Word problem of Linear inequalities Questions in English

(B) For the square root to be defined,we must have $x + 4 \geq 0$,which implies $x \geq -4$.
Given the inequality $x + 2 > \sqrt{x + 4}$.
If $x + 2 < 0$ (i.e.,$x < -2$),the inequality is not satisfied because the left side is negative and the right side is non-negative.
Thus,we must have $x + 2 \geq 0$,i.e.,$x \geq -2$.
Squaring both sides,we get $(x + 2)^2 > x + 4$.
$x^2 + 4x + 4 > x + 4$.
$x^2 + 3x > 0$.
$x(x + 3) > 0$.
This inequality holds when $x > 0$ or $x < -3$.
Combining this with the condition $x \geq -2$,we get $x > 0$.

(B) Case $I$: When $x + 2 \ge 0$ i.e.,$x \ge -2,$
The given inequality becomes ${x^2} - (x + 2) + x > 0 \implies {x^2} - 2 > 0 \implies |x| > \sqrt{2}.$
This implies $x < -\sqrt{2}$ or $x > \sqrt{2}.$
Since $x \ge -2,$ the solution set for this case is $[ -2, -\sqrt{2} ) \cup (\sqrt{2}, \infty ).$
Case $II$: When $x + 2 < 0$ i.e.,$x < -2,$
The given inequality becomes ${x^2} + (x + 2) + x > 0 \implies {x^2} + 2x + 2 > 0.$
Since the discriminant $D = 2^2 - 4(1)(2) = 4 - 8 = -4 < 0$ and the coefficient of ${x^2}$ is positive,the expression ${x^2} + 2x + 2$ is always positive for all real $x.$
Thus,the solution set for this case is $( -\infty, -2).$
Combining the two cases,the total solution set is $( -\infty, -2) \cup [ -2, -\sqrt{2} ) \cup (\sqrt{2}, \infty ) = ( -\infty, -\sqrt{2} ) \cup (\sqrt{2}, \infty ).$

(B) Step $1$: Solve the first inequality: $5x + 2 < 3x + 8 \implies 2x < 6 \implies x < 3$.
Step $2$: Solve the second inequality: $\frac{x + 2}{x - 1} < 4 \implies \frac{x + 2}{x - 1} - 4 < 0 \implies \frac{x + 2 - 4(x - 1)}{x - 1} < 0 \implies \frac{x + 2 - 4x + 4}{x - 1} < 0 \implies \frac{-3x + 6}{x - 1} < 0$.
Multiplying by $-1$,we get $\frac{3x - 6}{x - 1} > 0 \implies \frac{x - 2}{x - 1} > 0$.
Using the sign scheme (Wavy curve method),the solution is $x \in (-\infty, 1) \cup (2, \infty)$.
Step $3$: Find the intersection of $x < 3$ and $x \in (-\infty, 1) \cup (2, \infty)$.
The intersection is $(-\infty, 1) \cup (2, 3)$.

(C) Given the inequality: $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$
Factor the denominator: $\frac{2x}{(2x + 1)(x + 2)} > \frac{1}{x + 1}$
Subtract $\frac{1}{x + 1}$ from both sides: $\frac{2x}{(2x + 1)(x + 2)} - \frac{1}{x + 1} > 0$
Find a common denominator: $\frac{2x(x + 1) - (2x + 1)(x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$
Simplify the numerator: $\frac{2x^2 + 2x - (2x^2 + 5x + 2)}{(2x + 1)(x + 2)(x + 1)} > 0$
$\frac{-3x - 2}{(2x + 1)(x + 2)(x + 1)} > 0$
Multiply by $-1$ and reverse the inequality: $\frac{3x + 2}{(2x + 1)(x + 2)(x + 1)} < 0$
The critical points are $x = -2, -1, -\frac{2}{3}, -\frac{1}{2}$.
Testing the intervals,the inequality holds for $x \in (-2, -1) \cup (-\frac{2}{3}, -\frac{1}{2})$.

(B) To find the values of $n$ for which $10^{n - 2} > 81n$ holds true,we test integer values starting from $n = 1$:
For $n = 1$: $10^{-1} = 0.1$ and $81(1) = 81$. $0.1 > 81$ is false.
For $n = 2$: $10^{0} = 1$ and $81(2) = 162$. $1 > 162$ is false.
For $n = 3$: $10^{1} = 10$ and $81(3) = 243$. $10 > 243$ is false.
For $n = 4$: $10^{2} = 100$ and $81(4) = 324$. $100 > 324$ is false.
For $n = 5$: $10^{3} = 1000$ and $81(5) = 405$. $1000 > 405$ is true.
For $n = 6$: $10^{4} = 10000$ and $81(6) = 486$. $10000 > 486$ is true.
Since the inequality holds for $n \geq 5$,the correct option is $B$.

(C) The equation $|x + y| = 4$ represents two parallel lines: $x + y = 4$ and $x + y = -4$.
For the point $(a, a)$ to lie between these two lines,it must satisfy the inequality $-4 < a + a < 4$.
This simplifies to $-4 < 2a < 4$.
Dividing by $2$,we get $-2 < a < 2$.
This is equivalent to $|a| < 2$.

