisfy
$4\%$ of $(x+640)$ $<$ $2\%$ of $x$ + $8$ $\%$ of $640$ $<$ $6\%$ of $(x+640)$
$$4\%<\dfrac{\frac{2}{100}x+\dfrac{8}{100}\cdot640}{x+640}<6\%.$$
Multiply through by $100(x+640)$ to clear percentages and denominator:
$$4(x+640)<2x+8\cdot640<6(x+640).$$
Compute $8\cdot640=5120$ and simplify:
$$4x+2560<2x+5120<6x+3840.$$
Split and solve the two inequalities:
Left inequality:
$$4x+2560<2x+5120\implies 2x<2560\implies x<1280.$$
Right inequality:
$$2x+5120<6x+3840\implies 1280<4x\implies x>320.$$
Combine:
$$320<x<1280.$$
Thus the amount $x$ of $2\%$ solution to be added must satisfy
$$\boxed{320<x<1280}$$
litres — i.e. strictly more than $320$ L and strictly less than $1280$ L.
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Question 13. 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?
Solution:
Let the amount of water added be $x$ litres.
Total mixture $=x+1125$ litres. The amount of acid remains $45\%$ of $1125$, i.e. $\dfrac{45}{100}\cdot1125$.
We require
$25$ $\%$ of $(x+1125)$ $<$ $45$ $\%$ of $1125$ $<$ $30$ $\%$ $(x+1125)$
$$25\%<\dfrac{\dfrac{45}{100}\cdot1125}{x+1125}<30\%.$$
Equivalently,
$$25(x+1125)<45\cdot1125<30(x+1125).$$
Compute $45\cdot1125=50625$.
From the left inequality:
$$25(x+1125)<50625$$ $$ x+1125<\frac{50625}{25}=2025$$ $$ x<2025-1125=900.$$
From the right inequality:
$$50625<30(x+1125)$$ $$ \frac{50625}{30}<x+1125$$ $$ \frac{50625}{30}=1687.5$$ $$ x>1687.5-1125=562.5.$$
Therefore,
$$562.5<x<900.$$
Final Answer: the amount of water to be added must satisfy $$\boxed{562.5<x<900}\text{ litres.}$$
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NCERT Question 14 : IQ of a person is given by the formula:
$$IQ = \left(\frac{MA}{CA}\right) \times 100$$
where $MA$ is mental age and $CA$ is chronological age. If $80 \le IQ \le 140$ for a group of 12-year-old children, find the range of their mental age.
Solution :
Given:
Chronological age:
$$CA = 12 \text{ years}$$
IQ range:
$$80 \le IQ \le 140$$
Using the formula:
$$IQ = \left(\frac{MA}{CA}\right) \times 100$$
Substitute $CA = 12$:
$$80 \le \left(\frac{MA}{12}\right) \times 100 \le 140$$
Multiply both sides by 12:
$$80 \times \frac{12}{100} \le MA \le 140 \times \frac{12}{100}$$
Simplify:
$$\frac{96}{10} \le MA \le \frac{168}{10}$$
$$9.6 \le MA \le 16.8$$
✅ Final Answer
The mental age of the 12-year-old children must lie in the range:
$$9.6 \le MA \le 16.8$$
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