It costs ₹ $x$ each to make the first thousand copies of a compact disc and ₹ $y$ to make each subsequent copy. If $z$ is greater than 1000, how much will it cost to make $z$ copies of the compact disc?

### Concept & Logic Split the total quantity into two distinct tiers and calculate the cost for each tier separately. Then, combine the costs and factor the algebraic expression to match the given options. $$Total\ Cost = (Cost\ of\ First\ Tier) + (Cost\ of\ Remaining\ Tier)$$ ### Step-by-Step Solution * **Given:** * Cost for first 1000 copies = ₹ $x$ per copy * Cost for copies after 1000 = ₹ $y$ per copy * Total copies = $z$ (where $z > 1000$) * **Calculation:** 1. Calculate the total cost for the first 1000 copies: $Cost_1 = 1000 \times x = 1000x$ 2. Determine the number of subsequent copies made: $Remaining\ copies = z - 1000$ 3. Calculate the cost for these remaining copies: $Cost_2 = (z - 1000) \times y = yz - 1000y$ 4. Add the two costs together for the total cost: $Total\ Cost = 1000x + yz - 1000y$ 5. Rearrange and factor the expression to match the options: Factor out 1000 from the $x$ and $y$ terms: $1000(x - y) + yz$ ### Exam Strategy & Shortcut When algebraic manipulation gets confusing, use the substitution method. Pick simple, concrete numbers. Let $x = 10$, $y = 5$, and $z = 1001$. Cost = $(1000 \times 10) + (1 \times 5) = 10000 + 5 = 10005$. Now plug these values into the options: (a) $(1001\times10) - (1001\times5) = 10010 - 5005 = 5005$ (Incorrect) (b) $1000(10) + (5\times1001) = 10000 + 5005 = 15005$ (Incorrect) (c) $1000(10 - 5) + (5\times1001) = 5000 + 5005 = 10005$ (Matches perfectly) ### Common Pitfall Students often correctly formulate $1000x + y(z - 1000)$ but fail to recognize how it expands and regroups into option (c). Don't assume you made a mistake if your initial expression isn't listed; expand the options to check for equivalence. ### Final Answer **Therefore, the correct answer is $1000(x - y) + yz$.**