(D) Given $2\sin^2 x + 3\sin x - 2 > 0$.
Factoring the quadratic: $2\sin^2 x + 4\sin x - \sin x - 2 > 0$.
$2\sin x(\sin x + 2) - 1(\sin x + 2) > 0$.
$(\sin x + 2)(2\sin x - 1) > 0$.
Since $\sin x + 2 > 0$ for all real $x$,we must have $2\sin x - 1 > 0$,which implies $\sin x > 1/2$.
For $x$ in a reasonable range,$\sin x > 1/2$ implies $x > \pi/6$.
Given $x^2 - x - 2 < 0$.
Factoring the quadratic: $(x - 2)(x + 1) < 0$.
This inequality holds for $x \in (-1, 2)$.
Combining the two conditions: $x > \pi/6$ and $-1 < x < 2$.
Since $\pi/6 \approx 0.523$,the intersection is $x \in (\pi/6, 2)$.

(B) Since $\theta$ is an acute angle,we have $0 < \theta < 90^{\circ}$.
Therefore,$0 < \sin\theta < 1$.
Substituting the given expression: $0 < \frac{p - 6}{8 - p} < 1$.
Case $1$: $\frac{p - 6}{8 - p} \ge 0$. Since the denominator $8 - p$ must be positive for $\sin\theta$ to be defined and positive,$p < 8$. Thus,$p - 6 \ge 0$,which gives $p \ge 6$.
Case $2$: $\frac{p - 6}{8 - p} < 1$. Since $8 - p > 0$,we have $p - 6 < 8 - p$,which simplifies to $2p < 14$,or $p < 7$.
Combining these,we get $6 \le p < 7$.

(C) Let $f(x) = ||x - 2| - |3 - x||$.
By the triangle inequality,$||x - 2| - |3 - x|| \leq |(x - 2) - (3 - x)| = |2x - 5|$.
More precisely,the expression $||x - 2| - |3 - x||$ represents the absolute difference of the distances of $x$ from $2$ and $3$.
For $x < 2$,$f(x) = |(2 - x) - (3 - x)| = |-1| = 1$.
For $2 \leq x \leq 3$,$f(x) = |(x - 2) - (3 - x)| = |2x - 5|$,which ranges from $1$ to $0$ and back to $1$.
For $x > 3$,$f(x) = |(x - 2) - (x - 3)| = |1| = 1$.
Thus,the range of $f(x)$ is $[0, 1]$.
For the equation $f(x) = 2 - a$ to have a solution,we must have $0 \leq 2 - a \leq 1$.
This implies $a - 2 \leq 0$ and $2 - a \leq 1$,so $a \leq 2$ and $a \geq 1$.
The integral values of $a$ are $1$ and $2$.
The sum of these values is $1 + 2 = 3$.

(B) For $f(x)$ to be a decreasing function,the coefficient of $x$ must be less than or equal to $0$.
Let $k = [t]+1$. Then $\frac{|k|+a}{|k|+1-a} \leq 0$.
Since $a < 0$,we have $1-a > 0$.
For the fraction to be $\leq 0$,the numerator must be $\leq 0$ (since the denominator is always positive).
$|k| + a \leq 0 \implies |k| \leq -a$.
Since $-a > 0$,this implies $a \leq k \leq -a$.
Substituting $k = [t]+1$,we get $a \leq [t]+1 \leq -a$.
Subtracting $1$ from all sides,$a-1 \leq [t] \leq -a-1$.
Since $a \notin I$,the range for $t$ is $[[a], [-a])$.
Thus,the correct option is $B$.

(C) We test values for $n$ to check the inequality $10^{n-2} > 91n$:
For $n=1$: $10^{-1} = 0.1$,$91(1) = 91$. $0.1 > 91$ is false.
For $n=2$: $10^{0} = 1$,$91(2) = 182$. $1 > 182$ is false.
For $n=3$: $10^{1} = 10$,$91(3) = 273$. $10 > 273$ is false.
For $n=4$: $10^{2} = 100$,$91(4) = 364$. $100 > 364$ is false.
For $n=5$: $10^{3} = 1000$,$91(5) = 455$. $1000 > 455$ is true.
For $n=6$: $10^{4} = 10000$,$91(6) = 546$. $10000 > 546$ is true.
Since the exponential function $10^{n-2}$ grows much faster than the linear function $91n$,the inequality will hold for all $n \ge 5$.
Thus,the set of values is $\left\{ 5, 6, 7, 8, \dots \right\}$.

(D) Let $t = |x|$. Since $|x| \ge 0$,we have $t \ge 0$. The inequality becomes $\frac{1 - t}{2 - t} \ge 0$.
Multiplying by $-1$ in the numerator and denominator,we get $\frac{t - 1}{t - 2} \ge 0$.
Using the wavy curve method for $t$,the critical points are $t = 1$ and $t = 2$.
The inequality holds for $t \in [0, 1] \cup (2, \infty)$.
Substituting $t = |x|$ back,we have $|x| \in [0, 1] \cup (2, \infty)$.
Case $1$: $0 \le |x| \le 1 \implies x \in [-1, 1]$.
Case $2$: $|x| > 2 \implies x \in (-\infty, -2) \cup (2, \infty)$.
Combining these,the solution is $x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty)$.

