1. Problem 7: If $(x - 4)$ varies inversely as $(y + 3)$ and $x = 8$ when $y = 2$, find the value of $x$ when $y$ equals one of the options. Step 1: Write inverse variation formula: $x - 4 = \frac{k}{y + 3}$. Step 2: Substitute $x=8$ and $y=2$: $8 - 4 = \frac{k}{2 + 3} \Rightarrow 4 = \frac{k}{5} \Rightarrow k = 20$. Step 3: Rewrite formula with $k=20$: $x - 4 = \frac{20}{y + 3}$. Step 4: Find $x$ for each $y$ to match options: - For $x-4=\frac{20}{y+3}$, options are values to check but user only gave $x$ values; however, question likely asks for $x$ values corresponding to certain $y$. (Note: options a=20, b=18, c=16, d=14 likely refer to values to check for $x$ or $y$; assuming $x$ values). Check if $x=20$ is possible: - For $x=20$: $20-4=16=\frac{20}{y+3} \implies y+3=\frac{20}{16}=1.25 \implies y=-1.75$ (not given). - For $x=18$: $18-4=14=\frac{20}{y+3} \implies y+3=\frac{20}{14}=1.4286 \implies y=-1.5714$ (no match) - For $x=16$: $12=\frac{20}{y+3} \implies y+3=\frac{20}{12}=1.6667 \implies y=-1.3333$ - For $x=14$: $10=\frac{20}{y+3} \implies y+3=2 \implies y=-1$ None matches $y=2$ exactly; likely the problem expects calculation of $x$ for given $y$ values. So problem statement is ambiguous but key was to find $k=20$. 2. Problem 8: Gasoline usage varies jointly as distance and square root of speed. Given 25 liters for 100 km at 100 kph, estimate usage for 1000 km at 64 kph. Step 1: Write the joint variation formula: $G = k \times d \times \sqrt{s}$. Step 2: Given $G=25, d=100, s=100$, solve for $k$: $$25 = k \times 100 \times \sqrt{100} = k \times 100 \times 10 = 1000k \Rightarrow k = \frac{25}{1000} = 0.025$$ Step 3: Calculate $G$ for $d=1000$, $s=64$: $$G = 0.025 \times 1000 \times \sqrt{64} = 0.025 \times 1000 \times 8 = 200$$ Answer: 200 liters. 3. Problem 9: $y$ varies directly as the square of $x$. Find % change in $y$ if $x$ is increased by 20%. Step 1: Model: $y = kx^2$. Step 2: Let original $x$ be $x$, new $x$ is $1.2x$. Step 3: New $y$: $y_{new} = k(1.2x)^2 = k imes 1.44 x^2 = 1.44y$. Step 4: Percent increase in $y$ = $44\%$. Answer: 44% increase in $y$. 4. Problem 10: $w$ varies jointly as $l^2$ and $i$, and inversely as $g$. Given $h=50$ when $j=2, i=5, g=1/2$, find $h$ when $j=4, i=10, g=1/4$. Step 1: Model: $w = \frac{k l^2 i}{g}$. Step 2: Given $h=50$ when $j=2, i=5, g=1/2$, $$50 = \frac{k \times 2^2 \times 5}{1/2} = \frac{k \times 4 \times 5}{0.5} = k \times 40 \Rightarrow k = \frac{50}{40} = 1.25$$ Step 3: Find $h$ when $j=4, i=10, g=1/4$: $$h = \frac{1.25 \times 4^2 \times 10}{1/4} = \frac{1.25 \times 16 \times 10}{0.25} = \frac{200}{0.25} = 800$$ Answer: 800. Summary: - Problem 7 constant $k=20$. - Problem 8: 200 liters. - Problem 9: 44% increase. - Problem 10: 800.