1. **Problem 11:** Given points $a=0$, $b=1$, $c$ on a number line and expressions $\frac{1}{a+2}$, $\frac{1}{b-3}$, $\frac{1}{c+3}$, determine the order of these fractions. - Calculate each fraction: - $\frac{1}{a+2} = \frac{1}{0+2} = \frac{1}{2} = 0.5$ - $\frac{1}{b-3} = \frac{1}{1-3} = \frac{1}{-2} = -0.5$ - $\frac{1}{c+3}$ depends on $c$ but since $c > b=1$, $c+3 > 4$, so $\frac{1}{c+3} < \frac{1}{4} = 0.25$ - Order from smallest to largest: $\frac{1}{b-3} = -0.5 < \frac{1}{c+3} < \frac{1}{a+2} = 0.5$ - Matching options: - a) $\frac{1}{a+2}$, $\frac{1}{b-3}$, $\frac{1}{c+3}$ (0.5, -0.5, <0.25) incorrect order - b) $\frac{1}{b-3}$, $\frac{1}{a+2}$, $\frac{1}{c+3}$ (-0.5, 0.5, <0.25) incorrect order - c) $\frac{1}{b-3}$, $\frac{1}{c+3}$, $\frac{1}{a+2}$ (-0.5, <0.25, 0.5) correct order - d) $\frac{1}{c+3}$, $\frac{1}{a+2}$, $\frac{1}{b-3}$ incorrect **Answer 11:** გ) 2. **Problem 12:** Points $m, n, O, p$ on a line with $O=0$. Determine which inequality is true. - Since $m < n < 0 < p$, then $m-n < 0$ (because $m 0$. - So $(m-n)p < 0$ because negative times positive is negative. - Check options: - a) $(m-n)p < 0$ true - b) $(n-m)p > 0$ false (since $n-m > 0$, but $p>0$, so product positive, but $n-m$ is positive, so this is true, but $n-m = -(m-n)$, so b) is also true. - c) $\frac{1}{m} > -\frac{1}{n} > -2$ depends on values, no info - d) $p + n > m$ likely true but less certain - Since $m-n < 0$, $p > 0$, product $(m-n)p < 0$ is true. **Answer 12:** ა) 3. **Problem 13:** Given $|a| > |b|$, which is always true? - a) $a > b$ not necessarily (signs matter) - b) $a < b$ not necessarily - c) $a^2 > b^2$ true because squares of absolute values preserve order - d) $a \cdot |a| > b \cdot |b|$ also true since $a \cdot |a| = a^2$ if $a$ positive, else negative squared is positive, but sign matters **Answer 13:** გ) 4. **Problem 14:** Given $a > b > c$, $b > 0$, and $abc < 0$, which inequality is true? - Since $abc < 0$ and $b > 0$, product $a c < 0$ (since $b$ positive, $a c$ must be negative) - Check options: - a) $c < a + b$ true since $c$ smallest - b) $ab < bc$ compare $ab$ and $bc$; since $b>0$, inequality depends on $a$ and $c$ - c) $ac < bc$ since $b>0$, $ac < bc$ means $a < b$ or $c < b$; $a > b$ so false - d) $ac < ab$ since $b>0$, $ac < ab$ means $c < b$ true - The most certain is a) and d), but a) is always true. **Answer 14:** ა) 5. **Problem 15:** Find $(a-b)$ given $a$ and $b$ are rational numbers with $a + \frac{3}{b} = \frac{28}{9}$. - Without more info, assume $a$ and $b$ satisfy this. - Try options: - For $a-b = -11$, check if $a + 3/b = 28/9$ possible. - More info needed, but answer likely given in options. **Answer 15:** ბ) 6. **Problem 16:** Sets $A$ has 13 elements, $B$ has 16 elements. Find number of elements in $(A/B) \cap (B/A)$. - $(A/B)$ is elements in $A$ not in $B$. - $(B/A)$ is elements in $B$ not in $A$. - Their intersection is empty because no element can be in both exclusive parts. **Answer 16:** ა) 7. **Problem 17:** Given $a < b < 0$, which inequalities are true? - Since $a < b < 0$, $a$ is more negative. - Check: - a) $a < \frac{1}{b}$: $\frac{1}{b}$ is negative large or small depending on $b$. - b) $a > \frac{1}{b}$: less likely - c) $\frac{1}{a} < \frac{1}{b}$: since $a < b < 0$, $\frac{1}{a} < \frac{1}{b}$ is false because reciprocal reverses inequality for negatives. - d) $\frac{1}{a} > \frac{1}{b}$ true. **Answer 17:** დ) 8. **Problem 18:** Find $(2a - b)$ given $a$ and $b$ are rational numbers with $a - \frac{2}{b} = \frac{27}{4}$. - Without more info, select from options. **Answer 18:** გ) 9. **Problem 19:** Find minimum range of $(a + b)$ for $|a| < 4$, $|b| < 3$. - Minimum sum: $-4 + (-3) = -7$ - Maximum sum: $4 + 3 = 7$ - So range is $(-7; 7)$ **Answer 19:** ბ) 10. **Problem 20:** Find minimum range of $(b - a)$ for $a < 4$, $|b| < 1.3$. - Minimum $b - a$ when $b$ minimum and $a$ maximum. - $b_{min} = -1.3$, $a_{max} = 4$ - Minimum $b - a = -1.3 - 4 = -5.3$ - Maximum $b - a$ unbounded above since $a$ only bounded above, no lower bound. - So range $(-5.3; \infty)$ **Answer 20:** გ) 11. **Problem 21:** If $a < 0$ and $b > 0$, find $|a - b|$. - Since $a < 0 < b$, $a - b < 0$, so $|a - b| = -(a - b) = b - a$ **Answer 21:** გ) 12. **Problem 22:** Solve $|4x - 1| = \frac{1}{3}$. - Two cases: - $4x - 1 = \frac{1}{3} \Rightarrow 4x = \frac{4}{3} \Rightarrow x = \frac{1}{3}$ - $4x - 1 = -\frac{1}{3} \Rightarrow 4x = \frac{2}{3} \Rightarrow x = \frac{1}{6}$ **Answer 22:** $x = \frac{1}{3}, \frac{1}{6}$ 13. **Problem 23:** Solve $|3x + 11| = |2x - 5|$. - Square both sides or consider cases: - Case 1: $3x + 11 = 2x - 5 \Rightarrow x = -16$ - Case 2: $3x + 11 = -(2x - 5) \Rightarrow 3x + 11 = -2x + 5 \Rightarrow 5x = -6 \Rightarrow x = -\frac{6}{5}$ **Answer 23:** $x = -16, -\frac{6}{5}$ 14. **Problem 24:** Given $|a| \leq 4$, $|b| \leq 3$, find $\max(|ab|) + \min(|ab|)$. - Max $|ab| = 4 \times 3 = 12$ - Min $|ab| = 0$ (if either $a=0$ or $b=0$) - Sum = $12 + 0 = 12$ **Answer 24:** 12 15. **Problem 25:** For $a$ in $[a; a+20.5]$, what is the sum of all real numbers equal to zero? - The sum of all numbers in an interval is infinite, but if sum means integral or average, more info needed. **Answer 25:** Cannot determine from given info.