(D) The given inequality is $\frac{x^4 - 4x^3 + 3x^2}{(x^2 - 4)(x^2 - 7x + 10)} \ge 0$.
Factorizing the numerator: $x^2(x^2 - 4x + 3) = x^2(x - 1)(x - 3)$.
Factorizing the denominator: $(x - 2)(x + 2)(x - 2)(x - 5) = (x + 2)(x - 2)^2(x - 5)$.
The inequality becomes $\frac{x^2(x - 1)(x - 3)}{(x + 2)(x - 2)^2(x - 5)} \ge 0$.
The critical points are $x = -2, 0, 1, 2, 3, 5$.
Note that $x \neq -2, 2, 5$ as they make the denominator zero.
At $x = 0$,the expression is $0$,which satisfies the inequality.
Using the wavy curve method (sign scheme):
For $x > 5$,the expression is positive.
For $3 < x < 5$,the expression is negative.
For $2 < x < 3$,the expression is positive.
For $1 < x < 2$,the expression is positive.
For $0 < x < 1$,the expression is negative.
For $-2 < x < 0$,the expression is positive.
For $x < -2$,the expression is positive.
Combining these intervals and including $x=0$,we get $( - \infty, -2 ) \cup \{0\} \cup [1, 2) \cup (2, 3] \cup (5, \infty)$.

(D) Given the inequality: $\frac{x - 5}{(x + 7)(x - 2)} > 0$.
We use the wavy curve method (sign scheme) to find the intervals.
The critical points are $x = -7, 2, 5$.
Testing the intervals:
For $x > 5$,the expression is positive.
For $2 < x < 5$,the expression is negative.
For $-7 < x < 2$,the expression is positive.
For $x < -7$,the expression is negative.
The solution set is $x \in (-7, 2) \cup (5, \infty)$.
The integral values in the interval $(-7, 2)$ are $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.
The integral values in the interval $(5, \infty)$ are $\{6, 7, 8, \dots\}$.
The least integral value $\alpha$ is $-6$.
Checking the options for $\alpha = -6$:
$(A)$ $(-6)^2 + 3(-6) - 4 = 36 - 18 - 4 = 14 \neq 0$.
$(B)$ $(-6)^2 - 5(-6) + 4 = 36 + 30 + 4 = 70 \neq 0$.
$(C)$ $(-6)^2 - 7(-6) + 6 = 36 + 42 + 6 = 84 \neq 0$.
$(D)$ $(-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0$.
Thus,$\alpha = -6$ satisfies option $(D)$.

(D) The given inequalities are $|x - y| \leq 2$ and $|x + y| \leq 2$.
These can be written as $-2 \leq x - y \leq 2$ and $-2 \leq x + y \leq 2$.
This represents the region bounded by the lines $x - y = 2$,$x - y = -2$,$x + y = 2$,and $x + y = -2$.
The vertices of this region are the intersection points of these lines:
$1$) $x - y = 2$ and $x + y = 2$ gives $(2, 0)$.
$2$) $x + y = 2$ and $x - y = -2$ gives $(0, 2)$.
$3$) $x - y = -2$ and $x + y = -2$ gives $(-2, 0)$.
$4$) $x + y = -2$ and $x - y = 2$ gives $(0, -2)$.
The distance between consecutive vertices (e.g.,$(2, 0)$ and $(0, 2)$) is $\sqrt{(2-0)^2 + (0-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.
Since all sides are equal to $2\sqrt{2}$ and the diagonals are perpendicular (slopes are $1$ and $-1$),the region is a square.
The area of the square is $(\text{side})^2 = (2\sqrt{2})^2 = 8$ sq. units.
Thus,the region is a square of side length $2\sqrt{2}$ units.

(C) Let $x$ be the marks obtained by the student in the annual examination.
The average of the three examinations is given by $\frac{62 + 48 + x}{3}$.
According to the problem,the average must be at least $60$,so we have the inequality:
$\frac{62 + 48 + x}{3} \geq 60$
Multiplying both sides by $3$,we get:
$110 + x \geq 180$
Subtracting $110$ from both sides,we get:
$x \geq 70$
Thus,the student must obtain a minimum of $70$ marks to have an average of at least $60$ marks.

(A) Let $x$ be the smaller of the two consecutive odd natural numbers,so the other one is $x+2$.
According to the problem:
$x > 10$ $(1)$
$x + (x + 2) < 40$ $(2)$
Solving $(2)$:
$2x + 2 < 40$
$2x < 38$
$x < 19$ $(3)$
From $(1)$ and $(3)$,we have $10 < x < 19$.
Since $x$ is an odd natural number,the possible values for $x$ are $11, 13, 15, 17$.
Thus,the pairs $(x, x+2)$ are:
$(11, 13), (13, 15), (15, 17), (17, 19)$.

