[tex] \large \mathcal{ANSWER:} [/tex]
[tex] \boxed{\begin{array}{l} 1.\:\textsf{38.5 units, 39.5 units, 40.5 units, and} \\ \quad\textsf{41.5 units}\\ \\ 2.\: \textsf{Inzo = 20 years old, Migz = 10 years old} \\ \\ 3.\: \dfrac{675}{11}\:\textsf{kph or approx. 61.36 kph} \\ \\ 4.\:\textsf{Choose atleast 3 pairs from the list below.} \\ \\ 5.\: \textsf{The length of each side of the square is} \\ \quad \textsf{less than 38 meters.} \end{array}}[/tex]
[tex] \large \mathcal{SOLUTION:} [/tex]
[tex] \begin{array}{l} \textsf{1. Let}\:s, s+1, s+2, \:\textsf{and}\: s+3,\: \textsf{be the length} \\ \quad \textsf{of each side of the quadrilateral.} \\ \\ P = s + (s + 1) + (s + 2) + (s + 3) = 160 \\ \begin{aligned} 4s + 6 &= 160 \\ 4s &=160 - 6 \\ 4s &= 154 \\ \dfrac{\cancel{4}s}{\cancel{4}} &= \dfrac{154}{4} \\ \implies s &=38.5\:\textsf{units} \\ \implies s + 1 &=39.5\:\textsf{units} \\ \implies s + 2 &=40.5\:\textsf{units} \\ \implies s + 3 &=41.5\:\textsf{units} \end{aligned} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{2. Let}\: x\:\textsf{and}\:x + 10\:\textsf{be the ages of Migz and} \\ \quad \textsf{Inzo, respectively.} \\ \\ \begin{aligned} \ x - 5 &= \dfrac{1}{3}(x + 10 - 5) \\ 3(x - 5) &= \left [\dfrac{1}{\cancel{3}}(x + 5)\right] \cancel{3} \\ 3x - 15 &= x + 5 \\ 3x - x &=5 + 15 \\ \dfrac{\cancel{2}x}{\cancel{2}} &= \dfrac{20}{2} \\ \implies x &=10\:\textsf{years old}\quad(\textsf{Migz's age}) \\ \implies x + 10 &=20\:\textsf{years old}\quad(\textsf{Inzo's age}) \end{aligned} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{3. Let}\:d\:\textsf{be the distance from station A to} \\ \quad \textsf{station B.} \\ \\ \textsf{Given:} \\ \quad \textsf{Speed of express train} = 100\:\textsf{kph} \\ \quad \textsf{Speed of local train} = 45\:\textsf{kph} \\ \\ \textsf{Note: Time} = \dfrac{\textsf{Distance}}{\textsf{Speed}} \\ \\ \begin{aligned} \quad \dfrac{d}{45} - \dfrac{d}{100} &= \dfrac{3}{4} \\ \dfrac{20d - 9d}{900} &= \dfrac{3}{4} \\ 11d &=\dfrac{3}{4}(900) \\ \dfrac{\cancel{11}d}{11} &= \dfrac{675}{11} \\ \implies d &= \dfrac{675}{11} \approx 61.36\: \textsf{kph} \end{aligned} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{4. Let}\: 2n\:\textsf{and}\:2n + 2\:\textsf{be the two consecutive} \\ \quad \textsf{even integers.} \\ \\ \quad \quad 24 < 2n + (2n + 2) < 60 \\ \quad \quad \quad 24 < 4n + 2 < 60 \\ \quad \quad 24 - 2 < 4n < 60 - 2 \\ \quad \quad \quad \:\:22 < 4n < 58 \\ \quad\quad\quad \:\dfrac{22}{2} < \dfrac{4n}{2} < \dfrac{58}{2} \\ \quad\quad \quad \:\: 11 < 2n < 29 \\ \\ \implies \textsf{The smaller even integer is less than 29} \\ \quad\quad \:\:\textsf{but greater than 11.} \\ \\ \textsf{Possible pairs of two consecutive even integers} \\\textsf{which satisfy the given condition:} \\ \{(12, 14), (14, 16), (16, 18,), (18, 20), (20, 22), \\ \quad (22, 24), (24, 26), (26, 28), (28, 30)\} \end{array} [/tex]
[tex] \begin{array}{l} \textsf{5. Let}\:s\:\textsf{be the length of each side of the square.} \\ \\ \begin{aligned}\quad \quad \quad 4s &< 152 \\ \dfrac{\cancel{4}s}{4} &< \dfrac{152}{4} \\ \implies s &< 38 \:\textsf{m} \end{aligned} \end{array} [/tex]
[tex] \texttt \color{cyan} {\#CarryOnLearning} [/tex]
