1. Problem 210: Calculate the dealer's profit or loss as a percent of the initial cost for 60 cameras. - Let the dealer's initial cost per camera be $C$. - The selling price per camera is $250$, which is a 20% markup over $C$. Formula for markup: $$250 = C + 0.20C = 1.20C \implies C = \frac{250}{1.20} = 208.33$$ - Total initial cost for 60 cameras: $$60 \times 208.33 = 12500$$ - Cameras sold: 54 (since 6 were returned). - Revenue from sold cameras: $$54 \times 250 = 13500$$ - Refund for 6 returned cameras at 50% of initial cost: $$6 \times 0.50 \times 208.33 = 625$$ - Total revenue including refund: $$13500 + 625 = 14125$$ - Profit: $$14125 - 12500 = 1625$$ - Profit as a percent of initial cost: $$\frac{1625}{12500} \times 100 = 13\%$$ Answer: 13% profit (Option D). 2. Problem 211: Find the maximum possible length of the longest rope. - Seven ropes with average length 68 cm: $$\text{Total length} = 7 \times 68 = 476$$ - Median length is 84 cm, so the 4th rope length is 84 cm. - Let the shortest rope length be $x$ cm. - Longest rope length is $4x + 14$ cm. - To maximize the longest rope, minimize the other ropes while respecting the median. - The first three ropes must be $\leq 84$, so set them to $x, a, b$ with $x \leq a \leq b \leq 84$. - The 4th rope is 84. - The 5th and 6th ropes must be $\geq 84$, set them to 84 to minimize their sum. - Sum of all ropes: $$x + a + b + 84 + 84 + 84 + (4x + 14) = 476$$ - Simplify: $$x + a + b + 252 + 4x + 14 = 476$$ $$5x + a + b + 266 = 476$$ $$5x + a + b = 210$$ - To maximize $4x + 14$, minimize $a + b$ with $a, b \geq x$ and $a, b \leq 84$. - Minimum $a + b$ is $2x$. - Substitute: $$5x + 2x = 7x = 210 \implies x = 30$$ - Longest rope length: $$4(30) + 14 = 120 + 14 = 134$$ Answer: 134 cm (Option D). 3. Problem 212: Find the difference between the 6th and 5th terms of the sequence with nth term: $$a_n = n + 2^{n-1}$$ - Calculate $a_6$: $$6 + 2^{5} = 6 + 32 = 38$$ - Calculate $a_5$: $$5 + 2^{4} = 5 + 16 = 21$$ - Difference: $$38 - 21 = 17$$ Answer: 17 (Option E).