(A) Let $x$ be the marks obtained by Ravi in the third unit test.
Since the student should have an average of at least $60$ marks,we have:
$\frac{70+75+x}{3} \geq 60$
Multiplying both sides by $3$:
$145+x \geq 180$
Subtracting $145$ from both sides:
$x \geq 180-145$
$x \geq 35$
Thus,the student must obtain a minimum of $35$ marks to have an average of at least $60$ marks.

(A) Let $x$ be the marks obtained by Sunita in the fifth examination.
In order to receive grade $A$ in the course,she must obtain an average of $90$ marks or more in five examinations.
Therefore,the inequality is:
$\frac{87+92+94+95+x}{5} \geq 90$
$\Rightarrow \frac{368+x}{5} \geq 90$
$\Rightarrow 368+x \geq 450$
$\Rightarrow x \geq 450-368$
$\Rightarrow x \geq 82$
Thus,Sunita must obtain at least $82$ marks in the fifth examination to get grade $A$.

(A) Let $x$ be the smaller of the two consecutive odd positive integers. Then,the other integer is $x+2$.
Since both integers are smaller than $10$,we have:
$x+2 < 10$
$\Rightarrow x < 8$ ... $(i)$
Also,the sum of the two integers is more than $11$:
$x + (x+2) > 11$
$\Rightarrow 2x + 2 > 11$
$\Rightarrow 2x > 9$
$\Rightarrow x > 4.5$ ... $(ii)$
From $(i)$ and $(ii)$,we have $4.5 < x < 8$.
Since $x$ must be an odd positive integer,the possible values for $x$ are $5$ and $7$.
If $x = 5$,the pair is $(5, 7)$.
If $x = 7$,the pair is $(7, 9)$.
Thus,the required pairs are $(5, 7)$ and $(7, 9)$.

(A) Let $x$ be the smaller of the two consecutive even positive integers.
Then,the other integer is $x + 2$.
Since both integers are larger than $5$,we have $x > 5$ $(1)$.
Also,the sum of the two integers is less than $23$,so $x + (x + 2) < 23$.
$2x + 2 < 23$
$2x < 21$
$x < 10.5$ $(2)$.
From $(1)$ and $(2)$,we obtain $5 < x < 10.5$.
Since $x$ is an even number,$x$ can take the values $6, 8,$ and $10$.
Thus,the required possible pairs are $(6, 8), (8, 10),$ and $(10, 12)$.

(C) Let the length of the shortest side of the triangle be $x \, cm$.
Then,the length of the longest side is $3x \, cm$.
The length of the third side is $(3x - 2) \, cm$.
Since the perimeter of the triangle is at least $61 \, cm$,we have:
$x + 3x + (3x - 2) \geq 61$
$\Rightarrow 7x - 2 \geq 61$
$\Rightarrow 7x \geq 63$
$\Rightarrow x \geq 9$
Thus,the minimum length of the shortest side is $9 \, cm$.

(N/A) Let the length of the shortest piece be $x \, cm$. Then,the lengths of the second and third pieces are $(x+3) \, cm$ and $2x \, cm$ respectively.
Since the three lengths are cut from a single board of length $91 \, cm$:
$x + (x+3) + 2x \leq 91$
$4x + 3 \leq 91$
$4x \leq 88$
$x \leq 22$ ...... $(1)$
Also,the third piece is at least $5 \, cm$ longer than the second piece:
$2x \geq (x+3) + 5$
$2x \geq x + 8$
$x \geq 8$ ...... $(2)$
From $(1)$ and $(2)$,we obtain:
$8 \leq x \leq 22$
Thus,the possible length of the shortest board is greater than or equal to $8 \, cm$ but less than or equal to $22 \, cm$.

(A) It is given that the temperature $C$ in Celsius is $30 < C < 35$.
Substituting the conversion formula $C = \frac{5}{9}(F - 32)$,we get:
$30 < \frac{5}{9}(F - 32) < 35$
Multiplying the entire inequality by $\frac{9}{5}$:
$\frac{9}{5} \times 30 < F - 32 < \frac{9}{5} \times 35$
$54 < F - 32 < 63$
Adding $32$ to all parts of the inequality:
$54 + 32 < F < 63 + 32$
$86 < F < 95$
Thus,the required range of temperature in Fahrenheit is between $86^{\circ} \text{F}$ and $95^{\circ} \text{F}$.

Let $x$ litres of $30 \%$ acid solution be added to the mixture.
The total volume of the resulting mixture is $(x + 600) \text{ litres}$.
The amount of acid in the mixture is $0.30x + 0.12(600)$.
According to the problem,the acid content must be between $15 \%$ and $18 \%$:
$0.15(x + 600) < 0.30x + 0.12(600) < 0.18(x + 600)$
Solving the first inequality:
$0.15x + 90 < 0.30x + 72$
$18 < 0.15x$
$x > \frac{18}{0.15} = 120$
Solving the second inequality:
$0.30x + 72 < 0.18x + 108$
$0.12x < 36$
$x < \frac{36}{0.12} = 300$
Thus,the number of litres of the $30 \%$ solution must be greater than $120 \text{ litres}$ and less than $300 \text{ litres}$.

(A) Given that the temperature $F$ is between $68\,^{\circ} F$ and $77\,^{\circ} F$,we have the inequality:
$68 < F < 77$
Substituting the formula $F = \frac{9}{5} C + 32$ into the inequality:
$68 < \frac{9}{5} C + 32 < 77$
Subtracting $32$ from all parts:
$68 - 32 < \frac{9}{5} C < 77 - 32$
$36 < \frac{9}{5} C < 45$
Multiplying by $\frac{5}{9}$ throughout:
$36 \times \frac{5}{9} < C < 45 \times \frac{5}{9}$
$20 < C < 25$
Thus,the required range of temperature in degree Celsius is between $20\,^{\circ} C$ and $25\,^{\circ} C$.

(A) Let $x$ litres of $2 \%$ boric acid solution be added.
Total volume of the mixture $= (x + 640)$ litres.
Amount of boric acid in the mixture $= (0.02x + 0.08 \times 640) = (0.02x + 51.2)$ litres.
The concentration of the resulting mixture must be between $4 \%$ and $6 \%$.
Condition $1$: $\frac{0.02x + 51.2}{x + 640} > 0.04$
$0.02x + 51.2 > 0.04x + 25.6$
$51.2 - 25.6 > 0.04x - 0.02x$
$25.6 > 0.02x \Rightarrow x < 1280$.
Condition $2$: $\frac{0.02x + 51.2}{x + 640} < 0.06$
$0.02x + 51.2 < 0.06x + 38.4$
$51.2 - 38.4 < 0.06x - 0.02x$
$12.8 < 0.04x \Rightarrow x > 320$.
Thus,the amount of $2 \%$ solution to be added is between $320$ litres and $1280$ litres.

(N/A) Let $x$ litres of water be added to the solution.
Total volume of the mixture becomes $(1125 + x)$ litres.
The amount of acid remains constant at $45 \%$ of $1125$ litres,which is $0.45 \times 1125 = 506.25$ litres.
According to the problem,the acid content in the new mixture must be between $25 \%$ and $30 \%$.
So,$25 \% < \frac{506.25}{1125 + x} \times 100 < 30 \%$.
First,consider the inequality $\frac{506.25}{1125 + x} < 0.30$:
$506.25 < 0.30(1125 + x)$
$506.25 < 337.5 + 0.3x$
$168.75 < 0.3x$
$x > \frac{168.75}{0.3} = 562.5$.
Next,consider the inequality $\frac{506.25}{1125 + x} > 0.25$:
$506.25 > 0.25(1125 + x)$
$506.25 > 281.25 + 0.25x$
$225 > 0.25x$
$x < \frac{225}{0.25} = 900$.
Therefore,the amount of water to be added must be between $562.5$ litres and $900$ litres,i.e.,$562.5 < x < 900$.

(A) It is given that for a group of $12$ years old children,$CA = 12$ years.
The formula for $IQ$ is $IQ = \frac{MA}{12} \times 100$.
Given the inequality $80 \leq IQ \leq 140$,we substitute the expression for $IQ$:
$80 \leq \frac{MA}{12} \times 100 \leq 140$
To solve for $MA$,multiply the entire inequality by $\frac{12}{100}$:
$80 \times \frac{12}{100} \leq MA \leq 140 \times \frac{12}{100}$
$9.6 \leq MA \leq 16.8$
Thus,the range of mental age for the group of $12$ years old children is $9.6 \leq MA \leq 16.8$.

(77) Let the marks obtained by Ravi in the fifth examination be $x$.
The total marks for the five examinations are $94 + 73 + 72 + 84 + x = 323 + x$.
The average of the five examinations is $\frac{323 + x}{5}$.
According to the condition for grade $'B'$,the average must be at least $80$ and less than $90$:
$80 \le \frac{323 + x}{5} < 90$.
Multiplying the inequality by $5$:
$400 \le 323 + x < 450$.
Subtracting $323$ from all parts:
$400 - 323 \le x < 450 - 323$.
$77 \le x < 127$.
Since the maximum marks in an examination is $100$,the minimum marks required is $77$.

(B) Let the marks scored by Mohan in the third test be $x$.
Given that the marks in the first two tests are $62$ and $48$.
The average of the three tests is given by $\frac{62 + 48 + x}{3}$.
We want the average to be at least $60$,so we set up the inequality:
$\frac{62 + 48 + x}{3} \geq 60$
$110 + x \geq 180$
$x \geq 180 - 110$
$x \geq 70$
Thus,the minimum marks he should score in the third test is $70$.

(A) Let the shortest side of the triangle be $x \text{ cm}$.
According to the problem, the longest side is $2x \text{ cm}$ and the third side is $(x + 3) \text{ cm}$.
The perimeter of the triangle is the sum of its three sides:
$P = x + 2x + (x + 3) = 4x + 3$.
Given that the perimeter is at least $51 \text{ cm}$, we have the inequality:
$4x + 3 \geq 51$.
Subtracting $3$ from both sides:
$4x \geq 48$.
Dividing by $4$:
$x \geq 12$.
Thus, the minimum length of the shortest side is $12 \text{ cm}$.

(A) Let the two consecutive odd positive integers be $x$ and $x + 2$.
Given that both integers are smaller than $18$,we have $x + 2 < 18$,which implies $x < 16$.
Also,their sum is more than $20$,so $x + (x + 2) > 20$.
$2x + 2 > 20$
$2x > 18$
$x > 9$.
Since $x$ is an odd integer and $9 < x < 16$,the possible values for $x$ are $11, 13, 15$.
If $x = 11$,the pair is $(11, 13)$.
If $x = 13$,the pair is $(13, 15)$.
If $x = 15$,the pair is $(15, 17)$.
Thus,the pairs are $(11, 13), (13, 15), (15, 17)$.

(A) Given the temperature range in Celsius is $30 < C < 35$.
Using the conversion formula $F = \frac{9}{5} C + 32$:
For $C = 30$,$F = \frac{9}{5}(30) + 32 = 54 + 32 = 86^{\circ} F$.
For $C = 35$,$F = \frac{9}{5}(35) + 32 = 63 + 32 = 95^{\circ} F$.
Thus,the range of temperature in Fahrenheit is between $86^{\circ} F$ and $95^{\circ} F$.

(A) Given the inequality $\frac{|x-1|}{x+2} < 1$.
Case $1$: If $x+2 > 0$,i.e.,$x > -2$,then $|x-1| < x+2$.
If $x \ge 1$,then $x-1 < x+2 \implies -1 < 2$,which is true for all $x \ge 1$.
If $-2 < x < 1$,then $-(x-1) < x+2 \implies -x+1 < x+2 \implies 2x > -1 \implies x > -\frac{1}{2}$.
So,for $x > -2$,the solution is $x > -\frac{1}{2}$.
Case $2$: If $x+2 < 0$,i.e.,$x < -2$,then $|x-1| > x+2$.
Since $x < -2$,$x-1$ is negative,so $|x-1| = -(x-1) = 1-x$.
$1-x > x+2 \implies 2x < -1 \implies x < -\frac{1}{2}$.
Since we assumed $x < -2$,the solution is $x < -2$.
Combining both cases,the solution set is $(-\infty, -2) \cup (-\frac{1}{2}, \infty)$.

(D) For the first inequality: $\frac{x}{2x+1} - \frac{1}{4} \geq 0 \implies \frac{4x - (2x+1)}{4(2x+1)} \geq 0 \implies \frac{2x-1}{4(2x+1)} \geq 0$. The critical points are $x = -\frac{1}{2}$ and $x = \frac{1}{2}$. Testing intervals,we get $x \in (-\infty, -\frac{1}{2}) \cup [\frac{1}{2}, \infty)$.
For the second inequality: $\frac{6x}{4x-1} - \frac{1}{2} < 0 \implies \frac{12x - (4x-1)}{2(4x-1)} < 0 \implies \frac{8x+1}{2(4x-1)} < 0$. The critical points are $x = -\frac{1}{8}$ and $x = \frac{1}{4}$. Testing intervals,we get $x \in (-\frac{1}{8}, \frac{1}{4})$.
Since there is no common interval between $(-\infty, -\frac{1}{2}) \cup [\frac{1}{2}, \infty)$ and $(-\frac{1}{8}, \frac{1}{4})$,the solution set is $\emptyset$ (empty set).

(B) Case $1$: $x+2 \ge 0 \implies x \ge -2$. The inequality becomes $\frac{x+2-x}{x} < 2 \implies \frac{2}{x} < 2 \implies \frac{2-2x}{x} < 0 \implies \frac{1-x}{x} < 0$. This holds for $x \in (-\infty, 0) \cup (1, \infty)$. Considering $x \ge -2$ and $x \neq 0$,we get $x \in [-2, 0) \cup (1, \infty)$.
Case $2$: $x+2 < 0 \implies x < -2$. The inequality becomes $\frac{-(x+2)-x}{x} < 2 \implies \frac{-2x-2}{x} < 2 \implies \frac{-2x-2-2x}{x} < 0 \implies \frac{-4x-2}{x} < 0 \implies \frac{2x+1}{x} > 0$. This holds for $x \in (-\infty, -1/2) \cup (0, \infty)$. Considering $x < -2$,we get $x \in (-\infty, -2)$.
Combining both cases,the solution is $(-\infty, -2) \cup [-2, 0) \cup (1, \infty)$,which simplifies to $(-\infty, 0) \cup (1, \infty)$.

(A) Let $f(x) = |x-1|+|x-2|+|x-3|$.
Case $1$: $x < 1$,$f(x) = -(x-1)-(x-2)-(x-3) = -3x+6$. Setting $-3x+6 \geq 6$,we get $-3x \geq 0$,so $x \leq 0$.
Case $2$: $1 \leq x < 2$,$f(x) = (x-1)-(x-2)-(x-3) = -x+4$. Setting $-x+4 \geq 6$,we get $-x \geq 2$,so $x \leq -2$ (No solution in this interval).
Case $3$: $2 \leq x < 3$,$f(x) = (x-1)+(x-2)-(x-3) = x$. Setting $x \geq 6$ (No solution in this interval).
Case $4$: $x \geq 3$,$f(x) = (x-1)+(x-2)+(x-3) = 3x-6$. Setting $3x-6 \geq 6$,we get $3x \geq 12$,so $x \geq 4$.
Combining the results,the solution is $x \in (-\infty, 0] \cup [4, \infty)$.

Let $x$ be the number of litres of the $30 \%$ acid solution to be added.
The total volume of the mixture will be $(600 + x) \text{ litres}$.
The amount of acid in the mixture is $(0.12 \times 600 + 0.30 \times x) = (72 + 0.3x) \text{ litres}$.
According to the problem,the acid content must be more than $15 \%$ and less than $18 \%$ of the total volume:
$0.15(600 + x) < 72 + 0.3x < 0.18(600 + x)$.
Solving the first inequality: $90 + 0.15x < 72 + 0.3x \implies 18 < 0.15x \implies x > 120$.
Solving the second inequality: $72 + 0.3x < 108 + 0.18x \implies 0.12x < 36 \implies x < 300$.
Thus,the amount of $30 \%$ acid solution to be added must be more than $120 \text{ litres}$ and less than $300 \text{ litres}$.

(B) Given the temperature range in Fahrenheit is $86 < F < 95$.
Substituting the formula $F = \frac{9}{5} C + 32$ into the inequality:
$86 < \frac{9}{5} C + 32 < 95$
Subtracting $32$ from all parts:
$86 - 32 < \frac{9}{5} C < 95 - 32$
$54 < \frac{9}{5} C < 63$
Multiplying by $\frac{5}{9}$:
$54 \times \frac{5}{9} < C < 63 \times \frac{5}{9}$
$30 < C < 35$
Thus,the range of temperature in Celsius is between $30^{\circ} C$ and $35^{\circ} C$.

(N/A) To realize no profit or loss,the revenue must equal the cost,i.e.,$R(x) = C(x)$.
Substituting the given functions: $3x = 500 + \frac{5}{2}x$.
Subtracting $\frac{5}{2}x$ from both sides: $3x - 2.5x = 500$.
$0.5x = 500$.
$x = \frac{500}{0.5} = 1000$.
Thus,$1000$ items must be sold to realize no profit or loss.
Profit occurs when $R(x) > C(x)$,which implies $3x > 500 + 2.5x$,or $0.5x > 500$,which means $x > 1000$. Therefore,there is a profit if more than $1000$ items are sold.

Let $x$ be the amount of acid added in litres.
The initial amount of acid is $40 \% \text{ of } 1120 = 0.40 \times 1120 = 448 \text{ litres}$.
The initial amount of water is $1120 - 448 = 672 \text{ litres}$.
After adding $x$ litres of acid,the total volume becomes $(1120 + x) \text{ litres}$.
The amount of water remains $672 \text{ litres}$.
The percentage of water in the new mixture is $\frac{672}{1120 + x} \times 100$.
We are given that $40 < \frac{672}{1120 + x} \times 100 < 50$.
Dividing by $100$: $0.4 < \frac{672}{1120 + x} < 0.5$.
Taking the first inequality: $0.4 < \frac{672}{1120 + x} \implies 1120 + x < \frac{672}{0.4} = 1680 \implies x < 560$.
Taking the second inequality: $\frac{672}{1120 + x} < 0.5 \implies 1120 + x > \frac{672}{0.5} = 1344 \implies x > 224$.
Thus,the amount of acid to be added is between $224 \text{ litres}$ and $560 \text{ litres}$.

(N/A) Given the formula $I.Q. = \frac{MA}{CA} \times 100$ and $CA = 16$.
Substituting the values,we get $75 < \frac{MA}{16} \times 100 < 125$.
Divide the entire inequality by $100$: $0.75 < \frac{MA}{16} < 1.25$.
Multiply by $16$: $0.75 \times 16 < MA < 1.25 \times 16$.
Calculating the values: $12 < MA < 20$.
Thus,the range of their mental age is between $12$ and $20$ years.

(N/A) Consider the first part of the inequality: $\frac{4}{x+1} \leq 3$.
Since $x > 0$,$x+1 > 0$,so we can multiply by $(x+1)$ without changing the inequality sign:
$4 \leq 3(x+1)$
$4 \leq 3x + 3$
$1 \leq 3x$
$x \geq \frac{1}{3} \quad (i)$
Consider the second part of the inequality: $3 \leq \frac{6}{x+1}$.
Since $x+1 > 0$,we have:
$3(x+1) \leq 6$
$3x + 3 \leq 6$
$3x \leq 3$
$x \leq 1 \quad (ii)$
Combining $(i)$ and $(ii)$,we get:
$\frac{1}{3} \leq x \leq 1$
Thus,the solution set is $x \in [\frac{1}{3}, 1]$.

(N/A) Given the inequality: $\frac{1}{|x|-3} \leq \frac{1}{2}$.
Case $1$: If $|x|-3 > 0$,then $|x| > 3$. Multiplying both sides by $2(|x|-3)$,we get $2 \leq |x|-3$,which implies $|x| \geq 5$. Since $|x| \geq 5$ satisfies $|x| > 3$,the solution for this case is $x \in (-\infty, -5] \cup [5, \infty)$.
Case $2$: If $|x|-3 < 0$,then $|x| < 3$. Multiplying both sides by $2(|x|-3)$ reverses the inequality sign: $2 \geq |x|-3$,which implies $|x| \leq 5$. Since we must satisfy $|x| < 3$,the solution for this case is $x \in (-3, 3)$.
Combining both cases,the final solution is $x \in (-\infty, -5] \cup (-3, 3) \cup [5, \infty)$.

(NONE) We have $4x + 3 \geq 2x + 17$.
$\therefore 4x - 2x \geq 17 - 3$
$\therefore 2x \geq 14$
$\therefore x \geq 7$
$...(i)$
Now,we have $3x - 5 < -2$.
$\therefore 3x < -2 + 5$
$\therefore 3x < 3$
$\therefore x < 1$
$...(ii)$
From $(i)$ and $(ii)$,we need to find the values of $x$ that satisfy both $x \geq 7$ and $x < 1$ simultaneously.
Since there is no real number $x$ that is both greater than or equal to $7$ and less than $1$,the solution set is the empty set,denoted by $\emptyset$.

(A) Profit is defined as Revenue minus Cost,i.e.,$P(x) = R(x) - C(x)$.
Given $R(x) = 43x$ and $C(x) = 26000 + 30x$.
$P(x) = 43x - (26000 + 30x)$
$P(x) = 43x - 26000 - 30x$
$P(x) = 13x - 26000$
For the company to realize a profit,the profit must be greater than zero,so $P(x) > 0$.
$13x - 26000 > 0$
$13x > 26000$
$x > \frac{26000}{13}$
$x > 2000$
Therefore,the company must sell more than $2000$ cassettes to realize a profit.

(N/A) Let the third pH reading be $x$.
The average of the three readings must be between $8.2$ and $8.5$.
$\therefore 8.2 < \frac{8.48 + 8.35 + x}{3} < 8.5$
Multiply the entire inequality by $3$:
$\therefore 3 \times 8.2 < 16.83 + x < 8.5 \times 3$
$\therefore 24.6 < 16.83 + x < 25.5$
Subtract $16.83$ from all parts:
$\therefore 24.6 - 16.83 < x < 25.5 - 16.83$
$\therefore 7.77 < x < 8.67$
Hence,the third pH reading must be in the range $(7.77, 8.67)$.

(N/A) Let $x \text{ L}$ of $3 \%$ solution be added to $460 \text{ L}$ of $9 \%$ solution of acid.
Total quantity of mixture $= (460 + x) \text{ L}$.
Total acid content in the $(460 + x) \text{ L}$ of mixture $= 460 \times \frac{9}{100} + x \times \frac{3}{100}$.
It is given that acid content in the resulting mixture must be more than $5 \%$ but less than $7 \%$ acid.
$\therefore 5 \% \text{ of } (460 + x) < \left( 460 \times \frac{9}{100} + \frac{3x}{100} \right) < 7 \% \text{ of } (460 + x)$.
$\therefore (460 + x) \frac{5}{100} < \frac{4140 + 3x}{100} < (460 + x) \frac{7}{100}$.
$\therefore 5(460 + x) < 4140 + 3x < 7(460 + x)$.
$\therefore 2300 + 5x < 4140 + 3x < 3220 + 7x$.
Now,solving $2300 + 5x < 4140 + 3x$:
$2x < 1840 \implies x < 920 \quad \dots(i)$.
Solving $4140 + 3x < 3220 + 7x$:
$3x - 7x < 3220 - 4140 \implies -4x < -920 \implies 4x > 920 \implies x > 230 \quad \dots(ii)$.
From $(i)$ and $(ii)$,the number of litres of the $3 \%$ solution must be more than $230 \text{ L}$ and less than $920 \text{ L}$.

(A) Given that the temperature $C$ in Celsius lies between $40^{\circ} C$ and $45^{\circ} C$,we have the inequality: $40 < C < 45$.
The conversion formula is $F = \frac{9}{5} C + 32$.
To find the range in Fahrenheit,we substitute $C = \frac{5}{9}(F - 32)$ into the inequality:
$40 < \frac{5}{9}(F - 32) < 45$.
Multiply the entire inequality by $\frac{9}{5}$:
$40 \times \frac{9}{5} < F - 32 < 45 \times \frac{9}{5}$.
$72 < F - 32 < 81$.
Add $32$ to all parts of the inequality:
$72 + 32 < F < 81 + 32$.
$104 < F < 113$.
Thus,the range of temperature in degree Fahrenheit is between $104^{\circ} F$ and $113^{\circ} F$.